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Holt Physics Problem Bank

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Problem 1A Ch. 1–1

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 1AMETRIC PREFIXES

P R O B L E MThe scanning tunneling microscope (STM) has a magnifying ability of100 million and can distinguish between two objects that are separatedby only 3.0 × 10−10 m, or about one-hundredth the diameter of an atom.Express 3.0 × 10−10 m ina. nanometers.b. picometers.

S O L U T I O N

Given: distance = 3.0 × 10−10 m

Unknown: distance = ? nm distance = ? pm

Build conversion factors from the relationships given in Table 1-3.

1 ×

1

1

n

0

m−9 m

1 ×

1

10

p−m12 m

Convert from meters to nanometers by multiplying the distance by the first con-

version factor.

distance = 3.0 × 10−10 m × 1 ×

1

1

n

0

m−9 m = 3.0 × 10−1 nm =

Convert from meters to picometers by multiplying the distance by the second

conversion factor.

distance = 3.0 × 10−10 m × 1 ×

1

1

p

0−m

12m = 3.0 × 102 pm

0.30 nm

ADDITIONAL PRACTICE

1. One of the more unusual of world wonders is the Plain of Jars in Laos.

Several hundred huge stone jars, which do not seem to be made from

local rock, are scattered across the plain. The largest of these jars has a

mass of around 6.0 × 103 kg. Express this mass in

a. milligrams.

b. megagrams.

2. The French drink about 6.4 × 104 cm3 of mineral water per person per

year. Express this volume in

a. cubic meters.

b. cubic millimeters.

Holt Physics Problem BankCh. 1–2

NAME ______________________________________ DATE _______________ CLASS ____________________

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3. The explosive energy of powerful explosives is measured in terms of

“tons.” The ton referred to is a ton of TNT (trinitrotoluene), one of the

most powerful of chemical explosives. A ton of TNT will release 4.2 ×109 J (joules). Express this energy in

a. megajoules.

b. gigajoules.

4. A parsec, a distance measurement used by astronomers, is equal to

3.262 light years, where a light year is the distance light travels in one

year. In SI units, a parsec equals 3.086 × 1016 m. Express this distance in

a. kilometers.

b. exameters.

5. An acre is a common unit used to measure the area of a portion of

land. An acre is equal to about 4.0469 × 103 m2. Express this area in

a. square kilometers.

b. square centimeters.

6. Electric charge is measured in terms of the coulomb (1 C), although

this is a very large and not extremely practical unit of measurement.

For example, the charge in a bolt of lightning is about 15 C. Express the

charge in a lightning bolt in

a. millicoulombs.

b. kilocoulombs.

7. The wettest spot on Earth is generally considered to be Mt. Waialeale,

on the island of Kauai, Hawaii. In one year, this long-extinct volcano

receives 1.168 × 103 cm of rainfall. Express this quantity in

a. meters.

b. micrometers.

8. The United States Department of Defense is housed in the Pentagon,

one of the largest office buildings in the world. The entire floor area of

the Pentagon equals 0.344 279 km2. Express this area in

a. square meters.

b. square millimeters.

9. The Earth is approximately 4.50 billion years old. Setting 1 year equal

to 365.25 days, express the age of the Earth in

a. gigaseconds.

b. petaseconds.

10. One of the isotopes with the shortest half-life (the time it takes for half

of a sample of the element to decay) is beryllium-8. Its half life is mea-

sured as 6.7 × 10–17 s. Express this time in

a. microseconds.

b. attoseconds.

1ChapterThe Science of Physics

1. mass = 6.0 × 103 kga. mass = 6.0 × 103 kg ×

1

1

0

k

3

g

g ×

1

1

0

m−3

g

g =

b. mass = 6.0 × 103 kg × 1

1

0

k

3

g

g ×

1

10

M6 g

g = 6.0 Mg

6.0 × 109 mg

Additional Practice 1A

Givens Solutions

2. Volume = 6.4 × 104 cm3

a. volume = 6.4 × 104 cm3 × 10

12m

cm

3= 6.4 × 104 cm3 ×

10

16m

cm

3

3 =

b. volume = 6.4 × 104 cm3 × 11

0

c

m

m

m

3= 6.4 × 104 cm3 ×

10

1

3

c

m

m

m3

3

= 6.4 = 107 mm3

6.4 × 10−2 m3

3. energy = 4.2 × 109 J a. energy = 4.2 × 109 J × 1

10

M6 J

J =

b. energy = 4.2 × 109 J × 1

1

0

G9

J

J = 4.2 GJ

4.2 × 103 MJ

4. distance = 1 parsec = 3.086 × 1016 m a. distance = 1 parsec = 3.086 × 1016 m ×

1

1

0

k3m

m =

b. distance = 1 parsec = 3.086 × 1016 m × 1

1

01E8m

m = 3.086 × 10–2 Em

3.086 × 1013 km

5. area = 1 acre = 4.0469 × 103 m2 a. area = 1 acre = 4.0469 × 103 m2 × 1

1

0

k3m

m2

area = 4.0469 × 103 m2 × 1

1

0

k6m

m

2

2 =

b. area = 1 acre = 4.0469 × 103 m2 × 10

1

2

m

cm

2

area = 4.0469 × 103 m2 × 10

1

4

m

cm2

2

= 4.0469 × 107 cm2

4.0469 × 10−3 km2

6. electric charge = 15 Ca. electric charge = 15 C ×

10

1

3

C

mC =

b. electric charge = 15 C × 1

1

0

k3C

C = 1.5 × 10–2 kC

1.5 × 104 mC

Section Two—Problem Workbook Solutions V Ch. 1–1

V

Holt Physics Solution ManualV Ch. 1–2

V

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7. depth = 1.168 × 103 cma. depth = 1.168 × 103 cm ×

10

12m

cm =

b. depth = 1.168 × 103 cm × 10

12m

cm ×

1

1

0−m6

m

m = 1.168 × 107 mm

1.168 × 101 m = 11.68 m

Givens Solutions

8. area = 0.344 279 km2

a. area = 0.344 279 km2 × 110

k

3

m

m

2= 0.344 279 × 106 m2 =

b. area = 0.344 279 km2 × 110

k

3

m

m

2× 10

1

3

m

mm

2= 0.344 279 × 1012 mm2

area = 3.442 79 × 1011 mm2

3.442 79 × 105 m3

9. time = 4.50 × 109 years ×

365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

36

1

0

h

0 s

= 1.42 × 1017 s

a. time = 1.42 × 1017 s × 1

1

0

G9

s

s =

b. time = 1.42 × 1017 s × 1

1

01P5s

s = 1.42 × 102 Ps = 142 Ps

1.42 × 108 Gs

10. time = 6.7 × 10−17 s a. time = 6.7 × 10−17 s × 10

1

6

s

ms =

b. time = 6.7 × 10−17 s × 10

1

18

s

as = 6.7 = 101 as = 67 as

6.7 × 10−11 ms

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2AFINDING THE AVERAGE VELOCITY

P R O B L E MTo qualify for the finals in a racing event, a race car must achieve an aver-age speed of 2.50 × 102 km/h on a track with a total length of 1.60 km. If aparticular car covers the first half of the track at an average speed of 2.30× 102 km/h, what minimum average speed must it have in the second halfof the event to qualify?

S O L U T I O NGiven: vtot, avg = 2.50 × 102 km/h

v1, avg = 2.30 × 102 km/h

∆xtot = 1.60 km

∆x1 = ∆x2 = 12

∆xtot

Unknown: v2, avg = ?

Use the definition of average velocity, and rearrange it to solve for time.

vavg = ∆∆

x

t

∆t1 = time required to travel ∆x1 = v1

∆

,

x

a

1

vg

∆t2 = time required to travel ∆x2 = v2

∆

,

x

a

2

vg

∆ttot = ∆t1 + ∆t2 = v

∆

1,

x

a

1

vg +

v

∆

2,

x

a

2

vg =

vt

∆

ot

x

,

t

a

o

v

t

g

Use the last two equations for ∆ttot to solve for v2, avg.

v

∆

to

x

t,

t

a

o

v

t

g =

v

∆

1,

x

a

1

vg +

vt

∆

2,

x

a

2

vg =

1

2 ∆xtot v1,

1

avg +

v2,

1

avg

Divide by 1

2 ∆xtot on each side.

vtot

2

, avg =

v1,

1

avg +

v2,

1

avg

Rearrange the equation to calculate v2, avg.

v2,

1

avg =

vtot

2

, avg −

v1,

1

avg

Invert the equation.

v2, avg =

v2, avg =1

2.50 × 1

2

02 km/h − 2.30 × 1

2

02 km/h

1

vtot

2

, avg −

v1,

1

avg


NAME ______________________________________ DATE _______________ CLASS ____________________

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v2, avg =

v2, avg = 3.65

1

×k

1

m

0−3 h

v2, avg = 274 km/h

18.00 × 10−3 h/km − 4.35 × 10−3 h/km

ADDITIONAL PRACTICE

1. The fastest helicopter, the Westland Lynx, can travel 3.33 km in the forward

direction in just 30.0 s. What is the average velocity of this helicopter? Ex-

press your answer in both meters per second and kilometers per hour.

2. The fastest airplane is the Lockheed SR-71 Blackbird, a high-altitude spy

plane first built in 1964. If an SR-71 is clocked traveling 15.0 km west in

15.3 s, what is its average velocity in kilometers per hour?

3. At its maximum speed, a typical snail moves about 4.0 m in 5.0 min.

What is the average speed of the snail?

4. The arctic tern migrates farther than any other bird. Each year, the Arctic

tern travels 3.20 × 104 km between the Arctic Ocean and the continent of

Antarctica. Most of the migration takes place within two four-month pe-

riods each year. If a tern travels 3.20 × 104 km south in 122 days, what is

its average velocity in kilometers per day?

5. Suppose the tern travels 1.70 × 104 km south, only to encounter bad

weather. Instead of trying to fly around the storm, the tern turns around

and travels 6.00 × 102 north to wait out the storm. It then turns around

again immediately and flies 1.44 × 104 km south to Antarctica. What are

the tern’s average speed and velocity if it makes this trip in 122 days?

6. Eustace drives 20.0 km to the east when he realizes he left his wallet at

home. He drives 20.0 km west to his house, takes 5.0 min to find his wal-

let, then leaves again. Eustace is 40.0 km east of his house exactly

60.0 min after he left the first time.

a. What is his average velocity?

b. What is his average speed?

7. Emily takes a trip, driving with a constant velocity of 89.5 km/h to the

north except for a 22.0 min rest stop. If Emily’s average velocity is

77.8 km/h to the north, how long does the trip take?

8. Laura is skydiving when at a certain altitude she opens her parachute and

drifts toward the ground with a constant velocity of 6.50 m/s, straight down.

What is Laura’s displacement if it takes her 34.0 s to reach the ground?

9. A tortoise can run with a speed of 10.0 cm/s, and a hare can run exactly 20

times as fast. In a race, they both start at the same time, but the hare stops to

rest for 2.00 min. The tortoise wins by 20.0 cm. How long does the race take?

10. What is the length of the race in problem 9?

Problem 2B Ch. 2–3

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2BAVERAGE ACCELERATION

P R O B L E MA basketball rolling with a velocity of +1.1 m/s comes to a stop in 8.5 s.What is the basketball’s average acceleration?

S O L U T I O NGiven: vi = +1.1 m/s

vf = 0 m/s

∆t = 8.5 s

Unknown: aavg = ?

Use the definition of average acceleration to find aavg.

aavg = ∆∆

v

t =

vf

∆−t

vi

aavg = 0 m/s

8

−.5

1

s

.1 m/s =

−1

8

.1

.5

m

s

/s

aavg = −0.13 m/s2

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ADDITIONAL PRACTICE

1. With a time of 6.92 s, Irina Privalova of Russia holds the women’s

record for running 60 m. Suppose she ran this distance with a constant

acceleration, so that she crossed the finish line with a speed of

17.34 m/s. Assuming she started at rest, what was the magnitude of

Privalova’s average acceleration.

2. The solid-fuel rocket boosters used to launch the space shuttle are able

to lift the shuttle 45 kilometers above Earth’s surface. During the

2.00 min that the boosters operate, the shuttle accelerates from rest

to a speed of nearly 7.50 × 102 m/s. What is the magnitude of the shut-

tle’s average acceleration?

3. A type of firework consists of a cardboard tank mounted on plastic

wheels and driven forward by a small rocket. Once the rocket ignites,

the tank rolls from rest to a maximum velocity of 0.85 m/s forward, at

which point the rocket burns out. If the total time that the rocket re-

mains ignited is 3.7 s, what is the average acceleration of the tank?

4. A handball is hit toward a wall with a velocity of 13.7 m/s in the for-

ward direction. It returns with a velocity of 11.5 m/s in the backward

direction. If the time interval during which the ball is accelerated is

0.021 s, what is the handball’s average acceleration?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A certain type of rocket sled is used to measure the effects of extreme

deceleration. The sled reaches a velocity of +320 km/h, then comes to a

complete stop in 0.18 s. What is the average acceleration that takes

place in this time interval?

6. In 1970, Don “Big Daddy” Garlits set what was then the world record

for drag racing. With an average acceleration of 16.5 m/s2, Garlits

started at rest and reached a speed of 386.0 km/h. How much time was

needed for Garlits to reach his final speed?

7. A freight train traveling with a velocity of −4.0 m/s begins backing into

a train yard. If the train’s average acceleration is −0.27 m/s2, what is the

train’s velocity after 17 s?

8. A student on in-line roller skates travels at a speed of 4.5 m/s along the

top of a hill. She then skates downhill with an average acceleration of

0.85 m/s2. If her final speed is 10.8 m/s, how long does it take her to

skate down the hill?

9. The Impact is the first commercial electric car to be developed in over

60 years. During tests in 1994, the car reached a top speed of nearly

296 km/h. Suppose the car started at rest and then underwent an aver-

age acceleration of 1.60 m/s2. How long did it take the Impact to reach

its top speed?

10. A bicyclist accelerates –0.87 m/s2 during a 3.8 s interval. What is the

change in the velocity of the bicyclist and bicycle?

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Problem 2C Ch. 2–5

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2CDISPLACEMENT WITH UNIFORM ACCELERATION

P R O B L E MThe arrow on a crossbow undergoes uniform acceleration over a distanceof 38.1 cm. If the acceleration takes place over 8.93 × 10-3 s and the arrowis initially at rest, what is the arrow’s final speed?

S O L U T I O NGiven: ∆x = 38.1 cm

∆t = 8.93 × 10−3 s

vi = 0 m/s

Unknown: vf = ?

Use the equation for displacement with uniform acceleration.

∆x = 1

2 (vi + vf ) ∆t

Rearrange the equation to solve for vf.

vf = 2

∆∆t

x − vi

vf = − 0 m/s

vf = 85.3 m/s − 0 m/s

vf = 85.3 m/s

(2)(38.1 cm)10

1

0

m

cm

8.93 × 10−3 s

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ADDITIONAL PRACTICE

1. A device at Sandia Laboratories in Albuquerque, New Mexico, uses

highly compressed air to accelerate small metal disks to supersonic

speeds. Suppose the disk, which is initially at rest, undergoes a uniform

acceleration for 0.910 s, at which point it reaches its top speed. If the disk

travels 7.19 km in that time, what is its final speed?

2. Despite their size and awkward appearance, polar bears can run at re-

spectable speeds for short distances. Suppose a polar bear running with

an initial speed of 4.0 m/s accelerates uniformly for 18 s. What is the

bear’s maximum speed if the bear travels 135 m during the 18 s of accel-

eration? Give the answer in both meters per second and kilometers per

hour?

3. A hockey puck slides 55.0 m along the length of the rink in just 1.25 s.

The slight friction between the puck and the ice provides a uniform ac-

celeration. If the puck’s final speed is 43.2 m/s, what is its initial speed?


NAME ______________________________________ DATE _______________ CLASS ____________________

4. A child sleds down a snow-covered hill with a uniform acceleration.

The slope of the hill is 38.5 m long. If the child starts at rest and

reaches the bottom of the hill in 5.5 s, what is the child’s final speed?

5. The longest stretch of straight railroad tracks lies across the desolate

Nullarbor Plain, between the Australian cities of Adelaide and Perth.

The tracks extend a distance of 478 km without a curve. Suppose a

train with an initial speed of 72 km/h travels along the entire length of

straight track with a uniform acceleration. The train reaches the end of

the straight track in 5 h, 39 min. What is the train’s final speed?

6. A golf ball at a miniature golf course travels 4.2 m along a carpeted

green. When the ball reaches the hole 3.0 s later, its speed is 1.3 m/s. As-

suming the ball undergoes constant uniform acceleration, what is the

ball’s initial speed?

7. A speedboat uniformly increases its velocity from 25 m/s to the west to

35 m/s to the west. How long does it take the boat to travel 250 m west

while undergoing this acceleration?

8. Airplane racing, like horse and auto racing, uses a “track” of a specific

length. Unlike the horse or auto tracks, the racing area for airplanes is

bounded on the inside by tall columns, or pylons, around which the pi-

lots must fly, and by altitude limitations that the pilots must monitor

using their instruments. Different types of races use different arrange-

ments of pylons to make the length of the race longer or shorter. In one

particular race, a pilot begins the race at a speed of 755.0 km/h and ac-

celerates at a constant uniform rate for 63.21 s. The pilot crosses the

finish line with a speed of 777.0 km/h. From this data, calculate the

length of the course.

9. A hovercraft, also known as an air-cushion vehicle, glides on a cushion

of air, allowing it to travel with equal ease on land or water. Suppose a

hovercraft undergoes constant uniform acceleration, which causes the

hovercraft to move from rest to a speed of 30.8 m/s. How long does the

hovercraft accelerate if it travels a distance of 493 m?

10. A spaceship travels 1220 km with a constant uniform acceleration.

How much time is required for the acceleration if the spaceship in-

creases its speed from 11.1 km/s to 11.7 km/s?

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Problem 2D Ch. 2–7

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2DVELOCITY AND DISPLACEMENT WITH UNIFORM ACCELERATION

P R O B L E MA barge moving with a speed of 1.00 m/s increases speed uniformly, sothat in 30.0 s it has traveled 60.2 m. What is the magnitude of the barge’sacceleration?

S O L U T I O N

Given: vi = 1.00 m/s

∆t = 30.0 s

∆x = 60.2 m

Unknown: a = ?

Use the equation for displaement with constant uniform acceleration.

∆x = vi∆t + 12

a∆t2

Rearrange the equation to solve for a.

12

a∆t2 = ∆x − vi∆t

a = 2(∆x

∆−t2

vi∆t)

a =

a =

a = 9

(

.

2

0

)

0

(3

×0

1

.2

02m

s

)2

a = 6.71 × 10−2 m/s2

(2)(60.2 m − 30.0 m)

9.00 × 102 s2

(2)[60.2 m − (1.00 m/s)(30.0 s)]

(30.0 s)2

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ADDITIONAL PRACTICE

1. The flight speed of a small bottle rocket can vary greatly, depending on

how well its powder burns. Suppose a rocket is launched from rest so that

it travels 12.4 m upward in 2.0 s. What is the rocket’s net acceleration?

2. The shark can accelerate to a speed of 32.0 km/h in a few seconds. As-

sume that it takes a shark 1.5 s to accelerate uniformly from 2.8 km/h to

32.0 km/h. What is the magnitude of the shark’s acceleration?

3. In order for the Wright brothers’ 1903 flyer to reach launch speed, it had

to be accelerated uniformly along a track that was 18.3 m long. A system

of pulleys and falling weights provided the acceleration. If the flyer was

initially at rest and it took 2.74 s for the flyer to travel the length of the

track, what was the magnitude of its acceleration?


NAME ______________________________________ DATE _______________ CLASS ____________________

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4. A certain roller coaster increases the speed of its cars as it raises them to

the top of the incline. Suppose the cars move at 2.3 m/s at the base of

the incline and are moving at 46.7 m/s at the top of the incline. What is

the magnitude of the net acceleration if it is uniform acceleration and

takes place in 7.0 s?

5. A ship with an initial speed of 6.23 m/s approaches a dock that is 255 m

away. If the ship accelerates uniformly and comes to rest in 82 s, what is

its acceleration?

6. Although tigers are not the fastest of predators, they can still reach and

briefly maintain a speed of 55 km/h. Assume that a tiger takes 4.1 s to

reach this speed from an initial speed of 11 km/h. What is the magni-

tude of the tiger’s acceleration, assuming it accelerates uniformly?

7. Assume that a catcher in a professional baseball game catches a ball that

has been pitched with an initial velocity of 42.0 m/s to the southeast. If

the catcher uniformly brings the ball to rest in 0.0090 s through a dis-

tance of 0.020 m to the southeast, what is the ball’s acceleration?

8. A crate is carried by a conveyor belt to a loading dock. The belt speed

uniformly increases slightly, so that for 28.0 s the crate accelerates by

0.035 m/s2. If the crate’s initial speed is 0.76 m/s, what is its final speed?

9. A plane starting at rest at the south end of a runway undergoes a uni-

form acceleration of 1.60 m/s2 to the north. At takeoff, the plane’s ve-

locity is 72.0 m/s to the north.

a. What is the time required for takeoff?

b. How far does the plane travel along the runway?

10. A cross-country skier with an initial forward velocity of +4.42 m/s ac-

celerates uniformly at −0.75 m/s2.

a. How long does it take the skier to come to a stop?

b. What is the skier’s displacement in this time interval?

Problem 2E Ch. 2–9

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2EFINAL VELOCITY AFTER ANY DISPLACEMENT

P R O B L E MA radio-controlled toy car increases speed over a distance of 15.2 m. If thecar starts at rest and has a final speed of 0.76 m/s, what is the magnitudeof its acceleration?

S O L U T I O NGiven: ∆x = 15.2 m

vi = 0 m/s

vf = 0.76 m/s

Unknown: a = ?

Choose the equation(s) or situation: Use the equation for the final velocity after

any displacement.

vf2 = vi

2 + 2a∆x

Rearrange the equation(s) to isolate the unknown(s):

a = vf

2

2∆−

x

vi2

Substitute the values into the equation(s) and solve:

a =

a =

a = 0

3

.

0

5

.

8

4 m/s2

a =

The magnitude of the acceleration suggests that the car increases speed very slowly.

This is confirmed by using the difference in speeds and the acceleration to calculate

the time interval for the acceleration. The car reaches its final speed in 40 s.

1.9 × 10−2 m/s2

0.58 m2/s2 − 0 m2/s2

30.4 m

(0.76 m/s)2 − (0 m/s)2

(2)(15.2 m)

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A dumptruck filled with sand moves 1.8 km/h when it begins to acceler-

ate uniformly at a constant rate. After traveling 4.0 × 102 m, the truck’s

speed is 24.0 km/h. What is the magnitude of the truck’s acceleration?

2. One of the most consistent long-jumpers is Jackie Joyner-Kersee of the

United States. Her best distance in this field and track event is 7.49 m. To

achieve this distance, her speed at the point where she started the jump

was at least 8.57 m/s. Suppose the runway for the long jump was


NAME ______________________________________ DATE _______________ CLASS ____________________

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19.53 m, and that Joyner-Kersee’s initial speed was 0 m/s. What was

the magnitude of her acceleration if it was uniform acceleration?

3. Although ungraceful on land, walruses are fine swimmers. They nor-

mally swim at 7 km/h, and for short periods of time are capable of

reaching speeds of nearly 35 km/h. Suppose a walrus accelerates from

7.0 km/h to 34.5 km/h over a distance of 95 m. What would be the

magnitude of the walrus’s uniform acceleration?

4. Floyd Beattie set an unofficial speed record for a unicycle in 1986. He

rode the unicycle through a 2.00 × 102 m speed trap, along which his

speed was measured as being between 9.78 m/s and 10.22 m/s. Suppose

that Beattie had accelerated at a constant rate along the speed trap, so

that his initial speed was 9.78 m/s and his final speed was 10.22 m/s.

What would the magnitude of his acceleration have been?

5. A fighter jet lands on an aircraft carrier’s flight deck. Although the deck

is 300 m long, most of the jet’s acceleration occurs within a distance of

42.0 m. If the jet’s velocity is reduced uniformly from +153.0 km/h to

0 km/h as it moves through 42.0 m, what is the jet’s acceleration?

6. Most hummingbirds can fly with speeds of nearly 50.0 km/h. Suppose

a hummingbird flying with a velocity of 50.0 km/h in the forward di-

rection accelerates uniformly at 9.20 m/s2 in the backward direction until

it comes to a hovering stop. What is the hummingbird’s displacement?

7. A thoroughbred racehorse accelerates uniformly at 7.56 m/s2, reaching

its final speed after running 19.0 m. If the horse starts at rest, what is its

final speed?

8. A soccer ball moving with an initial speed of 1.8 m/s is kicked with a

uniform acceleration of 6.1 m/s2, so that the ball’s new speed is 9.4 m/s.

How far has the soccer ball moved?

9. A dog runs with an initial velocity of 1.50 m/s to the right on a waxed

floor. It slides to a final velocity of 0.30 m/s to the right with a uniform

acceleration of 0.35 m/s2 to the left. What is the dog’s displacement?

10. A hippopotamus can run up to 30 km/h, or 8.33 m/s. Suppose a hip-

popotamus uniformly accelerates 0.678 m/s2 until it reaches a top

speed of 8.33 m/s. If the hippopotamus has run 46.3 m, what is its ini-

tial speed?

Problem 2F Ch. 2–11

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 2FFALLING OBJECT

P R O B L E MWhen it is completed in 2002, the International Financial Center inTaipei, Taiwan, will be the tallest building in the world. Suppose a con-struction worker on the top-most floor of the building accidentallyknocks a wrench off a ledge. The wrench hits the ground below 9.56 slater. What is the distance between the top-floor of the International Fi-nancial Center and the ground. Assume there is no air resistance.

S O L U T I O NGiven: ∆t = 9.56 s

a = −9.81 m/s2

vi = 0 m/s

Unknown: ∆x = ?

Choose the equation(s) or situation: Displacement is unknown, as is the final

velocity. Because time, acceleration, and initial velocity are known, the equation

for displacement with constant acceleration can be used.

∆x = vi∆t + 12

a∆t2


∆x = (0 m/s)(9.56 S) + 12

(−9.81 m/s2)(9.56 s)2

∆x = (0 m) + (−448 m)

∆x = −448 m

∆x =

From the value for ∆x the wrench’s final speed can be determined as 93.8 m/s, or

nearly 340 km/h.

distance from top of building to ground = 448 m

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. Suppose a safety net at one of the floors of the International Financial

Center catches the wrench in Problem 2F. The wrench falls into the net

with a velocity of 49.5 m/s downward. How far above the ground is the

safety net located?

2. A gumdrop is released from rest at the top of the Empire State Building,

which is 381 m tall. Disregarding air resistance, calculate the displace-

ment of the gumdrop after 1.00, 2.00, and 3.00 s.

3. A small sandbag is dropped from rest from a hovering hot-air balloon.

After 2.0 s, how far below the balloon is the sand bag?


NAME ______________________________________ DATE _______________ CLASS ____________________

4. A physics student throws a softball straight up into the air with a speed

of 17.5 m/s. The ball is in the air for a total of 3.60 s before it is caught

at its original position. How high does the ball rise?

5. A surface probe lands on a highland region of the planet Mercury. A

few hours later the ground beneath the probe gives way and the probe

falls, landing below its original position with a velocity of 11.2 m/s

downward. If the free-fall acceleration near Mercury’s surface is

3.70 m/s2 downward, what is the probe’s displacement?

6. A ball thrown vertically is caught by the thrower after 5.1 s. Find the

maximum height the ball reaches.

7. Find the initial velocity with which the ball in problem 6 is thrown.

8. An archer fires an arrow directly upward, then quickly runs from the

launching spot to avoid being struck by the returning arrow. If the

arrow’s initial velocity is 85.1 m/s upward how long does the archer

have to run away before the arrow lands?

9. A popular scene in recent action films shows a character in free-fall

speed up to catch a freely falling parachute. Suppose a packed para-

chute is dropped from rest from an airplane and that a daredevil is

launched straight down from the plane 3.00 s later. Neglecting air

resistance, the daredevil catches up to the parachute 4.00 s after the

daredevil leaves the plane. What are the daredevil’s initial and final

velocities?

10. The elevators in the Landmark Tower in Yokohama, Japan, are among

the fastest in the world. They accelerate upward at 3.125 m/s2 for 4.00 s

to reach their maximum speed. Suppose an empty elevator is moving

upward with its maximum speed when the cable breaks, so that the ele-

vator slows down, comes to a stop, and then begins to fall freely. What

will the elevator’s velocity be 0.00 s, 1.00 s, 2.00 s, and 3.00 s after the

cable breaks?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3AFINDING RESULTANT MAGNITUDE AND DIRECTION

P R O B L E MA hummingbird flies 9.0 m horizontally and then flies up for 3.0 m. Whatis the bird’s resultant displacement?

S O L U T I O NGiven: ∆x = 9.0 m ∆y = 3.0 m

Unknown: d = ? q = ?

Diagram:

∆x = 9.0 m

∆y = 3.0 m

y

d

xθ

Choose the equation(s) or situation: The Pythagorean theorem can be used to

find the magnitude of the hummingbird’s displacement. The direction of the dis-

placement can be found using the tangent function.

d2 = ∆x 2 + ∆y 2

tan q =


d =√

∆x2+ ∆y2

q = tan−1 Substitute the values into the equation(s) and solve:

d =√

(9.0 m)2 + (3.0m)2 =√

81 m2+ 9.0 m2 =√

9.0× 101 m2

d =

q = tan−1 39..00

mm

q =

The resultant displacement (d) is only slightly larger than the largest component

(∆x), as is the case for small angles (q 20°).

18° above horizonal

9.5 m

∆y∆x

∆y∆x

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. A tiger paces back and forth along the front of its cage, which is 8 m

wide. The tiger starts from the right side of the cage, paces to the left

side, then back to the right side, and finally back to the left.

a. What total distance has the tiger paced?

b. What is the tiger’s resultant displacement?

2. A particular type of rubber ball is able to bounce to 0.90 times the height

from which it is dropped. The ball is dropped from a height of 0.91 m,

but it falls slightly away from the vertical, so that by the time it has

bounced to its new height it has undergone a horizontal displacement of

0.11 m. What is the ball’s resultant displacement from its initial height to

its maximum height after one bounce?

3. A helicopter flies 165 m horizontally and then moves downward to land

45 m below. What is the helicopter’s resultant displacement?

4. A toy parachute is dropped from an open window that is 13.0 m above

the ground. If the parachute travels 9.0 m horizontally, what is the resul-

tant displacement?

5. An octopus swims 36.0 m east, 42.0 m north, and then rises 17.0 m to-

ward the surface of the water. What is the octopus’s displacement?

(TWO-DIMENSIONAL METHOD: Visualize a horizontal and a vertical

triangle. Find the horizontal resultant; use that with the vertical distance

to calculate the final resultant. Studying this method can lead to under-

standing the easier three-dimensional solution in the solutions manual.)

6. An airplane taxis to the end of a runway before taking off. The magni-

tude of the plane’s total displacement is 599 m. If the northern compo-

nent of this displacement is 89 m, what is the displacement’s eastern

component? What is the direction of the total displacement?

7. The straightest stretch of railroad tracks in the world extends for 478 km

in southwestern Australia. A train traveling along these tracks is dis-

placed to the south by about 42 km. What is the train’s displacement to

the west? What is the direction of the total displacement?

8. Before the widespread use of steamships, sailing from Europe to North

America was accomplished by use of the “trade winds.” The trade winds

move from the northeast to the southwest between 30° and 60° latitude

in the northern hemisphere. A ship sailing from Europe to the Caribbean

Sea would first travel southward to the Canary Islands, off the coast of

North Africa, and then use the trade winds to sail west. Suppose a ship

travels south from Iceland to the Canary Islands, and then west to

Florida. The ship’s total displacement is 7400 km at 26° south of west. If

the ship sails 3200 km south from Iceland to the Canary Islands, how

large is the western component of its journey?


NAME ______________________________________ DATE _______________ CLASS ____________________

9. The Palm Springs Aerial Tramway extends 3.88 km from the Valley Sta-

tion, which is located 0.8 km above sea level, to the Mountain Station

atop San Jacinto Peak. The actual path of the tramway’s cables is not

along a straight line, but if it were, the horizontal displacement of the

tramway would be 3.45 km. How far is San Jacinto Peak above sea

level?

10. The islands that form the Tristan da Cunha Group in the South At-

lantic Ocean are considered to be the most remote places in the world:

the next nearest inhabited island is 2400 km away. If you sail from

Capetown, South Africa, in a south by southwest direction, you must

travel 2.9 × 103 km before reaching the Tristan da Cunha islands. If the

western component of your displacement is 2.8 × 103 km, what is your

displacement south? In what direction is the resultant displacement?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3BRESOLVING VECTORS

P R O B L E MThe straight stretch of Interstate Highway 5 from Mettler, California, to apoint near Buttonwillow, California, is 53.0 km long and makes an angleof 48.7° north of west. What are the northern and western components ofthis highway segment?

S O L U T I O NGiven: d = 53.0 km q = 48.7° north of west

Unknown: ∆x = ? ∆y = ?

Diagram:

1. DEFINE

2. PLAN

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∆x

∆y

θ = 48.7°

N

d = 53.0 km

Choose the equation(s) or situation: Because the axes are perpendicular, the

sine and cosine functions can be used to find the components.

sin q = ∆dy

cos q = ∆dx


∆x = d (cos q)

∆y = d (sin q)


∆x = (53.0 km)(cos 48.7°)

∆x =

∆y = (53.0 km)(sin 48.7°)

∆y =

Using the Pythagorean theorem to check the answers confirms the magnitudes of

the components.

d2 = ∆x2 + ∆y2

(53.0 km)2 = (35.0 km)2 + (39.8 km)2

39.8 km, north

35.0 km, west

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

1. The distance from an observer on the plain to the top of a nearby

mountain is 5.3 km, and the angle between this line and the horizontal

is 8.4°. How tall is the mountain?

2. A bowling ball is released at the near right corner of a bowling lane and

travels 19.1 m at an angle of 3.0° with respect to the lane’s length. The

ball reaches the far left corner of the lane, where it knocks over the “7”

pin. What is the width of the lane?

3. A skyrocket travels 113 m at an angle of 82.4° with respect to the ground

and toward the south. What is the rocket’s horizontal displacement?

4. A hot-air balloon descends with a velocity of 55 km/h at an angle of

37° below the horizontal. What is the vertical velocity of the balloon?

5. A billiard ball travels 2.7 m at an angle of 13° with respect to the long

side of the table. What are the components of the ball’s displacement?

6. One hole at a certain miniature golf course extends for about 60 m. A golf

ball on this hole travels with a velocity of 1.20 m/s at 14.0° east of north.

What are the eastern and northern components of the ball’s velocity?

7. The Very Large Array in western New Mexico consists of several radio

telescopes that can be rearranged along railroad tracks. The largest of

these arrangements has the telescopes positioned in a “Y” pattern for

18 km along three separate tracks. Suppose an electrician inspects the

instruments in each antenna from the end of the northern track to the

end of the southwestern track. If the electrician’s resultant displace-

ment is 31.2 km at 30 0° west of south, what are the southern and west-

ern components of the displacement?

8. Barnard’s Star is the closest star to Earth after the sun and the triple

star Alpha Centauri. Barnard’s Star has a velocity of 165.2 km/s at an

angle of 32.7° away from its forward motion. What are the forward and

side components of this velocity?

9. A tiger leaps with an initial velocity of 55.0 km/h at an angle of 13.0°

with respect to the horizontal. What are the components of the tiger’s

velocity?

10. A certain type of balloon is designed to ascend rapidly. Suppose this

balloon has a velocity 13.9 m/s at 26.0° above the horizontal and 24.0°

east of north. What are the upward, northern, and eastern components

of the balloon’s velocity? (HINT: Draw horizontal and vertical right tri-

angles whose sides represent the velocity’s components.)

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ADDITIONAL PRACTICE

2.80 × 103 km2 = 1.22 × 103 km2 + 1.58 × 103 km2

2.80 × 103 km2 = 2.80 × 103 km2


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3CADDING VECTORS ALGEBRAICALLY

P R O B L E MThe southernmost point in the United States is called South Point, and islocated at the southern tip of the large island of Hawaii. A plane designedto take off and land in water leaves South Point and flies to Honolulu, onthe island of Oahu, in three separate stages. The plane first flies 83.0 km at22.0° west of north from South Point to Kailua Kona, Hawaii. The planethen flies 146 km at 21.0° west of north from Kailua Kona to Kahului, onthe island of Maui. Finally, the plane flies 152 km at 17.5° north of westfrom Kahului to Honolulu. What is the plane’s resultant displacement?

S O L U T I O NGiven: d1 = 83.0 km q1 = 22.0° west of north

d2 = 146 km q2 = 21.0° west of north

d3 = 152 km q3 = 17.5° north of west

Unknown: d = ? q = ?

Diagram:

1. DEFINE

2. PLAN

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θ

θ

d

d2 = 146 km

d1 = 83.0 km

1 = 22.0°

θ2 = 21.0°

θ3 = 17.5°

d3 = 152 km N

Choose the equation(s) or situation: Express the components of each vector in

terms of sine or cosine functions.

∆x1 = d1 (sin q1) ∆y1 = d1 (cos q1)

∆x2 = d2 (sin q2) ∆y2 = d2 (cos q2)

∆x3 = d3 (cos q3) ∆y3 = d3 (sin q3)

Note that the angles q1 and q2 are with respect to the y axis (north), and so the x

components are in terms of sin q . Write the equations for ∆xtot and ∆ytot , the

components of the total displacement.


NAME ______________________________________ DATE _______________ CLASS ____________________

∆xtot = ∆x1 + ∆x2 + ∆x3

= d1 (sin q1) + d2(sin q2) + d3 (cos q3)

∆ytot = ∆y1 + ∆y2 + ∆y3

= d1 (cos q1) + d2(cos q2) + d3 (sin q3)

Use the components of the total displacement, the Pythagorean theorem, and the

tangent function to calculate the total displacement.

d =√

(∆xtot)2 + (∆ytot)2

q = tan−1 Substitute the values into the equation(s) and solve:

∆xtot = (83.0 km)(sin 22.0°) + (146 km)(sin 21.0°) + (152 km)

(cos 17.5°)

= 31.1 km + 52.3 km + 145 km

= 228 km

∆ytot = (83.0 km)(cos 22.0°) + (146 km)(cos 21.0°) + (152 km)

(sin 17.5°)

= 259 km

d =√

(228 km)2 + (259km)2 =√5.20 × 104 km2+ 6.71× 104 km2 =

√11.91× 104 km2

=

q = tan−1 =

If the diagram is drawn to scale, compare the calculated results to the drawing.

The length of the drawn resultant is fairly close to the scaled magnitude for d,

while the angle appears to be slightly greater than 45°.

48.6° north of west259 km228 km

345.1 km

∆ytot∆xtot

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3. CALCULATE

4. EVALUATE

1. U.S. Highway 212 extends 55 km at 37° north of east between Newell and

Mud Butte, South Dakota. It then continues for 66 km nearly due east

from Mud Butte to Faith, South Dakota. If you drive along this part of

U.S. Highway 212, what will be your total displacement?

2. Wrigley Field is one of only three original major-league baseball fields

that are still in use today. Suppose you want to drive to Wrigley Field

from the corner of 55th Street and Woodlawn Avenue, about 14 miles

south of Wrigley Field. Although not the fastest or most direct route, the

most straightforward way to reach Wrigley Field is to drive 4.1 km west

on 55th Street to Halsted Street, then turn north and drive 17.3 km on

Halsted until you reach Clark Street. Turning on Clark, you will reach

Wrigley Field after traveling 1.2 km at an angle of 24.6° west of north.

What is your resultant displacement?

ADDITIONAL PRACTICE

3. A bullet traveling 850 m ricochets from a rock. The bullet travels an-

other 640 m, but at an angle of 36° from its previous forward motion.

What is the resultant displacement of the bullet?

4. The cable car system in San Francisco is the last of its kind that is still

in use in the United States. It was originally designed to transport large

numbers of people up the steep hills on which parts of the city are

built. If you ride seven blocks on the Powell Street cable car from the

terminal at Market Street to Pine Street, you will travel 2.00 × 102 m on

level ground, then 3.00 × 102 m at an incline of 3.0° to the horizontal,

and finally 2.00 × 102 m at 8.8° to the horizontal. What will be your re-

sultant displacement?

5. An Arctic tern flying to Antarctica encounters a storm. The tern

changes direction to fly around the storm. If the tern flies 46 km at 15°

south of east, 22 km at 13° east of south, and finally 14 km at 14° west

of south, what is the tern’s resultant displacement?

6. A technique used to change the direction of space probes, as well as to

give them additional speed, is to use the gravitational pull of nearby

planet. This technique was first used with the Voyager probes. Voyager 2

had traveled about 6.3 × 108 km when it reached Jupiter. Jupiter’s grav-

ity changed Voyager’s direction by 68°. The probe then traveled about

9.4 × 108 km when it reached Saturn, and its direction was changed by

94°. Voyager 2 was now redirected; it encountered Uranus after traveling

3.4 × 109 km from Saturn. Use this information to calculate the resul-

tant displacement of Voyager 2 as it traveled from Earth to Uranus.

7. The city of Amsterdam, in the Netherlands, has several canals that con-

nect different sections of the city. Suppose you take a barge trip to the

harbor, starting at a point near the northwest corner of the Vondel-

park. You would sail 2.50 × 103 m at 58.5° north of east, 375 m at 21.8°

north of east, and 875 m at 21.5° east of north. What would be your re-

sultant displacement?

8. The elevated train, or “L,” in Chicago is a major source for mass transit

in that city. One of the lines extends from Jefferson Park, in the north-

west part of town, to the Clark Street station downtown. The route of

this line runs 5.0 km at 36.9° south of east, 1.5 km due south, 8.5 km at

42.2° south of east, and 0.8 km due east. What is the resultant displace-

ment of an “L” train from Jefferson Park to Clark Street?

9. A billiard table is positioned with its long side parallel to north. A cue

ball is then shot so that it travels 1.41 m at an angle of 45.0° west of

north, is deflected by the table’s left side, and continues to move 1.98 m

east of north at an angle of 45.0°. The ball is then deflected by the

table’s right side, so that it moves 0.42 m west of north at an angle of

45.0°. After a reflection on the north end of the table, the ball travels

1.56 m at an angle of 45.0° south of west. Determine the resultant dis-

placement of the cue ball.


NAME ______________________________________ DATE _______________ CLASS ____________________

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NAME ______________________________________ DATE _______________ CLASS ____________________

10. Hurricane Iniki was the most destructive cyclone to have crossed the

Hawaiian Islands in the twentieth century. It’s path was also unusual: it

moved south of the islands for 790 km at an angle of 18° north of west,

then moved due west for 150 km, turned north and continued for

470 km, and finally turned back 15° east of north and moved 240 km

to cross the island of Kauai. What was the resultant displacement of

Hurricane Iniki?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3DPROJECTILES LAUNCHED HORIZONTALLY

P R O B L E MAlthough not the fastest or tallest or steepest roller coaster in the world,the “High Roller” roller coaster atop the Stratosphere Tower, in Las Vegas,Nevada, is the highest. Suppose that during construction of the ride ametal bolt was accidentally knocked horizontally off the edge of theStratosphere. If the bolt’s initial speed was 0.80 m/s, it would have trav-eled 6.76 m in the horizontal direction before hitting the ground. Use thisinformation to calculate how tall the Stratosphere Tower is.

S O L U T I O NGiven: vx = 0.80 m/s

∆x = 6.76 m

g = 9.81 m/s2

Unknown: ∆y = ?

Diagram:

1. DEFINE

2. PLAN

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∆x = 6.76 m

∆y

vx

Choose the equation(s) or situation: The magnitude of the vertical displace-

ment is given by the equation for falling bodies with no ititial vertical velocity.

∆y = − 12

g∆t 2

The magnitude for horizontal displacement is given by the equation for displace-

ment at constant velocity.

∆x = vx ∆ t

Rearrange the equation(s) to isolate the unknown(s): Substitute for ∆ t in the

falling-body equation.

∆t = ∆vx

x

∆y = − 12

g ∆vx

x2


∆y = − 12

(9.81 m/s2) 2

= 350 m

height of building = 350 m

6.76 m0.80 m/s

3. CALCULATE


NAME ______________________________________ DATE _______________ CLASS ____________________

The solution can be checked by using both equations to solve for ∆t. From the

equation for falling bodies, ∆t, is found to be 8.4 s. From the equation for horizon-

tal displacement, ∆t is 8.4 s. Both times are the same, so ∆y is correctly calculated.

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4. EVALUATE

ADDITIONAL PRACTICE

1. Lookout Mountain, which overlooks the Tennessee River Valley near

Chattanooga, Tennessee, was of great strategic importance during the

Civil War. Today, some of the artillery used in the war remain at the park

located on top of the mountain. Suppose one of these cannons fired a

projectile horizontally with a speed of 430 m/s, so that the projectile

landed at a horizontal distance of 4020 m from the cannon. How high

would the ridge of the mountain be with respect to the valley below?

2. In 1977, a helicopter at the heliport atop the 59-story Pan Am building in

New York fell over, causing the rapidly-turning rotor blades to splinter.

One of these fragments landed about 101 m away, near the corner of

Madison Avenue and 43rd Street. Suppose the fragment moved off the

building horizontally with a speed of 14.25 m/s. Use this information to

find the height of the Pan Am building.

3. The LZ N07 is a newly designed airship in the manner of the old Zep-

pelin airships built in Germany between 1908 and 1940. New technology

has made the LZ N07 more efficient and safe, as well as speedier. This

airship can travel with a horizontal speed of up to 1.30 × 102 km/h. If a

parcel is dropped from this airship, so that it lands 135 m in front of the

spot over which it was released, how far above the ground is the airship?

4. The shape of Sugarloaf mountain, in Rio de Janeiro, Brazil, is such that, if

you were to kick a soccer ball hard enough, it could land near the base of

the mountain without hitting the mountain’s side. Suppose the ball is

kicked horizontally with an initial speed of 9.37 m/s. If the ball travels a

horizontal distance of 85.0 m, how tall is the mountain?

5. Although many structures taller than 500 m have been designed, few

have been built due to practical limitations, such as cost and safety. In

light of this, the Bionic Tower in Hong Kong may never be more than a

design. If it is built, the Bionic Tower will provide working space for

100,000 people, and transport them using over 300 elevators. Suppose a

plate-glass window falls out of place from the top floor of the Bionic

Tower. Although the window’s speed is only 6.32 cm/s in the horizontal

direction, the window will still have a horizontal displacement of 1.00 m

once it hits the street below. Use this information to calculate the pro-

posed height of the Bionic Tower.

6. A squirrel on a limb near the top of a tree loses its grip on a nut, so that

the nut slips away horizontally at a speed of 10.0 cm/s. If the nut lands at a

horizontal distance of 18.6 cm, how high above the ground is the squirrel?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. A lunch pail is accidentally kicked off a steel beam on a building under

construction. Suppose the initial horizontal speed is 1.50 m/s. How far

does the lunch pail fall after it travels 3.50 m horizontally?

8. If the building in problem 7 is 2.50 × 102 m tall, and the lunch pail is

knocked off the top floor, what will be the horizontal displacement of

the lunch pail when it reaches the ground?

9. What is the velocity of the lunch pail in problem 8 when it reaches the

ground?

10. What is the range of an arrow shot horizontally at 85.3 m/s if it is ini-

tially 1.50 m above the ground?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3EPROJECTILES LAUNCHED AT AN ANGLE

P R O B L E MA flying fish leaps out of the water with a speed of 15.3 m/s. Normallythese fish use winglike fins to glide about 40 m before reentering theocean, but in this case the fish fails to use its “wings” and so only travelshorizontally about 17.5 m. At what angle with respect to the water’s sur-face does the fish leave the water? Use the trigonometric identity2(sinq)(cosq) = sin (2q) to solve for q .

S O L U T I O NGiven: vi = 15.3 m/s

∆x = 17.5 m

g = 9.81 m/s2

Unknown: q = ?

Diagram:

1. DEFINE

2. PLAN

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∆x = 17.5 m

vi θ

Choose the equation(s) or situation: The horizontal component of the fish’s

initial velocity, vx, is equal to the horizontal displacement divided by the time of

the jump.

vx = vi (cos q) =

The vertical displacement of the fish is given by the equation for falling bodies,

with the vertical component of the initial velocity, vy , used.

∆y = vy∆t − 12

g∆t 2

Because the fish lands at the same vertical position from which it started, ∆y = 0.

∆y = 0

vy = vi(sin q) = 12

g∆t

Rearrange the equation(s) to isolate the unknowns: Substitute for ∆t using the

equation for horizontal velocity.

∆t =

vi (sin q) = 12

g (sin q)(cos q) =

Using the trigonometric identity allows a solution for q to be found.

(sin q)(cos q) = 12

[sin (2q)]

sin (2q) = g∆xvi

2

g∆x2vi

2

∆xvi (cos q)

∆xvi (cos q)

∆x∆t


NAME ______________________________________ DATE _______________ CLASS ____________________

q =

q =

=

Substituting the value for q into the original equations and solving for ∆t pro-

duces a time of 1.25 s for both, thus confirming the result for q .

23.6° above the horizontal

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sin−1 2

g∆xvi

2

sin−1 2

(9.81 m/s2)(17.5 m)

(15.3 m/s)2

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A baseball is thrown with an initial speed of 15.0 m/s. If the ball’s hori-

zontal displacement is 17.6 m, at what angle with respect to the ground is

the ball pitched? Use the trigonometric identity 2(sin q)(cos q) = sin

(2q) to solve for q .

2. A football is kicked so that its initial speed is 23.1 m/s. If the football

reaches a maximum height of 16.9 m, at what angle with respect to the

ground is the ball kicked?

3. Jackie Joyner-Kersee’s record long jump is 7.49 m. Suppose she ran

9.50 m/s to jump this horizontal distance. At what angle above the hori-

zontal did she jump? Use the trigonometric identity 2(sin q)(cos q) =sin (2q) to solve for q .

4. The small jumping spiders make up for their size by their ability to leap

relatively large distances. Some can jump fifty times the length of their

bodies. Suppose a jumping spider leaps a horizontal distance of 18.5 cm

with an initial speed of about 141 cm/s. At what angle above the horizon-

tal would a spider with this speed have to leap in order to travel a range

of 18.5 cm? Use the trigonometric identity 2(sin q)(cos q) = sin (2q) to

solve for q .

5. Olympic platform divers jump from a diving board that is 10.0 m above

the water. Suppose a diver jumps from the board with an initial speed of

6.03 m/s. The diver reaches a maximum height of 11.7 m above the

water, and lands in the water at a horizontal distance of 3.62 m from the

end of the board. At what angle with respect to the board does the diver

leave the board?

6. A ball is thrown from a roof with a speed of 10.0 m/s and an angle of

37.0° with respect to the horizontal. What are the vertical and horizontal

components of the ball’s displacement 2.5 s after it is thrown?

7. A downed pilot fires a flare from a flare gun. The flare has an initial

speed of 250 m/s and is fired at an angle of 35° to the ground. How long

does it take for the flare to reach its maximum altitude?


NAME ______________________________________ DATE _______________ CLASS ____________________

8. In the sport of ski jumping, a skier travels down the slope of a hill until

he or she reaches the takeoff. The takeoff is slanted slightly below the

horizontal, so that the skier is able to travel in the air just above the

ground. Suppose a skier leaves the takeoff and lands 73.0 m horizon-

tally beyond the takeoff and −52.8 m below the takeoff. If the takeoff

angle is −8.00° below the horizontal, what is the skier’s initial speed?

9. A shingle slides down a roof having a 30.0° pitch and falls off with a

speed of 2.0 m/s. How long will it take to hit the ground 45 m below?

10. A hole at a miniature golf course requires the ball to roll up a ramp, fly

over a small stream, and then land on the green beyond the stream.

The stream is 0.46 m wide, and the cup is 4.00 m beyond the stream’s

edge. The ramp makes an angle of 41.0° with the horizontal, and its

upper edge is 0.35 m above the green. What must the ball’s initial speed

be in order for the ball to fly over the water and land directly in the

cup?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 3FRELATIVE VELOCITY

P R O B L E MA polar bear swims 2.60 m/s south relative to the water. The bear is swim-ming against a current that moves 0.78 m/s at an angle of 40.0° north ofwest, relative to Earth. How long will it take the polar bear to reach theshore, which is 5.50 km to the south?

S O L U T I O NGiven: vbc = 2.60 m/s due south (velocity of the bear, b, with respect to

the current, c)

vce = 0.78 m/s at 40.0° north of west (velocity of the current, c,

with respect to Earth, e)

∆y = 5.50 km, south

Unknown: ∆t = ?

Diagram:

1. DEFINE

2. PLAN

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N

θce = 40.0°vce

vbcvbe

Choose the equation(s) or situation: To find vbe , write the equation so that the

subscripts of the vectors on the right begin with b and end with e.

vbe = vbc + vce

Because vectors vbc and vce are not perpendicular, their x and y components

must be calculated. Aligning the positive y axis with north and treating west as

the positive x direction for convenience, the following equations apply for the

magnitude of the components of vbe .

vx,be = vx,bc + vx,ce = vx,ce = vce (cos q ce)

vy,be = vy,bc + vy,ce = −vbc + vce (sin q ce)

From these components the magnitude and direction of vbe could be found

from the Pythagorean theorem and the tangent function, respectively. However,

only the component vy,be is needed to calculate the time required for the bear to

swim in the negative y direction.

∆t = −∆yvy,be


NAME ______________________________________ DATE _______________ CLASS ____________________


∆t = =


∆t =

= =

∆t =

Without the current, the polar bear would arrive about 500 s or 8.3 min sooner.

The 500 s delay is about one fourth (25%) of the bear’s swimming time without

the current. This proportion is equal to the ratio of the current’s northern com-

ponent to the bear’s velocity to the south.

2.62 × 103 s, or 43 min 40 s

−5.50 × 103 m−2.10 m/s

−5.50 km(−2.60 m/s + 0.50 m/s)

−5.50 km(−2.60 m/s + 0.78 m/s)(sin 40.0°)]

−∆y[−vbc + vce (sin q ce)]

−∆yvy,be

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A bird flies directly into a wind. If the bird’s forward speed relative to the

wind is 58.0 km/h and the wind’s speed in the opposite direction is

55.0 km/h, relative to Earth, how long will it take the bird to fly 1.4 km?

2. A moving walkway at an airport has a velocity of 1.50 m/s to the west. A

man rushing to catch his flight runs down the walkway with a velocity of

4.20 m/s to the west relative to the walkway. If the walkway if 8.50 ×102 m long, how much time does the man save by running on the walk-

way as opposed to running on a non-moving surface?

3. The greatest average speed for a race car in the Daytona 500 is 286 km/h,

which was achieved in 1980. Suppose a race car moving at this speed is in

second place, being 0.750 km behind a car that is moving at a speed of

252 km/h. How long will it take the second-place car to catch up to the

first-place car?

4. A mosquito can fly with a speed of 1.10 m/s with respect to the air. Sup-

pose a mosquito flies east at this speed across a swamp. The mosquito is

flying into a breeze that has a velocity of 5.0 km/h with respect to Earth

and moves 35° west of south. If the swamp is 540 m across, how long will

it take the mosquito to cross the swamp?

5. A glider descends with a velocity relative to the air of 150 km/h at an

angle of 7.0° below the horizontal. Suppose that the glider encounters an

updraft with a velocity relative to Earth of 15 km/h upward. How long

will it take the glider to reach the ground if it encounters the updraft at

166 m? How long would it take for the glider to land without the updraft?

6. A flare gun is mounted on an automobile and fired perpendicular to the

car’s motion. The car’s velocity with respect to Earth is 145 km/h to the

north. The flare’s velocity with respect to the car is 87 km/h to the west.

What are the components of the flare’s displacement with respect to

Earth 0.45 s after the flare is launched?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. An airship moving north at 55.0 km/h with respect to the air encoun-

ters a wind from 17.0° north of west. If the wind’s speed with respect to

Earth is 40.0 km/h, what is the airship’s velocity with respect to Earth?

8. How far to the north and west does the airship in problem 7 travel after

15.0 minutes?

9. A torpedo fired at an anchored target moves against a current. Suppose

the torpedo’s velocity with respect to the current is 51 km/h east, and

the current’s velocity with respect to the target is 4.0 km/h south. If the

torpedo hits the target in 14 s, how far away is the target from the point

where the torpedo is launched? How far north of the target must the

torpedo be launched in order to hit the target?

10. A sailboat travels south with a speed of 12.0 km/h with respect to the

water. Suppose the boat encounters a current that has a velocity with

respect to Earth of 4.0 km/h at 15.0° south of east. What is the sail-

boat’s resultant velocity with respect to Earth?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4ANET EXTERNAL FORCE

P R O B L E MTwo soccer players kick a ball at the same instant. One player kicks with aforce of 65 N to the north, while the other player kicks with a force of88 N to the east. In what direction does the ball travel?

S O L U T I O NGiven: F1 = 65 N north

F2 = 88 N east

Unknown: q = ?

Diagram: F1 = 65 Ν

F2 = 88 Ν

Ν

Select a coordinate system and apply it to the free-body diagram. Choose the

positive x-axis to align with east and the positive y-axis to align with north.

Find the x and y components of all vectors.

F1,x = 0 N F1,y = 65 N

F2,x = 88 N F2,y = 0 N

Find the net external force in both the x and y directions.

Fx,net = ΣFx = F1,x + F2,x = 0 N + 88 N = 88 N

Fy,net = ΣFy = F1,y + F2,y = 65 N + 0 N = 65 N

Find the direction of the net external force. Use the tangent function to find

the angle q of Fnet.

q = tan−1FFx

y

,

,n

n

e

e

t

t = tan−1685

8

N

N = 36°

q =

The direction is about three-fourths of the way to the midpoint (45°) between

north and east. This corresponds closely to the ratio of 65 N to 88 N (0.74).

1. Two tugboats pull a barge across the harbor. One boat exerts a force of

7.5 × 104 N north, while the second boat exerts a force of 9.5 × 104 N at

15.0° north of west. Precisely, in what direction does the barge move?

2. Three workers move a car by pulling on three ropes. The first worker ex-

erts a force of 6.00 × 102 N to the north, the second a force of 7.50 × 102 N

to the east, and the third 6.75 × 102 N at 30.0° south of east. In what pre-

cise direction does the car move?

36° north of east

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

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3. Four forces are acting on a hot-air balloon: F1 = 2280.0 N up, F2 = 2250.0 N

down, F3 = 85.0 N west, and F4 = 12.0 N east. What is the precise direction

of the net external force on the balloon?

4. What is the magnitude of the largest net force that can be produced by com-

bining a force of 6.0 N and a force of 8.0 N? What is the magnitude of the

smallest such force?

5. Two friends grab different sides of a videotape cartridge and pull with forces

of 3.0 N to the east and 4.0 N to the south, respectively. What force would a

third friend need to exert on the cartridge in order to balance the other two

forces? What would be that force’s precise direction?

6. A four-way tug-of-war has four ropes attached to a metal ring. The forces on

the ring are as follows: F1 = 4.00 × 103 N east, F2 = 5.00 × 103 N north,

F3 = 7.00 × 103 N west, and F4 = 9.00 × 103 N south. What is the net force on

the ring? What would be that force’s precise direction?

7. A child pulls a toy by exerting a force of 15.0 N on a string that makes an

angle of 55.0° with respect to the floor. What are the vertical and horizontal

components of the force?

8. A shopper pushes a grocery cart by exerting a force on the handle. If the

force equals 76 N at an angle of 40.0° below the horizontal, how much force

is pushing the cart in the forward direction? What is the component of force

pushing the cart against the floor?

9. Two paramedics are carrying a person on a stretcher. One paramedic exerts a

force of 350 N at 58° above the horizontal and the other paramedic exerts a

force of 410 N at 43° above the horizontal. What is the magnitude of the net

upward force exerted by the paramedics?

10. A traffic signal is supported by two cables, each of which makes an angle

of 40.0° with the vertical. If each cable can exert a maximum force of

7.50 × 102 N, what is the largest weight they can support?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4BNEWTON’S SECOND LAW

P R O B L E MTwo students reach for a jar of mustard at the same time. One studentpulls to the left with a force of 13.2 N, while the other student pulls to theright with a force of 12.9 N. If the jar has a net acceleration of 0.44 m/s2 tothe left, what is the mass of the jar?

S O L U T I O NGiven: F1 = 13.2 N to the left

F2 = 12.9 N to the right

anet = 0.44 m/s2 to the left

Unknown: m = ?

Use Newton’s second law and solve for m.

ΣF = m a = m anet

ΣF = F1 + F2 = F1 − F2 to the left

F1 − F2 = 13.2 N − 12.9 N = 0.30 N

m = a

Σ

n

F

et =

0.

0

4

.

4

30

m

N

/s2

m =

1. A house is lifted from its foundations onto a truck for relocation. The

house is pulled upward by a net force of 2850 N. This force causes the

house to move from rest to an upward speed of 15 cm/s in 5.0 s. What

is the mass of the house?

2. Suppose an empty grocery cart rolls downhill in a parking lot. The cart

undergoes a constant increase in speed of 1.0 m/s over a 5.0 s interval.

If the downhill force acting on the cart is 18.0 N and the uphill force

due to friction is 15.0 N, what is the cart’s mass?

3. A certain cable of an elevator is designed to exert a force of 4.5 × 104 N. If

the maximum acceleration that a loaded car can withstand is 3.5 m/s2

(the current fastest elevators in the world undergo an acceleration of less

than 3.2 m/s2), what is the combined mass of the car and its contents?

4. An 2.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, start-

ing from rest. Assuming the acceleration is constant, find the magnitude

and direction of the net force acting on the fish during this interval.

0.68 kg

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

5. An 8.0-kg bag of coins is being pulled upward by a rope rises 20.0 cm

in 0.50 s, starting from rest. Assuming the acceleration is constant, cal-

culate the net force on the bag. What is the upward force on the bag ex-

erted by the rope?

6. A pedestrian with a mass of 75 kg accelerates at 0.15 m/s2 to the west.

A high wind comes up, blowing toward the east. The wind is capable of

giving the pedestrian an acceleration of 2 × 10−2 m/s2. What are the

magnitude and direction of the net force acting on the pedestrian?

7. Assume that a catcher in a professional baseball game exerts a force of

−65.0 N to stop the ball. If the baseball has a mass of 0.145 kg, what is

its net acceleration as it is being caught?

8. A 214 kg boat is sinking in the ocean. The boat’s weight is partially off-

set by the 790 N buoyant force of the water. What is the net accelera-

tion of the boat?

9. The Goliath beetle, which is found in Africa, can reach a mass of

0.080 kg. Suppose a Goliath beetle is placed on a slope that makes an

angle of 37.0° with the horizontal. Find the acceleration of the beetle

along the slope, assuming the slope to be frictionless.

10. If an force of 1.40 N upward along the slope is applied to the beetle in

problem 9, what is the beetle’s acceleration?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4CCOEFFICIENTS OF FRICTION

P R O B L E MA cabinet initially at rest on a horizontal surface requires a 115 N hori-zontal force to set it in motion. If the coefficient of static friction betweenthe cabinet and the floor is 0.38, what is the normal force exerted on thecabinet? What is the mass of the cabinet?

S O L U T I O NGiven: Fs,max = 115 N

ms = 0.38

g = 9.81 m/s2

Unknown: Fn = ? m = ?

Use the equation for the coefficient of static friction to find Fn.

ms = Fs

F, m

n

ax

Fn = Fs,

mm

s

ax = 1

0

1

.

5

38

N = 3.0 × 102 N

Fn =

Use the definition for the normal force to find m.

Fn = mg, for a horizontal surface

m = F

gn =

3

9

.0

.8

×1

1

m

0

/

2

s2N

m =

1. A ship launched from a dry-dock slides into the water at a constant ve-

locity. Suppose the force of gravity that pulls the ship downward along

the dry-dock is 4.26 × 107 N. If the coefficient of kinetic friction be-

tween the ship’s hull and the dry-dock is 0.25, what is the magnitude of

the normal force that the dry-dock exerts on the ship’s hull?

2. If the incline of the dry-dock in problem 1 is 10.0°, what is the ship’s

mass?

3. A frictional force of 2400 N keeps a crate of machine parts from sliding

down a ramp. If the coefficient of static friction between the box and

the ramp is 0.20 and the incline of the ramp is 30.0°, what is the nor-

mal force of the ramp on the box? What is the mass of the crate of ma-

chine parts?

31 kg

3.0 × 102 N upward

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

4. A passenger with a mass of 60.0 kg is standing in a subway car that is

accelerating at 3.70 m/s2. If the coefficient of static friction between the

passenger’s shoes and the car floor is 0.455, will the passenger be able

to stand without sliding?

5. A 90.0 kg skier glides down a slope with an incline of 17.0°. What fric-

tional force is needed for the skier to move at a constant velocity?

6. A dogsled with a mass of 47 kg is loaded with 33 kg of supplies. If the

coefficient of kinetic friction between the sled’s runners and the snow

is 0.075, what is the magnitude of the frictional force on the sled as it

moves across flat ground? What is the magnitude of the frictional force

on the sled as it moves up a hill with a 15° incline?

7. A car with a mass of 1.8 × 103 kg is parked on a hill in San Francisco. Sup-

pose the hill makes a 15.0° incline with the horizontal. If the frictional

force required to keep the car from sliding down the hill is 1.25 × 104 N,

what is the coefficient of static friction between the pavement and the

car’s tires?

8. A metal disk with a mass of 15.0 g slides along a metal sheet. Both the

disk and sheet have been coated with a substance that reduces fric-

tional forces. If the sheet needs to be tilted only 2.3° for the disk to slide

down the sheet with a constant velocity, what is the coefficient of ki-

netic friction between the disk and sheet?

9. In 1994, a commercial automobile accelerated from rest to 88.0 km/h

in 3.07 s. Cars accelerate because of traction, which in turn depends on

the force of static friction between the rubber of their tires and the

road. If the force of acceleration is entirely provided by static friction

between the tires and pavement (an overly simplified assumption), cal-

culate the coefficient of static friction between the tires and the road.

10. A snowboarder slides down a 5.0° slope at a constant speed. What is

the coefficient of kinetic friction between the snow and the board?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 4DFINAL VELOCITY AFTER ANY DISPLACEMENT

P R O B L E MA bicyclist riding in the rain suddenly applies the brakes and slides to astop. If the acceleration is −9.5 m/s2, what is the coefficient of kinetic fric-tion between the bicycle’s rubber tires and the wet concrete?

S O L U T I O N

Given: anet = −9.5 m/s2

g = 9.81 m/s2

Unknown: mk = ?

Choose the equation(s) or situation: Use Newton’s second law to describe the

forces acting on the bicycle.

Fnet = m anet = −Fk

Use the definition of frictional force to express Fk in terms of the coefficient of

friction.

Fk = mk Fn = mk (mg)


m anet = −mkmg

mk = − an

get


mk = −(

9

−.8

9

1

.5

m

m

/s

/2s2)

mk =

The coefficient of static friction for rubber and most surfaces is high. This is indi-

cated by the value for rubber and wet concrete. Even under these conditions, ms is

nearly 1.

1. Blocks of ice are slid down a metal chute with an incline of 12.0° above

the horizontal. The blocks undergo a constant acceleration of 1.22 m/s2.

What is the coefficient of kinetic friction between the ice and the chute?

2. A force of 1760 N is required to start moving a bundle of wooden

planks up a ramp. If the ramp’s incline is 17° and the mass of the

planks is 266 kg, what is the coefficient of static friction between the

planks and the ramp?

0.97

1. DEFINE

2. PLAN

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

3. A bundle of bricks is pulled up a ramp to a construction site. The bun-

dle has a mass of 5.1 × 102 kg, and the incline of the ramp is 14°. If the

minimum force needed to move the bricks up the ramp is 4.1 × 103 N,

what is the coefficient of static friction between the bricks and the

ramp?

4. A force of 5.00 N to the left causes a 1.35 kg book to have a net acceler-

ation of 0.76 m/s2 to the left. What is the frictional force acting on the

book?

5. A jar is slid horizontally across a smooth table. If the coefficient of ki-

netic friction between the jar and the table is 0.20, what is the magni-

tude of the jar’s acceleration?

6. A skier is pulled by an applied force of 2.50 × 102 N up a slope with an

incline of 18.0°. If the combined mass of the skier and skis is 65.0 kg

and the net acceleration uphill is 0.44 m/s2, what is the frictional force

between the skis and the snow?

7. If the skier in problem 6 skis down the same hill, what will the skier’s

acceleration be?

8. A crate is pushed across a level floor by a force of 3.00 × 102 N exerted

at an angle of 20.0° below the horizontal. The coefficient of kinetic

friction between the crate and floor is 0.250. If the crate’s velocity is

constant, what is the magnitude of the normal force exerted on the

crate by the floor? What is the mass of the crate?

9. A horse must exert a force of 590 N just to keep a sleigh from sliding

down a snowcovered hill. The component of the sleigh’s weight down

the slope of the hill is 950 N, and the coefficient of static friction be-

tween the sleigh’s runners and the snow is 0.095. What is the normal

force exerted by the ground on the sleigh? What is the sleigh’s mass if

the slope of the hill is 14.0°?

10. A freight elevator accelerates upward at 1.20 m/s2. A crate is lifted in-

side the elevator. In order to move the crate along the floor of the ele-

vator, a worker must exert a force of 1.50 × 103 N at an angle of 10.0°

below the horizontal on the upper corner of the crate. If the coefficient

of static friction is 0.650, what is the normal force that the elevator

floor exerts on the crate? What is the crate’s mass?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5AWORK

P R O B L E MA girl playing tug-of-war with her dog pulls the dog a distance of 8.0 m byexerting a force at an angle of 18° with the horizontal. If the amount ofwork the girl does in pulling the dog is 190 J, what is the magnitude of theforce?

S O L U T I O NGiven: W = 190 J

d = 8.0 m

q = 18°

Unknown: F = ?

Use the equation for work done by a constant force, and rearrange it to solve for F.

W = Fd (cos q)

F = d(c

W

os q) =

(8.0 m

1

)

9

(

0

co

J

s 18°)

F = 25 N

1. A roller coaster must do work raising its cars to the highest point on the

ride. From there, the cars coast at varying speed until they return to the

starting point. Suppose a loaded roller coaster car must be pulled 3.00 ×102 m from the ride’s starting point to the top of the first rise. If 2.13 ×106 J of work must be done on the car during this stage of the ride, how

large is the force exerted on the car by the raising mechanism?

2. A building under construction requires building materials to be raised

to the upper floors by cranes or elevators. An amount of cement is

lifted 76.2 m by a crane, which exerts a force on the cement that is

slightly larger than the weight of the cement. If the net work done on

the cement is 1.31 × 103 J, what is the magnitude of the net force

exerted on the cement?

3. Two workers load identical refrigerators into identical trucks by differ-

ent methods. One worker has the refrigerator lifted upward onto the

back of the truck, which is 1.5 m above the ground. The other worker

uses a ramp to slide the refrigerator onto the back of the truck. The

ramp is 5.0 m long, and raises the refrigerator 1.5 m above the ground.

The amount of work done by both workers is the same: 1800 J. What

are the magnitudes of the forces each worker must exert to load the

refrigerators?

ADDITIONAL PRACTICE

Holt Physics Problems BankCh. 5–2

NAME ______________________________________ DATE _______________ CLASS ____________________

4. A sunken treasure has a mass of 2140 kg, most of which is due to silver

and gold coins. In order to make it easier to raise the treasure, a diver

descends 17 m to where the treasure is located and attaches balloon-

like bladders to each corner of the treasure chest. The diver then in-

flates these bladders, so they provide buoyancy to the chest. The chest

is still too heavy to float upward, but its weight has been largely

counteracted by the inflated bladders, so that now it can be easily lifted

by 4.27 × 103 J of work. What is the magnitude of the net force that is

exerted on the treasure in order to raise it to the water’s surface?

5. A wrench slides off a tilted shelf, although if a force of 1.6 N is applied

opposite the wrench’s motion the wrench will slide down the shelf with

a constant velocity. If the shelf is 1.2 m long, what is the work done by

the applied force on the wrench?

6. A girl pulls a wagon along a level path for a distance of 15.0 m. The

handle of the wagon makes an angle of 20.0° with the horizontal, and

the girl exerts a force of 35.0 N on the handle. Friction provides a force

of 24.0 N. Find the net work that is done on the wagon.

7. In 1947, a deceleration sled was built to test the effects of extreme

forces on humans and equipment. In this sled, a test pilot undergoes a

sudden negative acceleration of about 50.0 times free-fall acceleration

(g). In 0.181 s, through a distance of 8.05 m, the pilot’s speed decreases

from 88.9 m/s to 0 m/s. If the pilot’s mass is 70.0 kg, how much work is

done against the pilot’s body during the deceleration?

8. A car has run out of gas. Fortunately, there is a gas station nearby. You

must exert a force of 715 N on the car in order to move it. By the time

you reach the station, you have done 2.72 × 104 J of work. How far have

you pushed the car?

9. A catcher picks up a baseball from the ground. If the net upward force

on the ball is 7.25 × 10–2 N and the net work done lifting the ball is

4.35 × 10–2 J, how far is the ball lifted?

10. At the 1996 Summer Olympics in Atlanta, Georgia, a mass of 260 kg

was lifted for the first time ever in a clean-and-jerk lift. The lift, per-

formed by Russian weightlifter Andrei Chemerkin, earned him the un-

official title as “the world’s strongest man.” If Chemerkin did 6210 J of

work in exerting a force of 2590 N, how high did he lift the mass?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5BKINETIC ENERGY

P R O B L E M

A 2.00 g projectile has a speed of 3.00 102 m/s. What is its kinetic energy?

S O L U T I O NGiven: m = 2.00 g

v = 3.00 × 102 m/s

Unknown: KE = ?

Use the kinetic energy equation to solve for KE.

KE = 12

mv 2 = 12

(2.00 × 10–3 kg)(3.00 × 102 m/s)2

KE = 90.0 J

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ADDITIONAL PRACTICE

1. The Queen Mary was one of the largest ocean liners of the mid-twentieth

century, crossing the Atlantic Ocean 1000 times. The ship is now a

tourist attraction at Long Beach, California. Given that the mass of the

Queen Mary is 7.5 × 107 kg and her maximum cruising speed was

57 km/h, what would be the kinetic energy of the ship at maximum speed?

2. The fastest speed achieved on Earth for any object, with the exception

of sub-atomic particles in particle accelerators, is 15.8 km/s. A device at

Sandia Laboratories in Albuquerque, New Mexico, uses highly com-

pressed air to accelerate a small metal disk to supersonic speeds. Sup-

pose the disk has a mass of 0.20 g. What is the maximum kinetic

energy of the disk?

3. Although ungraceful on land, walruses are fine swimmers. They nor-

mally swim at 7 km/h, and for short periods of time are capable of

reaching speeds of nearly 35 km/h. If a walrus swimming at a speed of

35.0 km/h has a mass of 9.00 × 102 kg, what is its kinetic energy?

4. The Shinkansen, Japan’s high-speed trains, have been in service since

1964. Since that time, several train designs have been developed. Most

of these trains travel between 240 km/h and 285 km/h. The exceptions

are the “0” series, which began service in 1964, and the “500” series,

which began service in 1997. Series 0 trains travel up to 220.0 km/h

and have a total mass of about 8.84 × 105 kg. The lighter, streamlined

series 500 trains travel up to 320.0 km/h, and have an estimated total

mass of about 4.80 × 105 kg. What are the maximum kinetic energies

that can be achieved by each of these trains?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. The most massive of the Shinkansen are the series 200 trains, yet they

are among the fastest. Series 200 trains can reach speeds of 275 km/h. If

a 16-car series 200 train has a maximum kinetic energy of 2.78 × 109 J,

what is its mass?

6. The largest airplane built that has flown more than once is the Ukrainian-

built Antonov-225 Mriya. With a length of 85 m and a wingspan of 88 m,

the Mriya (Dream) was designed to carry the space shuttle of the Soviet

Union’s space program. Unloaded, the top speed of Mriya is 850 km/h, at

which point its kinetic energy is 9.76 × 109 J. What is its mass?

7. Though slow on land, the leatherback turtle holds the record for the

fastest water speed of any reptile: 9.78 m/s. It is also among the largest

of reptiles. Suppose the largest leatherback yet discovered were to swim

at the top leatherback speed. If its kinetic energy was 6.08 × 104 J, what

was its mass?

8. At the time a 55.0 kg skydiver jumps from a plane, her speed steadily

increases until air resistance provides a force that balances that due to

free-fall. How fast is the skydiver falling if her kinetic energy at the

moment is 7.81 × 104 J?

9. The kinetic energy of a golf ball is measured to be 1433 J. If the golf

ball has a mass of about 47.0 g, what is the ball’s speed?

10. A running student has half the kinetic energy that his younger brother

has. The student speeds up by 1.3 m/s, at which point he has the same

kinetic energy as his brother. If the student’s mass is twice as large as

his brother’s mass, what were the original speeds of both the student

and his brother? (See Appendix A of the text for hints on solving

quadratic equations.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5CWORK-KINETIC ENERGY THEOREM

P R O B L E MA forward force of 11.0 N is applied to a loaded cart over a distance of15.0 m. If the cart, which is initially at rest, has a final speed of 1.98 m/s,what is the combined mass of the cart and its contents?

S O L U T I O NGiven: Fapplied = 11.0 N

d = 15.0 m

q = 0°vi = 0 m/s

vf = 1.98 m/s

Unknown: m = ?

Diagram:

Choose the equation (s) or situation: The net work done on the cart can be ex-

pressed by using the definition of work in terms of net force. Because the force is

in the same direction as the cart’s displacement (q = 0°), the net work is simply

the product of the net force and the distance the cart is pushed. The net work can

also be explained in terms of changing kinetic energy by using the work-kinetic

energy theorem.

Wnet = Fnetd(cos q) = Fnetd

Wnet = ∆KE = KEf − KEi = 12

mvf2 − 1

2mvi

2

The net force on the cart is equal to the applied force. By inserting Fapplied into

the equation for Wnet and using the work-kinetic energy equivalence, the follow-

ing equation is obtained.

Fappliedd = 12

m(vf2 − vi

2)

Rearrange the equation(s) to isolate the unknown (s):

m = 2

v

F

f

a2p

−pli

ve

i

d2d


m =

m =

m = 84.2 kg

(2)(11.0 N)(15.0 m)

(1.98 m/s)2

(2)(11.0 N)(15.0 m)(1.98 m/s)2 − (0 m/s)2C

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1. DEFINE

2. PLAN

3. CALCULATE

Fapplied

v

d


NAME ______________________________________ DATE _______________ CLASS ____________________

1. A hockey puck with an initial speed of 8.0 m/s coasts 45 m to a stop

across the ice. If the force of friction on the puck has a magnitude of

0.12 N, what is the puck’s mass?

2. A meteoroid is a small fragment of rock that orbits a planet or the sun.

When a meteoroid enters a planet’s atmosphere, it most likely will burn

up entirely, glowing brilliantly as it does so. It is then referred to as a

meteor. Consider a meteoroid that has an initial speed of 15.00 km/s

when it enters the thin upper region of Earth’s atmosphere. Suppose

this meteoroid encounters a force of resistance with a magnitude of

9.00 × 10−2 N, so that after it travels 500.0 km parallel to Earth’s surface

the meteoroid’s speed is 14.97 km/s. Assume that the meteoroid does

not lose any mass as its temperature increases, and that the change in

the gravitational potential energy is negligible. What is the mass of the

meteoroid?

3. A car moving at a speed of 48.0 km/h accelerates 100.0 m up a steep

hill, so that at the top of the hill its speed is 59.0 km/h. If the car’s mass

is 1100 kg, what is the magnitude of the net force acting on it?

4. A 450 kg compressor slides down a loading ramp that is 7.0 m long. Ini-

tially at rest, the compressor’s speed at the bottom of the ramp is 1.1 m/s.

What is the magnitude of the net force acting on the compressor?

5. The force that stops a fighter jet as it lands on the flight deck of an

aircraft carrier is provided by a series of arresting cables. These cables

are attached to large springs that stretch enough to keep the plane from

slowing down too suddenly. Suppose a Hornet jet traveling with an

initial speed of 2.40 × 102 km/h lands on the flight deck, where it is

brought to rest by a net acceleration of magnitude 30.8 m/s2. If the jet’s

mass is 1.30 × 104 kg, how far does the jet travel during its deceleration?

6. A 50.0 kg parachutist falls at a speed of 47.00 m/s when the parachute

opens. The parachutist’s speed upon landing is 5.00 m/s. How much

work is done by the air to reduce the parachutist’s speed?

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ADDITIONAL PRACTICE

Note that the form of the equation is equivalent to Newton’s second law, with ac-

celeration given by the kinematic equation.

a = vf

2

2

−d

vi2

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

7. The giant sequoia redwood trees of the Sierra Nevada in California are

said never to die from old age. Instead, an old tree dies when its shallow

roots become loosened and the tree falls over. Removing a dead mature

redwood from a forest is no easy feat, as the tree can have a mass of

nearly 2.0 × 106 kg. Suppose a redwood with this mass is lifted 7.5 m

with a net upward acceleration of 7.5 × 10–2 m/s2. If the tree’s initial ki-

netic energy is zero, what is the final kinetic energy?

8. An applied force of 92 N is exerted horizontally on an 18 kg box of

books. The coefficient of kinetic friction between the floor and the box

is 0.35. If the box is initially at rest with zero kinetic energy, what is the

final kinetic energy after it has been moved 7.6 m across the floor?

9. A 2.00 × 102 kg iceboat is propelled across the horizontal surface of a

frozen lake by the wind. The wind exerts a constant force of 4.00 × 102 N while the boat moves 0.90 km. Assume that frictional forces are

negligible and that the boat starts from rest. What is the boat’s final

speed?

10. A certain firework is made of a small cardboard tube with a mass of

about 20.0 g. When lit, the tube slides 2.5 m across a smooth surface. If

the forward force on the tube is 7.3 × 10−2 N and the coefficient of fric-

tion between the tube and the ground is 0.20, what is the tube’s final

speed? Assume the tube is initially at rest.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5DPOTENTIAL ENERGY

P R O B L E MA 70.0 kg stuntman jumps from a bridge that is 50.0 m above the water.Fortunately, a bungee cord with an unstretched length of 15.0 m is at-tached to the stuntman, so that he breaks his fall 12.0 m above the water’ssurface. If the total potential energy associated with the stuntman andcord is 3.43 104 J, what is the force constant of the cord?

S O L U T I O NGiven: m = 70.0 kg h = 12.0 m

x = 50.0 m − 12.0 m − 15.0 m = 23.0 m

PEg = 0 J at river level

PEtot = 3.43 × 104 J

g = 9.81 m/s2

Unknown: k = ?

Diagram:

Choose the equation(s) or situation: The zero level for gravitational potential

energy is chosen to be at the water’s surface. The total potential energy is the sum

of the gravitational and elastic potential energies.

PEtot = PEg + PEelastic = mgh + 12

kx2


12

kx2 = PEtot − mgh

k = 2(PEto

xt2− mgh)]


k =

k = =

k =

The situation in this problem is basically the same as that in Sample Problem 5D

in the textbook. In order for the stuntman to have a greater height above the water,

the bungee cord must store more elastic potential energy with less stretching.

This occurs when the spring constant is larger (98.7 N/m > 71.8 N/m).

98.7 N/m

(2)(2.61 × 104 J)

(23.0 m)2(2)(3.43 × 104 J − 8.24 × 103 J)

(23.0 m)2

(2)[3.43 × 104 J − (70.0 kg)(9.81 m/s2)(12.0 m)]

(23.0 m)2

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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50.0 m

12.0 m

Relaxed length = 15.0 m

Stretched length = 38.0 m


NAME ______________________________________ DATE _______________ CLASS ____________________

1. A highway guardrail is designed so that it can be distorted as much as

5.00 cm when struck by an automobile. What is the minimum force

constant of the guardrail if it is to withstand the impact of a car with

1.09 × 104 J of energy?

2. An arresting cable helps to slow jet planes as they land on an aircraft

carrier. This is accomplished by two springs, each of which is attached

to one end of the cable. Suppose the elastic potential energy stored in

the springs while a jet is landing is 5.78 × 107 J. If each spring is

stretched 102 m, what is the force constant of each spring?

3. A produce scale at a supermarket uses a stretched spring to indicate the

weight of fruits and vegetables. If five oranges with a total mass of

0.76 kg are placed in the scale, the spring will be stretched 2.3 cm.

What is the force constant of the spring?

4. A 5.0 kg stone is slid up a frictionless ramp that has an incline of 25.0°.

How long is the ramp if the gravitational potential energy associated

with the stone is 2.4 × 102 J?

5. A pogo stick contains a spring with a force constant of 1.5 × 104 N/m.

Suppose the elastic potential energy stored in the spring as the pogo

stick is pushed downward is 120 J. How far is the spring compressed?

6. A 1750 kg weather satellite moves in a circular orbit with a gravita-

tional potential energy of 1.69 × 1010 J. At the satellite’s altitude above

Earth’s surface, the free-fall acceleration is only 6.44 m/s2. How high

above Earth’s surface is the satellite?

7. An automobile to be transported by ship is raised 7.0 m above the

dock. If its gravitational potential energy is 6.6 × 104 J, what is the

automobile’s mass?

8. One of the largest planes ever to fly, and the largest to fly frequently, is

the Ukrainian-built Antonov An-124 Ruslan. Its wingspan is 73.2 m

and its length is 69.2 m. The gravitational potential energy associated

with the plane at an altitude of 1.45 km is 3.36 × 109 J. What is the

airplane’s mass?

9. The force constant of the spring in a child’s toy car is 550 N/m. How

much elastic potential energy is stored in the spring if the spring is

compressed a distance of 1.2 cm?

10. With an elevation of 5334 m above sea level, the village of Aucanquilca,

Chile, is the highest inhabited town in the world. What would be the

gravitational potential energy associated with a 64.0 kg person in

Aucanquilca? Assume that the free-fall acceleration at Aucanquilca is

equal to that at sea level.

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5ECONSERVATION OF MECHANICAL ENERGY

P R O B L E MA raindrop with a mass of 0.500 g falls to Earth from a height of 1.50 km.The raindrop reaches Earth’s surface with a speed of 6.67 m/s. How muchof the raindrop’s mechanical energy is lost because of air resistance? As-sume free-fall acceleration to be the same at 1.5 km above Earth’s surfaceas it is at Earth’s surface.

S O L U T I O NGiven: m = 0.500 g

h = 1.50 km

vf = 6.67 m/s

vi = 0 m/s

g = 9.81 m/s2

Unknown: ∆ME = ?

Choose the equation(s) or situation: Use the conservation of mechanical energy

to account for energy dissipated through air resistance.

MEi + ∆ME = MEf

The zero level for gravitational potential energy is the ground. Because the rain-

drop starts at altitude h, the initial potential energy is its maximum value. Be-

cause the raindrops’ initial velocity is zero, the initial kinetic energy is zero.

MEi = PEi + KEi = PEi = mgh

When the raindrop reaches the ground, the gravitational potential energy is zero.

MEf = PEf + KEf = KEf = 12

mvf2

Substituting the last two equations into the first yields the following equation:

mgh + ∆ME = 12

mvf2


∆ME = 12

mvf2 − mgh


∆ME = 12

(0.500 × 10−3 kg)(6.67 m/s)2

− (0.500 × 10−3 kg)(9.81 m/s2) (1.50 × 103 m)

∆ME = 1.11 × 10−2 J − 7.36 J

∆ME =

Most of the gravitational potential energy is given up through the interaction of

the raindrop with the surrounding air. Only 0.15 percent of the initial energy

remains as kinetic energy.

− 7.35 J

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALULATE

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NAME ______________________________________ DATE _______________ CLASS ____________________

1. What would be the kinetic energy of a 0.500 g raindrop if it fell

0.250 km without any resistance provided by air?

2. Angel’s Flight, which has been called “the world’s shortest railway,” is a

hill-climbing cable car, or funicular, that is located on Bunker Hill in

downtown Los Angeles, California. The funicular consists of two small

railway cars that are connected to each other by a cable. The cable is, in

turn, wrapped around a large pulley that is attached to a motor. As one

car rises up the hill, the other car descends to the street below.

a. The tracks of Angel’s Flight extend 96.0 m along the side of the

hill at an angle of 18.4° with respect to the horizontal. If each car

has a passenger with a mass of 70.0 kg, what is the total mechani-

cal energy associated with the two passengers when the cars are

about to leave the boarding platforms?

b. What is the total mechanical energy associated with the two

passengers when the cars arrive at their destinations?

c. Except for a brief initial and final acceleration, the cars move in

opposite directions at a constant speed of about 1.0 m/s. Suppose

the ascending car is somewhere between street level and the mid-

level. If the descending car is 20.0 m above street level, what is the

gravitational potential energy associated with the passenger in

the ascending car?

3. A toy rocket is at a height of 75.0 m and is moving upward with a

speed of 1.2 m/s when it ejects a payload with a mass of 20.0 g. The

payload has an initial upward speed relative to the rocket of 3.5 m/s.

What is the height reached by the payload when its upward velocity is

zero?

4. A 25.0 kg falling trunk strikes the ground with a speed of 12.5 m/s. As-

suming that there is no loss of energy due to air resistance, what is the

height from which the trunk falls?

5. If you were to neglect air resistance, a projectile fired straight up into

the air would land again with the speed with which it was fired. (This is

why it is dangerous to shoot a bullet directly upward.) However, some

of the energy is lost because of air resistance. Suppose a 50.0 g projec-

tile is fired upward with an initial speed of 3.00 × 102 m/s. If it lands

with a speed of 89.0 m/s, how much mechanical energy is given up

because of air resistance?

6. The air resistance that slows the projectile in problem 5 affects it both

as it rises and falls. How high would the projectile rise if there were no

air resistance? How high does it rise because of air resistance? (HINT:

Use the work–kinetic energy theorem to describe the forces on the pro-

jectile as it goes up and again as it comes down. Then use these equa-

tions to solve for the projectile’s maximum height. Use the data from

problem 5 for the final calculation.)

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

7. A 50.0 kg circus performer jumps from a platform into a safety net

below. The net, which has a force constant of 3.4 × 104 N/m, is

stretched by 0.65 m. If the unstretched net is positioned 1.00 m above

the ground, what is the height of the platform? Ignore the effects of air

resistance.

8. A miniature golf course has a hole in which the fairway is 3.0 m above

the green. If you hit the ball into the middle hole in a row of three, the

ball will be directed to the green by a connecting pipe. Suppose the ball

falls down most of the length of the pipe and slides the rest of the way

without any loss of energy to friction. What is the ball’s speed as it

emerges from the pipe onto the green?

9. A 100.0 g arrow is pulled back 30.0 cm against a bowstring. If the

spring constant of the bowstring is 1250 N/m, at what speed will the

arrow leave the bow?

10. Because the wind speeds in Dayton, Ohio, are lower than at Kitty

Hawk, North Carolina, the Wright brothers improved their launch

mechanism for their 1904 flyer. Weights with a total mass of 546 kg

were dropped 5.64 m from a derrick. These weights pulled a rope that

was attached to the flyer, causing it to accelerate along a track for

5.64 m. This acceleration gave the flyer enough speed in addition to

that provided by its propellers to take off. If the flyer, which had a mass

of 273 kg, was initially at rest, what would its speed due to the falling

masses be when the masses reached the ground? (Note that your an-

swer is for a frictionless device; in reality, the speed of the flyer due to

the weights was somewhat lower.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 5FPOWER

P R O B L E MThe engines of the Queen Mary could deliver 174 MW to propel the mas-sive ship. How long does it take for the engines to do 7.31 1010 J of workon the ship?

S O L U T I O NGiven: P = 174 MW

W = 7.31 × 1010 J

Unknown: ∆t = ?

Use the equation for power and rearrange it to solve for time.

P = ∆W

t

∆t = W

P =

1

7

7

.3

4

1

××

1

1

0

06

10

W

J =

∆t = (4.20 × 102 s)(1 min/60 s) = 7.00 min

4.20 × 102 s

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1. The engine that moves the cables for the San Francisco cable cars de-

livers 380.3 kW of power for each line. How long does it take for 4.5 ×106 J of work (about the amount of work needed to raise a partially

loaded cable car up Nob Hill) to be done by this engine?

2. If the stairs of the Sears Tower in Chicago, Illinois, can be climbed by

an athlete with a power output of 331 W, how long does it take the ath-

lete to climb the building’s 442 m height? Assume the athlete has a

mass of 55 kg and that all of the power goes toward doing work against

gravity.

3. A runner exerts a force of 334 N against the ground while using 2100 W

of power. How long does it take the runner to run a distance of 50.0 m?

4. A ship’s diesel engine has a power output of 13.0 MW. How much

work is done by this engine in 15.0 min?

5. One horsepower (1 hp) is the unit of power based on the work that a

horse can do in one second. This is defined, in English units, as a force

of 550 lb that can move an object 1 foot in 1 s. In SI, 1 hp equals

745.7 W, or 745.7 J of work delivered in 1 s. Suppose you have a horse

that delivers 745.7 W of power, but that it does the work in only 0.55 s.

How much work has this horse done?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

6. The 300-series Shinkansen train of Japan has aluminum cars, so that it

can reach high speeds more easily. Ten of the sixteen cars of a 300-

series train have their own 300.0 kW motors, one for each of their four

axles. What is the work done by one car’s four motors during 25 s?

7. When it is completed in 2002, the International Financial Center in

Taipei, Taiwan, will be the tallest building in the world. The Interna-

tional Financial Center will also have the fastest elevators in the world.

Two of the 63 elevators will travel from the ground floor to the eighty-

ninth floor in just 39 s. Suppose the power output of each elevator

motor is 158 kW. How much work will these motors do in lifting the

elevator to the eighty-ninth floor?

8. The space shuttle, which was first launched on April 12, 1981, is the

world’s first reusable space vehicle. The shuttle is placed in orbit by

three engines that do 1.4 × 1013 J of work in 8.5 min. What is the

power output of these engines?

9. Borax was mined in Death Valley, California, during the nineteenth

century. It was transported from the valley by massive wagons, each

pulled by a team of mules. Suppose the team does 2.82 × 107 J of work

on the wagon for 30.0 min? How much power is delivered, on the aver-

age, by the team? Express your answer in both watts and horsepower

(1 hp = 745.7 W).

10. James Watt did not invent the steam engine, but by adding a condenser

to an existing engine he discovered how to make steam engines more

efficient and practical. His own engine of 1778 was able to do 3.0 ×106 J of work in 5.0 min. How much power was delivered by this

engine?

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Problem 6A Ch. 6-1

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6AMOMENTUM

P R O B L E MAn ostrich with a mass of 146 kg is running with a momentum of2480 kg•m/s to the right. What is the velocity of the ostrich?

S O L U T I O NGiven: m = 146 kg

p = 2480 kg•m/s to the right

Unknown: v = ?

Use the equation for momentum to solve for v.

p = mv

v = m

p

v = = 17.0 m/s to the right2480 kg•m/s

146 kg

1. If a blue whale has a mass of 1.46 × 105 kg and momentum of 9.73 ×105 kg•m/s to the south, what is its velocity?

2. The highest land speed for a rail-guided vehicle was set in 1982 by a

rocket sled at Holloman Air Force Base in southern New Mexico. The

sled was unmanned, but if it had a payload with a mass of 25 kg, the

magnitude of the payload’s momentum would have been 6.8 ×104 kg•m/s. What was the speed, in m/s and km/h, of the payload and

sled?

3. Thoroughbred horses are among the fastest horses in the world and are

used in famous racing events such as the Kentucky Derby. The mass ofa thoroughbred is about 5.00 × 102 kg. If a horse with this mass is gal-

loping with a momentum of 8.22 × 103 kg•m/s to the west, what is its

velocity?

4. The World Solar Challenge in 1987 was the first car race in which all

the vehicles were solar powered. The winner was the GM Sunraycer,

which had a mass of 177.4 kg, not counting the driver’s mass. Assume

that the driver had a mass of 61.5 kg, so that the total momentum of

the car and driver had a magnitude of 4.416 × 103 kg•m/s. What was

the car’s speed in m/s and km/h?

5. The current holder of the men’s world record for running 200 m is

Michael Johnson, who in 1996 ran 200.0 m in 19.32 s. Johnson’s mass

at the time of his record-breaking run was about 77 kg. What was the

magnitude of his momentum at his average speed?

ADDITIONAL PRACTICE

Holt Physics Problem BankCh. 6-2

NAME ______________________________________ DATE _______________ CLASS ____________________

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6. Although it cannot sustain its top speed for more than 8.65 s, the chee-

tah can run a distance of 274 m during that time. If a cheetah with a

mass of 50.0 kg is moving north at its top speed, what is its momentum?

7. The high-speed 300-series Shinkansen trains of Japan consist of 16 alu-

minum cars with a combined mass of 7.10 × 105 kg. The reduction in

mass from the 100-series trains enables the 300-series trains to reach a

top speed of 270 km/h. What is the magnitude of a 300-series train’s

momentum at its top speed?

8. The largest species of hummingbird is the Patagonia gigas, or the Giant

Hummingbird of the Andes. This bird has a length of 21 cm and can

fly with a speed of up to 50.0 km/h. Suppose one of these humming-

birds flies at this top speed. If the magnitude of its momentum is

0.278 kg•m/s, what is the hummingbird’s mass?

9. A hovercraft, or air-cushion vehicle, glides on a cushion of air, allow-

ing it to travel with equal ease on land or water. The first commercial

hovercraft to cross the English Channel, the V. A-3, had an average

velocity of 96 km/h to the southeast. Its average momentum was

4.8 × 104 kg•m/s to the southeast. What was the mass of the V. A-3?

10. The brightest, hottest, and most massive stars (over 10 times as massive

as the sun) are the brilliant blue stars designated as spectral class O. If a

class O star moves with a speed of 255 km/s and has a momentum of

8.62 × 1036 kg•m/s, what is the star’s mass?

Problem 6B Ch. 6-3

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6BFORCE AND MOMENTUM

P R O B L E MA student with a mass of 55 kg rides a bicycle with a mass of 11 kg. A netforce of 125 N to the east accelerates the bicycle and student during a timeinterval of 3.0 s. What is the final velocity of the bicycle and student? As-sume the student and bicycle are initially at rest.

S O L U T I O NGiven: ms = 55 kg

mb = 11 kg

F = 125 N to the east

∆t = 3.0 s

vi = 0 m/s

Unknown: vf = ?

Use the impulse-momentum theorem to solve for vf.

F∆t = ∆p = mvf − mvi

vf = F∆t

m

+ mvi

m = ms + mb = 55 kg + 11 kg = 66 kg

vf =

vf = (125

6

N

6

)

k

(

g

3.0 s)

vf = 5.7 m/s to the east

(125 N)(3.0 s) − (66 kg)(0 m/s)

66 kg

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ADDITIONAL PRACTICE

1. A net force of 10.0 N to the right pushes a 3.0 kg book across a table. If

the force acts on the book for 5.0 s, what is the book’s final velocity?

Assume the book to be initially at rest.

2. A 60.0 g egg dropped from a window is caught by a student. If the stu-

dent exerts a net force of −1.5 N over a period of 0.25 s to bring the egg

to a stop, what is the egg’s initial speed?

3. A child riding a sled is pulled down a snowy hill by a force of 75 N. If

the child and sled have a combined mass of 55 kg, what is their speed

after 7.5 s? Assume the child and sled are initially at rest.

4. A billiard ball with a mass of 0.195 kg and a velocity of 0.850 m/s to the

right is deflected by the cushioned edge of the billiard table. The cush-

ion exerts a force of 3.50 N to the left for 0.0750 s. What is the ball’s

final velocity?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A 5.00 g projectile has a velocity of 255 m/s to the right. What force is

required to stop this projectile in 1.45 s?

6. The Pacific walrus has an average mass of 1.1 × 103 kg and can swim

with a speed of about 9.7 m/s. Suppose a walrus starting from rest

takes 19 s to reach a velocity of 9.7 m/s to the east. What net force acts

upon the walrus?

7. With a mass of 3.000 × 103 kg, the Russian-made Zil-41047 is the most

massive automobile to have been manufactured on a regular basis.

Suppose one of these cars accelerates from rest to a velocity of 8.9 m/s

to the right in 5.5 s. Calculate the net force acting on the Zil-41047.

8. How much time would it take for a 0.17 kg ice hockey puck to decrease

its speed by 9.0 m/s if the coefficient of kinetic friction between the ice

and the puck is 0.050?

9. A girl pulls a 12.0 kg wagon along by exerting a force of 15.0 N on the

wagon’s handle, which makes an angle of 20.0° with the horizontal.

Friction provides a force of 11.0 N in the opposite direction. How long

does it take for the wagon, which is initially at rest, to reach a speed of

4.50 m/s?

10. The compressed-air device at Sandia Laboratories in Albuquerque,

New Mexico, accelerates small metal disks to a speed of 15.8 km/s. Sup-

pose the compressed air exerts a force of 12.0 N on a 0.20 g disk that is

initially at rest. How long will it take the disk to reach its maximum

speed?

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Problem 6C Ch. 6-5

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6CSTOPPING DISTANCE

P R O B L E M

A high-speed train with a total mass of 9.25 105 kg travels north at aspeed of 220 km/h. Suppose it takes 16.0 s of constant acceleration for thetrain to come to rest at a station platform. Calculate the force acting onthe train during this time. What is the train’s stopping distance?

S O L U T I O N

Given: m = 9.25 × 105 kg

vi = 220 km/h to the north

vf = 0 km/h

∆t = 16.0 s

Unknown: F = ? ∆x = ?

Use the impulse-momentum theorem to solve for F. Use the kinematic equation

for ∆x in terms of initial velocity, final velocity, and time to solve for ∆x.

F∆t = ∆p

F = ∆∆

p

t =

mvf

∆−t

mvi

F =

F =

F = −3.5 × 106 N =

∆x = 12

(vi + vf)∆t

∆x = 12

(220 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(16.0 s)

∆x = 490 m to the north

3.5 × 106 N to the south

−(9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)

16.0 s

(9.25 × 105 kg)(0 km/h) − (9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)

16.0 s

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ADDITIONAL PRACTICE

1. A race car has a velocity of 382 km/h to the right. If the car’s mass is

705 kg and the driver’s mass is 65 kg, what force is needed to bring the

car and driver to a stop in 12.0 s? What is the car’s stopping distance?

2. The danger that space debris poses to spacecraft can be understood in

terms of momentum. At 160 km above Earth’s surface, any object will

have a speed of about 7.82 × 103 m/s. Consider a meteoroid (a small

orbiting rock) with a mass of 42 g. Suppose this meteoroid collides

with a space shuttle and is brought to a full stop in 1.0 × 10–6 s. How

large is the force that stops the meteoroid? By how much would the

meteoroid dent the side of the shuttle?


NAME ______________________________________ DATE _______________ CLASS ____________________

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3. A 63 kg astronaut drifting with 7.0 m/s to the right with respect to a

spacecraft uses a jet pack to slow down. If it takes 14.0 s to come to a

stop with respect to the spacecraft, what is the force exerted by the jet

pack? How far does the astronaut travel before stopping?

4. A polar bear with a mass of 455 kg slides for 12.2 s across the surface of

a frozen lake. If the coefficient of kinetic friction between the bear and

the ice is 0.071, what is the change in the bear’s momentum as it comes

to a stop? How far does the polar bear slide?

5. A skyrocket that has consumed all of its fuel continues to move up-

ward, slowed mostly by the force of gravity. If the rocket’s mass is

75.0 g and it takes 1.2 s for the rocket to stop, what is the change in the

rocket’s momentum? What is the rocket’s stopping distance?

6. A 4400 kg sailboat drifts in calm waters. The force of resistance that the

water exerts on the ship is 2200 N to the left. What is the change in the

ship’s momentum after 8.0 s? If the ship’s initial speed is 6.5 m/s, how

far has the ship drifted?

7. A jet of water exerts a 25.0 N force on a type of sail that is attached to a

small wagon. What is the magnitude of the change in the wagon’s mo-

mentum after 7.00 s? If the wagon’s mass is 14.0 kg and there are no

other forces acting on it, how far will the wagon travel during the

7.00 s? Assume the wagon is initially at rest.

8. How long will it take a 2.30 × 103 kg truck to go from 22.2 m/s to a

complete stop if acted on by a force of –1.26 × 104 N? What would be

its stopping distance?

9. Consider a Hornet jet landing on the flight deck of an aircraft carrier.

The jet’s mass is 1.35 × 104 kg and its initial velocity is 66.1 m/s to the

west. If the force required to bring the jet to a stop is 4.00 × 105 N to

the east, how long does it take the jet to come to rest? How far does the

jet travel during its deceleration?

10. The surface of Jupiter’s moon Europa is covered with ice. Suppose an

ice boat moving with a speed of 14.5 m/s drifts to a stop. The boat’s

mass is 1.50 × 103 kg and the coefficient of kinetic friction between the

boat’s runners and the ice is 0.065. However, free-fall acceleration on

Europa is only 1.305 m/s2. How long will it take the boat to stop? How

far does the ice boat glide before stopping?

Problem 6D Ch. 6-7

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6DCONSERVATION OF MOMENTUM

P R O B L E M

A 20.0 kg cannonball is fired from a 2.40 × 103 kg. If the cannon recoils with a velocity of 3.5 m/s backwards, what is the velocity of thecannonball?

S O L U T I O NGiven: m1 = mass of cannonball = 20.0 kg

m2 = mass of cannon = 2.40 × 103 kg

v1,i = initial velocity of cannonball = 0 m/s

v2, i = initial velocity of cannon = 0 m/s

v2,f = final velocity of cannon = 3.5 m/s backwards = −3.5 m/s

Unknown: v1, f = final mass of cannonball = ?

Choose the equation(s) or situation: Because the momentum of the cannon-

cannonball system is conserved and therefore remains constant, the total initial

momentum of the cannon and cannonball will equal the total final momentum

of the cannon and cannonball.

m1v1,i + m2v2,i = m1v1,f + m2v2,f

Because the cannon and cannonball are initially at rest, the initial momentum for

each is zero. From momentum conservation it follows that the total final mo-

mentum is also zero.

m1v1,f + m2v2,f = 0

Rearrange the equation(s) and isolate the unknown(s):

v1,f = −m

m2

1

v2,f


v1,f = = 420 m/s

v1,f =

The velocity is positive, indicating the forward direction. The cannonball’s mass

is less than one-hundredth the mass of the cannon, so its speed should be over a

hundred times greater than the recoil speed of the cannon.

420 m/s forward

−(2.40 × 103 kg)(−3.5 m/s)

20.0 kg

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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ADDITIONAL PRACTICE

1. A student stumbles backward off a dock and lands in a small boat. The

student isn’t hurt, but the boat drifts away from the dock with a

velocity of 0.85 m/s to the west. If the boat and student each have a

mass of 68 kg, what is the student’s initial horizontal velocity?


NAME ______________________________________ DATE _______________ CLASS ____________________

2. A coal barge with a mass of 1.36 × 104 kg drifts along a river. When it

passes under a coal hopper, it is loaded with 8.4 × 103 kg of coal. What

is the speed of the unloaded barge if the barge after loading has a speed

of 1.3 m/s?

3. A child jumps from a moving sled with a speed of 2.2 m/s and in the

direction opposite the sled’s motion. The sled continues to move in the

forward direction, but with a new speed of 5.5 m/s. If the child has a

mass of 38 kg and the sled has a mass 68 kg, what is the initial velocity

of the sled?

4. A swimmer with a mass of 58 kg and a velocity of 1.6 m/s to the north

climbs onto a 142 kg raft. The combined velocity of the swimmer and

raft is 0.32 m/s to the north. What is the raft’s velocity before the swim-

mer reaches it?

5. A 50.0 g shell fired from a 3.00 kg rifle has a speed of 400.0 m/s. With

what speed does the rifle recoil in the opposite direction?

6. Momentum conservation often assumes that the mass of an object re-

mains constant throughout a process or event. However, a change in

momentum can also occur when mass changes. Consider an automo-

bile with a full tank of gasoline traveling at a velocity of 88.0 km/h to

the east. The mass of the car when the fuel tank is full is 1292 kg. Sup-

pose that the car travels along a highway that extends eastward for

600 km. By the time the car has traveled this distance, its mass is

1255 kg. What is the car’s velocity at the end of the journey?

7. In 1976, Comet West was observed to break into four smaller parts as it

orbited near the sun. Suppose a comet with a mass of 5.0 × 1014 kg and

moving with a speed of 74.0 km/s breaks into two equal parts. One

part moves 15.0° above the original orbit with a speed of 105 km/s,

while a second fragment moves 30.0° below the original orbit. What is

the velocity of the second comet fragment?

8. A twig floating in a small pond is initially at rest. On the twig is a snail,

which begins moving along the length of the twig with a speed of

1.2 cm/s. The twig moves in the opposite direction with a speed of

0.40 cm/s. If the snail’s mass is 2.5 g, what is the mass of twig?

9. A toy that is initially at rest consists of three parts that are held together

by spring-loaded clips. At a given instant, the toy “explodes.” Two of the

pieces, which each have a mass of 25.0 g, travel with velocities of

7.0 cm/s to the south and 7.0 cm/s to the west, respectively. The third

piece has a velocity of 3.3 cm/s at an angle of 45° north of east. What is

the mass of the third piece?

10. An ice skater at rest catches a bag of sand moving to the north with a

speed of 5.4 m/s. This causes both the skater and the bag to move to

the north at a speed of 1.5 m/s. If the skater’s mass is 63 kg, what is the

mass of the bag of sand?

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Problem 6E Ch. 6-9

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6EPERFECTLY INELASTIC COLLISIONS

P R O B L E MAn arrow is fired into a small target at rest on a frictionless surface. Thearrow’s mass is 20.0 g and the target’s mass is 2.50 kg. If the speed of thearrow and target combined is 0.67 m/s, what is the arrow’s initial speed?

S O L U T I O N

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Given: m1 = mass of arrow = 20.0 g

m2 = mass of target = 2.50 kg

v2,i = initial speed of target = 0 m/s

vf = final speed of target and arrow = 0.67 m/s

Unknown: v1,i = initial speed of arrow = ?

Use the equation for a perfectly inelastic collision and rearrange it to solve for v1,i .

m1v1,i + m2v2,i = (m1 + m2)vf

v1,i =

v1,i =

v1,i = (2.

2

5

0

2

.0

kg

×)(

1

0

0

.6−3

7

k

m

g

/s)

v1,i = 84 m/s

(20.0 × 10−3 kg + 2.50 kg)(0.67 m/s) − (2.50 kg)(0 m/s)

20.0 × 10−3 kg

(m1 + m2)vf − m2v2,imi

ADDITIONAL PRACTICE

1. A 1550 kg torpedo strikes a 770 kg target that is initially at rest. If the

combined torpedo and target move forward with a speed of 9.44 m/s,

what is the initial velocity of the torpedo? Assume that no resistance is

provided by the water.

2. An ice hockey puck with a mass of 0.17 kg collides inelastically with a

0.75 kg snowball that is sliding to the left with a speed of 0.50 m/s. The

combined puck and snowball slide along the ice with a velocity of

4.2 m/s to the right. What is the velocity of the hockey puck before the

collision?

3. A clay ball with a mass of 45 g is attached to a long string to make a

pendulum. The ball is pulled back so that the string is horizontal to the

ground, and is then released. At the bottom of the ball’s path is another

clay ball that has a mass of 75 g and is at rest. The two balls collide in-

elastically, so that they follow the path of the first ball beyond the point

of collision. What must the speed of the first ball be just before the col-

lision so that the combined balls rise to a height of 8.0 cm above the


NAME ______________________________________ DATE _______________ CLASS ____________________

point of collision? How high must the first ball be raised for it to have

this speed at the bottom of its path?

4. A 5.00 × 102 kg log collides inelastically with a second log with the

same mass. These combined logs then collide with a third log with a

mass of 5.00 × 102 kg. The final speed of the three combined logs is

3.67 m/s. If the speed of the third log before collision was 3.00 m/s, and

the speed of the second log before collision was 3.50 m/s, what was the

speed of the first log before collision?

5. A railway car with a mass of 8500 kg and a velocity of 4.5 m/s to the

right collides inelastically with a railway car with a mass of 9800 kg and

a velocity of 3.9 m/s to the left. What is the final velocity of the com-

bined cars?

6. A 1400 kg automobile heading north at 45 km/h collides inelastically

with a 2500 kg truck traveling east at 33 km/h. What is the final veloc-

ity of the combined vehicles?

7. Four velcro-lined air-hockey disks collide with each other in a perfect

inelastic collision. The first disk has a mass of 50.0 g and a velocity of

0.80 m/s to the west, the second disk has a mass of 60.0 g and a velocity

of 2.50 m/s to the north, the third disk has a mass of 100.0 g and a ve-

locity of 0.20 m/s to the east, and the fourth disk has a mass of 40.0 g

and a velocity of 0.50 m/s to the south. What is the final velocity of the

disks after the collision?

8. A 25.0 kg sled carrying a 42.0 kg child is moving with a speed of

3.50 m/s when it collides with a snowman that is initially at rest. If the

speed of the snowman, sled, and child is 2.90 m/s, what is the snow-

man’s mass?

9. A remora is a type of fish that uses suckers underneath its head to at-

tach itself to other fish, notably sharks (for this reason it is also called

the “sharksucker”). Suppose a remora swimming with a velocity of

5.0 m/s to the right attaches itself to a 150.0 kg shark that is swimming

to the left with a speed of 7.00 m/s. If the remora collides inelastically

with the shark, the velocity of the two fish combined is 6.25 m/s to the

left. From this information, calculate the mass of the remora.

10. A proposed method for removing small but hazardous debris in orbit

around Earth involves a large ball composed of a type of gum or putty.

This soft ball would orbit Earth and collide inelastically with small par-

ticles of debris, sweeping them up in the process. Suppose this putty

ball moves in an orbit containing a stream of debris. The ball has a ve-

locity of 8.0 × 103 m/s to the right, while the particles of debris have a

velocity of 8.0 × 103 m/s to the left. Each particle of debris has an aver-

age mass of 2.5 g. If the putty ball sweeps up 5,000 particles before the

velocity of the ball and debris is 90.0 percent of the ball’s initial veloc-

ity, what is the ball’s mass?

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Problem 6F Ch. 6-11

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 6FKINETIC ENERGY IN PERFECTLY INELASTIC COLLISIONS

P R O B L E M

A ship with a mass of 4.50 107 kg and a velocity of 2.30 m/s to the northcollides with another ship whose mass is 2.30 107 kg. If the speed of thesecond ship is 3.40 m/s to the south, what is the change in the kinetic energy after the two ships undergo a perfectly inelastic collision?

S O L U T I O N

Given: m1 = mass of first ship = 4.50 × 107 kg

m2 = mass of second ship = 2.30 × 107 kg

v1,i = initial velocity of first ship = 2.30 m/s to the north

= +2.30 m/s

v2,i = initial velocity of second ship = 3.40 m/s to the south

= −3.40 m/s

Unknown: vf = ? ∆KE = ?

Use the equation for a perfectly inelastic collision to find vf.

m1v1,i + m2v2,i = (m1 = m2)vf

vf = m1v

m1

1

,i

++

m

m

2

2v2,i

vf =

vf =

vf =

vf = 0.38 m/s

Use the equation for kinetic energy to calculate the kinetic energy of each ship

before the collision and the final kinetic energy of the two ships combined.

Initial kinetic energy:

KEi = KE1,i + KE2,i = 12

m1v1,i2 + 1

2m2v2,i

2

KEi = 12

(4.50 × 107 kg)(2.30 m/s)2 + 12

(2.30 × 107 kg)(−3.40 m/s)2

KEi = 1.19 × 108 J + 1.33 × 108 J = 2.52 × 108 J

Final kinetic energy:

KEf = KE1,f + KE2,f = 12

(m1 + m2)vf

KEf = 12

(4.50 × 107 kg + 2.30 × 107 kg)(0.38 m/s)2

KEf = 12

(6.80 × 107 kg)(0.38 m/s)2

KEf = 4.9 × 107J

2.6 × 107 kg•m/s6.80 × 107 kg•m/s

1.04 × 108 kg•m/s − 7.82 × 107 kg•m/s

6.80 × 107 kg

(4.50 × 107 kg)(2.30 m/s) + (2.30 × 107 kg)(−3.40 m/s)

4.50 × 107 kg + 2.30 × 107 kg

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1. A ball of clay with a mass of 55 g and a speed of 1.5 m/s collides with a

55 g ball of clay that is at rest. By what percent has the kinetic energy

decreased after the inelastic collision?

2. An unconventional artist creates paintings by sloshing buckets of paint

onto large canvases. Suppose the canvas and easel on which it is placed

have a combined mass of 4.5 kg and are initially at rest when the artist

throws 1.3 kg of paint onto the canvas. The canvas, easel, and paint to-

gether slide back on the smooth floor with a speed of 0.83 m/s. What is

the change in the kinetic energy after the inelastic collision?

3. The farthest source of comets is called the Oort cloud. This is a volume

of space ranging from 1.5 × 1010 km to 7.4 × 1012 km away from the

sun in which comets are loosely held by the sun’s gravitational force.

Suppose a comet in the Oort cloud has a mass of 1.50 × 1013 kg and a

speed of 250 m/s. This comet collides inelastically with another comet

that has a mass of 6.5 × 1012 kg and a velocity of 420 m/s in the same

direction as the first comet. What is the change in the kinetic energy of

the comets after the collision?

4. Two flying fish have an inelastic collision while in mid-flight. One fish

has a mass of 0.650 kg and a velocity of 15.0 m/s to the right; the other

has a mass of 0.950 kg and a velocity of 13.5 m/s to the left. Find the

change in their kinetic energy after the collision.

5. A 75.0 kg log floats downstream with a speed of 1.80 m/s. Eight frogs

hop onto the log in a series of inelastic collisions. If each frog has a

mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change

in kinetic energy for this system?

6. What is the change in kinetic energy for the inelastic collision between

the two railway cars described in problem 5 of the previous section?


the clay balls in problem 3 of the previous section?

8. What is the change in kinetic energy in the collision between the ball of

putty and space debris described in problem 10 of Section 6E?

9. What is the change in the kinetic energy for the colliding logs in prob-

lem 4 of the previous section?


the four disks in problem 7 of the previous section?


NAME ______________________________________ DATE _______________ CLASS ____________________

Change in kinetic energy:

∆KE = KEf − KEi = 4.9 × 107 J − 2.52 × 108 J =

By expressing ∆KE as a negative number, ∆KE indicates that the energy has left

the system to take a form other than mechanical energy.

−2.03 × 108 J

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ADDITIONAL PRACTICE

Problem 6G Ch. 6-13

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Holt Physics

Problem 6GELASTIC COLLISIONS

P R O B L E MIn the game of marbles, a shooter is a large marble about 2 cm in diame-ter that is used to knock smaller marbles out of the ring. Suppose ashooter with a speed of 0.80 m/s hits a 4.8 g marble that is at rest in thering. The shooter continues forward with a speed of 0.51 m/s while thesmaller marble moves forward with a speed of 1.33 m/s. What is the massof the shooter?

S O L U T I O NGiven: v1,i = initial velocity of shooter = 0.80 m/s forward

v2,i = initial velocity of marble = 0 m/s

v1,f = final velocity of shooter = 0.51 m/s forward

v2,f = final velocity of marble = 1.33 m/s forward

m2 = mass of marble = 4.8 g

Unknown: m1 = mass of shooter = ?

Choose the equation(s) or situation: Use the equation for the conservation of

momentum to determine the mass of the shooter.

m1v1,i + m2v2,i = m1v1,f + m2v2,f


m1(v1,i − v1,f) = m2v2,f − m2v2,i

m1 = m2

v

v

1

2

,

,

i

f

−−

v

m

1

2

,f

v2,i


m1 =

m1 =

m1 =

Confirm your answer by making sure that kinetic energy is also conserved.

12

m1v,i2 + 1

2m2v2,i

2 = 12

m1v1,f2 + 1

2m2v2,f

2

12

(22 × 10−3 kg)(0.80 m/s)2 + 12

(4.8 × 10−3 kg)(0 m/s)2

= 12

(22 × 10−3 kg)(0.51 m/s)2 + 12

(4.8 × 10−3 kg)(1.33 m/s)2

7.0 × 10−3 J + 0 J = 2.9 × 10−3 J + 4.2 × 10−3 J

7.0 mJ =

The slight difference arises from rounding.

7.1 mJ

22 g

6.4 g•m/s0.29 m/s

(4.8 g)(1.33 m/s) − (4.8 g)(0 m/s)

0.80 m/s − 0.51 m/s

1. DEFINE

2. PLAN

3. CALCULATE

3. EVALUATE

1. A highly elastic rubber ball is tossed at a moveable wooden wall panel

that is initially at rest. The ball’s initial velocity is 6.00 m/s to the right.

After the elastic collision, the ball returns with a velocity of 4.90 m/s to

the left, while the panel moves 1.09 m/s to the right. If the panel’s mass

is 1.25 kg, what is the mass of the ball?

2. A 2.0 kg block of ice with a speed of 8.0 m/s makes an elastic collision

with another block of ice that is at rest. The first block of ice proceeds

in the same direction as it did initially, but with a speed of 2.0 m/s.

What is the mass of the second block? (Hint: Use the conservation of

kinetic energy to solve for the second unknown variable.)

3. A golf ball slides down a pipe from the upper level of a miniature golf

course and heads directly for the hole on the green. Unfortunately, an-

other player’s ball is directly in the way. The second ball, which is ini-

tially at rest, moves forward with a speed of 3.0 m/s, causing it to land

in the cup. The first ball comes to a complete stop after the collision.

Both balls have a mass of 45 g. What is the first ball’s speed before the

collision.

4. Suppose two ships, one with a mass of 3.0 × 107 kg and the second with

a mass of 2.5 × 107 kg, are equipped with bumpers, so that they undergo

a completely elastic collision. Before the collision, the second ship

moves north with a speed of 4.0 km/h. After the collision, the first ship

moves 3.1 km/h to the north while the second ship moves 6.9 km/h to

the south. Assume that the ships move over the ocean without friction.

Given this information, calculate the initial velocity of the first ship.

5. A basketball player throws a ball at the same time a ball from a nearby

court is thrown. The two balls collide elastically, so that the final veloc-

ity of the first ball is 4.0 m/s to the west and the final velocity of the

second ball is 3.0 m/s to the north. If the first ball’s initial velocity is

3.0 m/s to the north, what is the initial velocity of the second ball. As-

sume both balls have identical masses.

6. A red ball with a mass of 0.75 kg strikes two croquet balls, a 0.50 kg

green ball and a 0.50 kg blue ball, that are at rest next to each other on

a smooth wood floor. After the collision, the red ball has a velocity of

0.80 m/s to the east, the green ball moves 45° north of east with a speed

of 3.4 m/s, and the blue ball moves 45° south of east with a speed of

3.4 m/s. What is the red ball’s initial velocity?

7. An elevator is moving upward at a speed of 2.000 m/s. At the instant

the elevator is 20.4 m from the top of the shaft, a ball is dropped down

the shaft. The ball collides elastically with the elevator, so that it rises

up the shaft again. The elevator’s velocity immediately after the colli-

sion is 1.980 m/s upward. If the ball has a mass of 0.150 kg and the ele-

vator’s mass is 325.0 kg, what is the velocity of the ball after the colli-

sion with the elevator? How high above the elevator shaft does the

returning ball bounce?


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

Problem 6G Ch. 6-15

NAME ______________________________________ DATE _______________ CLASS ____________________

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8. Suppose the initial conditions are the same as for problem 7, except

that the elevator’s initial velocity is 2.000 m/s downward and its final

velocity is 2.017 m/s downward. What is the velocity of the ball after

the collision with the elevator? How far below the top of the elevator

shaft does the returning ball bounce?

9. A steel ball with a mass of 0.50 kg is fastened to a cord that is 40.0 cm

long and is released from a height of 40.0 cm. At the bottom of its path

the ball strikes a 2.5 kg block that is initially at rest on a frictionless sur-

face. The collision is elastic. What are the final velocities of the ball and

block? (Hint: Use the conservation of kinetic energy to solve for the

second unknown variable.)

10. A bowling ball of mass 7.00 kg moves east with a speed of 2.00 m/s.

The ball collides with an identical ball at rest, which causes the first ball

to move 30.0° north of east at a speed of 1.73 m/s. What is the velocity

of the second ball after the collision?

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Problem 7A Ch. 7-1

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7AANGULAR DISPLACEMENT

P R O B L E MA woman on vacation admires the murals on the inner wall of Coit Towerin San Francisco, California. If the woman walks 10.0 m clockwise alongthe curved wall, what will her angular displacement be? Assume the innerradius of Coit Tower is 4.20 m.

S O L U T I O NGiven: ∆s = −10.0 m

r = 4.20 m

Unknown: ∆q = ?

Use the angular displacement equation to solve for ∆q.

∆q = ∆r

s =

−4

1

.2

0

0

.0

m

m

∆q = −2.38 rad

1. At Tivoli, Italy, a circular ruin stands on the grounds of the emperor

Hadrian’s villa. A tourist walks 24.0 m counterclockwise along the edge

of the ruin. If the ruin’s radius is 3.50 m, what is the tourist’s angular

displacement?

2. In terms of volume, the largest tree is the General Sherman at Sequoia Na-

tional Park in California. Assuming the tree is perfectly circular, its radius

at the base is 5.55 m. If you were allowed to walk 31.3 m counterclockwise

along the side of the tree, what would your angular displacement be?

3. There was once a comedy skit that had a character who boasted of a stor-

age tank “the size of Rhode Island.” Suppose a circular storage tank with

this area did exist. The area of Rhode Island is 2730 km2. If you were to

drive 545 km clockwise along the perimeter of this tank, what would

your angular displacement be?

4. One of the largest stars in our galaxy is Betelgeuse, a red super giant that

has expanded as it has evolved. Betelgeuse is so large that, were it placed

in our solar system where the sun is, its surface would lie beyond Jupiter’s

orbit. Suppose a spacecraft orbits the sun at a distance equal to the radius

of Betelgeuse. The ship must travel through an arc length with a magni-

tude of 4.3 × 1011 m to have an angular displacement of magnitude

0.39 rad. From this information, determine the average radius of

Betelgeuse.

5. The only circular border in the United States is between northern

Delaware and Pennsylvania. The border has a length of about 35.0 km. If

a vehicle traveling through an arc length of this size has an angular dis-

placement with a magnitude of 1.75 rad, what is the border’s radius?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

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6. A laser etches a line with an arc length of 36.6 mm clockwise along the sur-

face of a spherical grain of sand. If this distance corresponds to a clockwise

angular displacement of p6

rad, what is the sand grain’s radius?

7. Casa Rinconada is the name given to one of the largest ceremonial struc-

tures, or kivas, in the Chaco Canyon area of northwestern New Mexico.

Built nearly a thousand years ago, Casa Rinconada’s inner wall has a radius

of 10.0 m. If you were to walk counterclockwise along this wall, so that

your angular displacement was 5.7 rad, what arc length would you travel

through?

8. Of the solar system’s planets, the one whose orbit around the sun is most

circular is Venus. The mean radius of Venus’s orbit is 1.08 × 108 km. What

arc length does Venus travel through when its angular displacement is p3

rad counterclockwise?

9. After Venus, the next most circular orbit belongs to Neptune. The mean

radius of Neptune’s orbit is 4.48 × 109 km. What arc length does Neptune

travel through when its angular displacement is p3

rad counterclockwise?

10. The Pantheon in Rome is the oldest surviving domed building. The dome

is a hemisphere with a radius of 21.8 m. Suppose you were to walk inside

the Pantheon in a circular path just underneath the dome’s rim. What arc

length would you have traveled through if your angular displacement was

7.50 rad clockwise?

Problem 7B Ch. 7-3

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7BANGULAR SPEED

P R O B L E MThe spin of a flying disk toy gives the disk stability while gliding. If one ofthese disks has an angular displacement with a magnitude of 60.0 rev in12.0 s, what is the disk’s average angular speed in rev/s and rad/s?

S O L U T I O NGiven: ∆q = 60.0 rev

∆t = 12.0 s

Unknown: wavg = ?

Use the angular speed equation to solve for waug.

wavg = ∆∆

qt

= 6

1

0

2

.0

.0

re

s

v

wavg =

wavg = (5.00 rev/s)(2π rad/rev) = 31.4 rad

5.00 rev/s

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ADDITIONAL PRACTICE

1. The propellers, or screws, of a steamship push water away from the ship,

causing the ship to move forward. Suppose a ship’s screw turns 106 rad

clockwise in 7.5 s. What is the average angular speed of the screw?

2. Devil’s Tower National Monument in northeastern Wyoming is an outcrop-

ping of volcanic rock that is 264 m tall. Suppose a golden eagle soars counter-

clockwise in a circular path around the top of the tower. If the eagle circles the

monument once in 4.56 min, what is the bird’s average angular speed in rad/s?

3. The speed of a bullet can be determined in the following way. Each end

of a long metal rod is passed through the center of a paper disk. The rod

is rapidly rotated, so that the two disks spin at the same angular speed.

The bullet, which is fired parallel to the rod, penetrates the first disk near

the rim and then travels the distance between the disks. During this time,

the disks rotate through a certain angular displacement. After the bullet

has penetrated the second disk, the angle between the two holes in the

disks can be measured. With the known distance between the disks and

the angular speed of the disks, the bullet’s speed can be calculated. Sup-

pose the bullet’s speed is 280 m/s, the disks are separated by 2.0 m, and

the counterclockwise angular displacement between the holes is

0.54 rad. What is the average angular speed of the disks in rev/s?

= 0.54 radθ

d

vω


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4. The average angular speed of a merry-go-round is 0.75 rad/s. How long

does it take the merry-go-round to rotate through an angular displace-

ment with a magnitude of 3.3 rad?

5. Located on a 21 km2 island in the western Pacific Ocean, the Republic of

Nauru is the smallest—and in terms of its population one of the wealthi-

est—republics in the world. Suppose a tuna, which can swim at 80 km/h,

swims in a circle just off the shore of Nauru. If the tuna’s average angular

speed is 8.6 × 10–3 rad/s, how long will it take to circle the island exactly

6 times?

6. With a diameter nearly 1 m, Rafflesia arnoldii of Malaysia is considered

the world’s largest flower. Suppose a mosquito flies around the rim of a

Rafflesia with an average angular speed of 2.75 rad/s. How long will it

take the mosquito to circle the flower exactly 3 times?

7. Suppose you were to stand on Earth’s equator for 4.2 h. What would your

angular displacement be in radians? Consider Earth’s rotation to be

counterclockwise.

8. Just as the planets in the solar system move in elliptical orbits around the

sun, the solar system itself moves in an elliptical orbit around the gravita-

tional center of the Milky Way galaxy, making one revolution every 212 ×106 years. What would the magnitude of the solar system’s angular dis-

placement be over its approximate lifetime of 4.50 × 109 years? Give your

answer in radians.

9. Venus is the only planet in the solar system to rotate in a clockwise (retro-

grade) direction with respect to its north pole. It also rotates slower than

any other planet: once every 243 days. Express this average angular speed

in terms of rad/day, then calculate the angular displacement of Venus

during one Earth year (365.25 Earth days) and one Venus year (224.7

Earth days). (Note that one Venus year is shorter than one Venus day.)

10. Imagine the following experiment: you are standing at the top rim

looking down into a amusement park centrifuge ride in which a motion-

picture camera has been mounted to the wall. The camera and ride are

started, and at a point where the camera is on the side of the ride oppo-

site you, you toss a marble into the ride. Although from your viewpoint

the marble moves in a straight line, the image the camera records shows

the marble to be deflected through a certain angular displacement (actu-

ally the camera’s angular displacement). If the diameter of the ride is

6.0 m, the marble has a horizontal speed of 5.0 m/s, and the average an-

gular speed of the ride is 6.00 rad/s, what is the magnitude of the cam-

era’s actual (or marble’s apparent) angular displacement?

Problem 7C Ch. 7-5

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7CANGULAR ACCELERATION

P R O B L E MFor a certain high-speed elevator, the drive wheel on the motor under-goes an average angular acceleration of 52.0 rad/s2. If the wheel is ini-tially at rest, what is its angular speed after 4.00 s?

S O L U T I O N

Given: αavg = 52.0 rad/s2

∆t = 4.00 s

w1 = 0 rad/s

Unknown: w2 = ?Use the angular acceleration equation and rearrange it to solve for w2.

αavg = ∆∆wt =

w2

∆−t

w1

w2 = w1 + αaug ∆t = 0 rad/s + (52.0 rad/s2)(4.00 s)

w2 = 208 rad/s

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ADDITIONAL PRACTICE

1. A turntable has an average angular acceleration of 1.5 rad/s2 and an ini-

tial angular speed of 3.0 rad/s. What is the turntable’s angular speed 4.0 s

later?

2. The take-up reel of a cassette tape must pull the tape across the magnetic

head at a constant speed. To do this, the reel’s angular speed must vary

with the increasing amount of tape on the reel. An empty reel has an an-

gular speed of 9.5 rad/s and an average angular acceleration of –5.4 ×10–3 rad/s2. What is the reel’s angular speed after 22 min?

3. The gasoline engine for a power lawnmower has an average angular ac-

celeration of 32 rad/s2 when first started. What is the engine’s angular

speed after 1.5 s, assuming an initial angular speed of 0 rad/s?

4. A tire on an accelerating car goes from rest to an angular speed of

76 rad/s. How long does it take the tire to reach this angular speed if the

tire’s average angular acceleration is 9.5 rad/s2?

5. A rolling hula hoop undergoes an average angular acceleration of

3.91 rad/s2. How long will it take for the hula hoop to increase its angular

speed from 2.50 rad/s to 7.70 rad/s?

6. As it turns around, a steamship has an angular speed of 5.14 × 10–2 rad/s

at the beginning of the turn. Its speed a short time later is 3.09 × 10–2 rad/s.

How much time does it take the ship to change speed if the average angu-

lar acceleration of the ship is –1.75 × 10–3 rad/s2?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. The speed of a steam engine is automatically controlled by a device called

a governor. A governor consists of two metal balls attached by rods to the

top of a rotating shaft. The balls are also attached by rods to a lever that

controls the flow of steam to the engine. As the engine’s speed increases,

the shaft’s angular speed also increases, the balls spin outward, and the

control lever is raised. This reduces the amount of steam that enters the

engine, causing it to slow down. The reverse process causes the engine to

speed up again. If a governor increases its angular speed from 4.0 rad/s to

5.0 rad/s in 7.5 s, what is its average angular acceleration?

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8. A toy consists of a windup car and a circular metal base with a shallow

groove that guides the car in a circular path. At first the car has an angu-

lar speed of 7.14 rad/s, but after 9.00 s its angular speed is 2.38 rad/s.

What is the car’s average angular acceleration?

9. A driver turns a corner when she realizes that oil has been spilled on the

pavement. By letting the car’s angular speed slow from 2.07 rad/s to

1.30 rad/s over a period of 2.2 s, the driver is able to make the turn with-

out skidding. What is the average angular acceleration of the car and

driver?

10. Because of the gravitational forces between Earth and the moon, the an-

gular speed of Earth is decreasing at about 1.7 × 10–3 s each century. This

means that 70.0 million years ago, Earth made one complete rotation in

just 23.66 h, whereas now it takes 24.00 h. What is Earth’s average angular

acceleration over the last 70.0 × 106 years?

controllever

for steamflow

rotationarrow

Problem 7D Ch. 7-7

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7DANGULAR KINEMATICS

P R O B L E MThe screws of an ocean liner undergo a constant angular acceleration of0.656 rad/s2. How long does it take for the screws to reach a final angularspeed of 10.5 rad/s, assuming that they are initially at rest?

S O L U T I O N

Given: α = 0.656 rad/s2

wf = 10.5 rad/s

wi = 0 rad/s

Unknown: ∆t = ?

Use the angular kinematic equation for wf in terms of wi, α, and ∆t, and re-

arrange it to solve ∆t.

wf = wi + α∆t

∆t = wf

α− wi =

10.5

0.

r

6

a

5

d

6

/s

ra

−d

0

/s

r2ad/s

∆t = 16.0 s

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ADDITIONAL PRACTICE

1. A model airplane is kept in a circular path by means of a control line.

The airplane starts at rest and reaches an angular speed of 3.33 rad/s after

a constant angular acceleration of 0.183 rad/s2. How long does it take the

airplane to reach its final angular speed?

2. A millstone is a large stone wheel once used to grind grain for flour. Sup-

pose a stone in a windmill is moved from rest with a constant angular

acceleration of 0.13 rad/s2. How long does it take the millstone to move

through an angular displacement of 1.6 rad?

3. A ceiling fan’s angular speed increases from 5.2 rad/s to 20.9 rad/s. Dur-

ing this constant angular acceleration, the fan moves through an angular

displacement of 216 rad. How long does it take the fan to reach its final

angular speed?

4. A runner on a circular track increases his angular speed from

0.111 rad/s to 0.178 rad/s. If the runner’s angular acceleration is 1.1 ×10–2 rad/s2, what is his angular displacement?

5. Even rarer than phonograph turntables for long-playing (LP) records are

those turntables used in the first half of the twentieth century. Unlike the

LP turntables, which rotated at 33.3 rev/min, these older turntables

turned at 78.0 rev/min. Suppose a turntable with this angular speed is

shut off, so that it comes to a stop 30.0 s later. If the turntable has a con-

stant angular acceleration of –0.272 rad/s2, what is its angular displace-

ment? Give your answer in both radians and number of revolutions.


NAME ______________________________________ DATE _______________ CLASS ____________________

6. A rotary saw with an angular speed of 298 rad/s cuts through a piece of

wood. The saw undergoes a constant angular acceleration of –44.0 rad/s2

while it goes through an angular displacement of 276 rad. What is the

saw’s final angular speed?

7. The lumberjack sport of logrolling, or birling, involves two competitors

running along the circumference of a floating log. Each competitor tries

to alter the log’s speed in such a way that the other competitor will fall

into the water. Suppose a lumberjack is practicing logrolling skills alone

on a log. Initially at rest, the log undergoes a constant angular accelera-

tion, so that after 13.0 s the log has turned through 10.0 rev. What is the

final angular speed of the log in rad/s?

8. An automobile engine’s angular speed is increased from 1200 rev/min to

3600 rev/min in 12 s. What is the engine’s angular acceleration in rad/s2?

9. The drive wheel on a vacuum pump has an angular displacement of

158 rad as it accelerates from rest to 70.0 rad/s. What is the wheel’s angu-

lar acceleration?

10. A pet hamster runs inside its exercise wheel. The wheel has an initial an-

gular speed of 3.29 rad/s. Over the next 2.50 s, the hamster increases its

speed and the wheel undergoes a constant angular acceleration. If the an-

gular displacement of the wheel during this time is 12.3 rad, what is the

wheel’s angular acceleration?

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Problem 7E Ch. 7-9

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7ETANGENTIAL SPEED

P R O B L E MA pog disk rolls on its edge with an angular speed of 44.5 rad/s. If the ra-dius of the pog is 1.75 cm, what is the tangential speed of the pog’s rim?

S O L U T I O NGiven: w = 44.5 rad/s

r = 1.75 cm

Unknown: vt = ?

Use the tangential speed equation to solve for vt.

vt = rw = (1.75 cm)(44.5 rad/s)

vt = 77.9 cm/s

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ADDITIONAL PRACTICE

1. The Maelstrom is a whirlpool formed in a narrow strait among the Lo-

foten Islands, off the coast of northern Norway. It occurs during changes

in the tides, when currents of water move in opposite directions through

the strait. The angular speed of the waters in the Maelstrom is about 2.1 ×10–3 rad/s, though this varies with location. Suppose a boat is at a distance

of 1.5 km from the center of the Maelstrom and the angular speed of the

boat at that point is 2.07 × 10–3 rad/s. What is the boat’s tangential speed?

2. A professionally pitched baseball has an angular speed of 188.5 rad/s.

The radius of a baseball is 3.73 cm. What is the tangential speed of a

point on the baseball’s surface?

3. A wind turbine is a type of windmill that generates electricity. In order

for the turbine to be reasonably efficient, the blades must be very long.

Suppose one of the many turbines located near Palm Springs, California,

has blades that are 15.2 m long and rotate at an angular speed of

6.28 rad/s. What is the tangential speed at the tip of one of these blades?

4. A particular unicycle has a wheel with a radius of 0.30 m. If the unicycle is

ridden with a linear speed of 4.5 m/s, what is the wheel’s angular speed?

5. The main rotor of a certain helicopter consists of blades that extend

2.00 m from the center of the drive shaft. If the tangential speed at the

tips of these blades is 94.2 m/s, what is the rotor’s angular speed?

6. The “Barrel of Fun” is found in “fun houses” at certain amusement parks.

It consists of a cylindrical tunnel that rotates, so that anyone trying to

walk the length of the tunnel is pulled off to one side. Suppose the inner

wall of a “Barrel of Fun” has a tangential speed of 0.63 m/s. If the barrel

has a radius of 1.5 m, what is the barrel’s angular speed?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. The windmills of the Netherlands are not the only ones in Europe, nor

are they the oldest, but they are among the most famous. These mills

have been used for grinding grain and for draining water to reclaim land

from the sea. Suppose the sails of one of these mills turn with an angular

speed of 3.14 × 10–2 rad/s. If the tangential speed at the tips of the sails is

0.45 m/s, what is the radius of each sail?

8. A grindstone for sharpening knives and axes turns with an angular speed

of 10.0 rad/s. If the tangential speed of the grindstone’s rim is 4.60 m/s,

what is the radius of a grindstone?

9. A toy consists of a track made from two parallel metal wires that are sepa-

rated by a distance of 3 cm. A wheel with a magnetic shaft is placed be-

tween the wires. The wheel spins when the track is tilted, and it remains

on the wires at the end of the track because of the magnetized shaft. In

this way, the angular speed of the wheel can be increased simply by

tilting the track back and forth when the wheel is at the ends of the track.

Suppose the wheel has an angular speed of 11 rad/s and a tangential

speed of 4.0 cm/s. What is the radius of the wheel’s shaft?

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10. An ice skater moves in a circle on a frozen pond. The skater has a tangen-

tial speed of 1.5 m/s and an angular speed of 0.33 rad/s. What is the ra-

dius of the circle in which the skater moves?

Problem 7F Ch. 7-11

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7FTANGENTIAL ACCELERATION

P R O B L E M

One of the fastest angular speeds for a laboratory centrifuge is 2.5 106 rev/min. Suppose the angular acceleration for this centrifuge is 5800 rad/s2. If the radius of the centrifuge is 0.15 m, what is the tangen-tial acceleration of a point on the rim of the centrifuge?

S O L U T I O N

Given: α = 5800 rad/s2

r = 0.15 m

Unknown: at = ?

Use the tangential acceleration equation to solve for at.

at = rα = (0.15 m)(5800 rad/s2)

at = 870 m/s2

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1. A compact disc has a radius of 6.0 cm. Suppose this disc undergoes an

angular acceleration of 35.2 rad/s2. What is the tangential acceleration of

a dust speck located on the rim of the disc?

2. The drive shaft of an automobile engine undergoes an angular accelera-

tion of 105 rad/s2. If the shaft’s radius is 1.75 cm, what is the tangential

acceleration along the side of the shaft?

3. The drive wheels of the river steamship Grand Republic each had a radius

of 5.87 m. Suppose the angular acceleration of each wheel was 1.40 ×10–2 rad/s2. What would the tangential acceleration at the rim of either

wheel have been?

4. Though it was built by an engineer named Basset rather than by Ferris,

one of the largest and oldest Ferris wheels in the world is located in Vi-

enna at the Prater amusement park. Suppose this wheel accelerates from

rest to its final angular speed of 1.23 × 10–2 rad/s in 10.0 s. The corre-

sponding tangential acceleration along the outer rim of the wheel is

7.50 × 10–2 m/s2. What is the wheel’s radius?

5. An airplane propeller undergoes an angular acceleration of 42 rad/s2. If

the tangential acceleration of the propeller’s tip is 64 m/s2, what is the

propeller’s radius?

6. A large carousel undergoes an angular acceleration of 6.25 × 10–2 rad/s2

before it reaches its final angular speed. During this time a rider on the

edge of the carousel experiences a tangential acceleration of 0.75 m/s2.

What is the carousel’s radius?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

7. An experimental bus in Switzerland used the kinetic energy of a massive

spinning flywheel as a means of propulsion. Eventually the flywheel

would slow down, and because there was no engine in the bus, it would

have to be driven periodically to a power station. At this station, the fly-

wheel’s angular speed would be increased by an electric motor. Suppose

that while the flywheel was being brought up to speed, a point on its rim

underwent a tangential acceleration of 0.157 m/s2. If the radius of the

flywheel was 0.90 m, what was its angular acceleration?

8. The largest tire ever made had a radius of 1.75 m. Suppose this tire un-

derwent a linear acceleration 0.83 m/s2. What would the angular acceler-

ation of the tire have been?

9. The old kinds of bicycles with the enormous front wheels and high seats

are called pennyfarthings. This is because the ratio of the big wheel’s area

to the small wheel’s area is similar to the ratio of the areas of an old En-

glish penny to an old English farthing. The radius of the big wheel on a

pennyfarthing is 0.50 m. Suppose a rider accelerates from rest to a linear

speed of 5.0 m/s in 8.5 s. What would the angular acceleration of the big

wheel be?

10.The small wheel of a pennyfarthing has a radius of 16 cm. Suppose the

tangential acceleration along the small wheel’s rim is the same as for the

big wheel in problem 9. What would the angular acceleration of the small

wheel be?

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Problem 7G Ch. 7-13

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7GCENTRIPETAL ACCELERATION

P R O B L E MThe most violent part of a hurricane is at the edge of the hurricane’s eye.This region, called the eyewall, can have winds with speeds of more than300 km/h. Suppose winds in a hurricane’s eyewall have a tangential speedof 82.2 m/s. If the eyewall is 25 km from the center of the hurricane, whatis the magnitude of the centripetal acceleration of particles in the eye-wall?

S O L U T I O NGiven: vt = 82.2 m/s

r = 25 km

Unknown: ac = ?

Use the centripetal acceleration equation for ac in terms of vt and r.

ac = v

rt2

= (

2

8

5

2.

×2

1

m

0

/3s

m

)2

ac = 0.27 m/s2

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ADDITIONAL PRACTICE

1. Because of the conditions that give rise to them, tornadoes do not have

the widespread destructive effects of hurricanes. Nevertheless, the winds

encountered in some tornadoes are even greater than those at the eyewall

of a hurricane. Suppose a small pebble is swept up in a tornado. The peb-

ble is 3.81 m from the center of the tornado and has a tangential speed

equal to that of the surrounding wind: 124 m/s. What is the magnitude

of the centripetal acceleration of the pebble?

2. A type of pinball machine uses a spinning rubber disk to deflect the ball.

The disk has a radius of 6.50 cm and an angular speed of 30.0 rad/s. Sup-

pose this machine is laid flat so that a ball on the edge of the disk isn’t

pulled to the side by gravity. How large is the ball’s centripetal accelera-

tion when the ball is at the edge of the disk?

3. A customer sits in a revolving restaurant 11 m from the center. If the cus-

tomer’s tangential speed is 1.92 × 10–2 m/s, how large a centripetal accel-

eration does the customer experience?

4. NASA uses large centrifuges to study the effects of large forces on astro-

nauts prior to their going into space. A subject in the 20-G centrifuge,

which has a radius of 8.9 m, can have a centripetal acceleration as large as

20.0g, where g equals 9.81 m/s2. What is the tangential speed of the subject?

5. A rotating furnace has been developed recently to give a rough parabolic

curve to molten glass. This makes the manufacture of very large telescope

mirrors easier and more economical than ever before. The largest mirror


NAME ______________________________________ DATE _______________ CLASS ____________________

made in this furnace to date has a radius of 4.2 m. While in the furnace,

the centripetal acceleration of the molten glass for this mirror was

2.13 m/s2. What was the tangential speed at the edge of the molten glass?

6. For several decades the idea of an orbiting space colony has been dis-

cussed. The colony would consist of a large hollow cylinder that rotates

at a constant angular speed. Colonists would live on the inner wall of the

cylinder, where centripetal acceleration would simulate free-fall accelera-

tion at Earth’s surface. If the structure has an inner radius of 150 m, what

would the tangential speed of a colonist standing on the cylinder’s inner

wall be?

7. The Indianapolis Motor Speedway has four banked curves, each of which

forms a quarter of a circle. Suppose a race car speeds along one of these

curves with a constant tangential speed of 75.0 m/s. Neglecting the

effects due to the banking of the curve, the centripetal acceleration on

the car is 22.0 m/s2. What is the radius of the curve?

8. A turntable spins with an angular speed of 3.5 rad/s. A quarter placed on

the turntable has a centripetal acceleration of 2.0 m/s2. How far is the

quarter from the center of the turntable?

9. A model electric train moves along a circular track. The train has a tan-

gential speed of 0.35 m/s and has a centripetal acceleration of 0.29 m/s2.

What is the radius of the track?

10.A roller-coaster has a loop-the-loop in which the centripetal acceleration

on the cars and passengers just equals 9.81 m/s2. If the tangential speed

of the roller-coaster cars is 15.7 m/s, what is the radius of the loop-the-

loop?

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Problem 7H Ch. 7-15

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 7HFORCE THAT MAINTAINS CIRCULAR MOTION

P R O B L E MA model airplane with a mass of 3.2 kg moves in a circular path with a ra-dius of 12 m. If the airplane’s speed is 45 m/s, how large is the force thatthe control line exerts on the plane to keep it moving in a circle?

S O L U T I O NGiven: m = 3.2 kg

r = 12 m

vt = 45 m/s

Unknown: Fc = ?

Use the equation for force that maintains circular motion that is expressed in

terms of vt.

Fc = m v

rt2

= 3.2 kg (45

12

m

m

/s)2

Fc = 540 N

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1. A 40.0 kg child takes a ride on a Ferris wheel that rotates with an angular

speed of 0.50 rad/s. Find the magnitude of the force that maintains the

circular motion of the child if the distance from the center of the Ferris

wheel to the child is 18.0 m.

2. A toy model of an amusement park ride has a central shaft that rotates

while carts attached to the top of the shaft by threads “fly” outward. The

force that keeps the carts in a circular path is provided by the tension in

the thread. When the carts are 0.25 m from the center of the shaft, the

largest tangential speed that the carts can have without the threads

breaking is 5.6 m/s. If the mass of a cart is 0.20 kg, how large is the maxi-

mum force that maintains circular motion?

3. An automobile with a tangential speed of 48.0 km/h follows a circular

road that has a radius of 35.0 m. The pavement is wet and oily, so the co-

efficient of kinetic friction between the car’s tires and the pavement is

only 0.500. How large is the force needed to maintain the car’s circular

motion? How large is the available frictional force? Is the available fric-

tional force large enough to maintain the automobile’s circular motion?

Assume the automobile has a mass of 1250 kg.

4. Another automobile with a mass of 1250 kg follows a curved road with a

radius of 35 m. This time, however, the road is banked, so that it tilts

toward the center of the curve with an angle of 9.5°. If the coefficient of

kinetic friction is 0.500, how large is the centripetal force from friction

and gravity? If this force is equal to the force needed to maintain the auto-

mobile’s circular motion, what is the auto’s maximum tangential speed?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A small asteroid with a mass of 2.05 × 108 kg is pulled into a circular

orbit around Earth. The distance from the asteroid to Earth’s center is

7378 km. If the gravitational force needed to keep the asteroid in orbit

has a magnitude of 3.00 × 109 N, what is the asteroid’s tangential speed?

6. The first of the great amusement parks on Coney Island, in Brooklyn,

New York, was Steeplechase Park. In 1905, one of the rides in the fun

house at Steeplechase Park was the “Human Roulette Wheel.” This ride

consisted of a large wooden wheel, nearly 6 m in diameter, on which sev-

eral people climbed. The wheel would then spin, causing all but the pas-

sengers closest to the center to slide off. Suppose a passenger with a mass

of 55 kg was able to stay on the wheel, which rotated with an angular

speed of 2.0 rad/s. If the magnitude of the frictional force that held the

passenger on the wheel was 135 N, how far was the passenger from the

center of the wheel?

7. The comets with the longest periods between appearances, as well as

comets that appear only once, come from a region of the solar system

called the Oort cloud. In the Oort cloud, comets have slow tangential

speeds as they orbit the distant sun. Suppose one of these comets has a

mass of 7.55 × 1013 kg and moves with a tangential speed of 0.173 km/s

relative to the sun. If the magnitude of the gravitational force that keeps

the comet in orbit is 505 N, how far is the comet from the sun?

8. The governor for a steam engine rotates with an angular speed of

36.7 rad/s. The balls of the governor are each 0.10 m from the rotation

axis. A force with a magnitude of 670 N is provided by a metal rod, so

that each ball is kept in a circular path. Determine the mass of each ball.

9. To encourage donations of loose change, a zoo has placed an interesting

type of coin well at its visitor’s center. The well is about 1 m tall and is

shaped like the flared bell of a trumpet, with the widest part at the top

and the hole perpendicular to the ground. A coin placed in a chute and

knocked into the well does not simply drop in, but rolls on its edge

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coinwell

coin

rotationvelocity

Problem 7H Ch. 7-17

NAME ______________________________________ DATE _______________ CLASS ____________________

around the rim of the well, gradually moving lower down it. If the coin is

placed correctly, it can still roll around the well’s wall even when the wall

is nearly vertical. Consider a half dollar rolling around the top rim of this

coin well. The radius of the top of the well is 35.0 cm and the coin rolls

around its edge with a tangential speed of 2.21 m/s. If the well’s inner wall

exerts a force of 0.158 N on the rim of the coin, what is the coin’s mass?

10.Since antiquity people have used the sling to increase the speed of a rock

and send it swiftly in a specific direction. While the rock is being spun

overhead, the force that keeps the rock moving in a circle is provided by

the tensile strength of the sling material. Leather has a fairly high tensile

strength, so that a strip of leather with a cross-sectional area of 0.25 cm2

can withstand a pulling force of 800 × 102 N. Assume that, for a certain

sling, 8.00 × 102 N is the largest force that can keep a rock in a circular

path. If the rock in the sling is 0.40 m from the center of rotation and has

a tangential speed of 6.0 m/s, what is the largest mass the rock can have?

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NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 7IGRAVITATIONAL FORCE

P R O B L E MOne of the more mysterious objects in the solar system is a large asteroidor comet called Chiron. Chiron’s orbit lies between the orbits of Jupiterand Uranus, and crosses the orbit of Saturn. Assuming a mass for Chironof 4.5 × 1019 kg, what is the gravitational force between the sun and Chi-ron when the distance between the two is 2.0 × 1012 m? The sun’s mass is1.99 × 1030 kg.

S O L U T I O N

Given: m1 = mass of Chiron = 4.5 × 1019 kg

m2 = mass of the sun = 1.99 × 1030 kg

r = distance between the sun and Chiron = 2.0 × 1012 m

G = 6.673 × 10−11 N

k

•

g

m2

2

Unknown: Fg = ?

Use the equation for Newton’s Universal Law of Gravitation to solve for Fg.

Fg = G m1

r 2m2 = 6.673 × 10−11

N

k

•

g

m2

2

Fg = 1.5 × 1015 N

(4.5 × 1019 kg)(1.99 × 1030 kg)

(2.0 × 1012 m)2

ADDITIONAL PRACTICE

1. The largest fish is the whale shark, which can have a mass of 2.04 × 104 kg.

The largest mammal, and indeed the largest animal ever to have lived on

Earth, is the blue whale, which can have a mass of 1.81 × 105 kg. If the

distance between these two creatures is 1.5 m, how large is the gravita-

tional force between them?

2. Although it contributes only 0.02 percent to Earth’s total mass, the water

in Earth’s oceans is still quite massive. Suppose the water of the oceans

could somehow be drained, kept in liquid form, and moved as far from

Earth as the moon is. How large would the gravitational force between

the water and Earth be? Assume the mass of the ocean’s water to be

1.4 × 1021 kg, the mass of Earth to be the same with water (5.98 × 1024 kg),

and the Earth-moon distance to be 3.84 × 108 m.

3. A gravimeter is a highly sensitive device that measures changes in the

gravitational force in a given area. These measurements reveal variations

in the density of underground rock. This information can be used to in-

dicate whether resources like oil are possibly present. Suppose a certain

type of gravimeter has a test mass of 0.500 kg inside it. How large is the

gravitational force between this mass and a mountain that is 10.0 km

away and that has a mass of 2.50 × 1012 kg?

Problem 7I Ch. 7-19

NAME ______________________________________ DATE _______________ CLASS ____________________

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4. The magnitude of the gravitational force between two Sumo wrestlers

just before they collide (r = 0.025 m) is 2.77 × 10–3 N. If one wrestler has

a mass of 157 kg, what is the mass of the other wrestler?

5. Jupiter’s largest moon, Ganymede, is also the eighth largest known object

in the solar system. The magnitude of the gravitational force between

Ganymede and Jupiter is 1.636 × 1022 N. Given that Jupiter’s mass is

1.90 × 1027 kg and the distance between Jupiter and Ganymede is

1.071 × 106 km, calculate Ganymede’s mass.

6. The largest known asteroid in the solar system is Ceres, which is located

in an orbit between Mars and Jupiter. Although the existence of Ceres

has been known for two hundred years, its mass is still not well deter-

mined. Suppose the magnitude of the gravitational force between the

sun and Ceres equals 1.17 × 1018 N. Given that the sun’s mass is 1.99 ×1030 kg and the distance between the sun and Ceres is 4.12 × 1011 m,

what is Ceres’ mass?

7. The nearest galaxy with a size similar to that of the Milky Way is the so-

called Andromeda galaxy. By all appearances, this galaxy is very similar to

our own. Suppose both the Milky Way and Andromeda galaxies have

identical masses equal to 500 billion solar masses, or 9.95 × 1041 kg. If

the gravitational force between the two galaxies has a magnitude of

1.83 × 1029 N, what is the distance separating the galaxies?

8. At the sun’s surface, which is called the photosphere, the gravitational

force between the sun and a 1.00 kg mass of hot gas has a magnitude of

274 N. Given that the sun’s mass equals 1.99 × 1030 kg and assuming that

the sun is spherical, what is the sun’s mean radius?

9. At the surface, or photosphere, of the red super giant star Betelgeuse, the

gravitational force between the star and a 1.00 kg mass of hot gas is only

2.19 × 10–3 N. This is because the mean radius of Betelgeuse is so large.

Given that the mass of Betelgeuse is 20 times that of the sun, or

3.98 × 1031 kg, what is the mean radius of Betelgeuse?

10.Comets from the Oort cloud do not frequently enter the solar system

close to the sun. This occurs only when they are perturbed by other

masses, such as other nearby comets, or even the closest stars. Suppose

two comets in the Oort cloud interact with a gravitational force of just

125 N. If the masses of the comets are 4.5 × 1013 kg and 1.2 × 1014 kg,

what is the distance separating them?


NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 8AROTATIONAL EQUILIBRIUM AND DYNAMICS

P R O B L E MWhile driving an automobile, the driver makes a left turn. To performthis maneuver, the driver exerts a torque with a magnitude of 3.5 N•m onthe rim of the steering wheel. If the radius of the wheel is 0.15 m, what isthe magnitude of the force applied by the driver?

S O L U T I O NGiven: t = 3.5 N•m

d = 0.15 m

Unknown: F = ?

Choose the equation(s) or situation: Apply the definition of torque. Note that

the force is applied to the rim of a wheel, so that the angle between the force and

the lever arm (radius) is always 90.0°.

t = Fd(sin q) = Fd


F = d

t


F =

F =

Note that the magnitude of torque is not a good indicator of the force that pro-

duces the torque. A small torque can be the product of a large force acting on a

small lever arm, or a small force acting on a large lever arm.

23 N

3.5 N•m0.15 m

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. A lever is used to lift a boulder. The fulcrum is placed 1.60 m away

from the end at which you exert a downward force, producing a torque

with a magnitude of 4.00 × 102 N•m. If the angle between the force

and the lever is 80.0°, what is the magnitude of the applied force? As-

sume that the lever is massless.

2. Suppose the applied force in problem 1 produces a counterclockwise

torque. If the net torque exerted on the lever in problem 1 is 14.0 N•m

counterclockwise, what is the weight of the boulder? Assume that the

lever arm between the boulder’s center of mass and the fulcrum is

0.200 m and that the angle between the boulder’s weight and the lever

arm is 80.0°.


3. Small windmills have been used for over a century to pump water on

farms and ranches in the United States. The rotors of these mills con-

sist of 18 metal blades called “sails.” Even a small wind can provide

enough torque for drawing water from underground wells. If the

length of a sail is 2.44 m and the torque exerted by the wind is

50.0 N•m counterclockwise, what is the magnitude of the wind’s force?

Assume that this force is exerted perpendicular to the blade and at the

blade’s tip.

4. A force is applied to a door at an angle of 60.0° and 0.40 m from the

hinge. What force produces a torque with a magnitude of 1.4 N•m?

How large is the maximum torque this force can exert?

5. The force exerted by the driving rods of a steam locomotive has a mag-

nitude of 2.27 × 105 N. Each rod is connected to one of the driving

wheels at a point halfway between the center and the rim of the wheel.

Suppose the driving wheel has a radius of 0.660 m. How large is the

maximum torque exerted on the driving wheels by the driving rods?

6. The world’s narrowest street, which is located in a small Italian village,

is only 43 cm wide. Suppose a fish with a mass of 1.6 kg is hung from a

string attached to a stick. The stick, slightly longer than the street is

wide, is placed horizontally across the narrow street with each end rest-

ing on a windowsill. The fish hangs a horizontal distance of 15 cm

from the windowsill on the right. If the axis of rotation for the stick is

taken to be the end farthest from the fish, what is the magnitude of the

torque produced by the fish? Assume the stick has negligible mass.

7. A golfer produces a torque with a magnitude of 0.46 N•m on a golf

club. If the club exerts a force with a magnitude of 0.53 N on a station-

ary golf ball, what is the length of the club?

8. In 1902, fresh water was provided to San Francisco, California, by two

large Dutch-style windmills on the western edge of the city. Though

not in use, both mills are still standing. Suppose a worker is restoring

one of these windmills when the ladder shifts to the side. The worker

grabs the end of one of the rotor vanes and hangs onto it until fellow

workers come to the rescue. The worker hangs at an angle of 65.0° to

the vane, exerting a counterclockwise torque of 8.25 × 103 N•m. If the

worker weighs 587 N, what is the length of the windmill vane?

9. A Foucault pendulum consists of a large balanced mass hanging on the

end of a long wire. The pendulum swings freely, so that the rotation of

Earth causes the pendulum’s surroundings to slowly shift position with

respect to the pendulum’s motion. The pendulum’s apparent path dur-

ing one rotation of Earth is an indication of both Earth’s rotation and

the location of the pendulum on Earth’s surface. At the point where a

28-kg pendulum has the greatest potential energy, the angle between

the pendulum’s weight and the wire is 89°. What is the wire’s length if

the magnitude of the torque exerted by the pendulum’s weight at this

position is 1.84 × 104 N•m?

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10. At the moment before a diver jumps from a diving board, a force of

1.200 × 103 N is exerted on the diver at an angle of 90.0° to the board.

This force produces a torque in the clockwise direction. At the same

time, the diver’s weight produces a torque in the counterclockwise di-

rection. The diver’s mass is 60.0 kg, and the angle between the diver’s

weight and the board is 87.7°. If the net torque acting on the diver is

2985 N•m clockwise, what is the length of the diving board?


Holt Physics

Problem 8BROTATIONAL EQUILIBRIUM

P R O B L E MThe narrowest navigable strait lies between the mainland of Greece andthe island of Euboea. Imagine that a uniform bridge spans the gap andthat a car with a mass of 8.2 102 kg is on the bridge. The car is 10.0 mfrom the Euboea side of the strait. The upward force exerted by the landsupporting the bridge on the Greece side is 2.40 104 N, while the up-ward force exerted by the land on the Euboea side is 2.80 104 N. What isthe length of the bridge?

S O L U T I O N

Given: mc = mass of car = 8.2 × 102 kg

dc = distance of car from Euboea = 10.0 m

F1 = upward force on Greece side of bridge = 2.40 × 104 N

F2 = upward force on Euboea side of bridge = 2.80 × 104 N

g = 9.81 m/s2

Unknown: l = length of bridge = ?

Diagram:

Greece Euboea

F1

mbg mcg

F2

dc

1. DEFINE

2. PLAN

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Choose the equation(s) or situation:

Apply the first condition of equilibrium: To find the length of the bridge, the

weight of the bridge must first be determined. All forces are in the vertical (y)

direction.

F1 + F2 − mc g − mbg = 0

Choose a point for calculating net torque: Either end of the bridge may be cho-

sen. Choosing the Euboea side, note that the torque produced by F2 is zero.

Apply the second condition of equilibrium: The torques produced by the

bridge’s and car’s weights are counterclockwise and therefore positive. The nor-

mal force F1 exerts a clockwise, and therefore negative, torque. The lever arm for

F1 is l, while the lever arm for the weight of the bridge is l /2.

tnet = mbg2

l + mcgdc − F1l = 0


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mbg = F1 + F2 − mcg

m

2bg − F1l + mcgdc = 0

l =


mbg = 2.40 × 104 N + 2.80 × 104 N − (8.2 × 102 kg)(9.81 m/s2)

mbg = 5.20 × 104 N − 8.0 × 103 N

mbg = 4.40 × 104 N

l =

l =

l =

l =

While the weights of the bridge and car remain constant, the normal forces vary

as the car moves across the bridge (F1 increases as F2 decreases by an equal

amount). The increase in F1, and the resulting increase in the clockwise torque, is

balanced by the increase in the counterclockwise torque produced by the car. The

clockwise torque increases because of an increase in force, while the counter-

clockwise torque increases because of the lengthened lever arm, dc.

4.0 × 101 m

(8.2 × 102 kg)(9.81 m/s2)(10.0 m)

0.20 × 104 N

(8.2 × 102 kg)(9.81 m/s2)(10.0 m)

2.40 × 104 N − (2.20 × 104 N

(8.2 × 102 kg)(9.81 m/s2)(10.0 m)

2.40 × 104 N − 4.40 ×2

104 N

mcgdcF1 −

m

2bg

3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. The largest measured ostrich egg—and therefore the largest bird egg—

had a mass of 2.3 kg. By contrast, an adult dwarf rabbit that was born

in France was reported to have a mass of only 0.40 kg. Suppose you

balance this egg and this rabbit on a twin-pan balance, shifting the ful-

crum of the balance to the point along the beam at which the torques

exerted by the two masses cancel. If the beam’s total length is 1.00 m,

and the pans holding the masses are placed at each end of the beam,

where should the fulcrum be placed? Assume that the beam and pans

have negligible mass.

2. A meter stick with a mass of 139 g is found to balance at the 49.7 cm

mark when placed on a fulcrum. Suppose a 50.0 gram mass is attached

at the 10.0 cm mark. To which mark on the meter stick must the ful-

crum be moved for balance?

3. The Firth of Forth Bridge, which spans the Forth River near Edin-

burgh, Scotland, is one of the oldest bridges with a span greater than 2

km. Built in the late nineteenth century, it is also one of the strongest,


safest, and most expensive bridges ever built. The Forth Bridge (as it is

often called) is a cantilever-type bridge, which means that the most of

the bridge spans are balanced over massive supports and anchored at

one end. Short spans are placed at the free ends of the cantilevers, thus

completing the bridge. Consider a single cantilever shown in the figure

below. The anchored arm of the bridge, A, is shorter and more massive

than the cantilever arm, C. For the Forth Bridge, A has a mass of

roughly 4.64 × 107 kg and a length of 3.00 × 102 m. If the connecting

span between cantilevers exerts a downward force of 3.22 × 107 N on

the end of the cantilever arm, and the support exerts an upward force

of 7.55 × 108 N, what is the length of the cantilever arm?

4. A special ladder used by firefighters is mounted on a hydraulic pump

that permits the ladder to be rotated, tilted, and extended from a set of

controls on the fire engine. When a 70.0 kg firefighter is at the end of

the fully-extended ladder, and the ladder makes an angle of 10.0° with

the vertical, a torque of 7.08 × 103 N•m must be exerted by the ma-

chinery at the base of the ladder in order to keep the ladder from rotat-

ing. If the upward force exerted on the ladder at its base is 3.14 × 103 N,

what is the length of the ladder?

5. A cable is attached 32.0 m from the base of a flagpole that is about to

be raised. The raising of the pole is temporarily halted when the pole is

at an angle of 60.0° with respect to the ground. If the cable exerts a ver-

tical force of 1.233 × 104 N downward and a horizontal force of 1.233 ×104 N to the left, what is the length of the flagpole?

6. Archimedes is supposed to have said, “Give me a place to stand on and

I will move the Earth.” While in principle the claim is valid, the prob-

lems of testing it are vast. Consider the following, highly unrealistic

conditions. Suppose you had a mass equal to that of the Earth—5.98 ×1024 kg— and that it rests on one end of a lever that is balanced on a

fulcrum. The fulcrum is placed on a vast, flat surface in a gravitational

field equal to that at Earth’s surface. Let Earth be placed 1.00 m from

the fulcrum, while the distance from the fulcrum to the applied force is

3.8 × 1016 m, or the approximate distance from the sun to the next

closest star. How large must the applied force be to balance the Earth’s

weight? How large is the upward force exerted by the fulcrum on the

lever? Assume the lever to be massless.

support(fulcrum)

cantileverarm (C)

anchorarm (A)

anchoredend

FcsmAg mcg

Fs

dA dC

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7. A storefront sign is hung from the end of a 2.00 m uniform massless

rod over the doorway of the business. Additional support for the sign is

provided by a guy wire that is anchored to the wall of the building and

is attached to the end of the rod. The angle between the rod and the

guy wire is 30.0°. If the guy wire provides a torque of 1.47 × 103 N•m

in the counterclockwise direction, what is the weight of the sign? What

is the magnitude of the force exerted by the wire on the sign? What is

the magnitude of the force exerted by the wall on the rod?

8. A garage door with a weight of 7.10 × 102 N is designed to slide side-

ways by being mounted on an overhead rail. Two small wheels on ei-

ther side at the top edge of the door roll along the rail. Suppose the

wheels have become rusted and no longer roll, and that the coefficient

of static friction is 0.75. The wheels are separated by a distance of

1.22 m. A force is applied on one side of the door at a distance of

1.00 m below the rail. How large can this force be without causing the

wheels to rise from the track or the door to slide sideways?

9. A pane of plate glass is tilted upward from the ground until the pane is

at an angle of 70.0° with respect to the ground. Suppose the mass of

the pane is 307 kg and its height is 2.44 m. If the force required to hold

the pane in place is applied perpendicular to the pane and 1.22 m from

the pane’s base, what is the magnitude of the lifting force? How large is

the force exerted on the pane by the ground?

10. Many vending machines have a label on them warning of the dangers

of trying to tip the machine forward. Fortunately, such machines are so

heavy that an average person cannot exert a large enough force to tilt

the machines. Suppose a hinge is placed along the front part of the ma-

chine’s base. With this an additional safety feature, the machine is an-

chored, so that in the event it tips over it rotates around the hinge

rather than tilting forward and sliding backwards. This increases the

force with which it falls forward. Now suppose that the machine is

tilted forward, so that the center of mass has a horizontal position of

0.250 m to the left of the hinge. A horizontal force is applied 1.50 m

above the hinge to prevent the machine from tilting further. If the ma-

chine weighs 1.96 × 103 N, what is the magnitude of the applied force?

What is the magnitude of the force exerted on the machine at the

hinge?

d1

F

Fg

d2


Holt Physics

Problem 8CNEWTON’S SECOND LAW FOR ROTATION

P R O B L E MA torque of –11 N •m is applied perpendicular to the radius of a potter’swheel. If the angular acceleration on this wheel equals –2.6 rad/s2, what isthe moment of inertia of the wheel?

S O L U T I O NGiven: t = −11 N•m

α = −2.6 rad/s2

Unknown: I = ?

Use the equation for Newton’s second law for rotating objects, and rearrange the

equation to solve for moment of inertia.

t = Ia

I = at

= = 4.2 kg•m2−11 N•m−2.6 rad/s2

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1. A wheel is made to “stand up” by using its diameter as the axis of rota-

tion and by spinning it rapidly. Suppose a torque of 2.98 N•m acts on

this wheel, so that its angular speed increases from 0 rad/s to 55 rad/s

in just 0.75 s. What is the wheel’s moment of inertia? If the wheel’s ra-

dius is 20.0 cm, what is the wheel’s mass, assuming that the wheel is a

thin hoop?

2. A torque of 1.7 N•m must be exerted on a croquet mallet in order for

the mallet to have an angular acceleration of 5.5 rad/s2. What is the

moment of inertia of the mallet?

3. A torque of 0.750 N•m is applied to a plastic flying disk, giving the disk

an angular acceleration of 499 rad/s2. What is the disk’s moment of

inertia?

4. Francis Johnson of Minnesota made a ball of string with a mass of

7.91 × 103 kg and a radius of 1.83 m. Suppose this ball rolls down an

incline with an angular acceleration of 6.13 rad/s2. What is the torque

acting on the ball?

5. The tidal force exerted on Earth by the moon causes Earth’s rotation

speed to slowly decrease. This angular acceleration is equal to –6.53 ×10–22 rad/s2. If the moment of inertia for Earth is described by the

equation 0.331MR2, where M is 5.98 × 1024 kg and R is 6.37 × 106 m,

what is the torque that causes the angular acceleration?

ADDITIONAL PRACTICE


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6. The largest centrifuge equipped for human testing at NASA is capable

of simulating free-fall acceleration up to 20 times that at Earth’s sur-

face. To reach the speeds necessary for a centripetal acceleration as

great as 12g, the centrifuge undergoes a constant angular acceleration

of 1.05 rad/s2. If the moment of inertia of the centrifuge is 8.14 × 104 kg•m2, what is the torque acting on the centrifuge?

7. A clutch consists of two disks whose flat, circular surfaces can be

moved together or apart. In this way, the rotational energy of one disk

can be transferred or removed from the other disk. Suppose the rotat-

ing disk applies a torque of 108 N•m to the stationary disk, which has a

moment of inertia of 5.40 kg•m2. What is the second disk’s angular ac-

celeration?

8. A gyroscope consists of a spinning wheel and shaft that is mounted in a

frame. The stability of the wheel resulting from its large angular mo-

mentum made gyroscopes useful for guiding rockets in the 1940s and

1950s. Suppose a small gyroscope has a moment of inertia of 3.85 ×10–5 kg•m2. If a torque of 1.01 N•m is applied to the rotor of this gy-

roscope, what is the rotor’s angular acceleration?

9. A toy catapult propels a 0.15 kg stone into the air. Assume that the

length of the catapult arm is 0.35 m and that its mass (and therefore its

moment of inertia) is negligible. What is the stone’s angular accelera-

tion if the net torque on the stone is 1.5 N •m? What is the stone’s tan-

gential speed if the angular acceleration lasts for 0.26 s?

10. A 15 kg grindstone with a radius of 0.25 m is initially rotating at an an-

gular speed of 9.5 rad/s. A lawn mower blade pressed against the grind-

stone’s outer edge produces a constant torque of −0.80 N•m. Assum-

ing that the grindstone is a uniform solid disk, what is the angular

acceleration of the grindstone? How long does it take for the grind-

stone to stop?


Holt Physics

Problem 8DCONSERVATION OF ANGULAR MOMENTUM

P R O B L E MAn experiment aboard a spacecraft consists of a cylindrical bucket, inwhich a disk of ice is placed. The bucket and ice are placed on a frictionlessturntable that rotates with an angular speed of 8.000 rad/s. The tempera-ture inside the bucket is then increased until the ice melts, at which pointthe liquid water forms a ring of water along the wall of the bucket. The an-gular speed of the turntable with the bucket and water is now 7.990 rad/s.If the moment of inertia of the turntable, bucket, and ice is 1.620 kg •m2,what is the moment of inertia of the turntable, bucket, and water?

S O L U T I O NGiven: w1 = 8.000 rad/s

w2 = 7.990 rad/s

I1 = 1.620 kg•m2

Unknown: I2 = ?

Choose The equation(s) or situation: Because there are no external torques, the

angular momentum of the turntable, bucket, and water/ice is conserved.

L1 = L2

I1w1 = I2w2


I2 = I

w1w

2

1


I2 =

I2 =

The liquid water redistributes itself because of the bucket’s motion. The ring of

water has a moment of inertia that is larger than a disk of ice with the same mass.

Because the total moment of inertia increases, the angular speed must decrease.

1.622 kg•m2

(1.620 kg•m2)(8.000 rad/s)

7.990 rad/s

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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ADDITIONAL PRACTICE

1. A bucket with an inner radius of 0.120 m is rotated on a frictionless

turntable. Inside the bucket are 25 osmium 22.0-g ball bearings. The

turntable, bucket, and bearings rotate with an angular speed of

50.00 rad/s. Suppose that the bucket has small holes underneath and

that these holes can be uncovered by remote control. At a certain point

the holes are opened and the ball bearings fly out. If the angular speed

of the bucket and turntable is now 50.24 rad/s, what is the combined

moment of inertia of the bucket and turntable?


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2. A type of satellite has a cylindrical shape when launched. In orbit, how-

ever, long arms are extended outward, so that the satellite ultimately

takes the form of a long rod. The satellite rotates about an axis that is

perpendicular to its length and passes through its center. Suppose that

before the arms are extended, the satellite’s mass is 755 kg, the distance

between the axis to the end of the satellite is 1.75 m, and its angular

speed of rotation is 1.25 rad/s. After the arms have been extended, the

angular speed of the satellite is 1.70 × 10–2 rad/s. What is the satellite’s

final moment of inertia?

3. The “Human Roulette Wheel” at Steeplechase Park on Coney Island

was a horizontal spinning wheel with a radius of 3.00 m. Passengers

would try to climb toward the center, where the centripetal accelera-

tion was smaller and the passenger would be more able to stay on the

wheel. Suppose the wheel is set to spinning at an angular speed of

2.00 rad/s and then is allowed to spin freely without any external

torque (including friction) applied to it. Four students, each 90° away

from the next, sit on the rim of the wheel, then crawl at the same speed

toward the center of the wheel. When they reach the wheel’s center,

where they are 0.20 m from the axis of rotation, the angular speed of

the wheel is 2.35 rad/s. If the students each have a mass of 55.0 kg,

what is the moment of inertia of the wheel?

4. A tetherball revolves around a pole with an angular speed of 2.1 rad/s

when the rope, which has a length of 1.2 m, is fully extended. What is

the angular speed of tetherball after the rope has wrapped around the

pole several times, so that the ball is only 0.50 m from the axis of

rotation?

5. Most asteroids in the solar system move in orbits outside that of Earth.

However, there are asteroids whose orbital paths cross Earth’s orbit,

causing concern over the possibility of collisions. Some of these aster-

oids move primarily outside Earth’s orbit, crossing it only when they

are closest to the sun. Others move within Earth’s orbit, crossing it only

when they are farthest from the sun. One of the largest of these second

types of asteroids has a tangential speed of 43.5 km/s when it is 7.00 ×107 km from the sun. What is its speed when it crosses Earth’s orbit at a

distance of 1.49 × 108 km from the sun?

6. A student standing at the center of a large turntable holds a spinning

bicycle wheel by a short axle that passes through its center. The bicycle

wheel is held so that it is over the student’s head, so that its axis of rota-

tion is the same as that for the turntable. The angular speed of the bicy-

cle wheel is 57.7 rad/s, while the turntable and student are at rest. The

bicycle wheel has a radius of 0.350 m and a mass of 3.81 kg, which is

concentrated in the wheel’s rim. The combined moment of inertia of

the turntable, student, and wheel is 2.09 kg•m2. If the student suddenly

stops the spinning wheel, what is the angular speed of the turntable,

student, and wheel?


7. Suppose a satellite identical to the one in problem 2 is launched. The

only difference between this second satellite and the first is that the ini-

tial angular speed is 1.50 rad/s, and the final angular speed is 2.04 ×10–2 rad/s. Use the values for mass and initial arm length from prob-

lem 2 to calculate the final length of the satellite’s arms after they are

completely extended.

8. The planet Pluto is, for most of its orbit, the farthest known planet in

the solar system. However, because its orbit is much more elliptical

than those of the other planets, Pluto is closer to the sun during a

twenty year period of its orbit. It is in the middle of this period of time

that Pluto is closest to the sun. Pluto’s tangential speed is 3.68 km/s

when it is farthest from the sun, at a distance of 7.35 × 109 km. Pluto’s

tangential speed is 6.14 km/s when it is closest to the sun. How close

does Pluto come to the sun?

9. A spacecraft for exploring Mars enters an elliptical orbit around the

planet. The craft travels with a tangential speed of 3403 m/s when it is

3593 km from Mars’s center of mass, or about 2.00 × 102 km from

Mars’s surface. How far is the craft from Mars’s center of mass when its

speed is 3603 m/s? How far is the spacecraft from Mars’s surface?

10. Two skaters, each with a mass of 55.0 kg, approach each other along

parallel paths separated by 5.00 m. They have equal and opposite

speeds of 5.00 m/s. The first skater carries a very light 5.00 m pole, the

end of which the second skater grabs when passing by the first skater.

Now the skaters pull themselves closer to the center of the pole. How

far apart are the skaters if they each have a tangential speed of

15.0 m/s?

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Holt Physics

Problem 8ECONSERVATION OF MECHANICAL ENERGY

P R O B L E MIn the sport of trapshooting, a small clay disk called a “clay pigeon” islaunched into the air by a device called a “trap.” There are many differentdesigns and levels of complexity to a trap, but a simple device consists of arack in which the clay pigeons are placed. The rack is hinged at one end,and is rapidly pulled forward by a large spring. Suppose this spring isused in a different device that provides a clay pigeon with kinetic energyto roll along a flat surface. The spring has a force constant of 1.05 104

N/m and is pulled 4.0 10–2 m from its equilibrium position. Assumingthat the clay pigeon is a uniform disk with a mass of 95.0 g and a radius of5.60 cm, what is its rotational kinetic energy upon launching?

S O L U T I O N

Given: k = 1.05 × 104 N/m

x = 4.0 × 10−2 m

m = 95.0 g

r = 5.60 cm

Unknown: KErot = ?

Choose the equation(s) or situation: Apply the principle of conservation of me-

chanical energy.

MEi = MEf

Initially, the system possesses only elastic potential energy.

MEi = PEelastic = 12

kx2

When the clay pigeon initially moves away from the spring, all of the potential

energy has been converted to translational and rotational kinetic energy.

MEf = KEtrans + KErot = 12

mvf2 + 1

2Iwf

2

Noting that vf = rwf, and for a uniform solid disk, I = 12

mr2, the equation for MEf

takes the following form.

MEf = 12

mr2wf2 + 1

2Iwf

2 = Iwf2 + 1

2Iwf

2

MEf = 32

Iwf2

The translational kinetic energy can now be expressed in terms of MEf.

KEtrans = Iwf2 = 2

3MEf

From conservation of mechanical energy, MEf can be equated to PEelastic.

MEf = MEi = 12

kx2

KEtrans = 23

MEf = 23

MEi = 23

(12

kx2)


12

kx2 = 23

(12

kx2) + KErot

KErot = (12

− 13

)kx2 = 16

kx2

1. DEFINE

2. PLAN



KErot = 16

(1.05 × 104 N/m)(4.00 × 10−2m)2 =

From the general treatment of the conservation of mechanical energy, it may be

noted that the rotational kinetic energy for a rolling disk is always one-third of

the total kinetic energy. Therefore, it is total potential energy.

2.8 J

NAME ______________________________________ DATE _______________ CLASS ____________________

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3. CALCULATE

4. EVALUATE

ADDITIONAL PRACTICE

1. Using the information given in the Sample problem, determine the

translational kinetic energy of the clay pigeon.

2. A child’s ball flies a few centimeters above the ground at a speed of

2.2 m/s. Suppose it lands on the ground without bouncing and rolls

forward. What is the rotational kinetic energy of the ball? Treat the ball

as a hollow sphere with a mass of 55 g.

3. An empty barrel falls off a wagon and rolls down a steep hill. If the hill

is 46.0 m high at the point where the barrel starts rolling, what is the

barrel’s translational kinetic energy when it reaches the bottom of the

hill? Assume the barrel is a thin hoop with a mass of 25.0 kg.

4. Suppose that the same arrangement described in the Sample problem

is used to set a metal hoop into motion. Assume a spring constant of

1.05 × 104 N/m and that the spring is stretched 4.0 cm. If the final an-

gular speed of the hoop is 43.5 rad/s, what is its moment of inertia?

5. A bowling ball has an initial kinetic energy of 45 J when it begins to roll

down a bowling lane. What is the moment of inertia of the ball if its an-

gular speed is 27 rad/s? What is the mass of the ball if its radius is 0.11 m?

6. A returning bowling ball rolls in a track between lanes until it reaches

the front of the lane. The ball then rolls up a sloped track; it rolls very

slowly to a circular track near the bowler. The height of the sloped

track is about 0.60 m. What linear speed must the returning bowling

ball have for it to come to a stop at the top of the sloped track?

7. A plunger surrounded by a spring is used in pinball machines to

launch the ball into the field of play. Suppose the force constant for this

spring is 150 N/m, that the spring is compressed 6.0 cm, and that the

ball has a mass of 67 g. Assuming that the pinball machine is level,

what is the linear speed of the ball after the spring has returned to its

equilibrium position?

8. Pinball machines are slightly tilted, so that gravity provides a constant

force on the ball. Suppose the ball has just been launched with the

speed calculated in problem 7. If the ball has a mass of 67 g and the

machine is tilted 3.5° above the horizontal, how far up the incline can

the ball go? Considering that most pinball machines are about 1.5 m

long, would the ball reach the back of the machine and still have ki-

netic energy?


NAME ______________________________________ DATE _______________ CLASS ____________________

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9. A uniform solid sphere rolls along a horizontal surface at 4.6 m/s. It

then encounters an inclined ramp. What maximum height will the

sphere reach as it rolls up the ramp?

10. A fun slide is constructed so that riders must climb a tower to reach the

top of the slide, at which point they slide down on protective mats.

Suppose a hollow ball with a radius of 15 cm is rolled down the slide.

Assuming that there is no slippage or air resistance, the ball will have

the remarkable angular speed of 102 rad/s when it reaches the end of

the slide. What is the height of the slide?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9ABUOYANT FORCE

P R O B L E MThe average density of Saturn is just 0.685 times that of water. Suppose acube with sides 1.00 m in length could be cut out of Saturn and deliveredto Earth. The cube is allowed to float in fresh water, which has a densityof 1.00 103 kg/m3. What would be the buoyant force acting on the cube?How much of the cube’s volume would be submerged?

S O L U T I O NGiven: rs = (0.685)rw

rw = 1.00 × 103 kg/m3

Vs = (1.00 m)3 = 1.00 m3

g = 9.81 m/s2

Unknown: FB = ?

Vw = ?

Choose the equation(s) or situation: For floating objects, the buoyant force

equals the floating object’s weight.

FB = msg = rsVsg

The volume of the object that is submerged can be calculated by recalling that the

weight of the displaced water equals the buoyant force. The volume of the dis-

placed water equals the submerged volume of the object.

FB = mwg = rwVwg


Vw = rF

w

B

g


FB = (0.685)(1.00 × 103 kg/m3)(1.00 m3)(9.81 m/s2) =

Vw = =

The fraction of a floating object’s volume that is submerged in a fluid is equal to

the ratio of the object’s density to the fluid’s density.

0.685 m36.72 × 103 N(1.00 × 103 kg/m3)(9.81 m/s2)

6.72 × 103 N

1. Now suppose the cube from Saturn described in the sample problem is

submerged in gasoline, which has a density of 675 kg/m3. What would

be the buoyant force acting on the cube?

2. One of the rarest non-radioactive elements is rhenium, which is part of

the same chemical group as manganese. The pure metal also has the

third highest melting point of any solid element and the fourth greatest

density of any element. Imagine a cube of rhenium, which has a den-

ADDITIONAL PRACTICE

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

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sity of 2.053 × 104 kg/m3, that is 10.0 cm long on each side. Suppose

this cube is partially submerged in fresh water. If the apparent weight

of the rhenium cube is 192 N, what is the buoyant force acting on

the cube?

3. The buoyant force on a floating object is equal to its weight as long as

the average density of the object is smaller than the density of the fluid

in which it is floating. Thus, by taking steel, which is nearly 8 times as

dense as water, and shaping it into a large hollow shell, the average den-

sity of the shell can be made less than that of the water and the shell

will float. This is the principle that allows steel ships to float. However,

the side of the ships shell, or hull, must be high enough so that the hull

does not fill with water, causing its density to increase and the buoyant

force supporting it to decrease. Consider a steel hull with a mass of

1.47 × 106 kg. What is the buoyant force exerted by the water on the

hull? Suppose the hull has a flat base that is 2.50 × 103 m2 in area.

Given that the density of sea water is 1.025 × 103 kg/m3, how deep does

the hull sink into the water?

4. Suppose the hull in problem 3 is made of gold instead of steel. Assum-

ing that the density of the steel in the hull of problem 3 was 7.86 ×103 kg/m3 and given the density of gold as 1.93 × 104 kg/m3, what

would the buoyant force be on the gold hull? If the flat base of the gold

hull has an area of 2.50 × 103 m2, how deep does the gold hull sink into

the sea water?

5. The element osmium has the greatest density of any element. Suppose

a cube of osmium with a volume of 166 cm3 is submerged in fresh

water. The apparent weight of the cube is 35.0 N. Given that fresh

water has a density of 1.00 × 103 kg/m3, what is the density of osmium?

6. Ebony is one of the densest of all woods. Suppose a block of ebony

with a volume of 2.5 × 10–3 m3 is submerged in fresh water. If the ap-

parent weight of the block is 7.4 N, what is the density of ebony? As-

sume the water has a density of 1.0 × 103 kg/m3.

7. Gallium is one of only three non-radioactive elements that melts be-

tween 0°C and 40°C, and the only one for which the solid form has a

lower density than the liquid form. Suppose that a cube of solid gal-

lium with a volume of 7.62 cm3, mass of 45.0 g floats in a container of

liquid gallium. Assuming that the measurement is made before any of

the solid gallium melts, the volume of liquid gallium displaced by the

solid is 7.38 × 10–6 m3. Use this information to calculate the densities

of solid and liquid gallium.

8. The composition Density 21.5 is so-called because it was written for a

musician playing a platinum flute (the density of platinum being

21.5 g/cm3). Suppose this flute is submerged in fresh water, which has a

density of 1.00 g/cm3. If the apparent weight of the submerged flute is

40.2 N, what is the flute’s mass?


NAME ______________________________________ DATE _______________ CLASS ____________________

9. With a density of only 534 kg/m3, lithium has the lowest density of any

metallic element. Suppose a block of lithium floats on a surface of

gasoline, which has a density of 675 kg/m3. If 5.93 × 10–4 m3 of gaso-

line is displaced by the lithium, what is the mass and volume of the

lithium block?

10. A boat floating in a large tank of fresh water contains a passenger and a

heavy object of uniform density. When the boat, passenger, and object

are placed in the water, the surface of the water rises 5.00 cm because of

the volume of water displaced by the buoyant force. The object is then

dropped into the water, where it sinks to the bottom of the tank. The

surface of the water drops from 5.00 cm above its original level to

4.30 cm above its original level. (The increased displacement of the

water due to the submerged object equals the mass of the object, not its

volume. Because the object is denser than the water, the overall volume

of displaced water is less than when the object was floating with the

passenger in the boat.) Use the equation for buoyant force on floating

objects to calculate the mass of the submerged object, given that the

area of the displaced water is 3.4 m2 and that the density of the water is

1.0 × 103 kg/m3.

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Holt Physics Problem WorkbookCh. 9–4

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9BPRESSURE

P R O B L E MThe surface of the planet Jupiter is believed to consist of liquid hydrogen.Above this surface lies a thick atmosphere that exerts a pressure of 1.00 107 Pa on Jupiter’s surface. If the total force exerted by this atmosphere is6.41 1023 N, what is the area of Jupiter’s surface?

S O L U T I O N

Given: P = 1.00 × 107 Pa

F = 6.41 × 1023 N

Unknown: A = ?

Use the equation for pressure, and rearrange it to solve for the area.

P = A

F A =

P

F

A = 6

1

.

.

4

0

1

0

××

1

1

0

0

2

7

3

P

N

a = 6.41 × 1016 m2

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ADDITIONAL PRACTICE

1. In a certain gasoline engine, the ignited gasoline and air mixture in a

single cylinder exert a pressure of 1.50 × 106 Pa on the upper surface of

the piston. If the force exerted by the gas on the piston is 1.22 × 104 N,

how large is the area on the piston’s upper surface?

2. The two Echo satellites, which were launched in 1960 and 1964, were

early examples of communications satellites. The Echo satellites con-

sisted of aluminum-coated Mylar balloons that reflected radio signals

from one point on Earth to another location on Earth’s surface. Suppose

the gas pressure inside Echo 1 was the same as the atmospheric pressure

at Earth’s surface. If the total force exerted on the inner surface of Echo 1

was 2.86 × 108 N, what was the surface area of the satellite? Given that

the surface area of a sphere is 4pr2, what was the radius of the satellite?

3. A ball strikes the pavement with a force of 5.0 N. If the pressure exerted

on the pavement is 9.6 × 103 Pa, what is the area of contact between the

ball and the pavement?

4. A hydraulic lift consists of a piston, on which is placed an automobile

with a mass of 1.40 × 103 kg. The other piston consists of a block of ice

with a uniform thickness of 0.076 m. Given that the density of ice is

917 kg/m3, what is the area of the piston holding the automobile? As-

sume that the weight of the automobile is uniformly distributed over

the piston’s surface. Assume the piston to be massless.


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A door with an airtight fit around its edge has an area of 1.54 m2. Sup-

pose the pressure on the inside of the door is 1.00 percent greater than

the pressure on the other side of the door. If the outside pressure is

1.013 × 105 Pa, how large a force must you exert on the door to pull it

open? How large a mass can just be lifted by this force?

6. A neutron star is a remnant of a massive star that has undergone an

explosion called supernova. The neutron star’s mass, which mainly con-

sists of neutrons, can be 1.5 times that of the sun. Yet the neutron star’s

radius is only about 6.0 km. Suppose the neutron star is spherical with a

radius of 6.0 km, its surface consists of closely packed neutrons, and the

pressure beneath the surface layer of neutrons is 1.2 × 1016 Pa. If the

surface area of a sphere is 4pr2, what force is exerted by the neutrons?

7. An engineer devises an elevator for tall buildings that does not rely on

cables suspending a car. The elevator consists of a piston that is raised

by a non-compressible fluid, which is pushed upward by another,

larger piston. Suppose the elevator piston is designed to rise 448 m

above street level and that the maximum load that can be lifted is

4.45 × 104 N. What force must be exerted on the larger piston if its

maximum displacement is 8.00 m? Assume the pistons to be massless.

8. Atmospheric pressure is often given in units of millimeters of mercury.

This refers to the height of a mercury column above the mercury’s sur-

face at the base of a barometer. The force exerted by the atmosphere on

the surface of the mercury in the reservoir equals the weight of the mer-

cury in the column. If the mercury column extends 760 mm above the

mercury’s surface in the reservoir, what is the atmosphere’s pressure

over the mercury? Use 13.6 × 103 kg/m3 for the density of mercury.

9 Osmium, the densest of all known elements, is rarely found in Earth’s

crust. This is partly because, being more dense than other elements,

most osmium present on Earth has drifted toward Earth’s center. Sup-

pose Earth’s supply of osmium was mined and formed into a single

massive cylinder. If the mass of this cylinder were 2.4 × 1013 kg and the

area of the cylinder were 3.14 km2, the cylinder would be 340 m high.

If this cylinder could be placed on a hydraulic lift with a piston area of

3.14 km2, what pressure would be necessary to support the osmium

cylinder? Assume the piston to be massless.

10. When a can of soda is shaken vigorously, carbon dioxide is temporarily

separated from the water. This causes the pressure inside the can to in-

crease. If the soda is opened before the carbon dioxide can be redis-

solved in the water, the gas—and soda—will rapidly escape through

the opening. Suppose the force that the carbon dioxide gas exerts on

the inside of the sealed can is 4.4 × 103 N. If the surface area of the can

is 2.9 × 10–2 m2, what is the absolute pressure exerted by the carbon

dioxide? What is the gauge pressure of the gas, assuming that the pres-

sure of the air outside the can is 1.0 × 105 Pa?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9CPRESSURE AS A FUNCTION OF DEPTH

P R O B L E MSaturn’s largest moon, Titan, is actually larger than the planet Mercury. Itis also one of the few bodies in the solar system that has a dense atmos-phere. Data from the Voyager 1 probe indicates that the atmosphericpressure on Titan’s surface is about 1.60 105 Pa, even though free-fallacceleration at Titan’s surface is only 1.36 m/s2. Assuming that Titan’s at-mosphere has a uniform density of 2.32 kg/m3, at what height above thesurface is Titan’s atmospheric pressure equal to 1.01 105 Pa?

S O L U T I O N

Given: Po = 1.60 × 105 Pa

gT = 1.36 m/s2

r = 2.32 kg/m3

P = 1.01 × 105 Pa

Unknown: h = ?

Use the equation for fluid pressure as a function of depth, and rearrange it to

solve for height, which yields a negative value for h.

P = Po + rgh

h = P

r−

g

Po

h =

h =

h = −1.9 × 104 m

h = 19 km above Titan’s surface

−5.9 × 104 Pa(2.32 kg/m3)(1.36 m/s2)

1.01 × 105 Pa − 1.60 × 105 Pa

(2.32 kg/m3)(1.36 m/s2)

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ADDITIONAL PRACTICE

1. The barheaded goose breeds in Tibet and spends the winter in India.

This migration requires the goose to fly at extremely high altitudes, so

that it can fly over the Himalayan mountains. The average atmospheric

pressure at the goose’s minimum altitude is 6.9 × 104 Pa. Because air is

not an incompressible fluid, its density is not uniform at all altitudes;

therefore assume that the air has a much lower average density of

0.55 kg/m3. From this data, calculate the altitude at which the bar-

headed goose flies. Use the value of 1.01 × 105 Pa for the atmospheric

pressure at sea level.


NAME ______________________________________ DATE _______________ CLASS ____________________

2. Throughout Yucatán, Mexico, there are sinkholes called cenotes. Filled

with water and often connected to underground caves, these cenotes

form some of the world’s deepest pools. Rumored to be the deepest of

these is the Cenote Azul, which has sheer walls and a true depth that

has never been measured. At the cenote’s greatest known depth, the

pressure is 1.47 × 106 Pa greater than the pressure at the water’s sur-

face. Using a density of 1.00 × 103 kg/m3 for the density of the water,

determine the maximum known depth of the Cenote Azul.

3. The surface of Lake Tahoe, California, lies 1899 m above sea level. At

this elevation, the average atmospheric pressure is about 9.0 × 104 Pa.

At the deepest part of the lake, the pressure is 5.00 × 106 Pa. If the den-

sity of the water in Lake Tahoe is 1.0 × 103 kg/m3, what is the maxi-

mum depth of the lake?

4. At the greatest depths that scuba divers can work (and then only

briefly), the pressure is 4.03 × 105 Pa. Noting that the density of sea

water is 1.025 × 103 kg/m3, what is the maximum depth for scuba

divers?

5. The world’s first underwater hotel was built in 1986 at Key Largo,

Florida. The hotel is located 9.1 below the surface of a mangrove

swamp. If the water had a density of 1.0 × 103 kg/m3, what would be

the gauge pressure across a window in the hotel at a depth of 9.1 m?

6. The lowest elevation in the western hemisphere is 86 m below sea level,

in Death Valley, California. Assuming the density of the air to be

1.29 kg/m3, what is the atmospheric pressure at this location?

7. Cinnabar is the mineral from which mercury is extracted. Consider a

pool of mercury in a cinnabar mine. Given that mercury’s density is

13.6 × 103 kg/m3 and that the atmospheric pressure above the pool is

1.01 × 105 Pa, what is the pressure 1.50 m below the mercury’s surface?

8. The planet Venus has one of the densest atmospheres in the solar sys-

tem. At the surface of Venus, the atmosphere’s pressure is nearly 9.10 ×106 Pa. Even at an elevation of 1.00 km, the atmospheric pressure is

8.60 × 106 Pa. Given that free-fall acceleration on the surface of Venus

is 8.87 m/s2, what is the density of Venus’ atmosphere?

9. Imagine a giant vat containing liquid bromine. The pressure at the sur-

face of the bromine is 1.01 × 105 Pa, but at a depth of 3.99 m below the

liquid’s surface the pressure is 2.23 × 105 Pa. What is the density of liq-

uid bromine?

10. Now imagine the same vat as in problem 9, but containing the organic

compound diethyl ether. If the pressure at a depth of 3.99 m below the

ether’s surface is 1.29 × 105 Pa, what is the density of the diethyl ether?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9DBERNOULLI’S EQUATION

P R O B L E MA feature of dams is a spillway, which provides for excess water from up-stream flooding to bypass the dam. At Hoover Dam on the border of Ari-zona and Nevada, the Arizona spillway is a tunnel with a 15 m diameterthat opens to Lake Mead near the top of the dam. Water levels in the lakehave not been high enough for the spillway to be used since it was firsttested, but if they were to reach that height they would be channeledthrough the spillway to a location downriver from the dam. Assumingthat the bottom of the spillway is on the same level as the bottom of thedam, if water enters the spillway with a speed of nearly 0 m/s, it will leavethe spillway with a speed of 65.83 m/s. Given that the pressure differencebetween the top and the bottom of the dam is 1.2 103 Pa and that freshwater has a density of 1.00 103 kg/m3, what is the height of Hoover dam?

S O L U T I O NGiven: v1 = 0 m/s

v2 = 65.83 m/s

∆P = P2 − P1 = 1.2 × 103 Pa

r = 1.00 × 103 kg/m3

g = 9.81 m/s2

Unknown: ∆h = h1 − h2 = ?

Choose the equation(s) or situation: Because this problem involves fluid flow, it

requires the application of Bernoulli’s equation.

P1 + 12

rvi2 + rgh1 = P2 + 1

2rv2

2 + rgh2


rg(h1 − h2) = P2 − P1 + 12

r(v22 − v1

2)

∆h = h1 − h2 =

∆h = ∆r

P

g +

v22

2

−g

v12

∆h = +

∆h = 0.12 m + 221 m

∆h =

The difference in atmospheric pressures on the flowing water does not contribute

significantly to the water’s motion. In this case, Bernoulli’s equation simplifies to

rgh = 12

rv22

This is simply the principle of mechanical energy conservation applied to fluid of

undetermined volume. The units on either side of the equation are J/m3.

221 m

(65.83 m/s)2 − (0 m/s)2

(2)(9.81 m/s2)

1.2 × 103 Pa(1.00 × 103 kg/m3)(9.81 m/s2)

P2 − P1 + 12

(v22 − v1

2)

g

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

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NAME ______________________________________ DATE _______________ CLASS ____________________

1. A pressure difference of 1.5 × 103 Pa draws water from a well. The

water emerges from the pump with a speed of 17.0 m/s. Given that the

density of the water is 1.00 × 103 kg/m3, what is the depth of the well?

2. A major tourist attraction in the town of Collinsville, Illinois, is the

“world’s largest catsup bottle.” The bottle is actually a water tower

shaped like the bottle of a once-locally produced catsup. The 100 000-

gallon tank is mounted on a tower that is about 69 m above the

ground. Imagine a water tank that is as tall as the catsup bottle, but is

cylindrical, open to the air, and filled to the top. A hole at the base of

the tank causes a horizontal stream of water to land 120 m horizontally

from the hole and 69 m below the base of the tank. If the difference in

atmospheric pressure is nearly the same at the top and base of the tank,

how tall is the tank?

3. A hydraulic ram is a device that uses the pressure created by a large

amount of water falling a small distance to raise a portion of that water

to a much greater height. It is suitable in valleys where springs feed a

stream below, and where farms needing irrigation are at the top of the

valley. Suppose a spring feeding a hydraulic ram pours water with a

speed of 2.00 m/s into a pipe, which descends a certain distance to the

base of the ram. At the ram’s base, the water has a speed of 7.93 m/s.

How far has the water descended? Assume that the pressure at the two

levels is the same.

4. Now water is raised up the hill by the hydraulic ram to a water storage

tank. The water leaves the ram with a speed of 24.45 m/s and arrives

with a speed of just 0.55 m/s. If there is no significant change in pres-

sure between the two levels, what is the height to which the water is

raised?

5. Rainwater flows through a drainage conduit with an initial speed of

2.50 m/s. How fast does it emerge from the other end of the conduit,

which is 3.00 m below the upper opening? Assume that the pressure is

the same at both ends of the conduit.

6. Sea water flows through a level pipe in a tidal power station. The initial

speed of the sea water is 0.90 m/s. If the water’s pressure increases by

311 Pa as it passes through another section of pipe, what is the water’s

speed in the second section? The density of sea water is 1.025 × 103 kg/m3.

7. A device called a Pitot tube uses the change in height a liquid’s height

to measure the speed a fluid passing through the tube. Suppose a Pitot

tube is attached to the wing of an airplane, which flies at a low altitude

above the ground. The air flowing through the Pitot tube has a density

of 1.3 kg/m3. The part of the tube used for measurement contains mer-

cury, which has a density of 1.36 × 104 kg/m3. If the displaced mercury

column is 3.5 cm high, what is the speed of the air flowing across the

airplane’s wing? (HINT: Solve Bernoulli’s equation for the change in

pressure of each fluid; then substitute one equation into the other.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

8. Oil transported along a level pipeline has a density of 950 kg/m3. The

oil moves with a speed of 10.00 m/s at one point along the pipeline, but

after traveling 1 km the oil has slowed to a speed of 9.90 m/s. What is

the difference in pressure between the two points in the pipeline?

9. Water leaves a fire hydrant pipe, which has an area of 7.8 × 10–3 m2,

with a speed of 2.0 m/s. The firehose attached to the hydrant has a noz-

zle with an area of 3.1 × 10–4 m2. If the nozzle is 10.5 m higher than the

hydrant, what is the gauge pressure of the water? Use 1.00 × 103 kg/m3

as the density of water. (HINT: Use the continuity equation to express

the Bernoulli equation in terms of the known fluid speed.)

10. A Venturi meter, like the Pitot tube, is a device that measures the

change in pressure in a moving fluid. The pressure change occurs when

the fluid passes through sections of the tube with different cross-

sectional areas. Suppose fresh water flows through a level Venturi

tube. At one section of the tube, where the cross-sectional area is

9.2 × 10–2 m2, the speed of the water is 8.3 m/s. At a second section

of the tube, the cross-sectional area of the tube is 4.6 × 10–2 m2. If

density of the water is 1.0 × 103 kg/m3, what is the pressure difference

between the two sections of the Venturi tube? (HINT: Use the continu-

ity equation to express the Bernoulli equation in terms of the known

fluid speed.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 9ETHE IDEAL GAS LAW

P R O B L E M

An ideal gas kept under constant pressure has a volume of 1.5 m3 whenthe gas is at a temperature of 380.0 K, which is the maximum tempera-ture of the two-week lunar day. The moon’s lowest temperature, whichoccurs during the two weeks of night, is 100.0 K. If the ideal gas is cooledto this temperature, what is its new volume?

S O L U T I O N

Given: V1 = 1.5 m3

T1 = 380.0 K

T2 = 100.0 K

Unknown: V2 = ?

Choose the equation(s) or situation: To solve for the volume under the new

conditions, apply the form of the ideal gas law that relates initial and final states.

P

T1V

1

1 = P

T2V

2

2

Because pressure remains constant, the equation simplifies to

V

I1

1 = V

I2

2


V2 = V

T1T

1

2


V2 = (1.5 m

38

3)

0

(

.0

10

K

0.0 K) =

The estimated answer, using T1 ≈ 4.0 K, is

V2 ≈ 1.5

4

m3

≈ 0.4 m3,

which corresponds closely to the result.

0.39 m3

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1. An ideal gas kept under constant pressure is heated from the lowest

recorded temperature on Earth, 184 K, to the highest recorded temper-

ature, 331 K. If the initial volume of the gas is 3.70 m3, what will its

volume be after heating?

ADDITIONAL PRACTICE

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE


NAME ______________________________________ DATE _______________ CLASS ____________________

2. An ideal gas held at constant temperature has a volume of 0.455 m3

and a pressure of 9.1 × 106 Pa, which is equal to the atmospheric pres-

sure on the surface of the planet Venus. Suppose this pressure is low-

ered to 5.0 × 104 Pa, which is equal to the atmospheric pressure on the

surface of Mars. What will be the volume of the gas after this reduction

of pressure?

3. A balloon filled with helium has a volume of 65.4 m3. The balloon is

insulated so that the temperature of the helium remains constant. The

balloon is launched from the Dead Sea, the lowest point on Earth’s sur-

face, where the pressure is 1.03 × 105 Pa. What will be the volume of

the balloon when it is at an elevation equal to the summit of Mt. Ever-

est, where the atmospheric pressure is 5.84 × 104 Pa?

4. The lowest level of Earth’s atmosphere is the troposphere. At Earth’s

surface in mid-latitude regions, the average temperature and pressure

are 295 K and 1.01 × 105 Pa, respectively. At the top of the troposphere,

at an altitude of about 10 km, the average temperature is about 225 K

and the pressure is 5.10 × 104 Pa. What would be the volume of 5.55 ×1022 particles of an ideal gas at the top and bottom of the troposphere?

5. An ideal gas at a pressure of 7.5 × 104 Pa (the approximate pressure at

the top of Saturn’s atmosphere) has a temperature of 250 K. Suppose

the pressure of the gas is increased to 2.0 × 106 Pa (the estimated pres-

sure on Saturn’s surface). If the volume of the gas is kept constant,

what will its new temperature be?

6. A volume of 1.00 m3 of helium is placed in a balloon. The temperature

of the gas is 295 K. If the gas is kept at constant pressure, what temper-

ature must the helium have for the volume of the balloon to expand to

65.4 m3?

7 Although the conditions are extreme, the ionized gas within the sun’s

core behaves like an ideal gas. The core contains about 2.1 × 1057 gas

particles. The pressure of the gas in the sun’s core is 2.1 × 1016 Pa, while

the volume of the core is 2.1 × 1025 m3. What is the temperature at the

sun’s core?

8. A tank of helium is kept at a constant volume with a pressure of 2.50 ×105 Pa and a temperature of 295 K. Suppose the warehouse where the

tank is stored catches fire, so that the tank and the helium reach a tem-

perature of 506 K. What is the pressure of the gas at this temperature?

9. Hydrogen is the most common element in the universe. Large clouds

of hydrogen are often found in regions between the stars. Some of

these clouds have high temperatures and pressures as stars form within

them. Others have lower temperatures near 1.0 × 102 K. Suppose the

volume of a hydrogen cloud at this temperature is 3.3 × 1043 m3, and

that the density of the hydrogen is 10.0 atoms/cm3. What would the

pressure of the hydrogen gas in the cloud be?

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NAME ______________________________________ DATE _______________ CLASS ____________________

10. The tallest mountain on Earth is Mauna Kea, which is the taller of the

two large volcanoes on the island of Hawaii. The island rises 5995 m

from the ocean floor, and then Mauna Kea rises another 4205 m above

sea level for a total height of 10 200 m (or nearly 2 km higher than

Mt. Everest). Suppose the oceans were drained and all of Mauna Kea

were exposed to air. At the base of the mountain, the air pressure

would be 1.42 × 105 Pa, assuming that air pressure at what was previ-

ously sea level is 1.01 × 105 Pa. If a balloon containing 1.00 m3 of he-

lium is launched from the base of the mountain, its volume will in-

crease to 1.83 m3 by the time it reaches the summit of the mountain. If

the balloon is insulated to keep the helium at constant temperature,

what is the pressure at the mountain’s summit?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 10ATEMPERATURE

P R O B L E MSummertime temperatures on Pluto’s surface average 45 K—the onlyplanet cold enough to keep methane solid. What is this temperature onthe Celsius and Fahrenheit scales?

S O L U T I O NGiven: T = 45 K

Unknown: TC = ? TF = ?


Use the Celsius-Kelvin equation and the Celsius-Fahrenheit equation.

TC = (T − 273)°C = (45 – 273)°C =

TF = 9T

5C + 32 =

9

5 (−228)°F + 32°F = (–4.10 + 102)°F + 32°F

TF = −378°F

−228°C

ADDITIONAL PRACTICE

1. When Mercury was farthest from the Sun in 1974 and 1975, the U.S.

probe Mariner 10 flew by Mercury three times to collect data. Surface

temperatures ranged from 463 K during the day to 93 K at night. Ex-

press this temperature range in degrees Fahrenheit and in degrees

Celsius.

2. On July 10, 1913, the temperature reached 330.0 K Death Valley, Cali-

fornia—the hottest temperature ever reached in the United States. Cal-

culate this temperature in degrees Fahrenheit and in degrees Celsius.

3. On January 21, 1918, Granville, North Dakota had a surprising change

in temperature. Within 12 hours, the temperature changed from 237 K

to 283 K. What is this change in temperature in the Celsius and

Fahrenheit scales?

4. In only 15 minutes, the temperature in Fort Assiniboine, Montana,

went from −5°F to +37°F on January 19, 1892. Calculate this change in

temperature in kelvins.

5. Hypothermia is a condition in which the body gives up to much energy

by heat to its colder surroundings. If a person’s body temperature

drops to 90.0°F, they can lose consciousness, and if their temperature

falls to 78°F, they can die. What is this second temperature in kelvins?


NAME ______________________________________ DATE _______________ CLASS ____________________

6. Just as the human body cannot survive if its temperature falls to too

low a temperature, it also cannot survive if its temperature is too high.

In a condition called hyperthermia, energy is transferred to the body

from its surroundings, causing the body’s temperature to increase. The

condition known as heat stroke is a severe form of hyperthermia. Nor-

mally, the human body cannot survive for long at a temperature of

about 42°C, although a recent survivor of heat stroke had a high tem-

perature of nearly 47°C. Express both of these temperatures in kelvins.

7. Air that slowly falls from high altitudes can result in cold fronts that

sweep across many states. At 30,000 feet above Earth’s surface, the air

temperature can be around –67°F. Find this temperature in kelvins.

8. The Hawaiian lavas at Kilauea Crater have the highest temperatures

measured on Earth’s surface—over 2192°F! Express this temperature in

degrees Celsius.

9. Much of the hot water in Reykjavík, Iceland comes from wells bored

into the hot springs of Reykir. The water temperature from the wells is

188.6°F. Express this temperature in degrees Celsius.

10. The present temperature of the background radiation in the universe is

2.7 K. What is this temperature in degrees Celsius?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 10BCONSERVATION OF ENERGY

P R O B L E MA bright fireball zoomed across four states on October 9, 1992. Supposethe meteorite had a speed of 35.0 km/s just before it smashed through thetrunk of a Chevrolet Malibu in Peekskill, New York. At the time the car’sowner discovered the 11.99 kg meteorite on the ground beneath the car,the meteorite was still warm. Suppose that, at the time of impact, the in-ternal energy of the meteorite had increased by an amount equal to 0.001times the kinetic energy before impact, while the rest of the kinetic en-ergy was transferred to the ground, car and atmosphere. If 5348 J of en-ergy were required to increase the meteorite’s temperature by a degreeCelsius, what was the meteorite’s temperature upon impact?

S O L U T I O NGiven: vi = meteorite speed = 35.0 km/s

m = meteorite mass = 11.99 kg

k = energy needed to raise meteorite’s temperature by 1°C

= 5348 J

∆Umeteorite = (0.00100)KEi

Unknown: ∆T = ?

Choose the equation(s) or situation: Use the equation for energy conservation,

including the term for the change in internal energy.

∆PE + ∆KE + ∆U = 0

∆U stands for the change in the internal energy of the meteorite, car, and

ground. The change in potential energy from just before to after impact and

the final kinetic energy are both zero.

∆PE + ∆KE + ∆U = 0 + KEf – KEi + ∆U = – KEi + ∆U = 0

∆U = KEi = 12

mvi2

The change in the meteorite’s internal energy is 0.00100 times the total

change in internal energy.

∆Umeteorite = (0.00100)∆U = (0.00100)KEi = (0.0010

2

0)mvi2

To find the change in the meteorite’s temperature, the change in the mete-

orite’s internal energy must be divided by the conversion constant, k.

∆T = ∆Um

keteorite =

(0.001

2

0

k

0)mvi2


∆T =

∆T =

The answer can be estimated by taking the ratio of m to k as 10

2

00. This

∆T ≈ (35)2 °C = 1200° C, which is close to the calculated value.

1370°C

(0.00100)(11.99 kg)(35.0 × 103 m/s)2

(2)(5348 J/°C)C

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1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

NAME ______________________________________ DATE _______________ CLASS ____________________

1. A football game begins by flipping a penny to decide which team will get

the ball first. The 5.25 g penny has a speed of 3.27 m/s just before it hits

the ground. If half of the increase in the internal energy of the ground

and penny is transferred to the penny’s internal energy, by how much

does the penny’s temperature increase after impact? Assume that the

penny’s temperature increases by 1.00° C as 2.03 J are added to its mass.

2. A squirrel drops an acorn from the top of a tree that is 9.5 m tall. When

the acorn lands, 85 percent of the increase in the internal energy of the

ground and acorn is transferred to the acorn’s internal energy. If it

takes 1200 J/kg to increase the acorn’s temperture by 1.0°C, by how

much does the acorn’s temperature increase upon landing?

3. On a hot summer’s day, a child’s triple-dip ice cream cone falls to the

sidewalk. The gravitational potential energy associated with the cone is

initially 6.2 J. After impact, the internal energy of the ice cream cone

increases by an amount equal to 10.0 percent of the increase in the in-

ternal energy of the ground and ice cream cone. What is the increase in

the ice cream cone’s temperature if after impact it takes 180 J to in-

crease the ice cream’s temperature by 1.0°C?

4. Although the peregrine falcon can fly faster in a dive, the fastest-flying

bird in general is the Asian spine-tailed swift, which can fly at 47.5 m/s.

Suppose this bird drops a 7.32-g twig she is carrying to build her nest

while she is flying 151 m above the ground. If after impact 10.0 percent

of the total internal energy of the twig and ground is transferred to the

twig’s internal energy and it takes 8.5 J to raise the twig’s temperature

by 1.0°C, by how much does the twig’s temperture increase when it

strikes the ground?

5. Balls Pyramid, a rock pinnacle off the eastern coast of Australia, is

561.7 m tall. Suppose erosion causes a stone to fall from the summit.

When the stone lands, its kinetic energy is transferred to the internal

energy of the ground and the stone, which increases by 105 J. What is

the mass of the stone?

6. Bones found from a Tyrannosaurus suggest that the dinosaur was 5.5 m

high and up to 14.3 m long. Suppose a Tyrannosaurus dropped its prey

while eating. When it lands, all of the prey’s kinetic energy, 2.77 × 103 J,

is transferred to the internal energy of the ground and prey. What was

the mass of the prey?

7. A person riding a bicycle on level ground reaches a velocity of 13.4 m/s

before stopping. While braking, the internal energy of the brakes,

wheels, and road increase by 5836 J. What is the mass of the cyclist? Cop

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ADDITIONAL PRACTICE

NAME ______________________________________ DATE _______________ CLASS ____________________

8. You can start a campfire by rubbing two dry sticks together. To create

sparks, rest the larger stick on the ground. Press the point of the

smaller stick into the larger stick so that the sticks are at right angles.

You then rotate the smaller stick between your hands. Suppose you

begin to see smoke when you rotate the stick so that you transfer a ki-

netic energy of 2.15 × 104 J. What is the internal energy at the point

where the sticks meet?

9. The fastest roller skater skated over 402 m, reaching a final velocity of

20.5 m/s. Suppose this roller skater has a mass of 61.4 kg. What is the

total change in the internal energy of the skates, wheels, and ground as

the skater skates past the finish line?

10. Suppose on a cold winter day, you forgot to bring your gloves. To warm

your hands, you rub them together vigorously with a kinetic energy of

7320 J. What is the change in the internal energy of the skin on your

hands if 30.0% of the total mechanical energy was transferred to the air

by heat?

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NAME ______________________________________ DATE _______________ CLASS ____________________

P R O B L E M

S O L U T I O N1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

1. DEFINE

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Holt Physics

Problem 10CCALORIMETRY

In 1906, a 636.73-g diamond was found at the Premier Mine in SouthAfrica—making it the world’s largest uncut diamond. After being cut, thediamond pieces were dropped into an insulated water bath containing1.00 kg of water. The water temperature increased by 1.30C, and the dia-mond’s temperature decreased by 15.54°C. What is the specific heat ca-pacity of diamond? The specific heat capacity of water is 4186 J/kg•C.

S O L U T I O NGiven: md = 636.73 g mw = 1.00 kg ∆Td = 15.54°C

∆Tw = 1.30°C cp,w = 4186 J/kg•°C

Unknown: cp,d = ?

Choose the equation(s) or situation: Use the equation for specific heat capacity to

equate the energy removed from the diamond to the energy absorbed by the water.

Energy removed from diamond = Energy absorbed by water

cp,dmd∆Td = cp,wmw∆Tw


cp,d =

Substitute values into the equation(s) and solve:

cp,d = =

The answer can be estimated using rounded values for cp,w (4200 J/kg•°C), md 23

kg, and

the ratio of ∆Tw to ∆Td 1

1

2. The resulting value for cp,d is then 530 J/kg•°C, which is close

to the calculated value.

5.50 × 102 J/kg•°C(4186 J/kg•°C)(1.00 kg)(1.30°C)

(0.63673 kg)(15.54°C)

cp,wmw ∆Twmd∆Td

ADDITIONAL PRACTICE

1. Mixing equal parts of hydrogen peroxide and water to use as a mouth-

wash disinfects the mouth and whitens teeth. Suppose you mix 15 g of

each into a plastic foam cup. The water’s temperature changes 1.0°C

and the hydrogen peroxide changes 1.6°C. Disregarding energy transfer

as heat to the solution’s surroundings, what is the specific heat capacity

of hydrogen peroxide?

NAME ______________________________________ DATE _______________ CLASS ____________________

2. Vinegar, which contains acetic acid, can be used as an effective and

environmentally-friendly household cleanser. Suppose you mix 0.340 kg

of vinegar at 21.0°C with 1.00 kg hot water at 90.0°C in a plastic bucket.

The solution of vinegar and water reaches a final equilibrium tempera-

ture of 73.7°C. Disregarding energy transfer as heat to the surrounding

air and bucket, what is the specific heat capacity of vinegar?

3. After eating a hearty stew you cooked over a campfire with your

0.250-kg aluminum-alloy pot, you place the pot in a plastic bucket

containing 1.00 kg of water. The water’s temperature increases 1.00°C

and the temperature of the pot decreases 17.5°C. Disregarding energy

transfer as heat to the surrounding air and bucket, what is the specific

heat capacity of the pot?

4. Suppose you bake cornbread in a 3.0-kg cast iron skillet. After remov-

ing the cornbread from the oven, you place the hot skillet in a sink

filled with 5.0 kg of dishwater. The water’s temperature increases

2.25°C, and the temperature of the skillet decreases 29.6°C. Disregard-

ing energy transfer as heat to the surrounding air and sink, what is the

specific heat capacity of the skillet?

5. The water in a swimming pool transfers 1.09 × 1010 J of energy as heat

to the cool night air. If the temperature of the water, which has a spe-

cific heat of 4186 J/kg•°C, decreases by 5.0°C, what is the mass of the

water in the pool?

6. Bismuth’s specific heat is 121 J/kg•°C, the lowest of any non-

radioactive metal. What is the mass of a bismuth sample if 25 J raises

its temperature 5.0°C?

7. The temperature of air in a foundry increases when molten metals cool

and solidify. Suppose 45 × 106 J of energy is added to the surrounding

air by the solidifying metal. The air’s temperature increases by 55 °C,

and the air has a specific heat capacity of 1.0 × 103 J/kg•°C. What is the

mass of the heated air?

8. A 0.190 kg piece of copper is heated and fashioned into a bracelet. The

amount of energy transferred as heat to the copper is 6.62 × 104 J. If

the specific heat of copper is 387 J/kg•°C, what is the change in the

temperature of the copper?

9. A 0.225 kg sample of tin, which has a specific heat of 2.2 × 103 J/kg•°C,

is cooled in water. The amount of energy transferred to the water is 3.9

× 104 J. What is the change in the tin’s temperature?

10. Tantalum is an element used, among other things, in making aircraft

parts. Suppose the properties of a tantalum part are being tested at

high temperatures. Tantalum has a specific heat of about 140 J/kg•°C.

The part, which has a mass of 0.23 kg, is cooled by being placed in

water. If 3.0 × 104 J of energy is transferred to the water, what is the

change in the part’s temperature?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 10D

P R O B L E MCarbon tetrachloride, a toxic compound used in dry cleaning, evaporatesat 76.7°C. It is therefore important to use it in well-ventilated areas. If ittakes 4.85 × 105 J to evaporate 2.50 kg of liquid carbon tetrachloride,what is the compound’s latent heat of vaporization?

S O L U T I O N

Given: Q = 4.85 × 105 J

m = 2.50 kg

Unknown: Lv = ?

Use the equation for latent heat, rearranging it to solve for Lv.

Q = mLv

Lv = m

Q =

4.8

2

5

.5

×0

1

k

0

g

5 J = 1.94 × 105 J/kg

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ADDITIONAL PRACTICE

1. Prior to the Bronze age, which began around 2200 B. C., the metal used

occasionally for tools was copper. Copper is not as hard or sturdy as

bronze, an alloy that combines copper with tin; but it is easy to locate

in mineral deposits by its color and the color of copper compounds. In

order to melt copper, it must be raised to a temperature of 1083°C. At

this temperture, 1.10 106 J of energy must be absorbed by 5.33 kg of

copper in order for the copper to completely melt. What is the latent

heat of fusion of copper?

2. Bronze, which is made by combining copper and tin in different pro-

portions, offers advantages that copper or tin alone lack. Not only is

bronze harder than either copper or tin, but bronze has a lower melting

point than copper, so that bronze is somewhat easier to manufacture. A

bronze alloy containing 15 percent tin and 85 percent copper has a

melting point approximately 950°C. At this temperature, 9.6 × 105 J of

energy must be transferred as heat to 5.33 kg sample of bronze in order

for the bronze to completely melt. What is the latent heat of fusion of

this particular bronze alloy?

NAME ______________________________________ DATE _______________ CLASS ____________________

3. At atmospheric pressure, solid carbon dioxide, or dry ice, undergoes a

process of evaporation called sublimation. In this process, at the tem-

perature of phase change, the solid dry ice goes directly into the gas

phase without becoming liquid. For carbon dioxide, the temperature of

sublimation is –78.44°C. At this temperture, 3.72 × 105 J of energy

must be transferred as heat to a 0.650 kg block of dry ice in order for

the carbon dioxide to completely sublime. What is the latent heat of

sublimation of carbon dioxide at atmospheric pressure?

4. A blacksmith has been hammering a piece of red-hot iron into the

shape of a horseshoe. To cool and temper the iron, he places the still

hot horseshoe into a barrel of water. This causes some of the water to

evaporate. Suppose the liquid water is initially at 100°C. The horseshoe

gives up 8.5 × 104 J of energy as heat to the water. Given that the latent

heat of vaporization for water a 100°C is 2.26 × 106 J/kg, what is the

mass of the water that evaporates?

5. A quantity of ethyl alcohol at a temperature of 78°C (or ethanol) ab-

sorbs 2.11 × 106 J from the surrounding air, causing it to completely

evaporate. If the latent heat of vaporization of ethyl alcohol at 78°C is

8.45 × 105 J/kg, what is the mass of the ethanol that is vaporized?

6. At a foundry, 250 kg of molten iron is poured into a mold. The iron so-

lidifies at a temperature near 1500°C and then further cools as a solid.

During this process, 1.380 × 108 J of energy is given up as heat by the

iron. What is the overall change in the temperature of the iron? The

specific heat capacity of solid iron at high temperatures is 605 J/kg•°C,

and the latent heat of fusion for iron is 2.47 × 105 J/kg.

7. A sample of lead with a mass of 1.45 kg is heated until it reaches its

melting point at 330°C. The lead is then heated further until it has en-

tirely melted. During this process the lead absorbs a total of 4.46 × 104 J

as heat. What is the temperature at which the lead is initially heated?

The specific heat capacity of solid lead over the range of temperatures

in question is 120 J/kg•°C, and the latent heat of fusion for lead is 2.45

× 104 J/kg.

8. Water vapor with a mass of 0.75 g and a temperature of 100.0°C con-

denses to form a raindrop with the same mass. By the time the rain-

drop reaches the ground, it has given up 2.0 × 103 J as heat. What is the

raindrop’s final temperature? Assume that the entire process takes place

at atmospheric pressure, that the specific heat capacity for liquid water

is 4200 J/kg•°C, and that the latent heat of vaporization for water is

2.26 × 106 J/kg.

9. Gallium melts at 29.78°C, a temperature that can be achieved by the at-

mosphere at Earth’s surface. In fact, gallium can melt in your hand.

Suppose a person holds 0.35 kg of gallium at its melting poiint. How-

much energy must be transferred as heat to the gallium in order for it

to completely melt? Gallium’s latent heat of fusion is 8.02 × 104 J/kg.

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NAME ______________________________________ DATE _______________ CLASS ____________________

10. A 55.0-g sample of mercury has an initial temperature of 20.0°C. It is

heated to 357°C, at which it changes phase from a liquid to a gas. How

much energy must be transferred as heat to the mercury for it to com-

pletely evaporate? The average specific heat capacity for mercury over

this temperature range is 130 J/kg•°C, and the latent heat of vaporiza-

tion for mercury is 2.95 × 105 J/kg.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 11AWORK DONE ON OR BY A GAS

P R O B L E MTo measure blood pressure, an inflatable rubber cuff is wrapped aroundthe upper arm and inflated by squeezing a rubber bulb connected to it by

a tube. Suppose the cuff is inflated from 0 m3 to 8.45 104 m3. If the airin the cuff does 16.9 J of work on an arm, what is the pressure of the air inthe cuff in excess of atmospheric pressure?

S O L U T I O NGiven: ∆V = 8.45 × 10–4 m3 W = 16.9 J

Unknown: P = ?


Use the equation that defines work in terms of changing volume, and rearrange

the equation to solve for pressure.

W = P∆V

Rearrange the equation(s) and isolate the unknown(s):

P = ∆W

V =

8.45

1

×6

1

.9

0−J4 m3 = 2.00 × 104 Pa

1. Three people inflate an empty hot-air balloon to a volume of 2190 m3.

If the air inside the balloon does 3.29 × 106 J of work on the balloon,

how much in excess of atmospheric pressure must the pressure of the

air be inside the balloon?

2. The air in a research balloon does 1.06 × 106 J of work to raise the bal-

loon to an altitude of 35 km. The balloon inflates from 2.80 × 103 m3

to 8.50 × 105 m3. How much in excess of atmospheric pressure must

the pressure of the air be inside the balloon?

3. Steam moves into the cylinder of a steam engine at a constant pressure

and does 1.3 J of work on a piston. The volume increases by 5.4 × 10–4 m3.

How much in excess of atmospheric pressure is the pressure of the

steam?

4. Suppose 472.5 J of work was required to fill an empty fire extinguisher

with compressed nitrogen gas. If the fire extinguisher was filled at a

constant pressure that was 25.0 kPa in excess of atmospheric pressure,

what was the change in the volume of the nitrogen gas?

5. Suppose you must change a flat inner tube in a bicycle tire. It requires

393 J of work to inflate the empty inner tube at a pressure of 655 kPa in

excess of atmospheric pressure. What is the change in the inner tube’s

volume?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

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6. A child blows on a wand dipped in a soapy solution to form bubbles.

Suppose 0.20 J of work is required to inflate a soap bubble. If the bub-

ble was filled at a constant pressure that was 39 Pa in excess of atmos-

pheric pressure, what is the change in the bubble’s volume?

7. After the combustion of fuel in a car’s engine, a piston in one of the

cylinders moves outward. The work done by the gas on the piston is 3.2

× 102 J. If the gas maintains a constant pressure of 4.0 × 105 Pa, what is

the change in the volume of the gas?

8. An empty scuba tank is filled with air at a constant pressure of 2.07 ×107 Pa in excess of atmospheric pressure. The gas expands to fill the

tank, which has a volume of 0.227 m3. How much work is done on the

gas?

9. You may have noticed your ears pop if you yawn at a high altitude.

Suppose the volume of air in your ear increases from 3.375 × 10–6 m3

to 5.694 × 10–6 m3 as the plane you’re in flies to a high altitude. Before

you yawn, the pressure in your middle ear is 101.33 kPa—the same as

it was on the ground. What is the work done by the air on the

eardrum? (Assume that you still haven’t yawned.)

10. When an athlete inhales during strenuous exercise, the volume of air in

the lungs increases from 2.0 × 10–3 m3 to 5.0 × 10–3 m3. During this in-

halation, the pressure of air in the lungs reaches 0.90 kPa in excess of

atmospheric pressure. How much work is done by the air on the lungs?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 11BTHE FIRST LAW OF THERMODYNAMICS

P R O B L E MThe internal energy of a gas decreases by 256 J. If the process is adiabatic,how much work is done on or by the gas?

S O L U T I O NGiven: ∆U = –256 J Q = 0 J

Unknown: W = ?

Choose the equation(s) or situation: Use the equation for the first law of ther-

modynamics.

∆U = Q − W

In an adiabatic process, no energy is transferred as heat, so Q = 0 J

W = Q − ∆U = 0 J − (–256 J) = 256 J

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ADDITIONAL PRACTICE

1. A 2.0 kg metal pipe in the center of a compost pile supports a load of

15 kg. During the day the pipe’s temperature increases from 28°C to

52°C, causing the rod to thermally expand and raise the load 2.3 mm.

What is the work done in this process?

2. A system’s initial internal energy is 39 J. Then 114 J of energy is trans-

ferred to the system as heat. If the final internal energy is 163 J, how

much work is done on or by the system?

3. A system’s initial internal energy is 8093 J. Then 6932 J of energy is

transferred to the system as heat. If the final internal energy is 2.092 ×104 J, how much work is done on or by the system?

4. A steam engine’s boiler converts water to steam by transferring 4.50 ×108 J of energy as heat to the water. If steam escaping through a safety

valve does 3.21 × 108 J of work expanding against the outside atmos-

phere, what is the net change in the internal energy of the water-steam

system?

5. A pressure cooker converts water to steam by transferring 632 kJ of en-

ergy as heat to the water. If steam escaping through a safety valve does

102 kJ of work expanding against the outside atmosphere, what is the

net change in the internal energy of the water-steam system?

6. A gas expands when 867 J of energy is added to it by heat. The expand-

ing gas does 623 J of work on its surroundings. What is the overall

change in the internal energy of the gas?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. The expanding steam from a geyser does 192 kJ of work, and the inter-

nal energy of the system increases by 786 kJ. How much energy is

transferred to the system as heat?

8. At a nuclear power plant, heat from radioactive rods in the reactor

causes water to vaporize into steam. The expanding steam does 602 kJ

of work, and the internal energy of the system increases by 1.09 × 105 J.

How much energy is transferred to the system as heat?

9. The internal energy of the air in a closed car increases by 873 J. How

much energy is transferred as heat into the car?

10. The internal energy of the gas in a closed greenhouse increases by

986 J. How much work is done on or by the gas? How much energy is

transferred as heat into the greenhouse?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 11CHEAT-ENGINE EFFICIENCY

P R O B L E MIf a gasoline engine has an efficiency of 33 percent and receives 660 J of en-ergy as heat from combustion during each cycle, how much work is doneby the engine?

S O L U T I O NGiven: eff = 0.33 Qh = 660 J

Unknown: Wnet = ?

Choose the equation(s) or situation: Rearrange the equation for efficiency.

Wnet = eff Qh = (0.33) (660 J) = 220 J

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ADDITIONAL PRACTICE

1. A steam engine of a locomotive has an efficiency of 17 percent and re-

ceives 5.5 × 109 J of energy by heat from its boiler. How much work is

done by the engine?

2. A coal-burning power plant has an efficiency of 35 percent. If the

power plant uses 7.37 × 108 J of energy as heat, how much work is

done by the power plant?

3. A geothermal power plant has an efficiency of 15 percent. If the power

plant takes in 9.36 × 108 J of energy as heat, how much work is done by

the power plant?

4. If a gasoline engine has an efficiency of 29 percent and receives 693 J of

energy by heat during each cycle, how much work is done by the engine?

5. Suppose an engine has an efficiency of 11 percent and performs 1150 J

of work each cycle. How much energy is taken in as heat?

6. Suppose a steam engine has an efficiency of 19 percent and performs

998 J of work each cycle. How much energy is received by the steam

engine as heat?

7. A certain propane engine performs 544 J of work in each cycle with an

efficiency of 22.25 percent. How much energy is received from the en-

gine to the exhaust and cooling system as heat?

8. Find the efficiency of a gasoline engine that, during one cycle, receives

365 J of energy from combustion and loses 223 J as heat to the exhaust.

NAME ______________________________________ DATE _______________ CLASS ____________________

9. Find the efficiency of an engine that, during one cycle, receives 571 J of

energy and loses 463 J of energy as heat.

10. A test model for an experimental engine that uses a new clean-burning

fuel does 128 J of work in one cycle and receives 581 J of energy as heat

from combustion. What is the engine’s efficiency?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 12AHOOKE’S LAW

P R O B L E MSome bathroom scales work by stepping on a spring. Suppose a personsteps on a scale, compressing the spring 1.5 cm. If the spring constant is650 N/m, what is the spring force acting on the scale when the personsteps off?

S O L U T I O NGiven: k = 650 N/m x = 1.5 cm × 1.5 10–2 m

Unknown: F = ?


Use the equation for Hooke’s law to determine the spring’s restoring force.

Felastic = −kx = −(650 N/m) (1.5 10–2 m) = −9.8 N

1. A shopper places some fruit in a spring scale at a supermarket. If the

spring has a spring constant of 420 N/m and is compressed from its

equilibrium position by 4.3 cm, what is the spring force on the scale at

the moment it is released?

2. A fuzzy ball attached to an elastic cord is suspended from a ceiling to

be a toy for a cat. As the cat plays, the toy is pulled 15 cm and released.

If the toy has a spring constant of 65 N/m, what is the spring force act-

ing on the toy at the moment it is released?

3. You see a pair of joke glasses at a toy store. Each lens is connected to a

loosely coiled spring which, in turn, is connected to a plastic “eyeball.”

One spring is pulled 12 cm from its equilibrium position and released.

If the spring constant is 49 N/m, what is the magnitude of the spring

force acting on the toy at the moment it is released?

4. A lock tight curly hair is pulled a distance of 5.0 cm from its equilib-

rium position and released. If the lock of hair has a spring constant of

26 N/m, what is the magnitude of the spring force acting on the lock of

hair at the moment it is released?

5. When a person weighing 669 N sits in a hanging chair, a giant spring

suspending the load expands 6.5 cm. What is the spring constant?

6. A 550 N jumper attached to a bungee cord dives off a precipice. The

bungee cord stretches 15 m beyond its equilibrium point before it

bounces back. What is the spring constant?

ADDITIONAL PRACTICE

Holt Physics Problems WorkbookCh. 12–2

NAME ______________________________________ DATE _______________ CLASS ____________________

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7. As a 620 N mountain biker rides across rough terrain, the spring in the

seat compresses a distance of 7.2 cm. What is the spring constant?

8. A child exerts a force of 12 N to shoot a rubber band across the room.

If the rubber band has a spring constant of 180 N/m, what is the rub-

ber band’s displacement?

9. An archer applies a force of 52 N on a bowstring to shoot an arrow. If

the bow string has a spring constant of 490 N/m, what is the bow

string’s displacement?

10. A mass of 3.0 kg is attached to a spring scale with a spring constant of

36 N/m. What is the spring’s displacement?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 12BSIMPLE HARMONIC MOTION OF A SIMPLE PENDULUM

P R O B L E MA simple pendulum with a length of 1.00 m would have a period of 13.3 son Saturn’s icy moon, Dione. Find the acceleration of gravity on Dione.

S O L U T I O N

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ADDITIONAL PRACTICE

1. A simple pendulum with a length of 3.0 × 10–1 m would have a period

of 1.16 s on Venus. Calculate the acceleration of gravity on Venus.

2. On Mars, a simple pendulum with a length of 65.0 cm would have a

period of 2.62 s. Calculate the acceleration of gravity on Mars.

3. On Mercury, a simple pendulum with a length of 1.14 m would have a

period of 3.55 s. What is the acceleration of gravity on Mercury?

4. A simple pendulum with a length of 50.0 cm would have a period of

2.99 s on Pluto. Calculate the acceleration of gravity on Pluto.

5. Find the length of a pendulum that oscillates with a frequency of 1.0 Hz.



8. Calculate the period and frequency of a 6.200 m long pendulum in

Oslo, Norway, where g = 9.819 m/s2.


Quito, Ecuador, where g = 9.780 m/s2.


Cairo, Egypt, where g = 9.793 m/s2.

Given: L = 1.00 m T = 13.3 s

Unknown: g = ?

Choose the equation(s) or situation: Use the equation for the period of a

simple pendulum and rearrange it to solve for g.

T = 2pT2 =

4pg

2L

g = 4

T

p2

2L

= 4p

(

2

13

(1

.3

.0

s

0

)2m)

= 0.223 m/s2

Lg

Holt Physics

Problem 12CSIMPLE HARMONIC MOTION OF A MASS SPRING SYSTEM

P R O B L E MThe antennae of male mosquitoes have many hairs that receive sound signals from female mosquitoes. Female mosquitoes emit a frequency ofabout 230 Hz. Suppose a mass is attached to a spring with a spring con-stant of 1.14 104 N/m. How large is the mass if its oscillation frequency isthe same as a mosquito’s?

S O L U T I O N

Given: f = 230 Hz k = 1.14 × 104 N/m

Unknown: m = ?


Use the equation for the period of a mass-spring system to solve for m:

T = 2p = 1

f

f

12 =

4pk

2m

m = 4p

k2f 2 =

(

1

4

.1

p42)

×(2

1

3

0

0

4 N

s–/1m

)2 = 5.46 × 10−3 kg = 5.46 g

mk


NAME ______________________________________ DATE _______________ CLASS ____________________

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ADDITIONAL PRACTICE

1. Honeybee scouts inform other honeybees where to find food by flap-

ping their wings and “waggle-dancing.” During part of the dance, a

scout bee’s wings flap with a maximum frequency of 3.00 102 Hz. Sup-

pose a mass is attached to a spring with a spring constant of 8.65 ×104 N/m. How large is the mass if its oscillation frequency is the same

as the wings of a waggle-dancing bee?

2. On Halloween, you see an “alien” that has one antenna made of a glit-

tery foam ball connected to a spring. The springs oscillate with a pe-

riod of 0.079 s, and have a spring constant of 63 N/m. Find the mass of

the ball.

3. A farmer rides over a bumpy field on his tractor. The tractor seat is sup-

ported by a spring with a spring constant of 2.03 103 N/m. As the

farmer drives over a bump, the seat oscillates at a frequency of 0.79 Hz.

For the first few seconds, the vibration approximates simple harmonic

motion. Find the farmer’s mass if the tractor seat acts like a spring scale.

4. A 32 N sack of potatoes vibrates with a period of 0.42 s placed on a

spring scale. What is the spring constant?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A 66 N pumpkin vibrates with a period of 2.9 s when attached to the

end of a spring scale. What is the spring constant?

6. As the wind moves the bough of a tree, it oscillates up and down. Dur-

ing the first few seconds, it approximates simple harmonic motion. If

the bough has a weight of 87 N and oscillates with a period of 0.64 s,

what is the spring constant of the bough?

7. A certain trampoline acts like a single spring with a spring constant of

364 N/m. If a 24 kg child jumps on the trampoline, what would be the

period of oscillation?

8. Two children jump on their parent’s bed (when the parents are not

looking). The combined mass of both kids is 55 kg. The mattress is

supported by 36 springs, each with a spring constant of 458 N/m.

If the children jump at the same time, what would be the period

of oscillation?

9. An 8.2 kg infant is placed in a jumper that is made of a seat that is sus-

pended from a door frame by a spring. If the spring has a spring con-

stant of 221 N/m. Calculate the period of oscillation.

10. Your friend’s key chain is coiled like a spring. Three keys, each with a

mass of 24 g, are on the chain. When your friend removes the keys

from a pocket, the keys bob up and down. If the key chain has a spring

constant of 99 N/m, what is the frequency of oscillation?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 12DWAVE SPEED

P R O B L E MEarthquakes generate shock waves that travel through Earth’s interior toother parts of the world. The fastest of these waves are longitudinalwaves, like sound waves, and are called primary waves, or just p-waves. Ap-wave has very low frequencies, typically around 0.050 Hz. If the wave-length of a p-wave is 160 km, what is its speed?

S O L U T I O N

Given: l = 160 km = 1.6 × 105 m f = 0.050 Hz

Unknown: v = ?

Choose the equation(s) or situation: Use the equation relating speed, wave-

length, and frequency for a wave.

v = fl = (1.6 × 105 m) (0.050 s–1) = 8.0 × 103 m/s

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1. Earthquakes also produce transverse waves that move more slowly than

the p-waves. These waves are called secondary waves, or s-waves. If the

wavelength of an s-wave is 2.3 × 104 m, and its frequency is 0.065 Hz,

what is its speed?

2. A dolphin can hear sounds with frequencies up to 280 kHz. What is the

speed of sound in water if a wave with this frequency has a wavelength

of 0.51 cm?

3. Waves in a lake are 6.0 m apart and pass a person on a raft every 2.0 s.

What is the speed of the waves?

4. Sonar is a device that uses reflected sound waves to measure underwa-

ter depths. If a sonar signal has a frequency of 288 Hz, and the wave-

length is 5.00 m, what is the speed of the sonar signal in water?

5. A buoy on the ocean bobs up and down. The waves have a wavelength

of 2.5 m, and a frequency of 1.6 Hz. What is the speed of the waves?

6. A dog whistle is designed to produce a sound with a frequency

beyond that which can be heard by humans (between 20 000 Hz and

27 000 Hz). If a particular whistle produces a sound with a frequency

of 2.5 × 104 Hz, what is the sound’s wavelength? Assume the speed of

sound in air to be 331 m/s.

7. The lowest pitch that the average human can hear has a frequency of

20.0 Hz. If sound with this frequency travels through air with a speed

of 331 m/s, what is its wavelength?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

8. A ship anchored at sea is rocked by waves whose crests are 14 m apart.

The waves travel at 7.0 m/s. How often do the wave crests reach the

ship?

9. One of the largest organ pipes is in the Auditorium Organ in the At-

lantic City Convention Hall, New Jersey. The pipe is 38.6 ft long and

produces a sound with a wavelength of about 10.6 m. If the speed of

sound in air is 331 m/s, what is the frequency of this sound?

10. A drum is struck, producing a wave with a wavelength of 110 cm and a

speed of 2.42 × 104 m/s. What is the frequency of the wave?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 13AINTENSITY OF SOUND WAVES

P R O B L E MYour friend whispers a secret to you with a power output of 2.05 10–10 W.If the whisper has a sound intensity 4.1 10–10 W/m2, how far are youfrom your friend?

S O L U T I O NGiven: Intensity = 4.1 × 10–10 W/m2 P = 2.05 × 10–10 W

Unknown: r = ?


Use the equation for the intensity of a spherical wave.

Intensity = 4p

P

r2

r = 4p(Int

P

ensity) = = 0.20 m

2.05 × 10–10 W(4p)(4.1 × 10–10 W/m2)

1. Your friend tells you about what happened last weekend with a power

output of 5.88 × 10–5 W and a sound intensity of 3.9 × 10–6 W/m2. How

far are you from your friend?

2. The power output of heavy street traffic is 1.57 × 10–3 W. At what dis-

tance is the sound intensity of the traffic 5.20 × 10–3 W/m2?

3. A subway train in New York City produces sound with a power output of

4.80 W and an intensity of 7.2 × 10–2 W/m2. How far are you from the

subway train?

4. A loud clap of thunder has a power output of 151 kW and a sound inten-

sity of 0.025 W/m2. How far are you from the thunder’s source?

5. What is the intensity of the sound waves produced by the jet engine of a

plane taking off at a distance of 32 m when the power radiated as sound

from the engine is 402 W? Assume that the sound waves are spherical.

6. Calculate the intensity of the sound waves from a car stereo at a distance

of 0.50 m when the sound has power output of 3.5 W.

7. At a maximum level of loudness, the power output of portable radio

headphones radiated as sound is 2.76 × 10–2 W. What is the intensity of

these sound waves to a jogger whose ear is 5.0 cm from the headphone’s

speaker?

ADDITIONAL PRACTICE

NAME ______________________________________ DATE _______________ CLASS ____________________

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8. If the intensity of a mosquito’s buzzing is 9.3 × 10–8 W/m2 at a distance

of 0.21 m, how much sound power does that mosquito generate?

9. How much power is radiated as sound from a vacuum cleaner whose in-

tensity is 4.5 × 10–4 W/m2 at a distance of 1.5 m?

10. To perforate an eardrum, an intensity of 1.0 × 104 W/m2 at a distance of

1.0 m is required. Calculate how much sound power must be generated.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 13BHARMONICS

P R O B L E MA piano wire vibrates with a fundamental frequency of 440 Hz when thespeed of sound on the wire is 550 m/s. What is the length of this wire?

S O L U T I O NGiven: v = 550 m/s n = 1 f1 = 440 Hz

Unknown: L = ?

Choose the equation(s) or situation: The fundamental frequency can be found

by using the equation for standing waves on a vibrating string:

fn = 2

n

L

v, n = 1, 2, 3, …

Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation

above to solve for the length of the wire.

L = 2

n

f

v

n =

(

(

1

2

)

)

(

(

5

4

5

4

0

0

m

H

/

z

s

)

) = 0.625 m

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ADDITIONAL PRACTICE

1. A saxophone plays a tune in the key of B-flat. The saxophone has a sec-

ond harmonic frequency of 466.2 Hz when the speed of sound in air is

331 m/s. What is the length of the pipe that makes up the saxophone?

Recall that a saxophone should be treated as a pipe closed at one end.

2. A clarinetist plays a clarinet on a cold day. At one point she produces the

sound of middle F sharp, which has a frequency of 370 Hz, by playing

the third harmonic of low B. If the speed of sound in the air is 331 m/s,

what is the length of the clarinet? Recall that a clarinet resembles a pipe

closed at one end.

3. A penny whistle plays a tune in the key of G with a fundamental fre-

quency of 392.0 Hz. The speed of sound in air is 331 m/s. What is the

length of the penny whistle? Treat the penny whistle as a pipe closed

at one end.

4. An organ pipe that is open at both ends has a fundamental frequency of

370.0 Hz when the speed of sound in air is 331 m/s. What is the length of

this pipe?

5. What is the fundamental frequency of a viola string that is 35.0 cm long

when the speed of waves on this string is 346 m/s?

6. What is the fundamental frequency of a mandolin string that is 42.0 cm

long when the speed of waves on this string is 329 m/s?

NAME ______________________________________ DATE _______________ CLASS ____________________

7. What is the fundamental frequency of a cello string that is 0.85 m long

when the speed of waves on this string is 499 m/s?

8. A pipe that is open at both ends has a fundamental frequency of 277.2 Hz.

If the pipe is 0.75 m long, what is the speed of the waves in the pipe?

9. A pipe that is closed on one end has a seventh harmonic frequency of

466.2 Hz. If the pipe is 1.53 m long, what is the speed of the waves in

the pipe?

10. A pipe that is open at both ends has a fundamental frequency of 125 Hz.

If the pipe is 1.32 m long, what is the speed of the waves in the pipe?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 14AELECTROMAGNETIC WAVES

P R O B L E MHow fast does light with a frequency of 5.4999 1014 Hz and a wave-length of 545.00 nm travel?

S O L U T I O NGiven: λ = 5.450 0 × 10–7 m f = 5.4999 1014 Hz

Unknown: c = ?


Use the wave speed equation for electromagnetic waves.

c = f λ = (5.4999 1014 s–1) (5.4500 × 10–7 m) = 2.9974 × 108 m/s

1. How fast does a radio wave travel that has a frequency of 7.6270 × 108

Hz and a wavelength of 39.296 cm? Does this radio wave travel

through Earth’s atmosphere or in space? Light travels in a vacuum at

2.997 924 58 × 108 m/s and in air at 2.997 09 × 108 m/s.

2. How fast does microwave radiation that has a frequency of 1.173 06 ×1011 Hz and a wavelength of 2.555 6 mm travel? Does this microwave

travel through Earth’s atmosphere or in space? Light travels in a vac-

uum at 2.997 924 58 × 108 m/s and in air at 2.997 09 × 108 m/s.

3. Scientists at Lucent Bell Labs use high-resolution microscopes to make

images of tiny organisms that provide a lot of information. By using

3.2 nm x-rays on human tissue, images can be made showing micro-

tubules in the nuclei of cells. What is the frequency of these x-rays?

4. In order to see objects, the wavelength of the light must be smaller than

the object. The Bohr radius of a hydrogen atom is 5.291 770 × 10–11 m.

a. What is the lowest frequency that can be used to locate a hydro-

gen atom?

b. The visible part of the spectrum ranges from 400 nm to 700 nm.

Why aren’t individual atoms visible?

5. Your skin tans when melanin within the skin oxidizes. Your skin sun-

burns when it receives more ultraviolet radiation than the protection

provided by the melanin. Generally, ultraviolet (UV) radiation has

been divided into two classes: UVA and UVB. You are more likely to be

sunburned if you are exposed to radiation in the UVB range (280 nm–

320 nm) than in the UVA range (320 nm–400 nm). To what range of

frequencies do these wavelength ranges correspond?

ADDITIONAL PRACTICE

NAME ______________________________________ DATE _______________ CLASS ____________________

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6. Suppose you must decide whether a pre-Columbian mask is genuine

before buying it for a museum. It looks genuine, but to ensure its au-

thenticity, you shine X rays of wavelength 1.67 × 10–10 m on it to see if

a certain element is present. What is the frequency of this radiation?

7. Suppose you use ultraviolet light of frequency 9.5 × 1014 Hz to deter-

mine whether a mineral is fluorescent. To what wavelength does this

correspond?

8. Meteorologists use Doppler radar to watch the movement of storms.

If a weather station uses electromagnetic waves with a frequency of

2.85 × 109 Hz, what is the length of the wave?

9. PCS cellular phones have antennas that use radio frequencies from

1800–2000 MHz. What range of wavelengths corresponds to these

frequencies?

10. Suppose the microwaves in your microwave oven have a frequency of

2.5 × 1010 Hz.

a. What is the wavelength of these microwaves?

b. The holes in the door of a microwave oven have a radius of

1.2 mm. Why don’t microwaves pass through these holes?

c. Visible light has a wavelength that ranges from 400 nm to

700 nm. Would visible light be able to pass through the holes?

Why or why not?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 14BCONCAVE MIRRORS

P R O B L E MSuppose you place a 5.0 cm tall spring in front of a concave mirror. Themirror has a focal length of 24 cm. The spring forms an image that ap-pears to be at the same position as the spring, but the image is inverted.Where did you place the spring? How tall is the spring’s image?

S O L U T I O NGiven: h = 5.0 cm f = 24 cm q = p

Unknown: h′ = ? p = ?

Choose the equation (s) or situation: Use the mirror equation to find the posi-

tion of the spring, and the equation for magnification to find the height of the

image.

1

f =

p

1 +

1

q =

p

1 +

p

1 =

p

2

M = h

h′= −

p

q


p = 2f = (2) (24 cm) =

q = p = 48 cm

h′ = − q

p

h = −

(48 cm

48

)(

c

5

m

.0 cm) = −5.0 cm

48 cm

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ADDITIONAL PRACTICE

1. You can appear to shake hands with yourself using a concave mirror.

Suppose you have a mirror with a focal length of 32.0 cm.

a. Where would you place your right hand so that you appear to be

shaking hands with yourself?

b. When your hand is in the shaking position, it is 7.5 cm tall. What

is the height of the image?

2. As you eat your soup, you look in the concave part of the spoon and

see your face. The focal length of the spoon is 9.5 cm, and the image of

your eye appears to be 15.5 cm from the mirror.

a. How far in front of the mirror is your eye?

b. If your eye is 3.0 cm tall, how tall is the image?

NAME ______________________________________ DATE _______________ CLASS ____________________

3. Suppose you bend a sheet of aluminized Mylar™ to form a reflective

surface that resembles a concave mirror when the axis is vertical. The

bent reflective sheet has a focal length of 17 cm.

a. Where must you stand so that the image of your eye appears at

23 cm?

b. If your eye is 2.7 cm tall, how tall will the image be?

4. A car’s headlamp is made of a light bulb in front of a concave spherical

mirror. If the bulb is 5.0 cm in front of the mirror, what is the radius

and focal length of the mirror?

5. Suppose you are 19 cm in front of the bell of your friend’s trumpet and

you see your image at 14 cm. Treating the trumpet’s bell as a concave

mirror, what would be its focal length and radius of curvature?

6. You look in to a metallic mixing bowl which resembles a spherical con-

cave mirror. When you are 35 cm in front of the bowl, you see an image

at 42 cm. What is the focal length and radius of curvature of the bowl?

7. You place an electric heater 3.00 m in front of a concave spherical mir-

ror that has a focal length of 30.0 cm.

a. Where would your hand feel warm?

b. If the heater is 15 cm tall, how tall is the image?

8. A new line of makeup offers concave spherical mirrors in their pressed

powder compacts. The focal length for one of these mirrors is 17.5 cm.

a. If someone used this mirror at a distance of 15.0 cm, where

would the image appear?

b. What is the magnification of the image?

9. A concave spherical mirror on an actor’s dressing table has a focal

length of 60.0 cm. Suppose the actor sits 35.0 cm in front of the mirror.

a. Where does the image appear?

b. What is the magnification of the image?

10. You place a book in front of a concave spherical mirror. The mirror has

a focal length of 23.0 cm.

a. Where does the image form when the book is 3.00 cm in front of

the mirror? What is the magnification of the image? Will you be

able to read the writing on the image?

b. Where does the image form when the book is 33.0 cm in front of

the mirror? What is the magnification of the image? Will you be

able to read the writing on the image?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 14CCONVEX MIRRORS

P R O B L E MYou have just received a silver key ring as a gift. The ring is connected to aspherical silver ball that acts like a convex spherical mirror. When youhold the ball 21 cm from your eye, your image forms 7.0 cm behind themirror. What is the magnification of the image? What is the mirror’sfocal length and radius of curvature?

S O L U T I O NGiven: p = 521 cm q = –7.0 cm

Unknown: M = ? f = ? R = ?

Choose the equation (s) or situation: Use the mirror equation to find the focal

length and radius of curvature, and the equation for magnification to find the

height of the image.

1

f =

p

1 +

1

q =

21

1

cm +

−70.

1

0 cm

f = [0.048 cm−1 − 0.143 cm−1]−1 = −10.5 cm

R = 2f = 2 (−11 cm) =

M = −p

q = −

(−(2

7

1

.0

c

c

m

m

)

) = +0.33

−22 cm

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ADDITIONAL PRACTICE

1. Your car has a side-view mirror with a convex spherical mirror on the

passenger’s side. When you pass a car, you see the car’s image in the mir-

ror. The image appears to be 9.0 cm tall, but the car is really 1.5 m tall.

a. What is the magnification of the mirror?

b. If the car is 3 m from the mirror, what is the focal length of the

mirror?

c. What is the mirror’s radius of curvature?

2. Sitting beside a Christmas tree, you notice your face is reflected in a

hanging spherical ornament. The image appears 5.2 cm behind the or-

nament when you are 17 cm in front of the ornament.

a. What is the ornament’s focal length and radius of curvature?

b. If your eye is 3.2 cm tall, how tall is the image?

3. You see an image of your hand as you reach for a polished brass door-

knob. The doorknob has a focal length of 6.3 cm. How far from the

doorknob is your hand when the image appears at 5.1 cm behind the

doorknob? What is the magnification of the image?


NAME ______________________________________ DATE _______________ CLASS ____________________

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4. As you turn the knob of a faucet to draw bath water, you see your re-

flection in the water spout. The focal length of the spout is −33 cm.

How far away from the spout are you if your image appears to be

16.1 cm behind the spout? What is the magnification of the image?

5. You see your reflection in your friend’s mirrored sunglasses. If each

lens has a focal length of −12 cm, and your image appears 9.0 cm be-

hind the sunglasses, how far from your friend are you standing? What

is the magnification of the image?

6. To supervise customers, many stores install spherical convex mirrors in

strategic locations. Suppose one store has a spherical convex mirror

with a magnification of 0.11. Suppose you are 1.75 m tall.

a. How tall is the image?

b. How far in front of the mirror are you when the image appears

42 cm behind the mirror?

7. A stainless-steel ladle, used to serve soup, is like a spherical convex mir-

ror. If the focal length of the ladle is 27cm and you are 43 cm in front

of the ladle, where does the image appear? What is the magnification of

the image?

8. Just after you dry a spoon, you look into the convex part of the spoon.

If the spoon has a focal length of −8.2 cm and you are 18 cm in front of

the spoon, where does the image appear? What is the magnification of

the image?

9. The base of an art deco lamp is made of a convex spherical mirror with

a focal length of −39 cm.

a. Where does the image appear when you are 16 cm from the base?

b. If your nose is 6 cm long, how long does the image appear?

10. The button on many electric hand dryers is a convex mirror. You see

the image of your hand as you reach to press the button. If the magnifi-

cation of the image is 0.24 and your hand is 12 cm away from the but-

ton, where does the image appear?


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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 15ASNELL’S LAW

P R O B L E MA ray of light traveling in air strikes the surface of a polished agate slab (n 1.544) on display in your friend’s home. If the ray in the agate makesan angle of 29.0° with the normal, what is the angle of incidence?

S O L U T I O NGiven: nr = 1.544 qr = 29.0° ni = 1.00

Unknown: qi = ?


Use the equation for Snell’s law.

ni (sin qi) = nr (sin qr)

qi = sin–1nr (s

n

in

i

qr) = sin−1 (1.544)

1

(

.

s

0

i

0

n 29.0°) = 48.5°

1. An old Greek superstition was that amethyst would protect those who

wore it from drunkenness—which is why they called it améthystos,

meaning “not drunken.” Until the discovery of the large Brazilian and

Uruguayan deposits at the end of the nineteenth century, deep-colored

amethyst was highly prized. Suppose a ray of light traveling in air

strikes the surface of an amethyst crystal (n = 1.553). If the ray in the

amethyst makes an angle of 35° with the normal, what is the angle of

incidence?

2. If you were to set a calcite crystal on top of this sentence, you would see

a double image of each word. This phenomenon is called “double re-

fraction.” Suppose a ray of light traveling in air strikes the surface of a

calcite crystal (n = 1.486) used to demonstrate this phenomenon. If the

ray in the calcite makes an angle of 41° with the normal, what is the

angle of incidence?

3. The Chinese have skillfully carved figurines made of a translucent

greenish material called serpentine. A ray of light traveling in air strikes

the flat surface of a serpentine figurine (n = 1.555). If the ray in the ser-

pentine makes an angle of 33° with the normal, what is the angle of

incidence?

4. When light in air enters an opal mounted on a ring, it travels at a speed

of 2.07 × 108 m/s. What is opal’s index of refraction?

5. When light enters a pearl in a necklace, it travels at a speed of 1.97 × 108

m/s. What is the pearl’s index of refraction?

ADDITIONAL PRACTICE

NAME ______________________________________ DATE _______________ CLASS ____________________

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6. When light enters albite, also called “moonstone”, it has a luminous

albedo—like a full moon. When light in air enters albite, it travels at a

velocity of 1.95 × 108 m/s. What is albite’s index of refraction?

7. Nephrite jade was once used virtually everywhere by Neolithic man for

polished stone weapons. Nephrite jade was also important in ancient

oriental art. Suppose light passes from air at an angle of incidence of

59.2° into a thin ornate handle of a nephrite jade vase (n = 1.61) on

display at a museum. Determine the angle of refraction in the jade.

8. Malachite is characterized by wavy light- and dark-green bands and

has double refraction (n =1.91 and n = 1.66). Suppose a ray of light

traveling in air strikes a malachite carving of an Aztec calendar at an

angle of 35.2° with the normal. What are the two angles of refraction?

9. Amber is a fossil resin of trees that lived tens of millions of years ago.

Sometimes insects were trapped by the resin and fossilized inside. Sup-

pose a ray of light traveling in air strikes a 2 mm thick clear amber pen-

dant (n = 1.54) at an angle of 17° with the normal. Find the angles of

refraction at each surface.

10. The Museo degli Argenti, in Florence, displays a plate that was carved

out of rock crystal in the sixteenth century. Suppose a ray of light trav-

eling in air strikes this plate (n = 1.544) at an angle of 22° with the nor-

mal. Trace the light ray through the plate, and find the angles of refrac-

tion at each surface.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 15BLENSES

P R O B L E MAn object is placed 49 cm in front of a converging lens. The image forms23 cm behind the lens and is 8.0 cm tall. Determine the focal length of thelens and the height of the object.

S O L U T I O NGiven: h′ = 8.0 cm p = 49 cm q = 23 cm

Unknown: f = ? h = ?

Choose the equation(s) or situation: Use the thin-lens equation to find the

focal length, and the equation for magnification to find the height of the object.

1

f =

p

1 +

1

q =

49

1

cm +

23

1

cm = 0.0204 cm−1 + 0.0435 cm−1

f =

M = − p

q =

h

h

′


h = – p

q

h′ = –

(49 cm

23

)(

c

8

m

.0 cm) = 17 cm

16 cm

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ADDITIONAL PRACTICE

1. An object is placed 13 cm in front of a converging lens. The image

forms 19 cm behind the lens and is 3.0 cm tall. Determine the focal

length of the lens and the height of the object.

2. An object that is 15 cm tall is placed 44 cm in front of a diverging lens.

A virtual image appears 14 cm in front of the lens. Determine the focal

length of the lens and the height of the image.

3. A magnifying glass has a diverging lens with a 13.0 cm focal length. At

what distance from a toothpick should you hold this lens to form an

image with a magnification of +5.00?

4. An object with a height of 18 cm is placed in front of a converging lens.

The image has a height of –9.0 cm.

a. What is the magnification of the lens?

b. If the focal length of the lens is 6.0 cm, how far in front of the

lens is the object?

c. Where does the image appear?

NAME ______________________________________ DATE _______________ CLASS ____________________

5. A lighthouse places a 1000-watt bulb 4 m in front of a converging lens.

The focal length of the lens is 4 m. What is the image distance and the

magnification?

6. A searchlight is constructed by placing a 500-watt bulb 0.5 m in front

of a converging lens. The focal length of the lens is 0.5 m. What is the

image distance and the magnification?

7. A microscope slide is placed in front of a converging lens with a focal

length of 3.6 cm. The lens forms a real image of the slide 15.2 cm be-

hind the lens. How far is the lens from the slide?

8. Where must an object be placed to form an image 12 cm in front of a

diverging lens with a focal length of 44 cm?

9. In the projection booth of a movie theatre, film is placed in front of a

converging lens with a focal length of 9.0 cm. The lens forms a magni-

fied real image on a screen 18 m behind the lens. How far is the lens

from the film?

10. An object is placed in front of a converging lens with a focal length of

5.5 m. A virtual image appears 5.5 cm in front of the lens. How far is

the object from the lens?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 15CCRITICAL ANGLE

P R O B L E MThe critical angle for light traveling from a red spinel gemstone into air is35.8. What is the index of refraction for red spinel?

S O L U T I O NGiven: qc = 35.8° nr = 1.00

Unknown: ni = ?

Choose the equation(s) or situation: Use the equation for critical angle.

sin qc = n

nr

i


ni = sin

nr

qc =

sin

1.

3

0

5

0

.8° = 1.71

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ADDITIONAL PRACTICE

1. The critical angle for light traveling from a green tourmaline gemstone

into air is 37.8°. What is tourmaline’s index of refraction?

2. The critical angle for light traveling from an aquamarine gemstone into

air is 39.18°. What is the index of refraction for aquamarine?

3. The critical angle for light traveling from almandine garnet into air

ranges from 35.3° − 33.1°. Calculate the range of almandine garnet’s

index of refraction.

4. Light moves from olivine (n =1.670) into onyx. If the critical angle for

olivine is 62.85°, what is the index of refraction for onyx?

5. Light moves from spessartite garnet (n = 1.80) —also called spessar-

tine—into obsidian. If the critical angle for spessartine is 57.0°, what is

the index of refraction for obsidian?

6. Light moves from a clear andalusite (n =1.64) crystal into ivory. If the

critical angle for andalusite is 69.9°, what is the index of refraction for

ivory?

7. Find the critical angle for light traveling from ruby (n = 1.766) into air.

8. Find the critical angle for light traveling from sapphire (n = 1.774) into

air.

9. Find the critical angle for light traveling from blue topaz (n = 1.61)

into air.

10. Find the critical angle for light traveling from emerald (n = 1.576) into air.

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Problem 16A Ch. 16-1

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 16AINTERFERENCE

P R O B L E MYou can make a hologram by splitting a laser beam into two beams thatconstructively interfere with each other. Each beam acts as a slit in a double-slit experiment—one beam is sent directly to a photographicfilm, and the other beam reflects off of an object before reaching the film. The two beams constructively interfere with each other on the film,making fringes. Suppose a hologram is made by using red light from a helium-neon laser ( = 632.8 nm). The first bright fringe is imprinted onthe film at an angle of 0.426 from the central maximum. What is the distance between the beams?

S O L U T I O N

Given: l = 632.8 nm = 6.328 × 10–7 m m = 1 q = 0.426°

Unknown: d = ?

Choose the equation(s) or situation:Use the equation for constructive interference, given on page 601.

d = s

m

in

lq

= 2 (6

s

.

i

3

n

28

(0

×.4

1

2

0

6

–

°

7

)

m) = 1.70 × 10–4 m = 0.170 mm

ADDITIONAL PRACTICE

1. A grating with 14 450 lines/cm is illuminated by light with a wavelength

of 625.0 nm. What is the highest order number that can be observed

with this grating? (Hint: Remember that sin q can never be greater than

1 for a diffraction grating.)

2. A grating with 12 660 lines/cm is illuminated by light with a wavelength

of 589.6 nm. What is the highest order number that can be observed

with this grating? (Hint: Remember that sin q can never be greater than

1 for a diffraction grating.)

3. A diffraction grating is calibrated by using the 546.1 nm line of hydro-

gen gas. The first-order maximum is found at an angle of 75.76°. Calcu-

late the number of lines per centimeter on this grating.

4. A diffraction grating is calibrated by using the 447.1 nm line of mercury

vapor. The first-order maximum is found at an angle of 40.25°. Calcu-

late the number of lines per centimeter on this grating.

5. A diffraction grating with 1950 lines/cm is used to examine the lithium

spectrum. Find the angles at which one would observe the first-order and

second-order maxima for the blue-green line of lithum (l = 497.3 nm).


NAME ______________________________________ DATE _______________ CLASS ____________________

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6. A diffraction grating with 7500 lines/cm is used to examine the mercury

spectrum. Calculate the second-order angular separation of the two

closely spaced yellow lines of mercury (579.0 nm and 577.0 nm).

7. Infrared light passes through a diffraction grating of 3600 lines/cm. If the

angle of the third-order maximum is 76.54°, what is the wavelength of

the light?

8. A diffraction grating of 9550 lines/cm is used to examine the carbon

spectrum. If the angle of the second-order maximum is 54.58°, what is

the wavelength of the light?

9. A diffraction grating of 12 500 lines/cm is used to examine the iron spec-

trum. If the angle of the first-order maximum is 38.77°, what is the

wavelength of the light?

10. Ultraviolet light passes through a diffraction grating of 2400 lines/cm. If

the angle of the second-order maximum is 26.54°, what is the wave-

length of the light?

Problem 16B Ch. 16-3

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 16BDIFFRACTION GRATINGS

P R O B L E M

A grating with 13 250 lines/cm is illuminated by light with a wavelengthof 725 nm. What is the highest order number that can be observed withthis grating? (Hint: Remember that sin q can never be greater than 1 for adiffraction grating.)

S O L U T I O N

Given: 13 250 lines/cm l = 725 nm = 7.25 × 10–7 m q < 90°

Unknown: m = ?

Choose the equation(s) or situation: Use the equation for a diffraction grating.

m = 1: q = sin−1[ml /d] = sin−1[(1)(7.25 × 10−7 m) ÷ (1 325 000 lines/m)−1] = 73.9°

m = 2: q = sin−1[ml /d] = sin−1[(2)(7.25 × 10−7 m) ÷ (132 500 lines/m)−1] = ∞

Therefore, 1 is the highest-order number that can be observed.

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1. A double-slit interference experiment is performed using red light from a

helium discharge tube (l = 587.5 nm). The second-order bright fringe in

the interference pattern is 0.130° from the central maximum. How far

apart are the two slits separated?

2. A double-slit interference experiment is performed using red light from a

laser pointer pen (l = 656.3 nm). The fourth-order bright fringe in the

interference pattern is 0.626° from the central maximum. How far apart

are the two slits separated?

3. A ruby laser was developed by T. H. Maiman in 1960. Suppose a double-

slit interference experiment is performed using red light from a ruby

laser (l = 693 nm). The third-order bright fringe in the interference pat-

tern is 0.578° from the central maximum. How far apart are the two slits

separated?

4. Light falls on a double slit with slit separation of 8.04 × 10−6 m, and the

third bright fringe is seen at an angle of 13.1° relative to the central max-

imum. Find the wavelength of the light.

5. A double-slit interference experiment is performed with green light from

a mercury discharge tube. The separation between the slits is 4.43 ×10−6 m, and the third-order maximum of the interference pattern is at an

angle of 21.7° from the center of the pattern. What is the wavelength of

argon laser light?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

6. Light shines on a double slit with slit separation of 3.92 × 10−6 m, and

the second bright fringe is seen at an angle of 20.4° relative to the central

maximum. What is the wavelength of the light?

7. If the distance between the two slits is 0.0220 cm, find the angle at which

a first-order bright fringe is observed for green light with a wavelength of

527 nm.

8. Blue light with a wavelength of 430.8 nm shines on two slits 0.163 mm

apart. What is the angle at which a first-order bright fringe is observed?

9. Orange light (l = 583 nm) passes through two slits 0.329 mm apart. Cal-

culate the angle at which a first-order bright fringe is observed.

10. If two slits are 0.267 mm apart, find the angle between the first-order

and second-order bright fringes for red light with a wavelength of

687 nm.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 17A

P R O B L E MTwo electrostatic point charges of +20.0 µC and –30.0 µC exert attractiveforces on each other of –145 N. What is the distance between the twocharges?

S O L U T I O N

Given: q1 = 2.00 × 10−5 C q2 = −3.00 × 10−5 C

Felectric = −145 N kC = 8.99 × 109 N•m2/C2

Unknown: r = ?


Use Coulomb’s law, given on page 634.

Felectric = kC

r

q21q2

Rearrange the equation(s) to solve for the unknown(s): Rearrange Coulomb’s

law to solve for the distance between the two charges.

r = k

FCel

qec

1t

q

r

i

2

c =

r = 0.193 m = 19.3 cm

(8.99 × 109 N•m2/C2)(−3.0 × 10–5 C)(2.0 × 10–5 C)

−145 N

ADDITIONAL PRACTICE

1. Two electrostatic point charges of –13.0 mC and –16.0 mC exert repul-

sive forces on each other of 12.5 N. What is the distance between the

two charges?

2. Two electrostatic point charges of 99.9 mC and 33.3 mC exert repulsive

forces on each other of 87.3 N. What is the distance between the two

charges?

3. Two electrostatic point charges of –43.2 mC and 22.4 mC exert attrac-

tive forces on each other of –6.5 N. What is the distance between the

two charges?

4. A glass rod rubbed against silk gains a charge of –5.3 mC. What is the

electric force between the rod and the silk when the two are separated

by a distance of 4.2 cm? (Assume that the charges are located at a

point.)

5. A glass rod rubbed against your hair gains a charge of –14.0 nC. What

is the electric force between the balloon and your hear when the two

are separated by a distance of 7.1 cm? (Assume that the charges are lo-

cated at a point.)


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6. A dog’s fur is combed and the comb gains a charge of 8.0 nC. What is

the electric force between the fur and the comb when the two are sepa-

rated by 2.0 cm?

7. Suppose two pions are separated by 8.3 × 10–10 m. If the magnitude of

the electric force between the charges is 3.34 × 10–10 N, what is the

value of q?

8. Suppose two muons having equal but opposite charge are separated by

6.4 × 10–8 m. If the magnitude of the electric force between the charges

is 5.62 ×10–14 N, what is the value of q?

9. Suppose two delta particles are separated by 9.3 × 10–11 m. If the mag-

nitude of the electric force between the charges is 2.66 × 10–8 N, what

is the value of q?

10. Suppose two equal charges are separated by 6.5 × 10–11 m. If the mag-

nitude of the electric force between the charges is 9.92 × 10–4 N, what

is the value of q?


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Holt Physics

Problem 17B

P R O B L E M

Consider three point charges on the x-axis: q1 = 4.92 10–9 C is at theorigin, q2 = –6.99 10–8 C is at x = –3.60 10–1 m, and q3 = 5.65 10–9 Cis at x = 1.44 m. Find the magnitude and direction of the resultant forceon q1.

S O L U T I O N

Given: q1 = 4.92 × 10–9 C r1,2 = –3.60 × 10–1 m

q2 = –6.99 × 10–8 C r1,3 = 1.44 m

q3 = 5.65 × 10–9 C kC = 8.99 × 109 N•m2/C2

Unknown: F1,tot = ?

Calculate the magnitude of the forces with Coulomb’s law:

F1,2 = kC

r1

q

,2

12q2 = = –2.39 × 10–5 N

F1,3 = kC

r1

q

,3

12q3 = = 1.21 × 10–5 N

The forces are all along the x-axis, so add up the x-components:

F1,tot = F1,2 + F1,3 = −2.39 × 10–5 N + 1.21 × 10–5 N = −1.18 × 10–5 N

(8.99 × 109 N•m2/C2)(4.92 × 10–9 C)(5.65 × 10–9 C)

(1.44 m)2

(8.99 × 109 N•m2/C2)(4.92 × 10–9 C)(–6.99 × 10–8 C)

(–3.60 × 10–1 m)2

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ed. 1. Suppose four protons were at the corners of a square. The length of

each side of the square is 1.52 × 10–9 m. If q1 is on the upper right cor-

ner, calculate the magnitude and direction of the resultant force on q1.

2. Consider three point charges, q1 = 4.50 C, q2 = 4.50 C, and q3 = 6.30 C,

located at the corners of an isosceles triangle. The charges q1 and q2 are

5.00 m apart and form the base. The triangle is 3.50 m high, and q3 is

located at the top. Calculate the magnitude and direction of the resul-

tant force on q3.

3. Imagine three point charges on the corners of a triangle: q1 = –9.00 nC

is at the origin, q2 = –8.00 nC is at x = 2.00 m, and q3 = 7.00 nC is at y =3.00 m. Find the magnitude and direction of the resultant force on q1.

4. Suppose three point charges are on the y-axis: q1 = –2.34 × 10–8 C is at

the origin, q2 = 4.65 × 10–9 C is at y = 0.500 m, and q3 = –2.99 × 10–10 C

is at y = 1.00 m. What is the magnitude and direction of the resultant

force on q1?

5. Consider four electrons at the corners of a square. Each side of the

square is 3.02 × 10–5 m. Find the magnitude and direction of the resul-

tant force on q3 if it is at the origin.

ADDITIONAL PRACTICEADDITIONAL PRACTICE

NAME ______________________________________ DATE _______________ CLASS ____________________

6. Imagine three point charges at the corners of an isosceles triangle: q1 =2.22 × 10–10 C, q2 = 3.33 × 10–9 C, and q3 = 4.44 × 10–8 C. The charges

q1 and q2 are 1.00 m apart and form the triangle’s base. The triangle is

0.250 m tall. If q3 is at the top, what is the magnitude and direction of

the resultant force on q3?

7. Consider three 2.0 nC point charges at the following locations: at (0 m,

0 m), at (1.0 m, 2.0 m), and at (1.0 m, 0 m). Find the magnitude and

direction of the resultant force on the charge at the origin.

8. Consider three point charges on the corners of a triangle, where q1 =–4.0 mC at the origin; q2 = –8.0 mC at (2.0 m, 0 m); and q3 = 2.0 mC

at (0 m, 2.0 m). Calculate the magnitude and direction of the resultant

force on q1.

9. Suppose three point charges are on the corners of a triangle: q1 =9.00 mC is at the origin, q2 = 6.00 mC is at the point (1.00 m, 1.00 m),

and q3 = 3.00 mC is at (–1.00 m, 1.00 m). Find the magnitude and di-

rection of the resultant force on q1.

10. Consider three equal point charges of 4.00 nC on a line. All charges are

4.00 m apart. Calculate the magnitude and direction of the resultant

force on the charge in the middle.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 17C

P R O B L E MThree charges lie along the x-axis. One charge, q1 = –12 µC is at the origin. Another charge, q2 = 24 µC is at x = 1.0 m. A third charge,q3 = –36 µC, is placed so that q2 is in electrostatic equilibrium with q1 and q3. How large must the electrostatic force between q2 and q3 be to balance the force of q1 and q2?

S O L U T I O NGiven: q1 = −12 mC r1,2 = 1.0 m

q2 = 24 mC kC = 8.99 × 109 N•m2/C2

q3 = −36 mC

Unknown: r3,2 = ? F3,2 = ?

Diagram: y

q1 q2 x

x = 0 x = 1.0 m

Choose the equation(s) or situation: To solve for the electrostatic force, r3,2 is

needed. To solve for r3,2, determine where q3 must be placed in order to achieve

electrostatic equilibrium. To have electrostatic equilibrium, the direction of the

force of the charge on q2 must be opposite the direction of the force of the first

charge. The electrostatic force due to q1 points to q2. If q3 were between q1 and

q2, both forces on q2 would point in the same direction, so it would not be in

electrostatic equilibrium. So q3 can not be between the charges. Because q3 is

greater than q1, we would expect the location of q1 to be closer to q2 in order for

the forces to balance. So q3 is opposite the position of q1. The electric forces must

be equal and opposite to have electrostatic equilibrium.

F3,2 = –F2,1 = (x

k

–C

1

q

.03q

m2

)2 = –kC

x

q22q1

q2 x2 = –q1(x – 1.0 m)2 = – q1x2 + 2q1x – (1.0 m2)q1

(q2 + q1)x2 – 2q1x + (1.0 m2)q1 = 0

Use the quadratic formula to solve for x.

x =2q1 ±

√(2q1)2 – 4(q2+ q1)(1.0 m2) q1

2(q2 + q1)

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x = = 2.0 m ± 1.4 m

To achieve electrostatic equilibrium, q3 must be opposite q1, so x = 3.4 m

from the origin.

2(24 mC) ±√

4(24 mC)2 – 4(–12 mC + 24mC)(24 mC)(1.0m2)2(–12 mC + 24 mC)

NAME ______________________________________ DATE _______________ CLASS ____________________

Substitute the values into the equation(s) and solve: Calculate the electro-

static force between q2 and q3.

r3,2 = x – r1,2 = 3.4 m – 1.0 m = 2.4 m

F3,2 = kC

r3

q

,

2

22q3 = = 0.67 m

(8.99 × 109 N•m2/C2)(–1.2 × 10–5 C)(–3.6 × 10–5 C)

(2.4 m)2

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Ch. 17–6

ADDITIONAL PRACTICE

1. Three charges are on the y-axis. One positive charge, q1 = 9.0 mC, is at

y = 3.0 m. A second charge of –19 mC is located at the origin. A third

charge of 9.0 mC is placed so that q2 is in electrostatic equilibrium with

q1 and q3. How large must the electrostatic force between q2 and q3 be

to balance the force between q1 and q2?

2. Three charges lie along the x-axis. One positive charge, q1 = 25 mC is at

x = 0.25 m. Another charge, q2 = –5.0 mC, is located at the origin. A

third charge, q3 = –35 mC, is placed so that q2 is in electrostatic equilib-

rium with q1 and q3. How large must the electrostatic force between q2

and q3 be to balance the force between q1 and q2?

3. Three charges are on the y-axis. One positive charge, q1 = 6.0 mC is

at y = 5.0 cm. Another charge, q2 = –12 mC, is at the origin. A third,

6.0 mC charge is placed so that q2 is in electrostatic equilibrium with

q1 and q3. How large must the electrostatic force between q2 and q3

be to balance the force between q1 and q2?

4. A charge of 7.2 nC and a charge of 6.7 nC are separated by 32 cm. Find

the equilibrium position for a –3.0 nC charge.

5. A charge of 5.5 nC and a charge of 11 nC are separated by 88 cm. Find

the equilibrium position for a –22 nC charge.

6. A charge of –2.5 nC and a charge of –7.5 nC are separated by 20.0 cm.

Find the equilibrium position for a 5.0 nC charge.

7. Three charges are on the y-axis. A –2.3 C charge is at the origin and an-

other, unknown charge is at y = 2.0 m. A third charge of –4.6 C is

placed at y = –2.0 m so that it is in electrostatic equilibrium with the

first two charges. What is the charge on q2?

8. Three charges are on the x-axis. At the origin is an 8.0 C charge. An un-

known charge, q2, is at x = 1.0 m. A third charge, q3 = –4.0 C, is placed

at x = –1.0 m, where it is in electrostatic equilibrium with the other two

charges. What is the charge on q2?

ADDITIONAL PRACTICE

Holt Physics Problem Bank


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9. Three charges are on the y-axis. At the origin is a charge, q1 = 49 C, an

unknown charge, q2, is at y = 7.0 m. A third charge, q3 = –7.0 C, is

placed at y = –18 m, where it is in electrostatic equilibrium with q1 and

q2. What is the charge on q2?

10. Three charges are on the y-axis. At the origin is a charge, q1 = 72 C, an

unknown charge, q2, is at y = 15 mm. A third charge, q3 = –8.0 C, is

placed at y = –9.0 mm, so that it is in electrostatic equilibrium with q1

and q2. What is the charge on q3?

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Holt Physics

Problem 17D

P R O B L E M

A typical thundercloud has an electric field of about 3.0 105 N/C. If theelectric field is directed toward the ground. What is the electric force onan 18 nC charge in this field?

S O L U T I O N

Given: E = 3.0 × 105 N/C q = 18 nC = 1.8 × 10–8 C

Unknown: Felectric = ?

Choose the equation(s) or situation: Use the definition of electric field and

rearrange it to solve for Felectric.

E = Fele

qctric

Felectric = Eq = (3.0 × 105 N/C)(1.8 × 10–8 C)

Felectric = 5.4 × 10–3 N, directed toward the ground

ADDITIONAL PRACTICE

1. An electric field of 9.0 N/C is directed along the positive x-axis. What is

the electric force on a –6.0 C charge in this field?

2. An electric field of 1500 N/C is directed along the positive y-axis. What

is the electric force on a 5.0 nC charge in this field?

3. Millikan’s experiment measures the charge of an electron by suspend-

ing charged oil droplets in an electric field. If an oil droplet with a mass

of 3.35 × 10–15 kg has the same charge as an electron. What electric

force is required to balance the weight of the oil droplet?

4. Two equal charges of 3.00 mC lie along the x-axis: one is at the origin,

and the other is 2.0 m from the origin. Find the magnitude and direc-

tion of the electric field at a point on the y-axis 0.25 m from the origin.

5. A charge, q1 = 15.0 mC, is at the origin, and a second charge, q2 =5.00 mC, is on the y-axis 0.500 m from the origin. Find the magnitude

and direction of the electric field at a point on the y-axis 1.00 m from

the origin.

6. An electric field of 1663 N/C is directed along the positive x-axis. If the

electric force on a charge is 8.42 × 10–9 N, what is the charge?

7. An electric field of 4.0 × 103 N/C is directed downward. If the electric

force on a charge is 6.43 × 10–9 N, what is the charge?

ADDITIONAL PRACTICE

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Problem17D Ch. 17–9

8. One electron is at the point (2.00 × 10–10 m, 0 m) and another electron

is at the point (3.00 × 10–10 m, 0 m). If the field point is at the origin,

how far from the origin (along the x-axis) could a proton be placed so

that the strength of the resultant electric field would be zero?

9. One charge, q1 = –9.00 C, is at the point (1.50 mm, 0 mm). Another

charge, q2 = 6.00 C, is at the point (–1.50 mm, 0 mm). If the field point

is at the origin, how far from the origin (along the x-axis) could a

3.00 C charge be placed so that the strength of the resultant electric

field would be zero?

10. One charge, q1 = –55.0 nC, is at the point (–5.00 × 10–7 m, 0 m). An-

other charge, q2 = 11.0 nC, is at the point (5.00 × 10–7 m, 0 m). If the

field point is at the origin, how far from the origin (along the x-axis)

could a 5.00 nC charge be placed so that the strength of the resultant

electric field would be zero?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 18AELECTRICAL POTENTIAL ENERGY

P R O B L E MWhat if, instead of an apple, Sir Isaac Newton noticed a charge of 6.0 nCfall on his head from a branch 2.5 m above him. What would have beenthe change in the electrical potential energy if Earth’s electric fieldstrength has a magnitude of 95 N/C and a downward direction?

S O L U T I O NGiven: q = 6.0 nC = 6.0 × 10–9 C E = 225 N/C d = 2.5 m

Unknown: PEelectric = ?


Use the equation for electrical potential energy in a uniform electric field, given

on page 667.

PEelectric = –qEd = – (6.0 × 10–9 C) (95 N/C) (2.5 m) = –1.4 10–6 J

1. As a child tries to attach a string to a helium-filled balloon, 14.5 nC of

charge collects on the balloon’s surface as it rubs against her wool

sweater. Suppose the child lets go of the balloon before the string is at-

tached. The yellow balloon rises 290 m. What is the change in the elec-

trical potential energy if Earth’s electric field strength has a magnitude

of 105 N/C and a downward direction?

2. After brushing your hair many times, you notice that your hair is at-

tracted to the brush. Suppose a charge of +64 nC is on the brush and a

charge of +64 nC is on your hair. What is the electrical potential energy

associated with your hair and the brush when you hold the brush

0.95 m from your hair?

3. Sparks fly when free atmospheric electrons move fast enough to knock

electrons off of an atom with which they collide, causing an electron

“avalanche.” Suppose the electric field in the air is 3.0 × 106 N/C. If the

electron moves 7.3 × 10–7 m, what is the electrical potential energy as-

sociated with the electron?

4. Sometimes wearing woolen socks while walking on carpet can cause

charge to accumulate. You might have noticed this if you felt a spark as

you tried to open a door. Suppose a charge of –42 µC exists on a door-

knob, and your hand has a charge of 63 µC. The electrical potential en-

ergy associated with the hand and the doorknob is –6.92 × 10–4 J. How

far is your hand from the knob?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

5. Suppose you have a charge of 16 nC and your friend has a charge of

14 nC. As you greet each other, the electrical potential energy associ-

ated with you and your friend is 2.1 × 10–6 J. How far are you standing

from your friend?

6. A magician holds a charged wand over scattered confetti and the con-

fetti moves to the wand. Suppose the confetti has a charge of −55 nC

and the wand has a charge of 77 nC. When the electrical potential en-

ergy associated with the wand and the confetti is −1.3 × 10–2 J, how far

is the wand from the confetti?

7. The largest electric field measured in a thundercloud is 3.4 × 105 N/C.

Suppose the airplane that measured this field flew into this thunder-

cloud at an altitude of 7.3 km. If the electrical potential energy associ-

ated with the cloud and the plane is –1.39 × 1011 J, what is the charge

on the plane?

8. Suppose you would like to develop a hat that wouldn’t fall off some-

one’s head by putting enough charge on the hat so that it would stay

no matter how much the wind blew. You begin by rubbing a silk hat

against your head so that the same amount of charge collects on the

hat as the charge on your head. When you wear the hat (r = 1.25 × 10–3 m), the electrical potential energy associated with the hat and

your head is 1.25 × 10–4 J. The hat stays as you walk into the wind.

a. How much charge was on the hat?

b. How much charge was on your head?

c. Suppose your friend approaches and puts the hat on. Is the hat

likely to stay on your friend’s head by charge alone? Why or why

not?

9. When you remove clothes from the dryer, a sock clings to a towel

(r = 9.4 × 10–4 m). The electrical potential energy associated with the

sock and the towel is 8.89 × 10–10 J. Suppose the sock and the towel

have equal and opposite charge. Calculate the charges on both.

10. After rubbing a balloon against your hair, the balloon sticks to the wall

because they have equal and opposite charge. The electrical potential

energy associated with your hair and the wall is 6.3 × 10–6 J when they

are 1.25 × 10–6 m apart. What are the charges on the balloon and your

hair?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 18BPOTENTIAL DIFFERENCE

P R O B L E MA giraffe is the tallest animal on Earth. With a height of 5.3 m, a giraffecan eat leaves high in acacia trees. Suppose an ion sits atop a giraffe’shead. If the potential difference across the giraffe is 90.0 V, what is thecharge on the ion?

S O L U T I O N

Given: ∆V = 90.0 V PEelectric = 1.18 × 10–15 J r = 5.3 m

Unknown: q = ?

Choose the equation(s) or situation: Use the equation for the potential differ-

ence near a point charge, given on page 672.

q = r∆kc

V = = 5.3 10–8 C

(5.3 m)(90.0 V)8.99 × 109 N•m2/C2

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1. The Nile is 6695 km long, making it the longest river in the world. Sup-

pose an ion placed at the Nile River’s remotest stream in Burundi. If

the potential difference across the Nile is 114.0 V, what is the charge on

the ion?

2. The Akashi Kaikyo Bridge is the world’s longest bridge. Built in 1998,

this bridge spans the Japanese islands of Honshu and Awaji—a total of

1991 m across the islands. Suppose a support at one end of the bridge

is charged. If the potential difference across the bridge is 18.6 kV, what

is the charge on the ion?

3. How far from an electron is the electric potential 1.0 V?

4. How far from a charge of 94 nC is the electric potential 9.0 V?

5. Find the potential difference between a point infinitely far away and a

point 3.95 cm from a carbon atom having a charge of 1.28 × 10−18 C.

6. Cars start by generating a 3.0 × 106 N/C electric field that causes a

spark to cross a gap in a spark plug. The gap of a standard spark plug is

6.25 × 10–4 m. What minimum potential difference must you apply to

the spark plug?

7. Gustave Alexandre Eiffel designed and built the 0.30 km-tall Eiffel

Tower in 1889 for the World’s Fair in Paris. Suppose the atmospheric

electric field of the Earth is 95 N/C directed downward. What is the

electrical potential difference between the ground and the tip of the

Eiffel Tower?

ADDITIONAL PRACTICE

8. A water molecule is V-shaped, with oxygen in the center, and hydrogen

on either side, as shown below. The oxygen nucleus is 9.58 × 10−11 m

away from each of the hydrogen nuclei. The angle between the two hy-

drogen atoms is 105°. Find the electrical potential produced by the

protons at the point P, the midpoint between the hydrogen nuclei.

[Hint: Oxygen has eight protons and hydrogen has one proton.]


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9. Some molecules, such as KBr, occur in a straight line in nature. There-

fore, you can imagine KBr as being on an imaginary x-axis. Potassium

has a charge of 3.04 × 10–18 C and is located 1.89 × 10–10 m from the

origin. Bromine has a charge of 5.60 × 10–18 C and is located –9.30 ×10–11 m from the origin. Calculate the electrical potential at the origin.

10. Salt forms when Cl– and Na+ ions combine into a cube. Suppose you

look at one face of the cube so that you see a square with charges on

each corner, as shown below. Each side of the square is 2.82 × 10–10 m

long. Calculate the electrical potential at the center of the square.

[Hint: Treat chlorine atoms as electrons and sodium atoms as

protons.]

P

0

H

H

1e

8e

1e

9.58

× 10

–ll m9.58 × 10 –

llm

105°

Problem 18C Ch. 18-5

NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 18CCAPACITANCE

P R O B L E MIce has a dielectric constant of 96.5. Suppose you were going to make a1.07 nF capacitor by placing an ice cube between two aluminum sheetswith an area of 6.25 cm2. What is the plate separation?

S O L U T I O N

Given: C = 1.07 × 10−9 F k = 96.5 A = 6.25 cm2 = 6.25 10−4 m2

Unknown: d = ?


Use the equation for constructive a parallel-plate capacitor in a vacuum, given on

page 677 and multiply it by the dielectric constant of ice, k.

d = κe

C0A = = 5.00 10–4 m2= 5.00 cm2(96.5)(8.85 × 10–12 C2/N•m2)(6.25 10–4 m2)

1.07 × 10–9 F

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ADDITIONAL PRACTICE

1. Suppose a high-voltage filter uses a parallel-plate capacitor where air

(κ = 1) is in the gap between the plates. The 18 m F capacitor is placed

across a potential difference of 9.9 kV. If each plate has an area of 4.8 ×10–3 m2, what is the distance between the plates? Assume that this ca-

pacitor is in air, with a dielectric constant of 1.

2. Suppose a transmitter uses a 4550 pF parallel-plate capacitor in a vac-

uum. Each plate has an area of 6.4 × 10–3 m2. The capacitor is placed

across a potential difference of 36 kV. What is the distance between the

plates?

3. The radius of the Earth is 6.4 × 106 m. What is the capacitance of the

Earth, regarded as a conducting sphere?

4. What is the capacitance of your head if it is a conducting sphere of ra-

dius 0.10 m?

5. A TV receiver contains a capacitor of 14 mF charged to a potential dif-

ference of 1.5 × 104 V. How much charge and electrical potential en-

ergy does this capacitor store?

6. Some glass capacitors can store as much as 1000 pF when placed across

a potential difference as high as 600 V. What is the maximum amount

of charge and electrical potential energy a glass capacitor can store?

7. A 0.50 pF capacitor is connected across a 1.5 V battery. How much

charge can this capacitor store?


NAME ______________________________________ DATE _______________ CLASS ____________________

8. Capacitors made of ceramic are very popular because of their small

size. Some ceramic capacitors can have a 1 m F capacity, allowing them

to store as much as 3 × 10–2 C. Across what potential difference can a

ceramic capacitor be placed?

9. Teflon, the same material used to coat some “non-stick” pots and pans,

is also used to build capacitors. Teflon capacitors can have a 2.0 m F ca-

pacitance, allowing them to store as much as 4.0 × 10–4 C. Across what

potential difference can a Teflon capacitor be placed?

10. Polyester capacitors can have a capacitance of 5.0 × 10–5 F. What po-

tential difference is required to store 6.0 × 10–4 C?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19ACURRENT

P R O B L E MA total charge of 55 mC passes through a cross-sectional area of a copperwire in 0.5 s. What is the current in the wire?

S O L U T I O NGiven: ∆Q = 5.5 × 10−2 C ∆t = 0.5 s

Unknown: I = ?


Use the equation for electric current, given on page 694.

I = ∆∆Q

t =

5.5 ×0.

1

5

0

s

−2 C = 0.11 A

1. A total charge of 76 C passes through a cross-sectional area of a copper

wire in 19 s. What is the current in the wire?

2. A total charge of 114 µC passes through a cross-sectional area of an

aluminum wire in 0.36 s. What is the current in the wire?

3. A total charge of 29 mC passes through a cross-sectional area of a

nichrome wire in 11 s. What is the current in the wire?

4. If a current in a wire of a telephone is 1.4 A, how long would it take for

98 C of charge to pass a point in this wire?

5. If a current in a wire of a vacuum cleaner is 9.65 A, how long would it

take for 30.9 C of charge to pass a point in this wire?

6. If a current in a wire of a blender is 7.8 A, how long would it take for

56 C of charge to pass a point in this wire?

7. A photocopy machine draws 9.3 A when it starts up. If the start-up

time is 15 s, how much charge passes a cross-sectional area of the cir-

cuit in this time?

8. A computer draws 3.0 A when it starts up. If the start-up time is

2.0 min, how much charge passes a cross-sectional area of the circuit in

this time?

9. A printer draws 0.70 A when it starts up. If the start-up time is

2.0 s, how much charge passes a cross-sectional area of the circuit in

this time?

10. A space heater draws 5.6 A when it starts up. If the start-up time is 4.3 s,

how much charge passes a cross-sectional area of the circuit in this

time?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19BRESISTANCE

P R O B L E MA clothes dryer is equipped with an electric heater. The heater works bypassing air across an electric wire that is heated by the electricity passingthrough it. If the wire’s resistance is 10.0 and the current through thewire equals 24 A, what is the potential difference across the heater wire?

S O L U T I O NGiven: R = 10.0 Ω I = 24 A

Unknown: ∆V = ?

Choose the equation(s) or situation: Rearrange the definition for resistance on

page 700 to solve for the potential difference across the wire.

∆V = IR = (24 A) (10.0 Ω) = 240 V

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ADDITIONAL PRACTICE

1. You have probably heard that high-voltages are more dangerous than

low voltages. To understand this, assume that your body has a resis-

tance of 1.0 × 105 Ω. What potential difference would have to be across

your body to produce a current of 1.0 mA (which would cause a tin-

gling feeling) and 15 mA (a fatal amount of current)?

2. A battery-powered electric lantern is used as a light source for camp-

ing. The light bulb in the lantern has a resistance of 6.4 Ω. Assume that

the rest of the lantern’s circuit has no resistance and that the current

through the circuit is 0.75 A. Calculate the potential difference across

the lantern’s battery.

3. Some kitchen sinks are equipped with electric garbage disposals. These

are units with rapidly rotating steel blades, which are able to crush and

chop up food so that it can be washed down the drain instead of taking

up space as solid garbage. The motor of a garbage disposal has a resis-

tance of about 25.0 Ω. If the current through the motor equals 4.66 A,

what is the potential difference across the motor’s terminals?

4. A washing machine motor carries a current through a circuit with a re-

sistance of 12.2 Ω. If the washing machine is plugged into a 120 V out-

let, what is the current in the motor?

5. If you were to swim in the Atlantic Ocean off the coast of Brazil, the re-

sistance of your body could drop as low as 1.0 × 102 Ω. An electric eel

in Brazil can have a potential difference of up to 650 V across it. If you

came into contact with this eel while swimming, what current would

be delivered to your body?


NAME ______________________________________ DATE _______________ CLASS ____________________

6. When traveling to another country, you should always find out the

voltage that is used in that country before you plug in an appliance. To

understand the reason for this precaution, calculate the current that a

laptop computer would draw from a 120 V outlet in the United States

if the computer has a resistance of 40.0 Ω. Then, calculate the current

that the same computer would draw if you plugged it into a 240 V out-

let in the United Kingdom.

7. A television set is plugged into a 120 V outlet. The television circuit

carries a current equal to 0.75 A. What is the overall resistance of the

television set?

8. An electric car is equipped with an electric motor that can deliver

50 horsepower. The voltage across the motor’s terminals equals 5.00 ×102 V, and the current through the motor is 0.89 A. What is the resis-

tance in the motor’s circuit?

9. A medium-sized household oscillating fan draws 545 mA of current

when the potential difference across its motor is 120 V. How large is the

fan’s resistance?

10. A refrigerator’s circuit carries a current equal to 0.65 A when the volt-

age across the circuit equals 117 V. How large is the resistance of the re-

frigerator’s circuit?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19CELECTRIC POWER

P R O B L E MIf an alarm clock is plugged into a 120 V outlet, the electric current in theclock’s circuit is 4.2 102 A. How much power does the alarm clock use?

S O L U T I O N

Given: ∆V = 120 V I = 4.2 × 10–2 A

Unknown: P = ?


Because the current and potential difference are given but the power is un-

known, use the third form of the power equation on page 709, which in-

cludes these three variables.

P = I∆V = (4.2 × 10−2 A) (120 V) = 5.0 W

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1. A generator at a central electric power plant produces electricity with a

potential difference of 2.5 × 104 V across power lines which carry a

current of 20.0 A. How much power does the generator produce?

2. An electric sports car was developed several years ago at Texas A&M

University in College Station, Texas. If the potential difference across

the car’s motor is 720 V and the resistance was 0.30 Ω, how much

power was needed for the car to run?

3. A light bulb has a filament with a resistance of 144 Ω, while a second

bulb has a filament with a resistance of 240 Ω. Both bulbs are con-

nected across a 120 V outlet. Which light bulb is brighter? [Hint: The

brightest bulb uses the most power.]

4. A microwave oven requires 1750 W of power to cook food. If the oven

is plugged into a 120 V outlet, what is the resistance in the oven’s

circuit?

5. A waffle iron requires 650 W of power to operate. If the waffle iron is

plugged into a 120 V outlet, what is the resistance in the waffle iron’s

circuit?

6. An electric kettle requires 370 W of power to boil water. If the kettle is

plugged into a 120 V outlet, what is the resistance in the kettle’s circuit?

7. A blender requires 350 W to power the rotating blades that chop food.

If the blender has a resistance of 75 Ω, how much current passes

through the blender’s circuit?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

8. A computer with a power input of 230.0 W has a resistance of

91.0 Ω. Find the current in the computer.

9. A laser was developed in 1995 at the University of Rochester, New York,

that produced a beam of light that lasted for about a billionth of a sec-

ond. The power output of this beam was 6.0 × 1013 W. If all of the elec-

trical power was converted into light and 8.0 × 106 A of current was

needed to produce this beam, what was the potential difference across

the circuit of the laser?

10. Fuel cells are chemical cells that combine hydrogen and oxygen gas to

produce electrical energy. In recent years, a fuel cell has been developed

that can generate 1.06 × 104 W of power. If this fuel cell has a current

of 16.3 A, what is the potential difference across the fuel cell?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 19DCOST OF ELECTRICAL ENERGY

P R O B L E MSuppose you woke up one morning and realized that you had forgotten toturn off your front porch light the night before. If you had used .540 kW•hof energy over a period of 12 h, what is power emitted by the light bulb?

S O L U T I O NGiven: Energy = 0.540 kW•h ∆t = 12 h

Unknown: P = ?


Use the equation relating energy and power on page 712.

Energy = P∆t


Rearrange the equation to solve for the total power.

P = En

∆er

t

gy =

0.54

1

0

2

k

h

W•h = 0.045 kW = 45 W

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ADDITIONAL PRACTICE

1. Suppose that you’ve just returned to work from your lunch hour.

When you reach your desk, you realize that you had forgotten to turn

off your computer. Fortunately, your computer was in its energy-

conserving “sleep” mode. What is the power of a computer which

consumed 2.7 × 108 J of energy?

2. Suppose you’ve just returned to the parking lot from a 3.0 h shopping

spree at the mall and realized that you had forgotten to turn off the

headlights to your car. If 4.86 × 108 J of energy was spent, what is the

power of the headlights?

3. A bread machine requires 1200 W to bake bread. How much time is re-

quired for it to use 1.512 × 1010 J of energy?

4. Calculate how much time is required for a 600 W air conditioner to use

8.64 × 109 J of energy?

5. As an incentive to conserve electricity, some electric companies charge

lower prices for electricity up to a certain number of kilowatt-hours,

and then raise the cost of electricity for each kilowatt-hour you use

over that number. Suppose your local electric company charges

$0.0650/kW•h for the first 200.0 kW•h, and then drastically raises the

price for every kilowatt-hour more. What is the maximum your family

would pay if they wanted to pay only the cheaper rate?

Problem 19D Ch. 19-7

NAME ______________________________________ DATE _______________ CLASS ____________________

6. Suppose you receive an electric bill which states the following:

Final Meter Reading 24422 kW•h

Previous Meter Reading 24204 kW•h

Cost of Electricity Used for 20 Days $0.078/kW•h

How much must you pay the electric company?

7. How much energy does a 125 W computer use in an 8.0 h workday?

8. Calculate how many joules of energy a 750 W refrigerator uses in 24 h.

9. Calculate how many joules of energy a 550.0 W hair dryer uses in

10.0 min.

10. Calculate how many joules of energy an 850 W toaster uses in 3.0 min.

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20ARESISTORS IN SERIES

P R O B L E MA 18.0-Ω resistor is connected in series with another resistor across a 1.55-V battery. The current in the circuit is 25 mA. Calculate the value ofthe unknown resistance.

S O L U T I O NGiven: ∆V = 1.55 V R1 = 18 Ω I = 0.025 A

Unknown: R2 = ?


Use the equation relating the potential difference across the load for resistors in

series, given on page 737.

∆V = IR1 + IR2

Rearrange the equation(s) to isolate the unknown(s): Rearrange to solve for R2.

R2 = ∆

I

V − R1 =

0

1

.

.

0

5

2

5

5

V

A – 18 Ω = 62 Ω – 18 Ω = 44 Ω

1. A 16- Ω resistor is connected in series with another resistor across a

12-V battery. The current in the circuit is 0.42 A. Calculate the value of

the unknown resistance.

2. A 24-Ω resistor is connected in series with another resistor across a 3.0-V

battery. The current in the circuit is 62 mA. Calculate the value of the

unknown resistance.

3. A 9-Ω resistor is connected in series with another resistor across a 9.0-V

battery. The current in the circuit is 0.33 A. Calculate the value of the

unknown resistance.

4. A string of holiday lights has 73 light bulbs in series. Each light bulb

has a resistance of 3.0 Ω. Calculate the equivalent resistance.

5. A movie theater has 25 surround-sound speakers wired in series. Each

speaker has a resistance of 12.0 Ω. What is the equivalent resistance?

6. In case of an emergency, a corridor on an airplane has 57 lights wired

in series. Each light bulb has a resistance of 2.0 Ω. Find the equivalent

resistance.

7. A quadraphonic car stereo operates on electricity provided by the car’s

12-V battery and is connected in series. Each channel of the stereo,

which feeds the electric signal to one of the stereo’s four speakers, has a

resistance of 4.1 Ω. How much current is in the circuit of each stereo

channel?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

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8. A chandelier has 10 sockets wired in series, each of which holds a 10 Ωlight bulb. If the voltage across the chandelier’s circuit is 100 V, what is

the current drawn by the chandelier?

9. A portable lamp has three bulbs wired in series: one bulb has a resistance

of 96 Ω, one bulb has a resistance of 48 Ω, and one bulb has a resistance

of 29 Ω. If the voltage across the lamp is 115 V, what is the current

through the lamp’s circuit?

10. Three bulbs are wired in series: one bulb has a resistance of 56 Ω, one

bulb has a resistance of 82 Ω, and one bulb has a resistance of 24 Ω. If

the voltage across the circuit is 9.0 V, what is the current through the

circuit?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20BRESISTORS IN PARALLEL

P R O B L E M

A 42.0 resistor is connected in parallel with another resistor across a9.0 V battery. The current in the circuit is 0.41 A. Calculate the value ofthe unknown resistance.

S O L U T I O NGiven: ∆V = 9.0 V R1 = 42.0 Ω I = 0.41 A

Unknown: R2 = ?

Choose an equation(s) or situation:

Use the equation relating potential difference and equivalent resistance for resis-

tors in parallel, given on page 742.

∆V = IReq

I = R

∆

e

V

q =

∆R

V

1 +

∆R

V

2

Rearrange the equation(s) to isolate the unknown(s): Rearrange to solve for R2.

∆R

V

2 = I –

∆R

V

1

R2 = = = = 45 Ω9.0 V

[0.41 A − 0.21 A]

9.0 V

0.41 A − 9

4

.

2

0

ΩV

∆V

I – ∆R

V

1

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ed. ADDITIONAL PRACTICE

1. A 3.3 Ω resistor is connected in parallel with another resistor across a

3.0 V battery. The current in the circuit is 1.41 A. Calculate the value of


2. A 56 Ω resistor is connected in parallel with another resistor across a

12 V battery. The current in the circuit is 3.21 A. Calculate the value of


3. An 18 Ω resistor is connected in parallel with another resistor across a

1.5 V battery. The current in the circuit is 103 mA. Calculate the value

of the unknown resistance.

4. A 39 Ω resistor, an 82 Ω resistor, a 12 Ω resistor and a 22 Ω resistor are

connected in parallel across a potential difference of 3.0 V. Calculate

the equivalent resistance.

5. A 10 Ω resistor, a 12 Ω resistor, a 15 Ω resistor and an 18 Ω resistor are

connected in parallel across a potential difference of 12 V. What is the

equivalent resistance?


NAME ______________________________________ DATE _______________ CLASS ____________________

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6. A 33 Ω resistor, a 39 Ω resistor, a 47 Ω resistor and a 68 Ω resistor are

connected in parallel across a potential difference of 1.5 V. Find the

equivalent resistance.

7. A refrigerator and an oven are wired in parallel across a potential differ-

ence of 120 V. The refrigerator has a resistance of 75 Ω and the oven has

a resistance of 91 Ω. How much current is in the circuit of each

appliance?

8. A computer and a printer are wired in parallel across a potential differ-

ence of 120 V. The computer has a resistance of 82 Ω and the printer

has a resistance of 24 Ω. How much current is in the circuit of each

machine?

9. A lamp and a stereo are wired in parallel across a potential difference of

120 V. The lamp has a resistance of 11 Ω and the stereo has a resistance

of 36 Ω. How much current is in the circuit of each load?

10. Two bulbs are wired in parallel: one bulb has a resistance of 3.3 Ω, and

the other bulb has a resistance of 4.3 Ω. If the voltage across the circuit

is 1.5 V, what is the current through each bulb?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20CANGULAR ACCELERATION

P R O B L E MDetermine the unknown resistance in thecomplex circuit shown at right. The currentin the circuit is 0.36 A.

S O L U T I O N1. Redraw the circuit as a group of resistors along one side of the circuit.

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12.0 V

8.0 ΩR = ?

9.0 Ω

5.0 Ω 3.0 Ω2.0 Ω

3.0 Ω5.0 Ω

5.0 Ω 3.0 Ω2.0 Ω

3.0 Ω5.0 Ω

9.0 Ω

(a)

(c)

(b)

R = ? 8.0 Ω

2. Identify components in series, and calculate their equivalent resistance.

Resistors in groups (a), (b), and (c) are in series.

For group (a): Req = 9.0 Ω + R

For group (b): Req = 2.0 Ω + 5.0 Ω + 3.0 Ω = 10.0 Ω = R1 below

For group (c): Req = 5.0 Ω + 3.0 Ω = 8.0 Ω = R2 below

3. Identify components in parallel, and calculate their equivalent resistance.

Groups (b) and (c) combine to become group (d).

For group (d):

R

1

eq =

R

1

1 +

R

1

2 =

10.

1

0 Ω +

8.0

1

Ω =

0

1

.1

Ω0

+ 0

1

.1

Ω3

= 0

1

.2

Ω3

R = 4.4 Ω

4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single

equivalent resistance. All resistors combine to become group (e). The remainder of

the resistors in group (e) are in series.

For group (e): Req = 9.0 Ω + R + 4.4 Ω + 8.0 Ω = R + 21.4 Ω

5. Choose the equation(s) or situation:

Use the equation relating equivalent resistance to potential difference and current.

Req = R − 21.4 Ω = ∆

I

V − 21.4 Ω =

1

0

2

.3

.0

6

V

A – 21.4 Ω = 33 Ω – 21.4 Ω = 12 Ω

NAME ______________________________________ DATE _______________ CLASS ____________________

2. Determine the un-

known resistance

in the complex cir-

cuit shown at right.

The current in the

circuit is 375 mA.

3. Determine the un-

known resistance in

the complex circuit

shown at right. The

current in the circuit

is 185 mA.

4. Determine the equivalent re-

sistance of the complex circuit

shown at right.

5. Determine the equivalent re-

sistance of the complex circuit

shown at right.

6. Determine the equiv-

alent resistance of the

complex circuit

shown at right.

7. What will be the net current

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18.0 Ω8.00 Ω

3.0 Ω 4.0 Ω

20.0 Ω

R = ?

22.0 Ω

6.0 Ω9.00 V

2.0 Ω 3.0 Ω8.00 Ω 8.00 Ω

3.0 Ω 4.0 Ω

2.0 Ω

R = ?8.0 Ω 8.0 Ω

3.0 Ω

12.0 V

3.0 Ω 4.0 Ω2.0 Ω

5.0 Ω 6.0 Ω 3.0 Ω 4.0 Ω4.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω10.0 Ω

10.0 Ω 10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

2.0 Ω 2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω 2.0 Ω

10.0 Ω 10.0 Ω

10.0 Ω

10.0 Ω10.0 Ω 10.0 Ω10.0 Ω

10.0 Ω

10.0 Ω 10.0 Ω 10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω 10.0 Ω 10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω

10.0 Ω 10.0 Ω

5.0 Ω

2.0 Ω

5.0 Ω

6.0 Ω4.0 Ω

1.5 Ω3.0 Ω

12.0 V

ADDITIONAL PRACTICE

1. Determine the unknown resistance

in the complex circuit shown at

right. The current in the circuit is

680 mA.

15.0 V

11 ΩR = ?12.0 Ω

15 Ω

7.0 Ω6.0 Ω

NAME ______________________________________ DATE _______________ CLASS ____________________

8. What will be the net current for

the circuit shown at right?





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5.0 Ω

5.0 Ω

5.0 Ω

3.0 Ω

5.0 Ω

5.0 Ω3.0 Ω

5.0 Ω

15.0 V

4.0 Ω 3.0 Ω 1.0 Ω

4.0 Ω 2.0 Ω 2.0 Ω

24.0 V

4.0 Ω 2.0 Ω2.0 Ω

4.0 Ω

4.0 Ω

8.0 Ω3.0 Ω

4.0 Ω

2.0 Ω

24.0 V

3.0 Ω

8.0 Ω

4.0 Ω5.0 Ω 2.0 Ω


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 20DCURRENT IN AND POTENTIAL DIFFERENCE ACROSS A RESISTOR

P R O B L E M

Determine the current in and the potential difference across the 5.0 re-sistor in the circuit diagram at right.

S O L U T I O N1. Determine the equivalent resistance in the circuit.

For group (a): Req,a = 5.0 Ω + 2.0 Ω = 7.0 Ω = R2

For group (b): Re

1

q,b =

R

1

1 +

R

1

2 =

4.0

1

Ω +

7.0

1

Ω

Re

1

q,b =

0

1

.2

Ω5

+ 0

1

.1

Ω4

= 0

1

.3

Ω9

Req,b = 2.6 Ω

For group (c): Req,c = 3.0 Ω + 2.6 Ω + 6.0 Ω = 11.6 Ω

2. Calculate the total current in the circuit, which is the current in group (c).

I = ∆R

V

e

t

q

ot = 1

1

1

2

.

.

6

0

ΩV

= 1.0 A

3. Determine a path from the equivalent resistance found in step 1 to the

5.0 resistor. Review the path taken to find the equivalent resistance and

work backward through this path.

4. Follow the path determined in step 3, and calculate the current in and the

potential difference across each equivalent resistance. Repeat this process

until the desired values are found.

Regroup, evaluate, and calculate. The circuit’s equivalent resistance is that of

group (c), as found in step 1 above. The resistors in group (c) are in series; there-

fore, the current in each resistor is the same as the current in the equivalent re-

sistance, which equals 1.0 A. The potential difference across group (b), which is

represented by the 2.6 Ω resistor in group (c), can be replaced with ∆V = IR.

Given: I = 1.0 A R = 2.6 Ω

Unknown: ∆V = ?

∆V = IR = (1.0 A)(2.6 Ω) = 2.6 V

Regroup, evaluate, and calculate. Replace the center resistor with group (b).

The resistors in group (b) are in parallel; therefore, the potential difference

across each resistor is the same as the potential difference across the 2.6 Ωequivalent resistance, which equals 2.6 V. The current in the 7.0 Ω resistor in

group (b) can be calculated using I = ∆V/R.

Given: ∆V = 2.6 V R = 7.0 Ω

Unknown: I = ?

I = ∆R

V =

7

2

.

.

0

6

ΩV

= 0.37 A

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4.0 Ω

2.0 Ω6.0 Ω

5.0 Ω3.0 Ω

(c)

(a)

(b)

12 V

NAME ______________________________________ DATE _______________ CLASS ____________________

Regroup, evaluate, and calculate. Replace the 7.0 Ω resistor with group (a).

The resistors in group (a) are in series; therefore, the current in each resistor

is the same as the current in the 7.0 Ω equivalent resistance, which equals

0.37 A.

The potential difference across the 5.0 Ω resistor can be calculated using

∆V = IR.

Given: I = 0.37 A R = 5.0 Ω

Unknown: ∆V = ?

∆V = IR = (0.37 A)(5.0 Ω) = 1.85 V

∆V = 1.85 V

I = 0.37 A

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ADDITIONAL PRACTICE

1. Determine the current in and the potential

difference across the 4.0 Ω resistor in the

circuit diagram at right.




3. Determine the current in and the

potential difference across the 6.0 Ωresistor in the circuit diagram at right.







3.0 Ω 12.0 V

10.0 Ω

10.0 Ω

10.0 Ω

4.0 Ω

1.5 V

2.0 Ω

9.0 Ω

8.0 Ω6.0 Ω 4.0 Ω

3.0 Ω7.0 Ω 6.0 Ω

4.0 Ω 5.0 Ω

2.0 Ω9.0 V

7.0 Ω 9.0 V

4.0 Ω 3.0 Ω5.0 Ω

10.0 Ω 6.0 Ω2.0 Ω

3.0 V

4.0 Ω 6.0 Ω2.0 Ω

5.0 Ω

3.0 Ω1.0 Ω

NAME ______________________________________ DATE _______________ CLASS ____________________


potential difference across the

3.0 Ω resistor in the circuit dia-

gram at right.


potential difference across the 2.0 Ωresistor in the circuit diagram at

right.


potential difference across the 7.0 Ωresistor in the circuit diagram at right.

9. Determine the cur-

rent in and the

potential difference

across the 12.0 Ωresistor in the circuit

diagram at right.


potential difference across the 15.0 Ωresistor in the circuit diagram at

right.

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2.0 Ω 3.0 Ω

2.0 Ω 4.0 Ω

3.0 V

2.0 Ω

2.0 Ω

5.0 Ω5.0 Ω

12.0 V

2.0 Ω

5.0 Ω

5.0 Ω

5.0 Ω5.0 Ω

3.0 Ω

2.0 Ω

7.0 Ω

3.0 Ω

12.0 V

4.0 Ω

5.0 Ω

3.0 Ω 3.0 Ω3.0 Ω5.0 Ω

1.5 V 4.0 Ω

4.0 Ω6.0 Ω

12.0 Ω 6.0 Ω6.0 Ω

15.0 Ω

6.0 Ω

30.0 Ω

5.0 Ω 12.0 V

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 21APARTICLE IN A MAGNETIC FIELD

P R O B L E MA bubble chamber is a device used to make the paths of charged particlesvisible in collision experiments. Suppose a proton travels in one of thesebubble chambers at a speed of 1.9 × 107 m/s. A magnetic force of 4.3 ×10–12 N acts on the proton. What is the magnitude of the magnetic field atright angles to the proton’s path?

S O L U T I O N

Given: q = 1.60 × 10–19 C v = 1.9 × 107 m/s

Fmagnetic = 4.3 × 10–12 N

Unknown: B = ?

Use the equation for the magnetic field from page 773.

B = Fma

qg

vnetic = = 1.4 Τ

4.3 × 10–12 N(1.60 × 10–19 C)( 1.9 × 107 m/s)

ADDITIONAL PRACTICE

1. An electron in a cathode ray tube (found in many televisions) moving at a speed of

1.2 × 106 m/s has a magnetic force of 1.2 × 10–17 N acting on it. What is the magnitude

of the magnetic field perpendicular to the electron’s path?

2. Humans emit a magnetic field because ions move through the blood. Suppose an elec-

tron whizzes horizontally by a standing person at a speed of 3.9 × 106 m/s. If the elec-

tron is affected by a magnetic force of 1.9 × 10–22 N, how much of a magnetic field does

this person emit?

3. An electron moves perpendicular to a sunspot at a speed of 7.8 × 106 m/s. A magnetic

force of 3.7 × 10–13 N is exerted on the electron. What is the magnitude of the magnetic

field emitted by the sunspot?

4. In the Fermilab accelerator, a proton maintains circular motion about a radius of

1.0 km. The proton travels at right angles to a uniform magnetic field of 3.3 T. What is

the speed of the proton? (Hint: The magnetic force exerted on the proton is the force

that maintains circular motion.)

5. Aurora Borealis, the “northern lights” found at high latitudes, happens when charged

particles fall into Earth’s atmosphere near the poles. The magnetic field near the poles is

5.0 × 10–5 T. Suppose the particle has a charge of 1.60 × 10–19 C. If the magnetic force

exerted on the particle is 6.1 × 10–17 N, at what speed is the particle moving? (Assume

that the particle moves perpendicular to Earth’s magnetic field.)

6. The magnetic field in the Crab Nebula is about 1 × 10–8 T. If an electron moving per-

pendicular to this field is affected by a magnetic force of 3.2 × 10–22 N, what is the elec-

tron’s speed?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. Suppose an electron moves to the right at a velocity of 6 × 106 m/s at an

angle of 45° to a uniform magnetic field. If the magnetic field is 3 × 10–4 T

upward, what is the magnitude and direction of the magnetic force ex-

erted on the electron? (Hint: Only the component perpendicular to the

magnetic field contributes to the magnetic force.)

8. A proton moves at right angles to a uniform magnetic field of 0.8 T. If

the speed of the proton is 3.0 × 107 m/s, what is the magnetic force ex-

erted on the proton?

9. Suppose an electron moves at a speed of 2.2 × 106 m/s on the surface of

the Sun. Suppose the Sun has a magnetic field of 1.1 × 10–2 T perpendic-

ular to the electron’s path. What is the magnetic force exerted on the

electron by the Sun?

10. Suppose an electron moves at a speed of 9.3 × 105 m/s. If it moves at

right angles to a uniform magnetic field of 4.1 × 10–10 T in interstellar

galactic space, what is the magnetic force exerted on the electron?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 21BFORCE ON A CURRENT CARRYING CONDUCTOR

P R O B L E MSuppose a wire carries a current of 7.32 A. If a magnetic field of 0.0038 Tproduces a force that has a magnitude of 4.2 × 10–3 N, how long is thewire?

S O L U T I O NGiven: I = 7.32 A B = 0.0038 T

Fmagnetic = 4.2 × 10–3 N

Unknown: l = ?

Use the equation for the force on a current-carrying conductor perpendicular to

a magnetic field, shown on page 776.

l = Fma

Bg

Inetic =

(0.0

4

0

.

3

2

8

×T

1

)

0

(

–

7

3

.3

N

2 A) = 0.15 m

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ADDITIONAL PRACTICE

1. Suppose a wire is placed perpendicular to a uniform magnetic field of 4.6 × 10–4 T. A

magnetic force of 2.9 × 10–3 N is exerted on the wire. If the current in the wire is 10.0 A,

how long is the wire?

2. An electric eel in Brazil can produce 1 A of current and up to 650 V. If the eel swims

perpendicular to the Earth’s magnetic field of 2.8 × 10–5 T and is affected by a magnetic

force of 5.6 × 10–5 N, how long is the eel?

3. A 12 m wire carrying a current of 12 A is placed at right angles to a uniform magnetic

field. A magnetic force of 7.3 × 10–2 N acts on this wire. What is the magnitude of the

magnetic field?

4. Suppose a lightning bolt strikes the ground in a straight path and a magnetic force of

7.8 × 105 N acts on the bolt. The bolt carries 1.8 × 104 A of current. The cloud from

which the lightning bolt came is 12 km above the ground. What is the magnitude of the

magnetic field perpendicular to this lightning bolt?

5. Suppose a 14.32 A current moves through a 15.0 cm wire. If a magnetic force of 6.62 ×10–4 N acts on the wire, what would be the magnitude of the magnetic field at right an-

gles to the wire?

6. A magician wants to build a device that will allow him to levitate for his magic act. His

device has a wire 10 m long bent into a loop. If the magician’s mass is 75 kg, and Earth’s

magnetic field (perpendicular to the wire) is 4.8 × 10–4 T, how much current must

move through the wire for the magician to levitate?


NAME ______________________________________ DATE _______________ CLASS ____________________

7. A printer is connected to a 1.0 m cable. If the magnetic force is 9.1 × 10–5 N,

and the magnetic field is 1.3 × 10–4 T, how much current passes through

the wire?

8. Suppose a high-voltage cable carries 1.5 × 103 A of current 15 km along a

straight city road. The Earth’s magnetic field at this latitude is at a 45°

angle to the power line and has a magnitude of 2.3 × 10–5 T. What is the

magnetic force exerted on the cable?

9. A fax machine requires 1.4 A of current to pass through a 2 m cable in

order to work. If the magnetic field perpendicular to the cable is 3.6 ×10–4 T, what is the magnetic force exerted on the cable?

10. Light bulb filaments can weaken partially because of the magnetic forces

acting on them. If a filament weakens enough, it breaks, and the light

bulb “burns out.” Suppose a current of 0.5 A moves through a 5 cm-long

wire filament and that a magnetic field of 1.3 × 10–4 T acts on the wire.

What is the magnetic force acting on the filament? (Assume that the fila-

ment is a straight wire.)

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 22AINDUCED EMF AND CURRENT

P R O B L E MA coil with 245 turns and an area of 0.51 m2 is placed 45° to a uniform ex-ternal magnetic field. The coil is 90.0° to the magnetic field 0.63 s later. Ifthe average induced emf is 45 V, what is the strength of the magnetic field?

S O L U T I O N

Given: N = 245 turns A = 0.51 m2 qi = 45°

qf = 90.0° ∆t = 0.63 s emf = 45 V

Unknown: B = ?

Choose the equation(s) or situation: Use Faraday’s law of magnetic induction

to find the induced emf in the coil.

emf = –N ∆[A

∆B

t

(cos q)]

Rearrange the equation(s) to isolate the unknown(s): In this example, only the

angle between the magnetic field and the coil changes with time. The other com-

ponents (the coil area and the magnetic field strength) remain constant.

B = –N

e

A

m

∆f

[c

∆o

t

s q] =

B =

B = = 0.32 T(45 V)(0.63 s)

–(245)(0.51 m2)(– 0.71)

(45 V)(0.63 s)– (245)(0.51 m2)[cos(90.0°) – cos(45°)]

emf ∆t–NA [cos qf – cos qi]

ADDITIONAL PRACTICE

1. A coil with 540 turns and an area of 0.016 m2 is placed parallel to a uniform external

magnetic field. The coil is perpendicular to the magnetic field 0.05 s later. If the average

induced emf is 3.0 V, what is the strength of the magnetic field?

2. A 320-turn coil with an area of 0.068 m2 is placed parallel to a uniform external mag-

netic field. Exactly 0.25 s later, the coil is at right angles to the magnetic field. This in-

duces an average emf of 4.0 V. How strong is the magnetic field?

3. A coil with an area of 0.93 m2 and 628 turns begins parallel to a uniform external mag-

netic field. It takes 0.30 s for the coil to rotate so that it makes an angle of 30.0° with the

magnetic field. If the average induced emf is 62 V, what is the strength of the magnetic

field?


NAME ______________________________________ DATE _______________ CLASS ____________________

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4. Credit cards have a magnetic tape along one edge of the card. The mag-

netic fields on the tape correspond with the unique number of the card.

When a credit card slides through a credit-card reader, the changes in the

magnetic fields on the tape induce an emf in a coil that are converted to

an electrical signal. Suppose a 550-turn coil with an area of 5.0 × 10–5 m2

is positioned parallel to a magnetic field that changes by 2.5 × 10–4 T in

2.1 × 10–5 s. What is the induced emf in the coil?

5. A VCR plays movies by placing a video head over a magnetic tape. The

changing magnetic fields in the tape induce a current in the coil, which is

converted into an electrical signal, which goes to a television set. Suppose

a VCR has a video head made of a 220-turn coil wrapped around an iron

ring with an area of 6.0 × 10–6 m2. The head is parallel to a magnetic

field that changes by 9.7 × 10–4 T in 1.7 × 10–6 s. What is the induced

emf in the coil?

6. Computers store memory by spinning a stack of magnetic-coated plates

beneath an electromagnetic head. Suppose a head consisting of a 148-

turn coil wrapped around an iron ring with an area of 1.25 × 10–8 m2.

The head is parallel to a magnetic field that changes by 5.2 × 10–4 T in

8.5 × 10–9 s. What is the induced emf in the coil?

7. Suppose an emf of 220 V is induced in a coil that has a resistance of

120 Ω. What is the induced current in the coil?

8. At your job, you use a security card that unlocks certain doors. A mag-

netic tape along the edge of the card induces a current in a coil when you

insert and remove the card from a slot. Suppose the coil designed to read

your card has 180 turns and an area of 5.0 × 10–5 m2. The coil is parallel

to the magnetic field that changes by 5.2 × 10–4 T in 1.9 × 10–5 s. What is

the induced current if the resistance in the coil is 1.0 × 102 Ω?

9. A 246-turn coil with an area of 0.40 m2 is parallel to a magnetic field. Ini-

tially, the magnetic field is 0.237 T, but it increases to 0.320 T in 0.9 s. If

this induces an emf of 9.1 V, what is the time interval necessary to induce

this emf?

10. To measure earthquakes, seismologists suspend a bar magnet (using very

sensitive springs) within a coiled wire housing that is parallel to the mag-

netic field. Suppose a coil has 785 turns of wire and an area of 7.3 ×10–2m2. Suppose an earthquake occurs, causing the magnetic field to

change by 6.9 × 10–3 T. Each time the magnet oscillates (moves up and

down), it induces an emf of 2.8 V. What is the time interval for each os-

cillation? If there are a total of 120 oscillations during this earthquake,

how long did the earthquake last?


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 22BINDUCTION IN GENERATORS

P R O B L E M

A generator is made of a 200-turn coil with an area of 0.040 m2 that ro-tates in a magnetic field of 6.9 × 10–3 T. What is the angular speed neededto produce a maximum emf of 25 V?

S O L U T I O NGiven: maximum emf = 25 V N = 200 turns

A = 0.040 m2 B = 6.9 × 10−3 T

Unknown: w = ?

Choose the equation(s) or situation: Use the maximum emf for a generator.

maximum emf = NABw


for maximum emf of a generator to solve for the angular frequency.

w = maxim

NA

um

B

emf


w = maxim

NA

um

B

emf = = 4.5 × 102 rad/s

25 V(200)(0.040 m2)((0.0069 T)

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ADDITIONAL PRACTICE

1. A generator consisting of a coil with 220 turns and an area of 0.080 m2

rotates in a magnetic field of 4.8 × 10–3 T. What is the angular speed

needed to produce a maximum emf of 12 V?

2. A generator is made of a 140-turn coil with a diameter of 0.33 m that ro-

tates in a magnetic field of 9.3 × 10–2 T. What is the angular speed needed

to produce a maximum emf of 150 V?

3. A generator is made of a 195-turn coil with an area of 0.052 m2 that ro-

tates in a magnetic field of 3.2 × 10–3 T. What is the angular speed needed

to produce a maximum current of 1.2 A on a 16 Ω load?

4. A generator is made of a 385-turn coil with an area of 0.38 m2 that ro-

tates in a magnetic field of 9.4 × 10–3 T. If the coil rotates at 45 Hz, what

is the maximum emf produced by the generator?

5. A generator produces a maximum current of 14 A on a 5 Ω load. What is

the maximum emf produced?


NAME ______________________________________ DATE _______________ CLASS ____________________

6. A generator is made of a 119-turn coil with an area of 4.9 × 10–2 m2 that

rotates in a magnetic field of 9.4 × 10–3 T. If the coil rotates at 345 rad/s,

what is the maximum emf produced by the generator?

7. A generator produces a maximum emf of 40 V on an 8 Ω load. What is

the maximum current produced by the generator?

8. A generator is made of a 425-turn coil with an area of 2.16 × 10–2 m2

that rotates in a magnetic field of 3.9 × 10–2 T. If the coil rotates at 33 Hz,

what is the maximum current produced by the generator if the load is

25 Ω?

9. A generator is made of a coil with an area of 1.20 × 10–2 m2 that rotates

in a magnetic field of 6.0 × 10–2 T. The coil rotates at 393 rad/s and in-

duces a maximum emf of 213 V. How many turns of wire are there in the

coil?

10. A generator is made of a coil with an area of 0.60 m2 that rotates in a

magnetic field of 0.012 T. The coil rotates at 44 Hz and induces a maxi-

mum emf of 320 V. How many turns of wire are there in the coil?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 22CRMS CURRENTS POTENTIAL DIFFERENCES

P R O B L E MA generator with a rms potential difference of 124 V is connected to a 85 resistor. Calculate the maximum output emf. Find the rms currentand the maximum ac current in the circuit.

S O L U T I O NGiven: ∆Vrms = 124 V R = 85 Ω

Unknown: ∆Vmax = ? Irms = ? Imax = ?

Choose the equation(s) or situation: Use the equation for the rms potential dif-

ference to find the rms potential difference.

∆Vrms = 0.707 ∆Vmax

Use the equation for rms current to find Imax.

Irms = 0.707 Imax

Rearrange the equation(s) to isolate the unknown(s): Rearrange the definition

for resistance to calculate the rms current.

Irms = ∆V

Rrms

Rearrange the equation relating rms current to maximum current so that maxi-

mum current is calculated.

Imax = 0

I

.

r

7

m

0

s

7

Rearrange the equation relating rms potential difference to maximum potential

difference so that maximum potential difference is calculated.

∆Vmax = ∆0

V

.7r

0m

7s


∆Vmax = ∆0

V

.7r

0m

7s =

1

0

2

.7

4

0

V

7 =

Irms = ∆V

Rrms =

1

8

2

5

4

ΩV

=

Imax = 0

I

.r

7m

0s

7 =

0

2

.

.

7

1

0

A

7 =

Evaluate: The maximum values for potential difference and current are a little

less than three-halves (1.5 times) the rms values, as expected.

3.0 A

2.1 A

175 V

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NAME ______________________________________ DATE _______________ CLASS ____________________

1. Suppose you build a hydroelectric turbine in your back yard along a

river. The turbine is connected to a generator which can provide a rms

potential difference of 320 V onto a 100 Ω load. Calculate the maximum

output emf. Find the rms current and the maximum ac current in the

circuit.

2. Some wind turbines in the United States can provide an rms current of

1.3 A. Calculate the maximum ac current.

3. In Iceland, people harness geothermal energy by placing pipes deep into

the Earth in order to tap hot-water aquifers. Steam from these aquifers

spins a turbine that is connected to a generator. Suppose such a generator

can provide a rms current of 2.5 A and an rms potential difference of

22 kV. Calculate the maximum ac current. Find the resistance of the

load.

4. In Denmark, farmers and people living in rural areas harness wind en-

ergy to spin a turbine that is coupled to a generator. The generator can

provide an rms current of 1.7 A and an rms potential difference of 220 V.

Calculate the maximum ac current and the maximum output emf.

5. A generator can provide a maximum ac current of 1.2 A and a maximum

output emf of 211 V. Calculate the rms current, the rms potential differ-

ence, and the resistance in the circuit.

6. A generator can provide a maximum output emf of 170 V. Calculate the

rms potential difference.

7. Suppose you measure a circuit with a voltmeter. You measure a rms po-

tential difference of 115 V and a resistance of 50.0 Ω. Find the maximum

ac current and the rms current in the circuit.

8. Off the western coast of Scotland, turbines are placed in deep ocean

water so that ocean waves can spin them. The turbines are connected to a

generator, which produces electricity for the locals. These turbines can

supply an rms current of 2.1 A on a 16 kΩ load. Calculate the maximum

ac current. How much power is supplied?

9. A wind turbine is built to service farmers in England. The turbine is con-

nected to a generator, which can supply a rms current of 1.3 A on a

12 kΩ load. If the rms potential difference is 15.6 kV, find the maximum

output emf provided by the wind turbine. Calculate the maximum ac

current. How much power is supplied?

10. The total world potential for hydroelectric generation is 1.2 million MW,

however, not all of it has been harnessed. Suppose the world’s rms cur-

rent is 2.2 × 1010 A and it supplies a load of 6.1 × 10–10 Ω. Calculate the

total maximum ac current that is harnessed. How much total power can

be supplied with this current?

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ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 22DTRANSFORMERS

P R O B L E MAn ac generator at a central power station uses a step-up transformer toprovide a potential difference of 68 kV across the secondary coil. If theprimary coil has 125 turns and the secondary coil has 625 turns, what isthe potential difference across the primary coil?

S O L U T I O N

Given: ∆V2 = 68 kV = 6.8 × 104 V N1 = 125 turns

N2 = 625 turns

Unknown: ∆V1 = ?

Choose the equation(s) or situation: Use the transformer equation.

∆∆

V

V1

2 =

N

N

2

1


for a transformer to solve for the potential difference in the primary coil.

∆V1 = ∆V

N2

2

N1


∆V1 = ∆V

N2

2

N1 = (6.8 × 1

(

0

6

4

25

V

)

)(125) =

Evaluate: The potential difference across the primary should be 14 kV. The step-

up factor for the transformer is 1:5.

1.4 × 104 V = 14 kV

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ADDITIONAL PRACTICE

1. A transmission line to a city has a potential difference of 6.9 kV across

the secondary coil. If the primary coil has 1400 turns and the secondary

coil has 140 turns, what is the potential difference across the primary

coil?

2. A power line has a potential difference of 3.4 kV across the secondary

coil. If the primary coil has 9.0 × 101 turns and the secondary coil has

2250 turns, what is the potential difference across the primary coil?

3. A transmission line to a city has a potential difference of 46 kV across the

primary coil. If the primary coil has 1250 turns and the secondary coil

has 250 turns, what is the potential difference across the secondary coil?

4. A high-voltage cable has a potential difference of 5.6 kV across the pri-

mary coil. If the primary coil has 140 turns and the secondary coil has

840 turns, what is the potential difference across the secondary coil?


NAME ______________________________________ DATE _______________ CLASS ____________________

5. A high-voltage cable has a potential difference of 9.2 kV across the pri-

mary coil. If the primary coil has 120 turns and the secondary coil has

1200 turns, what is the potential difference across the secondary coil?

6. A ac power station has a potential difference of 36 kV across the primary

coil and a potential difference of 7.2 kV across the secondary coil. If the

primary coil has 55 turns, how many turns does the secondary coil have?

7. The plug to your zip drive has a step-down transformer in it. The poten-

tial difference across the primary coil is 240 V and a potential difference

of 5.0 V across the secondary. What is the step-down ratio? (Hint: the

step-down ratio is the ratio of N1:N2.)

8. A transmission line has a potential difference of 3.6 kV across the pri-

mary coil and a potential difference of 1.8 kV across the secondary coil. If

the primary coil has 58 turns, how many turns does the secondary coil

have?

9. A central ac power station has a potential difference of 49 kV across the

primary coil and a potential difference of 4.9 kV across the secondary

coil. If the secondary coil has 480 turns, how many turns does the pri-

mary coil have?

10. A ac generator central power station can produce 1380 kW of power. The

secondary coil has a potential difference of 3.4 kV. If the primary coil has

340 turns, and the secondary coil has 17 turns, what is the potential dif-

ference across the primary coil? What is the current in the primary coil?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 23AQUANTUM ENERGY

P R O B L E MFind the energy of a photon of red light, having a wavelength of 690 nm.

S O L U T I O N

Given: l = 6.90 × 10–7 m

Unknown: f = ? E = ?

Choose the equation(s) or situation: Use the equation for the energy of a light

quantum, given on page 832.

E = hf

Use the equation relating frequency to wavelength, given on page 522.

c = lf


to solve for the frequency.

f = lc

= 6

3

.

.

9

0

0

××1

1

0

0

8

–m7 m

/s = 4.3 × 1014 Hz

E = hf = = 1.8 MeV(6.63 × 10–34 J • s)(4.3 × 1014 Hz)

1.60 × 10–19 J/eV

ADDITIONAL PRACTICE

1. Determine the energy of a photon of green light, having a wavelength of 527 nm.

2. What is the energy of a photon of blue light, having a wavelength of 430.8 nm?

3. Calculate the frequency of ultraviolet (UV) light, having the energy of 20.7 eV.

4. Microwave ovens work by vibrating the water molecules in food, causing the food to

warm up. Microwaves can reach energies as high as 1.24 × 10–3 eV. At what frequency is

this?

5. Calculate the frequency of infrared (IR) light, having the energy of 1.78 eV.

6. An X ray can have an energy of 12.4 MeV. To what wavelength does this correspond?

7. A neutron has 939.57 MeV of energy. If a photon had the same energy as a neutron,

what would be the photon’s wavelength? The visible part of the spectrum ranges from

700 nm – 400 nm. Would this wavelength lie within the visible spectrum?

8. Calculate the wavelength of a radio wave that has an energy of 3.1 × 10–6 eV.


NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 23BPHOTOELECTRIC EFFECT

P R O B L E M

Light of wavelength 3.5 × 10–7 m shines on a cesium surface. Cesium has awork function of 2.14 eV. What is the maximum kinetic energy of thephotoelectrons?

S O L U T I O N

Given: l = 3.5 × 10–7 m hft = 2.14 eV

Unknown: KEmax = ?

Choose the equation(s) or situation: Use the equation for the maximum kinetic

energy of a photoelectron, given on page 835.

KEmax = hf – hft = h

lc – hft

KEmax = – 2.14 eV

KEmax = 1.41 eV

(6.63 × 10–34 J • s)(3.0 × 108 m/s)(1.60 × 10–19 J/eV)(3.5 × 10–7 m)

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1. Light of wavelength 240 nm shines on a potassium surface. Potassium

has a work function of 2.3 eV. What is the maximum kinetic energy of

the photoelectrons?

2. Light of wavelength 519 nm shines on a rubidium surface. Rubidium has

a work function of 2.16 eV. What is the maximum kinetic energy of the

photoelectrons?

3. Light of frequency 6.5 × 1014 Hz illuminates a lithium surface. The

ejected photoelectrons are found to have a maximum kinetic energy of

0.20 eV. Find the threshold frequency of this metal.

4. Light of frequency 9.89 × 1014 Hz illuminates a calcium surface. The

ejected photoelectrons are found to have a maximum kinetic energy of

0.90 eV. Find the threshold frequency of this metal.

5. The threshold frequency of platinum is 1.36 × 1015 Hz. What is the work

function of platinum?

6. The threshold frequency of copper is 1.1 × 1015 Hz. What is the work

function of copper?

7. Manganese has a work function of 4.1 eV. What is the wavelength of the

photon that will just have the threshold energy for manganese?

ADDITIONAL PRACTICE


NAME ______________________________________ DATE _______________ CLASS ____________________

8. Cobalt has a work function of 5.0 eV. What is the wavelength of the pho-

ton that will just have the threshold energy for cobalt?

9. Light shines on a photoelectric metal and the maximum kinetic energy is

measured to be 0.6 eV. What is the speed of the photoelectrons?

10. Light shines on a photoelectric metal and the maximum kinetic energy is

measured to be 1.2 eV. What is the speed of the photoelectrons?

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NAME ______________________________________ DATE _______________ CLASS ____________________

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Holt Physics

Problem 23CDE BROGLIE WAVES

P R O B L E MA grain of sand blows along a seashore at a velocity of 5.2 m/s. If it has a deBroglie wavelength of 5.8 × 10–29 m, what is the mass of the sand grain?

S O L U T I O N

Given: v = 5.2 m/s l = 5.8 × 10–32 m

Unknown: m = ?

Choose the equation(s) or situation: Use the equation for the de Broglie wave-

length, given on page 849.

l = m

h

v


relating wavelength, mass, and velocity to solve for mass.

m = lh

v = = 2.2 × 10–6 kg

6.63 × 10–34 J • s(5.2 m/s)(5.8 × 10–29 m)

ADDITIONAL PRACTICE

1. A cheetah can run as fast as 28 m/s. If the cheetah has a de Broglie wavelength of

8.97 × 10–37 m, what is the cheetah’s mass?

2. A Boeing 747 jet airliner has a maximum airspeed of 7.1 × 102 m/s. If the airliner has a

de Broglie wavelength of 5.8 × 10–42 m, what is the mass of the jet?

3. The smallest known virus is a potato spindle. Suppose a potato spindle moves across a

Petri dish at 5.6 × 10–6 m/s and has a de Broglie wavelength of 2.96 × 10–8 m. What is

the mass of a potato spindle?

4. Suppose a raindrop falls from the sky at a velocity of 12 m/s and has a de Broglie wave-

length of 2.6 × 10–29 m. What is the mass of the raindrop?

5. Calculate the de Broglie wavelength of an electron orbiting the hydrogen atom at a

velocity of 2.19 × 106 m/s.

6. The ship Queen Elizabeth has a mass of 7.6 × 107 kg. Calculate the de Broglie wave-

length if this ship sails at 35 m/s.

7. Earth has a mass of 5.94 × 1024 kg and orbits the sun at a velocity of 3.0 × 104 m/s.

Calculate Earth’s de Broglie wavelength.

8. Our solar system is within the Milky Way galaxy. Astronomers estimate that our galaxy

has a mass of 4.0 × 1041 kg. Calculate the de Broglie wavelength of our galaxy if it were

to move at a velocity of 1.7 × 104 m/s.

9. What is the speed of an electron with a de Broglie wavelength of 9.87 × 10–14 m?

10.What is the speed of a neutron with a de Broglie wavelength of 5.6 × 10–14 m?

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NAME ______________________________________ DATE _______________ CLASS ____________________

Holt Physics

Problem 25ABINDING ENERGY

P R O B L E MGiven that the atomic mass of iron –55 is 54.938 297 u, calculate the bind-ing energy of 55

26Fe.

S O L U T I O NGiven: Z = 26 atomic mass of iron −55 = 54.938 297 u

N = 29 atomic mass of H = 1.007 825 u

mn = 1.008 665 u

Unknown: Ebind = ?


First find the mass defect with the following relationship:

∆m = Z(atomic mass of H) + Nmn − atomic mass

Then find the binding energy by converting the mass defect to rest energy.


∆m = 26(1.007 825 u) + 29(1.008 665 u) − 54.938 297 u

∆m = 0.51644 u

Ebind = (0.51644 u)(931.50 MeV/u) = 481.06 MeV

ADDITIONAL PRACTICE

1. Calculate the binding energy of 3919K.

2. Compute the binding energy of 12050Sn.

3. Determine the difference in the binding energy of 10747Ag and 63

29Cu.

4. Calculate the difference in the binding energy of 126C and 16

8O.

5. What is the binding energy of 3517Cl?

6. Calculate the binding energy of deuterium, 21H.

7. Find the mass defect of 5828Ni.

8. What is the mass defect of 6430Zn?

9. Calculate the binding energy of 9040Zr.

10. Find the mass defect of 3216S.

Holt Physics

Problem 25BNUCLEAR DECAY

P R O B L E MBromine-80 decays by emitting a positron and a neutino. Write the com-plete decay formula for this process.

S O L U T I O NGiven: The decay can be written symbolically as follows:

8035Br → X + 00

1e + v

Unknown: the daughter element (X)

The mass numbers and atomic numbers on the two sides of the expression must

be the same so that both charge and nucleon number are conserved during the

course of a particular decay.

Mass number of X = 80 − 0 = 80

Atomic number of X = 35 − (1) = 348035Br → 80

34X + 01e + v

The periodic table (Appendix F) shows that the nucleus with an atomic number

of 34 is selenium, Se. Thus, the process is as follows:

8035Br → 80

34Se + 01e + v


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ADDITIONAL PRACTICE

1. Complete this radioactive-decay formula: 21084Po → ? + 42He

2. Complete this radioactive-decay formula: 167N → ? + –1

0e + v

3. Complete this radioactive-decay formula: 14762Sm → 143

60Nd + ?

4. Complete this radioactive-decay formula: 1910Ne → ? + 01e + v

5. Complete this radioactive-decay formula: ? → 13154Xe + –1

0e + v

6. Complete this radioactive-decay formula: ? → 9039Y + –1

0e + v

7. Complete this radioactive-decay formula: 16074W → 156

72Hf + ?

8. Complete this radioactive-decay formula: ? → 10752Te + 42He

9. Complete this radioactive-decay formula: 15772Hf → 153

70Yb + ?

10. Complete this radioactive-decay formula: 14158Ce → ? + –1

0e + v


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Holt Physics

Problem 25CMEASURING NUCLEAR DECAY

P R O B L E MSuppose you start with 9.36 × 10–3 g of a pure radioactive substance anddetermine 4 h later that only 2.34 × 10–3 g of the substance is left unde-cayed. What is the half-life of this substance?

S O L U T I O N

Given: mi = 9.36 × 10–3 g mf = 2.34 × 10–3 g ∆t = 4 h

Unknown: T1⁄2 = ?

Choose the equation(s) or situation: Take the ratio of the final mass to the ini-

tial mass to find what fraction of the sample remains.

fraction of remaining sample = m

m

i

f = 2

9

.

.

3

3

4

6

××

1

1

0

0

–

–

3

3g

g =

1

4

If one-quarter of the sample remains after 4 h, then half of the sample must have

remained after 2.00 h, or T1⁄2 = .2.00 h

ADDITIONAL PRACTICE

1. Suppose you start with 5.25 × 10–3 g of a pure radioactive substance and

determine 12 h later that only 3.28 × 10–4 g of the substance is left unde-

cayed. What is the half-life of this substance?


determine 30.0 s later that only 8.22 × 10–4 g of the substance is left un-

decayed. What is the half-life of this substance?


determine 1.25 days later that only 2.07 × 10–4 g of the substance is left

undecayed. What is the half-life of this substance?

4. How long will it take a sample of lead-212 with a half-life of 10.64 h to

decay to one-eighth its original strength?

5. How long will it take a sample of cadmium-109 with a half-life of 462

days to decay to one-fourth its original strength?

6. The half-life of 5526Fe is 2.7 years. An iron-55 sample contains 3.2 × 109

nuclei. What is the decay constant for the decay?

7. A sample of a radioactive isotope is measured to have an activity of

765.3 mCi. If the sample has a half-life of 22 h, how many nuclei of the

isotope are there at this time?


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8. The half-life of 4824Cr is 21.6 h. A chromium-48 sample contains 6.5 × 106 nuclei. Calcu-

late the decay constant and the activity of the sample in mCi.

9. The half-life of 31H is 12.33 years. A tritium sample contains 4.8 × 108 nuclei. What is

the decay constant for the decay?

10. A sample of a radioactive isotope is measured to have an activity of 360.0 mCi. If the

sample has a half-life of 17.2 s, how many nuclei of the isotope are there at this time?

1ChapterThe Science of Physics

1. mass = 6.0 × 103 kga. mass = 6.0 × 103 kg ×

1

1

0

k

3

g

g ×

1

1

0

m−3

g

g =

b. mass = 6.0 × 103 kg × 1

1

0

k

3

g

g ×

1

10

M6 g

g = 6.0 Mg

6.0 × 109 mg


Givens Solutions

2. Volume = 6.4 × 104 cm3

a. volume = 6.4 × 104 cm3 × 10

12m

cm

3= 6.4 × 104 cm3 ×

10

16m

cm

3

3 =

b. volume = 6.4 × 104 cm3 × 11

0

c

m

m

m

3= 6.4 × 104 cm3 ×

10

1

3

c

m

m

m3

3

= 6.4 = 107 mm3

6.4 × 10−2 m3

3. energy = 4.2 × 109 J a. energy = 4.2 × 109 J × 1

10

M6 J

J =

b. energy = 4.2 × 109 J × 1

1

0

G9

J

J = 4.2 GJ

4.2 × 103 MJ

4. distance = 1 parsec = 3.086 × 1016 m a. distance = 1 parsec = 3.086 × 1016 m ×

1

1

0

k3m

m =

b. distance = 1 parsec = 3.086 × 1016 m × 1

1

01E8m

m = 3.086 × 10–2 Em

3.086 × 1013 km

5. area = 1 acre = 4.0469 × 103 m2 a. area = 1 acre = 4.0469 × 103 m2 × 1

1

0

k3m

m2

area = 4.0469 × 103 m2 × 1

1

0

k6m

m

2

2 =

b. area = 1 acre = 4.0469 × 103 m2 × 10

1

2

m

cm

2

area = 4.0469 × 103 m2 × 10

1

4

m

cm2

2

= 4.0469 × 107 cm2

4.0469 × 10−3 km2

6. electric charge = 15 Ca. electric charge = 15 C ×

10

1

3

C

mC =

b. electric charge = 15 C × 1

1

0

k3C

C = 1.5 × 10–2 kC

1.5 × 104 mC

Section Two—Problem Workbook Solutions V Ch. 1–1

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7. depth = 1.168 × 103 cma. depth = 1.168 × 103 cm ×

10

12m

cm =

b. depth = 1.168 × 103 cm × 10

12m

cm ×

1

1

0−m6

m

m = 1.168 × 107 mm

1.168 × 101 m = 11.68 m

Givens Solutions

8. area = 0.344 279 km2

a. area = 0.344 279 km2 × 110

k

3

m

m

2= 0.344 279 × 106 m2 =

b. area = 0.344 279 km2 × 110

k

3

m

m

2× 10

1

3

m

mm

2= 0.344 279 × 1012 mm2

area = 3.442 79 × 1011 mm2

3.442 79 × 105 m3

9. time = 4.50 × 109 years ×

365

1

.2

y

5

ea

d

r

ays ×

1

24

da

h

y ×

36

1

0

h

0 s

= 1.42 × 1017 s

a. time = 1.42 × 1017 s × 1

1

0

G9

s

s =

b. time = 1.42 × 1017 s × 1

1

01P5s

s = 1.42 × 102 Ps = 142 Ps

1.42 × 108 Gs

10. time = 6.7 × 10−17 s a. time = 6.7 × 10−17 s × 10

1

6

s

ms =

b. time = 6.7 × 10−17 s × 10

1

18

s

as = 6.7 = 101 as = 67 as

6.7 × 10−11 ms

Section Five—Problem Bank V Ch. 2–1

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2ChapterMotion in One Dimension

1. ∆x = 3.33 km forward∆t = 30.0 s vavg =

∆∆

x

t =

3.33

3

×0.

1

0

0

s

3 m =

vavg = (111 m/s)(3600 s/h)(10−3 km/m) = 4.00 × 102 km/h forward

111 m/s forward


Givens Solutions

2. ∆x = 15.0 km west∆t = 15.3 s

vavg = ∆∆

x

t = =

4.2

1

5

5

×.0

1

k

0

m−3 h

= 3.53 × 103 km/h west15.0 km

(15.3 s)36

1

0

h

0 s

3. ∆x = 4.0 m∆t = 5.0 min vavg =

∆∆

x

t = = 48 m/h

4.0 m

(5.0 min)60

1

m

h

in

4. ∆x = 3.20 × 104 km south∆t = 122 days vavg =

∆∆

x

t =

3.2

1

0

2

×2

1

d

0

a

4

ys

km = 262 km/day south

5. ∆x1 = 1.70 × 104 km south= +1.70 × 104 km

∆x2 = 6.0 × 102 km north= −6.0 × 102 km

∆x3 = 1.44 × 104 km south= +1.44 × 104 km

∆t = 122 days

d = total distance traveled = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3

d = 1.70 × 104 km + 6.0 × 102 km + 1.44 × 104 km

d = (1.70 + 0.060 + 1.44) × 104 km

d = 3.20 × 104 km

average speed = ∆d

t =

3.2

1

0

2

×2

1

d

0

a

4

ys

km =

vavg = ∆∆xt

tot =

∆x1 + ∆∆x

t2 + ∆x3

vavg =

vavg =

vavg = 3.0

1

8

2

×2

1

d

0

a

4

ys

km = +252 km/day = 252 km/day south

(1.70 − 0.060 + 1.44) × 104 km

122 days

(1.70 × 104 km) + (−6.0 × 102 km) + (1.44 × 104 km)

122 days

262 km/day

6. ∆x1 = 20.0 km east = + 20.0 km

∆x2 = 20.0 km west = − 20.0 km

∆x3 = 0 km

∆x4 = 40.0 km east = +40.0 km

∆t = 60.0 min

a. vavg = =

vavg =

vavg = = + 40.0 km/h = 40.0 km/h east40.0 km

(60.0 min) 60

1

m

h

in

(20.0 km) + (−20.0 km) + (0 km) + (40.0 km)

60.0 min

∆x1 + ∆x2 + ∆x3 + ∆x4∆t

∆xtot∆t


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b. d = total distance traveled

d = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3 + magnitude ∆x4

d = 20.0 km + 20.0 km + 0 km + 40.0 km = 80.0 km

average speed = = = 80.0 km/h80.0 km

(60.0 min)60

1

m

h

in

d∆t

Givens Solutions

7. v = 89.5 km/h north

vavg = 77.8 km/h north

∆trest = 22.0 min

∆x = vavg ∆t = v(∆t − ∆trest)

∆t(vavg − v) = −v∆trest

= =

∆t = 2.80 h = 2 h, 48 min

(89.5 km/h)(22.0 min)60

1

m

h

in

11.7 km/h

(89.5 km/h)(22.0 min)60

1

m

h

in

89.5 km/h − 77.8 km/h

∆t = v∆restv − vavg

8. v = 6.50 m/s downward = −6.50 m/s

∆t = 34.0 s

∆x = v∆t = (−6.50 m/s)(34.0 s) = −221 m = 221 m downward

9. vt = 10.0 cm/s

vh = 20 vt = 2.00 × 102 cm/s

∆trace = ∆tt

∆th = ∆tt − 2.00 min

∆xt = ∆xh + 20.0 cm = ∆xrace

∆xt = vt∆tt

∆xh = vh∆th = vh (∆tt − 2.00 min)

∆xt = ∆xrace = ∆xh + 20.0 cm

vt ∆tt = vh (∆tt − 2.00 min) + 20.0 cm

∆tt (vt − vh) = −vh (2.00 min) + 20.0 cm

∆tt =

∆trace = ∆tt =

∆trace = =

∆trace = 126 s

−2.40 × 104 cm−1.90 × 102 cm/s

20.0 cm − 2.40 × 104 cm

−1.90 × 102 cm/s

20.0 cm − (2.00 × 102 cm/s)(2.00 min)(60 s/min)

10.0 cm/s − 2.00 × 102 cm/s

20.0 cm − vh (2.00 min)

vt − vh

10. ∆xrace = ∆xt

vt = 10.0 cm/s

∆tt = 126 s

∆xrace = ∆xt = vt∆tt = (10.0 cm/s)(126 s) = 1.26 × 103 cm = 12.6 m

Additional Practice 2B

1. ∆t = 6.92 s

vf = 17.34 m/s

vi = 0 m/s

aavg = ∆∆

v

t =

vf

∆−t

vi = = 2.51 m/s217.34 m/s − 0 m/s

6.92 s

2. vi = 0 m/s

vf = 7.50 × 102 m/s

∆t = 2.00 min

aavg = = = = 6.25 m/s27.50 × 102 m/s − 0 m/s

(2.00 min)16

m

0

i

s

n

vf − vi∆t

∆v∆t


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Givens Solutions

3. vi = 0 m/s

vf = 0.85 m/s forward

∆t = 3.7 s

aavg = = = = 0.23 m/s2 forward0.85 m/s − 0 m/s

3.7 s

vf − vi∆t

∆v∆t

4. vi = 13.7 m/s forward = +13.7 m/s

vf = 11.5 m/s backward = −11.5 m/s

∆t = 0.021 s

aavg = = = =

aavg = −1200 m/s2, or 1200 m/s2 backward

−25.2 m/s

0.021 s

(− 11.5 m/s) − (13.7 m/s)

0.021 s

vf − vi∆t

∆v∆t

5. vi = +320 km/h

vf = 0 km/h

∆t = 0.18 saavg = = =

aavg = = −490 m/s2−89 m/s

0.18 s

(0 km/h − 320 km/h)36

1

0

h

0 s11

0

k

3

m

m

0.18 s

vf − vi∆t

∆v∆t

6. aavg = 16.5 m/s2

vi = 0 km/h

vf = 386.0 km/h∆t = = =

∆t = = 6.50 s107.2 m/s16.5 m/s2

(386.0 km/h − 0 km/h)36

1

0

h

0 s11

0

k

3

m

m

16.5 m/s2

vf − viaavg

∆vaavg

7. vi = −4.0 m/s

aavg = −0.27 m/s2

∆t = 17 s

vf = aavg ∆t + vi

vf = (−0.27 m/s2)(17 s) + (−4.0 m/s) = −4.6 m/s − 4.0 m/s = −8.6 m/s

8. vi = 4.5 m/s

vf = 10.8 m/s

aavg = 0.85 m/s2

∆t = a

∆

a

v

vg = =

10.8

0

m

.8

/

5

s −m

4

/s

.52

m/s =

0

6

.8

.3

5

m

m

/

/

s

s2 = 7.4 svf − viaavg

9. vf = 296 km/h

vi = 0 km/h

aavg = 1.60 m/s2∆t =

a

∆

a

v

vg = = =

1

8

.

2

6

.

0

2

m

m

/

/

s

s2 = 51.4 s

(296 km/h − 0 km/h)36

1

0

h

0 s11

0

k

3

m

m

1.60 m/s2

vf − viaavg

10. aavg = − 0.87 m/s2

∆t = 3.85

∆v = aavg ∆t = (−0.87 m/s2)(3.85 s) = –3.35 m/s

1. ∆t = 0.910 s

∆x = 7.19 km

vi = 0 km/s

vf = 2

∆∆t

x − vi =

(2)

0

(7

.9

.1

1

9

0

k

s

m) − 0 km/s = 15.8 km/s

Additional Practice 2C

Givens Solutions


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

2. vi = 4.0 m/s

∆t = 18 s

∆x = 135 m

vf = 2

∆∆t

x − vi =

(2)(

1

1

8

35

s

m) − 4.0 m/s = 15 m/s − 4.0 m/s =

vf = (11 m/s)36

1

0

h

0 s

1

1

0

k3m

m

vf = 4.0 × 101 km/h

11 m/s

3. ∆x = 55.0 m

∆t = 1.25 s

vf = 43.2 m/s

vi = 2

∆∆t

x − vf =

(2)

1

(5

.2

5

5

.0

s

m) − 43.2 m/s = 88.0 m/s − 43.2 m/s = 44.8 m/s

4. ∆x = 38.5 m

∆t = 5.5 s

vi = 0 m/s

vf = 2

∆∆t

x − vi =

(2)(

5

3

.

8

5

.5

s

m) − 0 m/s = 14 m/s

5. ∆x = 478 km

∆vi = 72 km/h

∆t = 5 h, 39 min

vf = 2

∆∆t

x − vi = − 72 km/h =

(

5

2

h

)(4

+7

0

8

.6

k

5

m

h

) − 72 km/h

vf = 9

5

5

.

6

65

km

h − 72 km/h = 169 km/h − 72 km/h = 97 km/h

(2)(478 km)

5 h + 39 min60

1

m

h

in

6. ∆x = 4.2 m

∆t = 3.0 s

vf = 1.3 m/s

vi = 2

∆∆t

x − vf =

(2)

3

(4

.0

.2

s

m) − 1.3 m/s = 2.8 m/s − 1.3 m/s = 1.5 m/s

7. vi = 25 m/s west

vf = 35 m/s west

∆x = 250 m west

∆t = v

2

i

∆+

x

vf =

25

(

m

2)

/

(

s

2

+50

35

m

m

)

/s =

6

5

.0

.0

××1

1

0

01

2

m

m

/s = 8.3 s

8. vi = 755.0 km/h

vf = 777.0 km/h

∆t = 63.21 s

∆x = 1

2 (vi + vf)∆t =

1

2 (755.0 km/h + 777.0 km/h)(63.21 s)36

1

0

h

0 s

∆x = 1

2 (1532.0 km/h)(1.756 × 10−2 h) = 13.45 km

9. vi = 0 m/s

vf = 30.8 m/s

∆x = 493 m

∆t = v

2

i

∆+

x

vf =

0 m

(2

/s

)(

+49

3

3

0.

m

8 m

)

/s =

3

9

0

8

.8

6

m

m

/s = 32.0 s

10. ∆x = 1220 km

vi = 11.1 km/s

vf = 11.7 km/s

∆t = v

2

i

∆+

x

vf = =

2

2

2

4

.

4

8

0

k

k

m

m

/s = 107 s

(2)(1220 km)11.1 km/s + 11.7 km/s


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.Givens Solutions

2. ∆t = 1.5 s

vi = 2.8 km/h

vf = 32.0 km/ha = =

a = = 5.4 m/s2

(29.2 km/h)36

1

0

h

0 s11

0

k

3

m

m

1.5 s

(32.0 km/h − 2.8 km/h)36

1

0

h

0 s10

k

3

m

m

1.5 s

vf − vi∆t

3. ∆x = 18.3 m

∆t = 2.74 s

vi = 0 m/s

Because vi = 0 m/s, a = 2

∆∆t

x2 =

(2

(

)

2

(

.

1

7

8

4

.3

s)

m2

) = 4.88 m/s2

4. vi = 2.3 m/s

vf = 46.7 m/s

∆t = 7.0 s

a = = 46.7 m/

7

s

.0

−s

2.3 m/s =

44

7

.4

.0

m

s

/s = 6.3 m/s2vf − vi

∆t

5. vi = 6.23 m/s

∆x = 255 m

∆t = 82 s

a = 2(∆x

∆−t2vi ∆t) =

a = = (

6

2

.

)

7

(−×2

1

5

0

53m

s2)

= −7.6 × 10−2 m/s2(2)(255 m − 510 m)

6.7 × 103 s2

(2)[255 m − (6.23 m/s)(82 s)]

(82 s)2

6. vi = 11 km/h

vf = 55 km/h

∆ = 4.1 sa = =

a = = 3.0 m/s2

(44 km/h)36

1

0

h

0 s11

0

k

3

m

m

4.1 s

(55 km/h − 11 km/h)36

1

0

h

0 s11

0

k

3

m

m

4.1 s

vf − vi∆t

7. vi = 42.0 m/s southeast

∆t = 0.0090 s

∆x = 0.020 m/s southeast

a = 2(∆x

∆−t2vi ∆t) =

a = =

a = −8.9 × 103 m/s2, or 8.9 × 103 m/s2 northwest

(2)(−0.36 m)8.1 × 10−5 s2

(2)(0.020 m/s − 0.38 m)

8.1 × 10−5 s2

(2)[0.020 m − (42.0 m/s)(0.0090 s)]

(0.0090 s)2

8. ∆t = 28 s

a = 0.035 m/s2

vi = 0.76 m/s

vf = a∆t + vi = (0.035 m/s2)(28.0 s) + 0.76 m/s = 0.98 m/s + 0.76 m/s = 1.74 m/s

Additional Practice 2D

1. ∆x = 12.4 m upward

∆t = 2.0 s

vi = 0 m/s

Because vi = 0 m/s, a = 2

∆∆t2x

= (2)

(

(

2

1

.0

2.

s

4

)2m)

= 6.2 m/s2 upward

Givens Solutions


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

4. ∆x = 2.00 × 102 m

vi = 9.78 m/s

vf = 10.22 m/s

a = vf

2

2∆−

x

vi2

= =

a = 4.

8

0

.

0

8

×m

1

2

0

/2s2

m = 2.2 × 10−2 m/s2

104.4 m2/s2 − 95.6 m2/s2

4.00 × 102 m

(10.22 m/s)2 − (9.78 m/s)2

(2)(2.00 × 102 m)

10. vi = +4.42 m/s

vf = 0 m/s

a = −0.75 m/s2

∆t = 5.9 s

a. ∆t = vf −

a

vi = 0 m

−/

0

s

.7

−5

4

m

.42

/s2m/s

= −−0

4

.

.

7

4

5

2

m

m

/

/

s

s2 =

b. ∆x = vi∆t + 1

2a∆t2 = (4.42 m/s)(5.9 s) +

1

2(−0.75 m/s2)(5.9 s)2

∆x = 26 m − 13 m = 13 m

5.9 s

1. vi = 1.8 km/h

vf = 24.0 km/h

∆x = 4.0 × 102 ma =

vf2

2∆−

x

vi2

=

a =

a = = 5.5 × 10−2 m/s2(573 km2/h2)36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

8.0 × 102 m

(576 km2/h2 − 3.2 km2/h2)36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

8.0 × 102 m

[(24.0 km/h)2 − (1.8 km/h)2]36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

(2)(4.0 × 102 m)

2. vf = 0 m/s

vf = 8.57 m/s

∆x = 19.53 m

a = vf

2

2

−∆x

vi2

= = 7

3

3

9

.4

.0

m

6

2

m

/s2

= 1.88 m/s2(8.57 m/s)2 − (0 m/s)2

(2)(19.53 m)

9. vi = 0 m/s

vf = 72.0 m/s north

a = 1.60 m/s2 north

∆t = 45.0 s

a. ∆t = vf −

a

vi = 72.0

1.

m

60

/s

m

−/

0

s2m/s

=

b. ∆x = vi∆t + 1

2a∆t2 = (0 m/s)(45.0 s) +

1

2 (1.60 m/s2)(45.0 s)2 = 0 m + 1620 m

∆x = 1.62 km

45.0 s

Additional Practice 2E

3. vi = 7.0 km/h

vf = 34.5 km/h

∆x = 95 ma =

vf2

2

−∆x

vi2

=

a =

a = = 0.46 m/s2

(1140 km2/h2)36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

190 m

(1190 km2/h2 − 49 km2/h2)36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

190 m

[(34.5 km/h)2 − (7.0 km/h)2]36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

(2)(95 m)


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

6. vi = 50.0 km/h forward= +50.0 km/h

vf = 0 km/h

a = 9.20 m/s2 backward= −9.20 m/s2

Givens Solutions

∆x = vf

2

2

−a

vi2

=

∆x =

∆x = 10.5 m = 10.5 m forward

−(2.50 × 103 km2/h2)36

1

0

h

0 s

2

110

k

3

m

m

2

−18.4 m/s2

[(0 km/h)2 − (50.0 km/h)2]36

1

0

h

0 s

2

110

k

3

m

m

2

(2)(−9.20 m/s2)

7. a = 7.56 m/s2

∆x = 19.0 m

vi = 0 m/s

vf =√

vi2+ 2a∆x =√

(0 m/s)2 + (2)(7.56m/s2)(19.0m)

vf =√

287m2/s2 = ±16.9 m/s = 16.9 m/s

8. vi = 1.8 m/s

vf = 9.4 m/s

a = 6.1 m/s2

∆x = vf

2

2

−a

vi2

= =

∆x = (2)

8

(

5

6.

m

1

2

m

/s

/

2

s2) = 7.0 m

88 m2/s2 − 3.2 m2/s2

(2)(6.1 m/s2)

(9.4 m/s)2 − (1.8 m/s)2

(2)(6.1 m/s2)

9. vi = 1.50 m/s to the right = +1.50 m/s

vf = 0.30 m/s to the right = +0.30 m/s

a = 0.35 m/s2 to the left = −0.35 m/s2

∆x = vf

2

2

−a

vi2

=

∆x =

∆x = −−2

0

.

.

1

7

6

0

m

m

2

/

/

s

s2

2

= +3.1 m = 3.1 m to the right

9.0 × 10−2 m2/s2 − 2.25 m2/s2

−0.70 m/s2

(0.30 m/s)2 − (1.50 m/s)2

(2)(−0.35 m/s2)

10. a = 0.678 m/s2

vf = 8.33 m/s

∆x = 46.3 m

vi =√

vf2− 2a∆x =√

(8.33m/s)2 − (2)(0.678 m/s2)(46.3m)

vi =√

69.4 m2/s2 − 62.8m2/s2 =√

6.6m2/s2 = ±2.6 m/s = 2.6 m/s

1. vi = 0 m/s

vf = 49.5 m/s downward= 49.5 m/s

a = −9.81 m/s2

∆tot = −448 m

∆xi = vf

2

2

−a

vi2

= = (2)

2

(

4

−5

9

0

.8

m

1

2

m

/s

/

2

s2) = −125 m

∆x2 = ∆xtot − ∆x1 = (−448 m) − (−125 m) = −323 m

distance from net to ground = magnitude ∆x2 = 323 m

(−49.5 m/s)2 − (0 m/s)2

(2)(−9.81 m/s2)

Additional Practice 2F

5. ∆x = +42.0 m

vi = +153.0 km/h

vf = 0 km/ha =

vf2

2∆−

x

vi2

=

a = = −21.5 m/s2

−(2.34 × 104 km2/h2)36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

(84.0 m)

[(0 km/h)2 − (153.0 km/h)2]36

1

0

h

0 s

2

1

1

0

k

3

m

m

2

(2) (42.0 m)

Givens Solutions


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

2. vi = 0 m/s

a = −9.81 m/s2

∆t1 = 1.00 s

∆t2 = 2.00 s

∆t3 = 3.00 s

Because vi = 0 m/s, ∆x1 = 1

2a∆t1

2 = 1

2(−9.81 m/s2)(1.00 s)2 =

∆x2 = 1

2a∆t2

2 = 1

2(−9.81 m/s2)(2.00 s)2 =

∆x3 = 1

2a∆t3

2 = 1

2(−9.81 m/s2)(3.00 s)2 = −44.1 m

−19.6 m

−4.90 m

3. vi = 0 m/s

∆t = 2.0 s

a = −9.81 m/s2

Because vi = 0 m/s, ∆x = 1

2a∆t2 =

1

2(−9.81 m/s2)(2.0 s)2 = −2.0 × 101 m

distance of bag below balloon = 2.0 × 101 m

4. vi = +17.5 m/s

a = −9.81 m/s2

∆ttot = 3.60 s

∆xtot = 0 m

To find the ball’s maximum height, calculate ∆xup. The time for this distance to betraveled is ∆tup = 1

2 ∆xtot

∆xup = vi ∆tup + 1

2a∆tup

2 = vi 1

2∆ttot +

1

2a

1

2∆ttot

2

∆xup = (17.5 m/s)3.6

2

0 s +

1

2(−9.81 m/s2)

3.6

2

0 s

2= (31.5 m) + (−15.9 m) = +15.6 m

softball’s maximum height = 15.6 m

5. vi = 0 m/s

vf = 11.4 m/s downward= −11.4 m/s

a = 3.70 m/s2 downward= −3.70 m/s2

∆x = vf

2

2

−a

vi2

= = = −17.6 m

∆x = 17.6 m downward

1.30 × 102 m2/s2

−7.40 m/s2

(−11.4 m/s)2 − (0 m/s)2

(2)(−3.70 m/s2)

6. ∆ttot = 5.10 s

∆tdown = 1

2 ∆ttop

vi = 0 m/s

a = −9.81 m/s2

To find the ball’s maximum height, calculate the displacement from that height to itsoriginal position. The time interval for this free-fall is 1

2 ∆ttot, and vi = 0 m/s.

∆xdown = 1

2 a∆tdown

2 = 1

2a

1

2∆ttot

2=

1

2(−9.81 m/s2)

5.1

2

0 s

2= −31.9 m

ball’s maximum height = 31.9 m

7. ∆ttot = 5.10 s

a = −9.81 m/s2

∆xtot = 0 m

∆ttot = vi ∆ ttot + 1

2a∆ttot

2

Because ∆xtot = 0,

vi = − 1

2a∆ttot = −

1

2(−9.81 m/s2)(5.10 s) = +25.0 m/s = 25.0 m/s upward

8. vi = 85.1 m/s upward= + 85.1 m/s

a = −9.81 m/s2

∆x = 0 m

Because ∆x = 0 m, vi ∆t + 1

2a∆t 2 = 0

∆t = − −2

a

vi = − −(

(

2

−)

9

(

.

8

8

5

1

.1

m

m

/s

/2s

)

) = 17.3 s


V

Cop

yrig

ht ©

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

Givens Solutions

10. aup = 3.125 m/s2, upward= +3.125 m/s2

∆tup = 4.00 s

vi = 0 m/s

vf = aup ∆tup + vi = (3.125 m/s2) (4.00 s) + 0 m/s = 2.5 m/s

When the cable brakes, the upward-moving elevator undergoes free-fall acceleration.

vi = +12.5 m/s

a = −9.81 m/s2

∆t1 = 0.00 s

∆t2 = 1.00 s

∆t3 = 2.00 s

∆t4 = 3.00 s

vf, 1 = a∆t1 + vi = (−9.81 m/s2)(0.00 s) + 12.5 m/s =

vf, 2 = a∆t2 + vi = (−9.81 m/s2)(1.00 s) + 12.5 m/s = −9.81 m/s + 12.5 m/s =

vf, 3 = a∆t3 + vi = (−9.81 m/s2)(2.00 s) + 12.5 m/s = −19.6 m/s + 12.5 m/s =

vf, 4 = a∆t4 + vi = (−9.81 m/s2)(3.00 s) + 12.5 m/s = −29.4 m/s + 12.5 m/s = − 16.9 m/s

−7.1 m/s

+2.7 m/s

+12.5 m/s

9. ∆td = 4.00 s

∆tp = ∆td + 3.00 s= 7.00 s

a = −9.81 m/s2

∆xd = ∆xp

vi, p = 0 m/s

∆xd = vi, d ∆td + 1

2a∆td

2

∆xp = vi, p ∆tp + 1

2a∆tp

2

vi, d ∆td + 1

2a∆td

2 = vi, p∆tp + 1

2a∆tp

2

vi, d = + vi,

∆p

t

∆

d

tp

Because vi, p = 0 m/s,

vi, d = =

vi, d = − = − =

vf, d = a ∆td + vi, d = (−9.81 m/s2)(4.00 s) + (−40.5 m/s) = (−39.2 m/s) + (−40.5 m/s)

vf, d = −79.7 m/s

−40.5 m/s(9.81 m/s2)(33.0 s2)

8.00 s

(9.81 m/s2)(49.0 s2 − 16.0 s2)

8.00 s

1

2(−9.81 m/s2)[(7.00 s)2 − (4.00 s)2]

4.00 s

1

2a(∆tp

2 − ∆td2)

∆td

1

2a(∆tp

2 − ∆td2)

∆td


Chapter 3Two -Dimensional Motion and Vectors

V

1. ∆x1 = 8 m to the left = +8 m

∆x2 = 8 m to the right = −8 m

∆x3 = 8 m to the left = +8 m

a. distance traveled = 8 m + 8 m + 8 m =

b. d = ∆x1 + ∆x2 + ∆x3 = 8 m + (−8 m) + 8 m = 8 m

24 m


Givens Solutions

Cop

yrig

ht ©

by H

olt,

Rin

ehar

t and

Win

ston

.All

right

s re

serv

ed.

2. hi = 0.91 m

hf = (0.90)hi

∆x = 0.11 m

∆y = hf − hi = (0.90 − 1.00)hi

∆y = (−0.10)(0.91 m) = −9.1 × 10−2 m

d =√

∆x2+ ∆y2 =√

(0.11m)2 + (−9.1× 10−2m)2 =√

1.2× 10−2m2+ 8.3 × 10−3m2

d = 2.0 × 10−2 m2 =

q = tan−1∆∆

x

y = tan−1 −9.1

0

×.11

10

m

−2 m

q = −4.0 × 101° = 4.0 × 101° below the horizontal

0.14 m

3. ∆x = 165 m

∆y = −45 m

d =√

∆x2+ ∆y2 =√

(165 m)2 + (−45 m)2√

2.72 × 104 m2+ 2.0 × 103 m2 =√

2.92 × 104 m2

d =

q tan−1 ∆∆

x

y = tan−1−16

4

5

5

m

m

q = −15° = 15° below the horizontal

171 m

4. ∆y = −13.0 m

∆x = 9.0 m

d =√

∆x2+ ∆y2 =√

(9.0 m)2 + (−13.0 m)2 =√

81 m2+ 169 m2 =√

2.50 × 102 m2

d =

q = tan−1 ∆∆

x

y tan−1−9

1

.

3

0

.0

m

m

q = −55° = 55° below the horizontal

15.8 m

5. ∆x = 36.0 m, east

∆y = 42.0 m, north

∆z = 17.0 m, up

d =√

∆x2+ ∆y2+ ∆z2 =√

(36.0m)2 + (42.0 m)2 + (17.0 m)2

d =√

1.30 × 103 m2+ 1.76× 103 m2+ 289 m2 =√

3.35 × 103 m2

d =

horizontal direction = qh = tan−1 ∆∆

x

y = tan−1 432

6

.

.

0

0

m

m

qh =

vertical direction = qv = tan−1 ∆x 2∆+z

∆y249.4° north of east

57.9 m

9. d = 3.88 km

∆x = 3.45 km

h1 = 0.8 km

d2 = ∆x2 + ∆y2

∆y =√

d2− ∆x2 =√

(3.88km)2 − (3.45 km)2 =√

15.1 km2− 11.9km2 =√

3.2km2

∆y = 1.8 km

height of mountain = h = ∆y + h1 = 1.8 km + 0.8 km

h = 2.6 km


V

qv = tan−1 = tan−1 = tan−1 √3.06

17

×.0

10

m3 m2

qv = 17.1° above the horizontal

17.0 m√

1.30 × 103 m2+ 1.76× 103 m217.0 m

√(36.0m)2 + (42.0 m)2

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6. d = 599 m

∆y = 89 m north

d2 = ∆x2 + ∆y2

∆x =√

d2− ∆y2 =√

(599 m)2 − (89 m)2 =√

3.59 × 105 m2− 7.9 × 103 m2

∆x =√

3.51 × 105 m2

∆x =

q = sin−1 ∆d

y = sin−1 58999m

m

q = 8.5 ° north of east

592 m, east

7. d = 478 km

∆y = 42 km, south = −42 km

d2 = ∆x2 + ∆y2

∆x =√

d2− ∆y2 =√

(478 km)2 − (−42 km)2√

2.28 × 105 km2− 1.8 × 103 km2

∆x =√

2.26 × 105 km2 = −475 km

∆x =

q = sin−1 ∆d

y = sin−1 −47

4

8

2

k

k

m

m

q = 5.0° south of west

475 km, west

8. d = 7400 km

q = 26° south of west

∆y = 3200 km, south =−3200 km

d2 = ∆x2 = ∆y2

∆x =√

d2− ∆y2 = √

(7400km)2 − (−3200 km)2 =√

5.5× 107 km2− 1.0 × 107 km2

∆x =√

4.5× 107 km2 = −6700 km

∆x = 6700 km, west


V

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10. d = 2.9 × 103 km

∆x = 2.8 × 103 km, west = −2.8 × 103 km

d2 = ∆x2 + ∆y2

∆y =√

d2− ∆x2 =√

(2.9 × 103 km)2 − (−2.8× 103)2

∆y =√

8.4× 106 km2− 7.8 × 106 km2 =√

0.6× 106 km2 = −800 km

∆y =

q = cos−1 ∆d

x = cos−1 −22.9.8×

×1

1

0

03

3

k

k

m

m

q = 15° south of west

800 km, south

1. d = 5.3 km

q = 8.4° above horizontal

∆y = d(sin q) = (5.3 km)(sin 8.4°)

∆y = 0.77 km = 770 m

the mountain’s height = 770 m

2. d = 19.1 m

q = 3.0° to the left

∆y = d(sin q) = (19.1 m)(sin 3.0°)

∆y = 1.0 m to the left

the lane’s width = 1.0 m


3. d = 113 m

q = 82.4° above the horizontal south

∆x = d(cos q) = (113 m)(cos 82.4°)

∆x = 14.9 m, south

4. v = 55 km/h

q = 37° below the horizontal= −37°

vy = v(sin q) = (55 km/h)[sin(−37°)]

vy = −33 km/h = 33 km/h, downward

5. d = 2.7 m

q = 13° from the table’slength

∆x = d(cos q) = (2.7 m)(cos 13°)

∆x =

∆y = d(sin q) = (2.7 m)(sin 13°)

∆y = 0.61 m along the table’s width

2.6 m along the table’s length

6. v = 1.20 m/s

q = 14.0° east of north

vx = v(sin q) = (1.20 m/s)(sin 14.0°)

vx =

vy = v(cos q) = (1.20 m/s)(cos 14.0°)

vy = 1.16 m/s, north

0.290 m/s, east

7. d = 31.2 km

q = 30.0° west of south

∆x = d (sin q) = (31.2 km)(sin 30.0°)

∆x =

∆y = d (cos q) = (31.2 km)(cos 30.0°)

∆y = 27.0 km, south

15.6 km, west


V

8. v = 165.2 km/s

q = 32.7°

vforward = v(cos q) = (165.2 km/s)(cos 32.7°)

vforward =

vside = v(sin q) = (165.2 km/s)(sin 32.7°)

vside = 89.2 km/s to the side

139 km/s, forward

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9. v = 55.0 km/h

q = 13.0° above horizontal

vy = v(sin q) = (55.0 km/h)(sin 13.0°)

vy =

vx = v(cos q) = (55.0 km/h)(cos 13.0°)

vx = 53.6 km/h, forward

12.4 km/h, upward

10. v = 13.9 m/s

qh = 24.0° east of north

qv = 26.0° above the horizontal

vz = v(sin qv) = (13.9 m/s)(sin 26.0°)

vz =

horizontal velocity = vh = v(cos qv)

vy = vh(cos qh) = v(cos qv)(cos qh) = (13.9 m/s)(cos 26.0°)(cos 24.0°)

vy =

vx = vh(sin qh) = v(cos qv)(sin qh) = (13.9 m/s)(cos 26.0°)(sin 24.0°)

vx = 5.08 m/s, east

11.4 m/s, north

6.09 m/s, upward

1. d1 = 55 km

q1 = 37 north of east

d2 = 66 km

q2 = 0.0° (due east)

∆x1 = d1(cos q1) = (55 km)(cos 37°) = 44 km

∆y1 = d1(sin q1) = (55 km)(sin 37°) = 33 km

∆x2 = d2(cos q2) = (66 km)(cos 0.0°) = 66 km

∆y2 = d2(sin q2) = (66 km)(sin 0.0°) = 0 km

∆xtot = ∆x1 + ∆x2 = 44 km + 66 km = 110 km

∆ytot = ∆y1 + ∆y2 = 33 km + 0 km = 33 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(110 km)2 + (33 km)2

=√

1.21 × 104 km2+ 1.1 × 103 km2=√

1.32 × 104 km2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 13130k

k

m

m

q = 17° north of east

115 km



V

2. d1 = 4.1 km

q1 = 180° (due west)

d2 = 17.3 km

q2 = 90.0° (due north)

d3 = 1.2 km

q3 = 24.6° west of north = 90.0° + 24.6° = 114.6°

∆x1 = d1(cos ∆1) = (4.1 km)(cos 180°) = −4.1 km

∆y1 = d1(sin q1) = (4.1 km)(sin 180°) = 0 km

∆x2 = d2(cos q2) = (17.3 km)(cos 90.0°) = 0 km

∆y2 = d2(sin q2) = (17.3 km)(sin 90.0°) = 17.3 km

∆x3 = d3(cos q3) = (1.2 km)(cos 114.6°) = −0.42 km

Dy3 = d3(sin q3) = (1.2 km)(sin 114.6°) = 1.1 km

∆xtot = ∆x1 + ∆x2 + ∆x3 = −4.1 km + 0 km + (−0.42 km) = −4.5 km

∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 17.3 km + 1.1 km = 18.4 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(−4.5km)2 + (18.4 km)2

=√

2.0× 101 km2+ 339 km2 =√

359km2

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−1 −18

4

.

.

4

5

k

k

m

m = −76° = 76° north of west

18.9 km

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3. d1 = 850 m

q1 = 0.0°

d2 = 640 m

q2 = 36°

∆x1 = d1(cos q1) = (850 m)(cos 0.0°) = 850 m

∆y1 = d1(sin q1) = (850 m)(sin 0.0°) = 0 m

∆x2 = d2(cos q2) = (640 m)(cos 36°) = 520 m

∆y2 = d2(sin q2) = (640 m)(sin 36°) = 380 m

∆xtot = ∆x1 + ∆x2 = 850 m + 520 m = 1370 m

∆ytot = ∆y1 + ∆y2 = 0 m + 380 m = 380 m

d = (∆xtot)2 + (∆ytot)2 = (1370m)2 + (380m)2 = 1.9× 106 m2+ 1.4 × 105 m2

= 2.0× 106 m2

d =

q = tan−1∆∆x

yt

t

o

o

t

t = tan−11338700m

m

q = 16° to the side of the initial displacement

1400 m


V

4. d1 = 2.00 × 102 m

q1 = 0.0°

d2 = 3.00 × 102 m

q2 = 3.0°

d3 = 2.00 × 102 m

q3 = 8.8°

∆x1 = d1(cos q1) = (2.00 × 102 m)(cos 0.02) = 2.0 × 102 m

∆y1 = d1(sin q1) = (2.00 × 102 m)(sin 0.0°) = 0 m

∆x2 = d2(cos q2) = (3.00 × 102 m)(cos 3.0°) = 3.0 × 102 m

∆y2 = d2(sin q2) = (3.00 × 102 m)(sin 3.0°) = 16 m

∆x3 = d3(cos q3) = (2.00 × 102 m)(cos 8.8°) = 2.0 × 102 m

∆y3 = d3(sin q3) = (2.00 × 102 m)(sin 8.8°) = 31 m

∆xtot = ∆x1 + ∆x2 + ∆x3 = 2.0 × 102 m + 3.0 × 102 m + 2.0 × 102 m = 7.0 × 102 m

∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 m + 16 m + 31 m = 47 m

d =√

(∆xtot)2 + (∆ytot)2 =√

(7.0 × 102 m)2 + (47 m)2 =√

4.9× 105 m2+ 2.2 × 103m2

=√

4.9× 105 m2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 7.0

4

×7

1

m

02 m

q = 3.8° above the horizontal

7.0 × 102 m

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5. d1 = 46 km

q1 = 15° south of east = −15°

d2 = 22 km

q2 = 13° east of south = −77°

d3 = 14 km

q3 = 14° west of south = −90.0° − 14° = −104°

∆x1 = d1(cos q1) = (46 km)[cos(−15°)] = 44 km

∆y1 = d1(sin q1) = (46 km)[sin(−15°)] = −12 km

∆x2 = d2(cos q2) = (22 km)[cos(−77°)] = 4.9 km

∆y2 = d2(sin q2) = (22 km)[sin(−77°)] = −21 km

∆x3 = d3(cos q3) = (14 km)[cos(−104°)] = −3.4 km

∆y3 = d3(sin q3) = (14 km)[sin(−104°)] = −14 km

∆xtot = ∆x1 + ∆x2 + ∆x3 = 44 km + 4.9 km + (−3.4 km) = 46 km

∆ytot = ∆y1 + ∆y2 + ∆y3 = −12 km + (−21 km) + (−14 km) = −47 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(46km)2 + (−47 km)2 =√

2.1× 103 km2+ 2.2 × 103 km2

=√

4.3× 103 km2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 −4467k

k

m

m = −46°

q = 46° south of east

66 km


V

6. d1 = 6.3 × 108 km

q1 = 0.0°

d2 = 9.4 × 108 km

q2 = 68°

d3 = 3.4 × 109 km

q3 = 94° + 68° = 162°

∆x1 = d1(cos q1) = (6.3 × 108 km)(cos 0.0°) = 6.3 × 108 km

∆y1 = d1(sin q1) = (6.3 × 108 km)(sin 0.0°) = 0 km

∆x2 = d2(cos q2) = (9.4 × 108 km)(cos 68°) = 3.5 × 108 km

∆y2 = d2(sin q2) = (9.4 × 108 km)(sin 68°) = 8.7 × 108 km

∆x3 = d3(cos q3) = (3.4 × 109 km)(cos 162°) = −3.2 × 109 km

∆y3 = d3(sin q3) = (3.4 × 109 km)(sin 162°) = 1.1 × 109 km

∆xtot = ∆x1 + ∆x2 + ∆x3 = 6.3 × 108 km + 3.5 × 108 km + (−3.2 × 109 km) = −2.2 × 109 km

∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 8.7 × 103 km + 1.1 × 109 km = 2.0 × 109 km

d =√

(xtot)2+ (ytot)2 =√

(−2.2× 109 km)2 + (2.0× 109 km)2

=√

4.8× 1018km2+ 4.0 × 1018km2 =√

8.8× 1018km2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 −22.0.2×

×1

1

0

0

9

9k

k

m

m = − 42° 180.0° − 42° = 138°

q = 138° from the probe’s initial direction

3.0 × 109 km

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7. di = 2.50 × 103 m

q1 = 58.5° north of east

d2 = 375 m

q2 = 21.8° north of east

d3 = 875 m

q3 = 21.5° east of north

∆x1 = d1(cos q1) = (2.50 × 103 m)(cos 58.5°) = 1310 m

∆y1 = d1(sin q1) = (2.50 × 103 m)(sin 58.5°) = 2130 m

∆x2 = d2(cos q2) = (375 m)(cos 21.8°) = 348 m

∆y2 = d2(sin q2) = (375 m)(sin 21.8°) = 139 m

∆x3 = d3(sin q3) = (875 m)(sin 21.5°) = 321 m

∆y3 = d3(cos q3) = (875 m)(cos 21.5°) = 814 m

∆xtot = ∆x1 + ∆x2 + ∆x3 = 1310 m + 348 m + 321 m = 1.98 × 103 m

∆ytot = ∆y1 + ∆y2 + ∆y3 = 2130 m + 139 m + 814 m = 3.08 × 103 m

d =√

(∆xtot)2 + (∆ytot)2 =√

(1.98× 103 m)2 + (3.08 × 103 m)2

=√

3.92 × 106 m2+ 9.49× 106 m2 =√

13.41× 106 m2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 31.

.

0

9

8

8

××

1

1

0

0

3

3m

m

q = 57.3° north of east

3.66 × 103 m


V

8. d1 = 5.0 km

q1 = 36.9° south of east = −36.9°

d2 = 1.5 km

q2 = 90.0° due south = −90.0°

d3 = 8.5 km

q3 = 42.2° south of east

= −42.2°

d4 = 0.8 km

q4 = 0° (due east)

∆x1 = d1(cos q1) = (5.0 km)[cos(−36.9°)] = 4.0 km

∆y1 = d1(sin q1) = (5.0 km)[sin(−36.9°)] = −3.0 km

∆x2 = d2(cos q2) = (1.5 km)[cos(−90.0°)] = 0 km

∆y2 = d2(sin q2) = (1.5 km)[sin(−90.0°)] = −1.5 km

∆x3 = d3(cos q3) = (8.5 km)[cos(−42.2°)] = 6.3 km

∆y3 = d3(sin q3) = (8.5 km)[sin(−42.2°)] = −5.7 km

∆x4 = d4(cos q4) = (0.8 km)(cos 0°) = 0.8 km

∆y4 = d4(sin q4) = (0.8 km)(sin 0°) = 0 km

∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = 4.0 km + 0 km + 6.3 km + 0.8 km = 11.1 km

∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (− 3.0 km) + (−1.5 km) + (−5.7 km) + 0 km = −10.2 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(11.1km)2 + (−10.2 km)2 =√

123km2+ 104 km2

=√

227km2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 −1110.1.2k

k

m

m = −42.6°

q = 42.6° south of east

15.1 km

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9. d1 = 1.41 m

q1 = 45.0° west of north = 90.0° + 45.0° = 135.0°

d2 = 1.98 m

q2 = 45.0° east of north = 45.0°

d3 = 0.42 m

q3 = 45.0° west of north = 135.0°

d4 = 1.56 m

q4 = 45.0° south of west = 180.0° + 45.0° = 225.0°

∆x1 = d1(cos q1) = (1.41 m)(cos 135.0°) = −0.997 m

∆y1 = d1(sin q1) = (1.41 m)(sin 135.0°) = 0.997 m

∆x2 = d2(cos q2) = (1.98 m)(cos 45.0°) = 1.40 m

∆y2 = d2(sin q2) = (1.98 m)(sin 45.0°) = 1.40 m

∆x3 = d3(cos q3) = (0.42 m)(cos 135.0°) = −0.30 m

∆y3 = d3(sin q3) = (0.42 m)(sin 135.0°) = 0.30 m

∆x4 = d4(cos q4) = (1.56 m)(cos 225.0°) = −1.10 m

∆y4 = d4(sin q4) = (1.56 m)(sin 225.0°) = −1.10 m

∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−0.997 m) + 1.40 m + (−0.30 m) + (−1.10 m)

= −0.997 m

∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 0.997 m + 1.40 m + 0.30 m + (−1.10 m) = 1.60 m

d =√

(∆xtot)2 + (∆ytot2 =√

(−0.997m)2 + (1.60 m)2 =√

0.994m2+ 2.56m2

=√

3.55 m2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 −1

0

.

.

6

9

0

97

m

m = −58.1°

q = 58.1° north of west

1.88 m


V

10. d1 = 79 km

q1 = 18° north of west 180.0° − 18° = 162°

d2 = 150 km

q2 = 180.0° due west

d3 = 470 km

q3 = 90.0° due north

d4 = 240 km

q4 = 15° east of north 90.0° − 15° = 75°

∆x1 = d1(cos q1) = (790 km)(cos 162°) = −750 km

∆y1 = d1(sin q1) = (790 km)(sin 162°) = 24 km

∆x2 = d2(cos q2) = (150 km)(cos 180.0°) = −150 km

∆y2 = d2(sin q2) = (150 km)(sin 180.0°) = 0 km

∆x3 = d3(cos q3) = (470 km)(cos 90.0°) = 0 km

∆y3 = d3(sin q3) = (470 km)(sin 90.0°) = 470 km

∆x4 = d4(cos q4) = (240 km)(cos 75°) = 62 km

∆y4 = d4(sin q4) = (240 km)(sin 75°) = 230 km

∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−750 km) + (−150 km) + 0 km + 62 km = −840 km

∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 240 km + 0 km + 470 km + 230 km = 940 km

d =√

(∆xtot)2 + (∆ytot)2 =√

(−840km)2 + (940km)2 =√

7.1× 105 km2+ 8.8 × 105 km2

=√

15.9 × 105 km2

d =

q = tan−1 ∆∆x

yt

t

o

o

t

t = tan−1 −984400k

k

m

m = −48°

q = 48° north of west

1260 km

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1. vx = 430 m/s ∆t = ∆vx

x

∆x = 4020 m∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)443

0

0

20

m

m

/s

2

= −430 m

2. ∆x = 101 m ∆t = ∆vx

x

vx = 14.25 m/s∆y = − 1

2g∆t2

g = 9.81 m/s2

y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)14

1

.

0

2

1

5

m

m/s

2

= 246 m

3. vx = 1.30 × 102 km/h ∆t = ∆vx

x

∆x = 135 m∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)1.30 ×13

1

5

0

m2 km/h

2

130630

m

0

/

s

k

/h

m

2

= −68.6 m

airship’s altitude = 68.6 m

height of building = 246 m

height of ridge = 430 m


V

4. vx = 9.37 m/s ∆t = ∆vx

x

∆x = 85.0 m∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)98.357

.0

m

m

/s

2

= − 404 m

5. vx = 6.32 cm/s ∆t = ∆vx

x

∆x = 1.00 m∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)6.32

1

×.0

1

0

0

m−2 m/s

2

= 1230 m

6. vx = 10.0 cm/s ∆t = ∆vx

x

∆x = 18.6 cm∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)1108.0.6c

c

m

m

/s

2

= −17.0 m

7. vx = 1.50 m/s ∆t = ∆vx

x

∆x = 3.50 m∆y = − 1

2g∆t2

g = 9.81 m/s2

∆y = − 12

g∆vx

x

2

= − 12

(9.81 m/s2)13.5.500m

m

/s

2

= −26.7 m

8. ∆y = −2.50 × 102 m ∆x = vx∆t

vx = 1.50 m/s ∆y = − 12

g∆t2

g = 9.81 m/s2

∆t = 2

−∆g

y

∆x = vx2

−∆g

y = (1.50 m/s) (2)(−

−2

9

..

5

80

1 ×m 1

/0

s22

m)

∆x = 10.7 m

the lunch pail falls 26.7 m

squirrel’s height = 17.0 m

building’s height = 1230 m

mountain’s height = 404 m

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V

9. vx = 1.50 m/s vy,f2 = −2g∆y + vy,i

2

∆y = −2.50 × 102 m vy,i = 0 m/s, so

g = 9.81 m/s2 vy,f = vy =√

−2g∆y =√

−(2)(9.81 m/s2)(−2.50 × 102 m)vy = 70.0 m/s

v =√

vx2+ vy2 =

√(1.50m/s)2 + (70.0 m/s)2 =

√2.25 m2/s2 + 4.90× 103 m2/s2

=√

4.90 × 103 m2/s2

v =

q = tan−1 v

vx

y = tan−117.

0

5

.

0

0

m

m

/

/

s

s

q =

10. vx = 85.3 m/s ∆t = 2

−∆g

y =

∆vx

x

∆y = −1.50 m

∆x = vx 2

−∆g

y = (85.3 m/s) (2

−)

9

(.

−81

1

. 5

m0

/m

s2) = 47.2 m

g = 9.81 m/s2

range of arrow = 47.2 m

1.23° from the vertical

70.0 m/s

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1. vi = 15.0 m/s ∆t = vi(c

∆o

x

s q)

∆x = 17.6 m ∆y = vi(sin q)∆t − 12

g∆t2 = 0

vi(sin q) = 12

g∆t = 12

gvi(c

∆o

x

s q)

2(sin q)(cos q) = g

v

∆

i2x

Using the identity 2 (sin q)(cos q) = sin (2 q),

sin (2 q) = g

v

∆

i2x

q = =

q =

2. vi = 23.1 m/s vy,f2 − vy,i

2 = −2g∆y

∆ymax = 16.9 m At maximum height, vy,f

g = 9.81 m/s2 vy,i = vi(sin q) =√

2g∆ymax

q = sin−1 = sin−1 q = 52.0°

√(2)(9.81 m/s2)(16.9m)

23.1 m/s

√2g∆ymax

vi

25.1°

sin − (9.81

(1

m

5.

/

0

s2

m

)(

/

1

s

7

)

.26 m)

2

sin−1g

v

∆

i2x

2



V

3. ∆x = 7.49 m Using the form of the equation derived in problem 1,

vi = 9.50 m/s 2(sin q)(cos q) = sin(2 q) = g

v

∆

i2x

g = 9.81 m/s2

q = sin−1 = sin−1

q =

4. vi = 141 cm/s Using the form of the equation derived in problem 1,

∆x = 18.5 cm 2(sin q)(cos q) = sin(2 q) = g

v

∆

i2x

g = 9.81 m/s2

q = sin−1

sin−1 2

q =

5. vi = 6.03 m/s vy,f2 − vy,i

2 = −2g∆y

hi = 10.0 m At maximum height, vy,f = 0.

hmax = hf = 11.7 m vy,i = vi(sin q) =√

2g∆ymax∆x = 3.62 m For the diver, hf is the maximum height above the diving board.

g = 9.81 m/s2 ∆y = hf − hi

q = sin−1 = sin−1 q =

6. vi = 10.0 m/s ∆x = vi(cos q)∆t = (10.0 m/s)(cos 37.0°)(2.5 s)

q = 37.0° ∆x =

∆t = 2.5 s ∆y = vi(sin q)∆t − 12

g∆t2 = (10.0 m/s)(sin 37.0°)(2.5 s) − 12

(9.81 m/s2)(2.5 s)2

g = 9.81 m/s2= 15 m − 31 m

∆y =

7. vi = 250 m/s At the maximum height

q = 35° vy,f = vy,i − g∆t = 0

g = 9.81 m/s2 vy,i = vi(sin q) = g∆t

∆t = vi(si

g

n q) =

(250

9

m

.8

/

1

s)

m

(s

/

i

s

n235°)

∆t = 15 s

−16 m

2.0 101 m

73.3°

√(2)(9.81)m/s2)(11.7m − 10.0m)

6.03 m/s

√2g(hf − hi)

vi

33.0°

(9.81 m/s2)(18.5 × 10−2 m)

(141 × 10−2 m/s)2

g

v

∆

i2x

2

27.3°

(9.81

(

m

9.5

/s

0

2)

m

(7

/s

.429 m)

2

g

v

∆

i2x

2

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V

8. ∆x = 73.0 m ∆t = vi(c

∆o

x

s q)

∆y = −52.8 m

∆y = vi(sin q)∆t − 12

g∆t 2 = vi(sin q)vi(c

∆o

x

s q) − 1

2gvi(c

∆o

x

s q)

2

q = −8.00°

∆y = ∆x(tanq) − 2 vi

2g

(

∆co

x

s

2

q)2g = 9.81 m/s2

2 vi

2g

(

∆co

x2

s q)2 = ∆x(tan q) − ∆y

2vi2(cos q)2 =

[∆x(ta

g

n

∆qx

)

2

− ∆y]

vi = vi = vi = vi = vi =

9. q = −30.0° ∆y = vi(sin q)∆t − 12

g∆t2

vi = 2.0 m/s 2

g∆t2 − [vi(sin q)]∆t + ∆y = 0

∆y = −45 mSolving for ∆t using the quadratic equation,

g = 9.81 m/s2

∆t =

∆t =

∆t = =

∆t =

∆t must be positive, so the positive root must be chosen.

∆t = 9.

2

8

9

1

m

m

/

/

s

s2 = 3.0 s

−1.0 m/s ± 3.0 × 101 m/s

9.81 m/s2

−1.0 m/s ±√

8.8× 102 m2/s29.81 m/s2

−1.0 m/s ±√

1.0m2/s2 + 8.8 × 102 m2/s29.81 m/s2

(2.0 m/s)[sin(−30.0°)] ±√

[(−2.0m/s)[sin(−30.0°)]2 − (2)(9.81m/s2)(−45 m)9.81 m/s2

vi(sin q) ± [−vi(sin q)]2− 42

g(∆y)

22

g

25.0 m/s

(9.81 m/s2)(73.0 m)2

(2)[−cos(–8.00°)]2 (42.5 m)

(9.81 m/s2)(73.0 m)2

(2)[cos(−8.00°)]2 (−10.3 m + 52.8 m)

(9.81 m/s2)(73.0 m)2

(2)[cos(−8.00°)]2 [(73.0 m)(tan[−8.00°]) − (−52.8 m)]

g∆x2

2(cos q)2 [∆x(tan q) − ∆y]

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V

10. ∆x1 = 0.46 m ∆xtot = ∆x1 + ∆x2

∆x2 = 4.00 m ∆t = vi(c

∆o

x

s q)

q = 41.0° ∆y = vi(sin q)∆t − 12

g∆t2 = vi(sin q)∆vi

x

(i

c

+os

∆qx

)2 − 1

2g∆vx

i(1

co

+s

∆qx

)2

2

∆y = − 0.35 m ∆y = (∆x1 + ∆x2)(tan q) − g

2

(∆v

x

i21

(c

+o

∆s

x

q2

)2)2

g = 9.81 m/s2 vi = vi = vi = vi = vi = 6.36 m/s

(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2 (4.23 m)

(9.81 m/s2)(4.46 m)2

(2)(cos 41.0°)2(3.88 m + 0.35 m)

(9.81 m/s2)(0.46 m + 4.00 m)2

(2)(cos 41.0°)2 [(0.46 m + 4.00 m)(tan 41.0°) − (−0.35 m)]

g(∆x1 + ∆x2)2

2(cos q)2[(∆x1 + ∆x2)(tan q) − ∆y]

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1. vbw = 58.0 km/h, forward vbe = vbw + vwe = + 58.0 km/h + (−55.0 km/h) = +3.0 km/h= +58.0 km/h

∆t = v

∆

b

x

e =

3

1

.0

.4

k

k

m

m

/h

vwe = 55.0 km/h, backward

∆t == −55.0 km/h

∆x = 1.4 km

0.47 h = 28 min


2. vwe = 1.50 m/s, west vme = vmw + vwe = 4.20 m/s + 1.50 m/s = 5.70 m/s, west

vmw = 4.20 m/s, west time of travel with walkway:

∆x = 8.50 × 102m∆t1 =

v

∆

m

x

e =

8.5

5

0

.7

×0

1

m

0

/

2

s

m = 149 s

time of travel without walkway:

∆t2 = v

∆

m

x

w =

8.5

4

0

.2

×0

1

m

0

/

2

s

m = 202 s

time saved = ∆t2 − ∆t1 = 202 s − 149 s =

3. v1e = 286 km/h, forward v12 + v2e = v1e

v2e = 252 km/h, forward v12 = v1e − v2e

∆x = 0.750 km v12 = v1e − v2e = 286 km/h − 252 km/h = 34 km/h

∆t = v

∆

1

x

2 −

0

3

.

4

75

k

0

m

k

/

m

h = 2.2 × 10−2 h

∆t = (2.2 × 10−2 h)36

1

0

h

0 s = 79 s

53 s


V

4. vma = 1.10 m/s, east vme = vma + vae

vae = 5.0 km/h at 35° Find the mosquito’s speed with respect to Earth in the x direction.west of south

vx, me = vx, ma + vx, ae = vma + vae(cos qae)∆x = 540 m

qae = –90.0° − 35° = −125°

vx, me = 1.10 m/s + (5.0 km/h)(103 m/km)36

1

0

h

0 s[cos(−125°)] = 1.10 m/s

+ (−0.80 m/s) = 0.30 m/s

∆t = vx

∆

,

x

me =

0

5

.3

4

0

0

m

m

/s

∆t =

5. vga = 150 km/h at 7.0° vge = vga + vaebelow horizontal

Find the glider’s speed with respect to Earth in the y (vertical) direction.vae = 15 km/h upward

vy, ge = vga + vy, ae = vga(sin qga) + vae∆y = −165 m

qga = −7.0°

vy, ge = (150 km/h)[sin(−7.0°)] + 15 km/h = −18 km/h + 15 km/h = −3 km/h

Time of descent with updraft:

∆t = v

∆

y,

y

ge =

∆t =

Time of descent without updraft:

∆t′ = v

∆

y,

y

ga =

∆t′ =

6. vfc = 87 km/h, west vfe = vfc + vce

vce = 145 km/h, north vx, fe = vx, fc + vx, ce = vfc = −87 km/h

∆t = 0.45 s vy, fe = vy, fc + vy, ce = vce = +145 km/h

∆x = vx, fe ∆t = (−87 km/h)(103 m/km)(1 h/3600 s)(0.45 s) = −11 m

∆x =

∆y = vy, fe ∆t = (145 km/h)(103 m/km)(1 h/3600 s)(0.45 s)

∆y = 18 m, north

11m, west

33 s

−166 m(−18 km/h)(103 m/km)(1h/3600 s)

200 s

−166 m(−3 km/h)(103 m/km)(1 h/3600 s)

1800 s = 3.0 × 101 min

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V

7. vaw = 55.0 km/h, north vae = vaw + vwe

vwe = 40.0 km/h at 17.0° vx, ae = vx, aw + vx, we = vwe (cos qwe)

north of west vy, ae = vy, aw + vy, we = vaw + vwe(sin qwe)

qwe = 180.0° − 17.0° = 163.0°

vx, ae = (40.0 km/h)(cos 163.0°) = −38.3 km/h

vy, ae = 55.0 km/h + (40.0 km/h)(sin 163.0°) = 55.0 km/h + 11.7 km/h = 66.7 km/h

vae =√

vx, ae)2+(vy, ae02 =√

(−38.3 km/h)2 + (66.7 km/h)2

vae =√

1.47 × 103 km2/h2+ 4.45× 103 km2/h2 =√

5.92 × 103 km2/h2

vae =

q = tan−1 v

v

x

y,

,

a

a

e

e = tan−1 = −60.1°

q =

8. vae = 76.9 km/h at 29.9° ∆x = vae(cos qae)∆twest of north ∆y = vae(sin qae)∆t

∆t = 15.0 min qae = 90.0° + 29.9° = 119.9°

∆x = (76.9 km/h)(cos 119.9°)(15.0 min)(1 h/60 min) = −9.58 km

∆y = (76.9 km/h)(sin 119.9°)(15.0 min)(1 h/60 min) = 16.7 km

∆x =

∆y =

9. vtc = 51 km/h, east vte = vtc + vce

vce = 4.0 km/h, south vx, te = vx, tc + vx, ce = vtc = 51 km/h

∆t = 14 s vy, te = vy, tc + vy, ce = vce = −4.0 km/h

∆x = vx, tc∆t = (51 km/h)(103 m/km)(1 h/3600 s)(14 s) = 2.0 × 102 m

∆y = vy, te∆t = (4.0 km/h)(103 m/km)(1 h/3600 s)(14 s) = 16 m

the torpedo must be fired 16 m north of the target

the target is 2.0 × 102 m away

16.7 km, north

9.58 km, west

60.1° west of north

66.7 km/h−38.3 km/h

76.9 km/h

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V

10. vbw = 12.0 km/h, south vbe = vbw + vwe

vwe = 4.0 km/h at 15.0° vx, be = vx, bw + vx, we = vwe(cos qwe)south of east vy, be = vy, bw + vy, we = vbw + vwe(sin qwe)

qwe = −15.0°

vx, be = (4.0 km/h)[cos(−15.0°)] = 3.9 km/h

vy, be = (−12.0 km/h) + (4.0 km/h)[sin(−15.0°)] = (−12.0 km/h) + (−1.0 km/h) = −13.0 km/h

vbe =√

(vx, be)2+ vy,be)2 =√

(3.9 km/h)2 + (−13.0 km/h)2

vbe =√

15 km2/h2+ 169 km2/h2 =√

184km2/h2

vbe =

q = tan−1vvx

y,

,

b

b

e

e = tan−1−3

1

.

3

9

.0

km

km

/h

/h = −73°

q = 73° south of east

13.6 km/h

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Section Five—Solution Manual V Ch. 4–1

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4ChapterForces and the Laws of Motion

1. F1 = 7.5 × 104 N north

F2 = 9.5 × 104 N at 15.0°north of west

q1 = 90.0°

q2 = 180.0° − 15.0° = 165.0°

Fx,net = Fx = F1(cos q1) + F2(cos q2)

Fx,net = (7.5 × 104 N)(cos 90.0°) + (9.5 × 104 N)(cos 165.0°)

Fx,net = −9.2 × 104 N

Fy,net = Fy = F1(sin q1) + F2(sin q2)

Fy,net = (7.5 × 104 N)(sin 90.0°) + (9.5 × 104 N)(sin 165.0°)

Fy,net = 7.5 × 104 N + 2.5 × 104 N = 10.0 × 104 N

q = tan−1 = tan−1 = −47°


10.0 × 104 N−9.2 × 104

Fy,netFx,net


Givens Solutions

2. F1 = 6.00 × 102 N north

F2 = 7.50 × 102 N east

F3 = 6.75 × 102 N at 30.0°south of east

q1 = 90.0°

q2 = 0.00°

q3 = −30.0°

Fx,net = Fx = F1(cos q1) + F2(cos q2) + F3(cos q3) = (6.00 × 102 N)(cos 90.0°)

+ (7.50 × 102 N)(cos 0.00°) + (6.75 × 102 N)[cos(−30.0°)]

Fx,net = 7.50 × 102 N + 5.85 × 102 N = 13.35 × 102 N

Fy,net = Fy = F1(sin q1) + F2(sin q2) + F3(sin q3) = (6.00 × 102 N)(sin 90.0°)

+ (7.50 × 102 N)(sin 0.00°) + (6.75 × 102 N)[sin (−30.0°)]

Fy,net = 6.00 × 102 N + (−3.38 × 102 N) = 2.62 × 102 N

q = tan−1 = tan−1 q = 11.1° north of east

2.62 × 102 N13.35 × 102 N

Fy,netFx,net

3. F1 = 2280.0 N upward

F2 = 2250.0 N downward

F3 = 85.0 N west

F4 = 12.0 N east

Fy,net = Fy = F1 + F2 = 2280.0 N + (−2250.0 N) = 30.0 N

Fx,net = Fx = F3 + F4 = −85.0 N + 12.0 N = −73.0 N

q = tan−1 = tan−1 = −22.3°

q = 22.3° up from west

30.0 N−73.0 N

Fy,netFx,net

4. F1 = 6.0 N

F2 = 8.0 N

Fmax = F1 + F2 = 6.0 N + 8.0 N

Fmax =Fmin = F2 − F1 = 8.0 N − 6.0 N

Fmin = 2.0 N

14.0 N


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Fx,net = F1 + F3(cos q) = 0F3(cos q) = −F1 = −3.0 N

Fy,net = F2 + F3(sin q) = 0F3(sin q) = −F2 = −(−4.0 N) = 4.0 N

F3 = √

[F3(cosq)]2+ [F3(sin q)]2 = √

(−3.0 N)2 + (4.0 N)2 = √

9.0N2+ 16N2 = √

25 N2

F3 =

q = tan−1 = tan−1 = −53°


4.0 N−3.0 N

F3(sin q)F3(cos q)

5.0 N

Givens Solutions

5. F1 = 3.0 N east

F2 = 4.0 N south

6. F1 = 4.00 × 103 N east

F2 = 5.00 × 103 N north

F3 = 7.00 × 103 N west

F4 = 9.00 × 103 N south

Fx,net = F1 + F3 = 4.00 × 103 N + (−7.00 × 103 N) = −3.00 × 103 N

Fy,net = F2 + F4 = 5.00 × 103 N + (−9.00 × 103 N) = −4.00 × 103 N

Fnet = √

(Fx,net)2+ (Fy,net)2 = √

(−3.00 × 103 N)2 + (−4.00 × 103 N)2

Fnet = √

9.00 × 106 N2+ 16.0× 106 N2 = √

25.0 × 106 N2

Fnet =

q = tan−1 = tan−1 q = 53.1° south of west

−4.00 × 103 N−3.00 × 103 N

Fy,netFx,net

5.00 × 103 N

7. F1 = 15.0 N

q = 55.0°

Fy = F(sin q) = (15.0 N)(sin 55.0°)

Fy =Fx = F(cos q) = (15.0 N)(cos 55.0°)

Fx = 8.60 N

12.3 N

8. F = 76 N

q = 40.0°

Fx = F(cos q) = (76 N)(cos 40.0°)

Fx =Fy = F(sin q) = (76 N)(sin 40.0°)

Fy = 49 N

58 N

9. F1 = 350 N

q1 = 58.0°

F2 = 410 N

q2 = 43°

Fy,net = F1(sin q1) + F2(sin q2) = (350 N)(sin 58°) + (410 N)(sin 43)

Fy,net = 3.0 × 102 N + 2.8 × 102 N

Fy,net = 580 N

10. F1 = 7.50 × 102 N

q1 = 40.0°

F2 = 7.50 × 102 N

q2 = −40.0°

Fy,net = Fg = F1(cos q1) + F2(cos q2)

Fy,net = (7.50 × 102 N)(cos 40.0°) + (7.50 × 102 N) [cos(−40.0°)]

Fg = 575 N + 575 N = 1.150 × 103 N


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1. Fnet = 2850 N

vf = 15 cm/s

vi = 0 cm/s

∆t = 5.0 s

anet = = = 3.0 cm/s2 = 3.0 × 10−2 m/s2

Fnet = m anet

m = =

m = 9.5 × 104 kg

2850 N3.0 × 10−2 m/s2

Fnetanet

15 cm/s − 0 cm/s

5.0 s

vf − vi∆t

2. ∆t = 1.0 m/s

∆t = 5.0 s

Fdownhill = 18.0 N

Fuphill = 15.0 N

anet = = = 0.20 m/s2

Fnet = m anet = Fdownhill − Fuphill = 18.0 N − 15.0 N = 3.0 N

m = = = 15 kg3.0 N0.20 m/s2

Fnetanet

1.0 m/s

5.0 s

∆v∆t

3. Fmax = 4.5 × 104 N

anet = 3.5 m/s2

g = 9.81 m/s2

Fnet = m anet = Fmax − mg

m(anet + g) = Fmax

m = = = = 3.4 × 103 kg4.5 × 104 N13.3 m/s2

4.5 × 104 N3.5 m/s2 + 9.81 m/s2

Fmaxanet + g

4. m = 2.0 kg

∆y = 1.9 m

∆t = 2.4 s

vi = 0 m/s

∆y = vi ∆t + anet ∆t2

Because vi = 0 m/s, anet = = = 0.66 m/s2

Fnet = m anet = (2.0 kg)(0.66 m/s) = 1.32 N

Fnet = 1.32 N upward

(2)(1.9 m)

(2.4 s)22∆y∆t2

12

5. m = 8.0 kg

∆y = 20.0 cm

∆t = 0.50 s

vi = 0 m/s

g = 9.81 m/s2

∆y = vi ∆t + anet ∆t2

Because vi = 0 m/s, anet = = = 1.6 m/s2

Fnet = m anet = (8.0 kg)(1.6 m/s2) = 13 N

Fnet =

Fnet = Fupward − mg

Fupward = Fnet + mg = 13 N + (8.0 kg)(9.8 m/s2) = 13 N + 78 N = 91 N

Fupward = 91 N upward

13 N upward

(2)(20.0 × 10−2 m)

(0.50 s)22∆y∆t2

12

6. m = 75 kg

aforward = 0.15 m/s2 west

abackward = 2 × 10−2 east

anet = aforward − abackward = 0.15 m/s2 − 2 × 10−2 m/s2

anet = 0.13 m/s2 west

Fnet = m anet = (75 kg)(0.13 m/s) = 9.8 N

Fnet = 9.8 N west


Givens Solutions

Givens Solutions


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7. Fnet = −65.0 N

m = 0.145 kganet = = = −448 m/s2−65.0 N

0.145 kg

Fnetm

8. m = 214 kg

Fbuoyant = 790 N

g = 9.81 m/s2

Fnet = Fbuoyant − mg = 790 N − (214 kg)(9.81 m/s2)

Fnet = 790 N − 2.10 × 103 N = −1310 N

anet = = = −6.12 m/s2−1310 N214 kg

Fnetm

9. m = 0.080 kg

q = 37.0°

g = 9.81 m/s2

Fnet = m anet = m g(sin q)

anet = g(sin q) = (9.81 m/s2)(sin 37.0°) = 5.90 m/s2

anet = 5.90 m/s2 down the incline (37.0° below horizontal)

10. m = 0.080 kg

Fupward = 1.40 N

adownward = 5.90 m/s2

Fnet = Fupward − m adownward = 1.40 N − (0.080 kg)(5.90 m/s2) = 1.40 N − 0.47 N = 0.93 N

Fnet = 0.93 N up the incline (37.0° above the horizontal)

anet = F

mnet =

0

0

.0

.9

8

3

0

N

kg = 12 m/s2

1. Fdownward = 4.26 × 107 N

mk = 0.25

Fnet = Fdownward − Fk = 0

Fk = mk Fn = Fdownward

Fn = = = 1.7 × 108 N4.26 × 107 N

0.25

Fdownwardmk

2. Fn = 1.7 × 108 N

q = 10.0°

g = 9.81 m/s2

Fn = mg (cos q)

m = = = 1.8 × 107 kg1.7 × 108 N

(9.81 m/s2)(cos 10.0°)

Fng (cos q)

3. Fs,max = 2400 N

ms = 0.20

q = 30.0°

g = 9.81 m/s2

Fs,max = ms Fn

Fn = = = 1.2 × 104 N

Fn =

Fn = mg(cos q)

m = = = 1400 kg1.2 × 104 N

(9.81 m/s2)(cos 30.0°)

Fng (cos q)

1.2 × 104 N perpendicular to and away from the incline

2400 N

0.20

Fs,maxms


4. m = 60.0 kg

a = 3.70 m/s2

ms = 0.455

g = 9.81 m/s2

For the passenger to remain standing without sliding,

Fs,max ≥ F = ma

Fs,max = ms Fn = ms mg

ms mg ≥ ma

ms g ≥ a

(0.455)(9.81 m/s2) = 4.46 m/s2 > 3.70 m/s2

The passenger will be able to stand without sliding.


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Givens Solutions

5. m = 90.0 kg

q = 17.0°

g = 9.81 m/s2

Fnet = mg(sin q) − Fk = 0

Fk = mg(sin q) = (90.0 kg)(9.81 m/s2)(sin 17.0°) = 258 N

Fk = 258 N up the slope

6. msled = 47 kg

msupplies = 33 kg

mk = 0.075

q = 15°

Fk = mk Fn = mk(msled + msupplies)g = (0.075)(47 kg + 33kg)(9.81 m/s2)

Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)

Fk =

Fk = mk Fn = mk(msled + msupplies)g(cos q) = (0.075)(47 kg + 33 kg)(9.81 m/s2)(cos 15°)

Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)(cos 15°) = 57 N

59 N

7. m = 1.8 × 103 kg

q = 15.0°

Fs,max = 1.25 × 104 N

g = 9.81 m/s2

Fs,max = ms Fn = ms mg(cos q)

ms = mg

F

(s

c,m

oa

sx

q) =

ms = 0.73

1.25 × 104 N(1.8 × 103 kg)(9.81 m/s2)(cos 15.0°)

8. m = 15.0 g

q = 2.3°

g = 9.81 m/s2

Fnet = mg(sin q) − Fk = 0

Fk = mk Fn = mk mg(cos q)

mk = m

m

g

g

(

(

c

si

o

n

s

qq

)

) = tan q = tan 2.3°

mk = 0.040

9. vf = 88.0 km/h

vi = 0 km/h

∆t = 3.07 s

g = 9.81 m/s2

Fnet = Fapplied − Fs,max = 0

Fs,max = ms Fn = ms mg

Fapplied = m a

m a = ms mg

ms = a

g =

vf

g∆−

t

vi =

ms = 0.812

(88.0 km/h − 0 km/h)(103 m/km)(1 h/3600 s)

(9.81 m/s2)(3.07 s)

10. q = 5.0° Fnet = mg(sin q) − Fk = 0


mk = m

m

g

g

(

(

c

si

o

n

s

qq

)

) = tan q = tan 5.0°

mk = 0.087


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3. m = 5.1 × 102 kg

q = 14°

Fapplied = 4.1 × 103 N

g = 9.81 m/s2

ms = =

ms = 0.40

1.00 × 103 N(266 kg)(9.81 m/s2)(cos 17°)

1760 − 760 N(266 kg)(9.81 m/s2)(cos 17°)

1. anet = 1.22 m/s2

q = 12.0°

g = 9.81 m/s2

Fnet = m anet = mg(sin q) − Fk


m anet + mk mg(cos q) = mg(sin q)

mk = g(s

g

in

(c

qo

)

s

−q

a

)net = =

mk = = 0.0850.82 m/s2

(9.81 m/s2)(cos 12.0°)

2.04 m/s2 − 1.22 m/s2

(9.81 m/s2)(cos 12.0°)

(9.81 m/s2)(sin 12.0°) − 1.22 m/s2

(9.81 m/s2)(cos 12.0°)

2. Fapplied = 1760 N

q = 17.0°

m = 266 kg

g = 9.81 m/s2

Fnet = Fapplied − mg(sin q) − Fs,max = 0


ms mg(cos q) = Fapplied − mg(sin q)

ms = Fappl

m

ied

g

−(c

m

os

g

q(s

)

in q) =

1760 − (266 kg)(9.81 m/s2)(sin 17°)

(266 kg)(9.81 m/s2)(cos 17°)

Fnet = Fapplied − mg(sin q) − Fs,max = 0


ms mg(cos q) = Fapplied − mg(sin q)

ms = Fappl

m

ied

g

−(c

m

os

g

q(s

)

in q) =

ms = =

ms = 0.60

2.9 × 103 N(5.1 × 102 kg)(9.81 m/s2)(cos 14°)

4.1 × 103 N − 1.2 × 103 N(5.1 × 102 kg)(9.81 m/s2)(cos 14°)

4.1 × 103 N − (5.1 × 102 kg)(9.81 m/s2)(sin 14°)

(5.1 × 102 kg)(9.81 m/s2)(cos 14°)


Givens Solutions

4. Fapplied = 5.0 N to the left

m = 1.35 kg

anet = 0.76 m/s2 to the left

Fnet = m anet = Fapplied − Fk

Fk = Fapplied − m anet

Fk = 5.0 N − (1.35 kg)(0.76 m/s2) = 5.0 N − 1.0 N = 4.0 N

Fk = 4.0 N to the right

5. mk = 0.20

g = 9.81 m/s2

Fnet = m anet = Fk

Fk = mkFh = mkmg

anet = mk

m

mg = mkg = (0.20)(9.81 m/s2)

anet = 2.0 m/s2


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6. Fapplied = 2.50 × 102 N

m = 65.0 kg

q = 18.0°

anet = 0.44 m/s2

Givens Solutions

Fnet = m anet = Fapplied − mg(sin q) − Fk

Fk = Fapplied − mg(sin q) − manet

Fk = 2.50 × 102 N − (65.0 kg)(9.81 m/s2)(sin 18.0°) − (65.0 kg)(0.44 m/s2)

Fk = 2.50 × 102 N − 197 N − 29 N = 24 N = 24 N downhill

7. m = 65.0 kg

g = 9.81 m/s2

Fk = 24 N

q = 18.0°

Fnet = m anet = mg(sin q) − Fk

anet = g(sin q) − F

mk = (9.81 m/s2)(sin 18.0°) −

6

2

5

4

.0

N

kg = 3.03 m/s2 − 0.37 m/s2 = 2.66 m/s2

anet = 2.66 m/s2 downhill

8. Fapplied = 3.00 × 102 N

q = 20.0°

mk = 0.250

Fx,net = Fapplied(cos q) − Fk = 0Fy,net = Fn − mg + Fapplied(sin q) = 0

Fk = mkFn

Fn = Fapplie

md(

k

cos q) = =

m = Fn + Fapp

glied(sin q) =

m = 113

9

0

.8

N

1

−m

1

/s

023 N

= 9.

1

8

0

1

3

m

0 N

/s2 = 105 kg

1130 N + (3.00 × 102 N)[sin(−20.0°)]

9.81 m/s2

1130 N(3.00 × 102 N)[cos(−20.0°)]

0.25°

9. Fapplied = 590 N

Fdownhill = 950 N

ms = 0.095

q = 14.0°

Fnet = Fapplied + Fs,max − Fdownhill = 0Fs,max = ms Fn = ms mg(cos q)

ms Fn = Fdownhill − Fapplied

Fn = Fdownhill

m−

s

Fapplied = 950 N

0.0

−9

5

5

90 N =

3

0

6

.0

0

9

N

5 = 3800 N

Fn =

m = g(c

F

on

s q) = = 4.0 × 102 kg

3800 N(9.81 m/s2)(cos 14.0°)

3800 N perpendicular to and up from the ground

10. anet = 1.20 m/s2

Fapplied = 1.50 × 103 N

q = −10.0°

ms = 0.650

g = 9.81 m/s2

Fx,net = Fapplied(cos q) − Fs,max = 0Fs,max = msFn

Fn = Fapplie

md(

s

cos q) = = 2.27 × 103 N

Fn =

Fy,net = m anet = Fn − mg + Fapplied(sin q)

m(anet + g) = Fn + Fapplied(sin q)

m = Fn + F

aa

n

p

e

p

t

li

+ed(

g

sin q) =

m = = 2

1

.0

1

1

.0

×1

1

m

0

/

3

s

N2 = 183 kg

2.27 × 103 N − 2.60 × 102 N

11.01 m/s2

2.27 × 103 N + (1.50 × 103 N)[sin(−10.0°)]

1.20 m/s2 + 9.81 m/s2

2.27 × 103 N, upward

(1.50 × 103 N)[cos(−10.0°)]

0.650


Chapter 5Work and Energy

V

1. d = 3.00 × 102 m

W = 2.13 × 106 J

q = 0°

F = d(c

W

os q) = = 7.10 × 3 N

2.13 × 106 J(3.00 × 102 m)(cos 0°)


Givens Solutions

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2. d = 76.2 m

Wnet = 1.31 × 103 J

q = 0°

Fnet = d(

W

con

set

q ) =

(76

1

.

.

2

31

m

×)(

1

c

0

o

3

s

J

0°) = 17.2 N

3. W = 1800 J

d1 = 1.5 m

d2 = 5.0 m

q = 0°

W = F1 d1 = (cos q) = F2 d2 (cos q)

F1 = d1 (c

W

os q) =

(1.5

1

m

8

)

0

(

0

co

J

s 0°) =

F2 = d2 (c

W

os q) =

(5.0

1

m

8

)

0

(

0

co

J

s 0°) = 3.6 × 102 N

1.2 × 103 N

4. Wnet = 4.27 × 103 J

d = 17 m

q = 0°

Fnet = d(

W

con

set

q) =

(1

4

7

.2

m

7

)

×(c

1

o

0

s

3

0

J

°) = 2.5 × 102 N

5. F = 1.6 N

d = 1.2 m

q = 180°

W = Fd(cos q) = (1.6 N) (1.2 m) (cos 180°) = −1.9 J

6. d = 15.0 m

Fapplied = 35.0 N

q1 = 20.0°

Fk = 24.0 N

q2 = 180°

Wnet = Fapplied d (cos q1) + Fk d (cos q2)

Wnet = (35.0 N) (15.0 m) (cos 20.0°) + (24.0 N) (15.0 m) (cos 180°)

Wnet = 493 J + (−3.60 × 102 J)

Wnet =

Alternatively,

Wnet = Fnet d (cos q ′)

q ′ = 0°

Fnet = Fapplied (cos q1) − Fk

Wnet = [Fapplied (cos q1) − Fk] d (cos q ′)

Wnet = [(35.0 N) (cos 20.0°) − 24.0 N] (15.0 m) (cos 0°)

Wnet = (32.9 N − 24.0 N) (15.0 m) = (8.9 N)(15.0 m) = 130 J

133J

1. m = 7.5 × 107 kg

v = 57 km/h

KE = 12

mv2 = 12

(7.5 × 107 kg) [(57 km/h)(103 m/km)(1h/3600 s)]2

KE = 9.4 × 109 J


V

7. v1 = 88.9 m/s

vf = 0 m/s

∆t = 0.181 s

d = 8.05 m

m = 70.0 kg

q = 180°

W = Fd(cos q)

F = ma

a = ∆∆

v

t =

vf

∆−t

vi

W = m (v

∆f

t

− vi) d (cos q) = (8.05 m)(cos 180°)

W =

W = 2.77 × 105 J

(70.0 kg)(88.9 m/s)(8.05 m)

(0.181 s)

(70.0 kg)(0 m/s − 88.9 m/s)

(0.181 s)

Givens Solutions

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8. F = 715 N

W = 2.72 × 104 J

q = 0°

d = F (c

W

os q) =

(7

2

1

.

5

72

N

×)(

1

co

0

s

4

0

J

°) = 38.0 m

9. Fnet = 7.25 × 10−2 N

Wnet = 4.35 × 10−2 J

q = 0°

d = Fnet

W

(cn

oet

s q) = = 0.600 m

4.35 × 10−2 J(7.25 × 10−2 N)(cos 0°)

10. W = 6210 J

F = 2590 N

q = 0°

d = F (c

W

os q) =

(2590

6

N

21

)

0

(c

J

os 0°) = 2.398 m


2. v = 15.8 km/s KE = 12

mv2 = 12

(0.20 × 10−3 kg)(15.8 × 103 m/s)2

m = 0.20 g KE =

3. v = 35.0 km/h KE = 12

mv2 = 12

(9.00 × 102 kg) [(35.0 km/h)(103 m/km)(1 h/3600 s)]2

m = 9.00 × 102 kg KE =

4. v1 = 220.0 km/h KE1 = 12

m1 v12 = 1

2 (8.84 × 105 kg)[(220.0 km/h)(103 m/km)(1 h/3600 s)]2

m1 = 8.84 × 105 kg KE1 =

v2 = 320.0 km/h KE2 = 12

m2 v22 = 1

2 (4.80 × 105 kg)[(320.0 km/h)(103 m/km)(1h/3600 s)]2

m2 = 4.80 × 105 kg KE2 =

5. KE = 2.78 × 109 J

v = 275 km/h

1.90 × 109 J

1.65 × 109 J

4.25 × 104 J

2.5 × 104 J

m = 2

v

K2E

=

m = 9.53 × 105 kg

(2)(2.78 × 109 J)[(275 km/h)(103 m/km)(1h/3600 s)]2


V

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6. v = 850 km/h

KE = 9.76 × 109 Jm =

2

v

K2E

=

m = 3.50 × 105 kg

(2)(9.76 × 109 J)[(850 km/h)(103 m/km)(1 h/3600 s)]2

7. v = 9.78 m/s

KE = 6.08 × 104 Jm =

2

v

K2E

= (2)

(

(

9

6

.

.

7

0

8

8

m

×/

1

s

0

)2

4 J) = 1.27 × 103 kg

8. KE = 7.81 × 104 J

m = 55.0 kgv =

2

m

K

E

= (2)(7

5

.8

5

1

.0

×kg104

J) = 53.3 m/s

10. KEA,i = 12

KEB

vA,f = vA,i + 1.3 m/s

KEA,f = KEB

mA = 2.0 mB

1. vi = 8.0 m/s

vf = 0 m/s

d = 45 m

Fk = 0.12 N

q = 180°


mvf2 − 1

2 mvi

2

Wnet = Fnetd(cos q) = Fkd(cos q)

12

m(vf2 − vi

2) = Fkd(cos q)

m = 2 F

vk

f2d(

−co

vi

s2q)

= = −(2)(

−0

6

.1

4

2

m

N2)

/

(

s

425 m)

m = 0.17 kg

(2)(0.12 N)(45 m)(cos 180°)

(0 m/s)2 − (8.0 m/s)2

9. KE = 1433 J

m = 47.0 gv =

2

m

K

E

= 47(

.

20

)(

×1 4

13

0

3−3

J)

kg = 247 m/s

KEA,i = 12

KEB

12

mAvA,i2 = 1

21

2 mBvB

212

(2mB)vA,i2 = 1

4 mBvB

2

vA,i2 = 1

4 vB

2, or vB

2 = 4vA,i2

vA,i = 12

vB

KEA,f = KEB

12

mAvA,f2 = 1

2 mBvB

2

12

(2mB)(vA,i + 1.3 m/s)2 = 12

mBvB2

(vA,i + 1.3 m/s)2 = 12

vB2 = 1

2 (4vA,i

2) = 2vA,i2

vA,i2 + (2.6 m/s) vA,i + 1.7 m2/s2 = 2vA,i

2

vA,i2 − (2.6 m/s) vA,i − 1.7 m2/s2 = 0

Using the quadratic equation,

vA,i = =

vA,i = 2.6 m/s ±√

2

13.6 m2/s2 = 2.6 m/s ±

2

3.69 m/s =

6.3

2

m/s =

vA,i = 12

vB

vB = vA,i = (2)(3.2 m/s) = 6.4 m/s

3.2 m/s

2.6 m/s ±√

6.8m2/s2 + 6.8 m2/s22

2.6 m/s ±√

(−2.6m/s)2 − 4(−1.7m2/s2)2



V

2. vi = 15.00 km/s

vf = 14.97 km/s

Fr = 9.00 × 10–2 N

d = 500.0 km

q = 180°

Wnet = ∆KE =KEf − KEi = 12

mvf2 − 1

2 mvi

2

Wnet = Fnetd(cos q) = Fr d (cos q)

12

m(vf2 − vi

2) = Fr d (cos q)

m = 2 F

vr

f

d2 −

(c

v

o

i

s2

q) =

m = = −−9

9

×.0

1

0

0

×8

1

m

02

4

/s

J2

m = 1.00 × 10−4 kg

−(2)(9.00 × 10−2 N)(500.0 × 103 m)2.241 × 108 m2/s2 − 2.250 × 108 m2/s2

(2)(9.00 × 10−2 N)(500.0 × 103 m)(cos 180°)

(14.97 × 103 m/s)2 − (15.00 × 103 m/s)2

Givens Solutions

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3. vi = 48.0 km/h

vf = 59.0 km/h

d = 100.0 m

m = 1100 kg

q = 0°


mvf2 − 1

2 mvi

2

Wnet = Fnet d (cos q)

Fnet d (cos q) = 12

m(vf2 − vi

2)

Fnet = m

2

(

d

vf

(

2

co

−s

v

qi2

)

) =

Fnet =

Fnet =

Fnet = 5.01 × 102 N

(1100 kg)(1.18 × 109 m2/h2)(1h/3600 s)2

(2)(100.0 m)

(1100 kg)(3.48 × 109 m2/h2 − 2.30 × 109 m2/h2)(1h/3600 s)2

(2)(100.0 m)

(1100 kg)[(59.0 km/h)2 − (48.0 km/h)2](103 m/km)2 (1h/3600 s)2

(2)(100.0 m)(cos 0°)

4. m = 450 kg

d = 7.0 m

vf = 1.1 m/s

vi = 0 m/s

q = 0°


mvf2 − vi

2

Wnet = Fnetd(cos q)

Fnetd(cos q) = 12

m(vf2 − vi

2)

Fnet = m

2

(

d

vf

(

2

co

−s

v

qi2

)

) = =

(450

(

k

2

g

)

)

(

(

7

1

.0

.2

m

m

)

2/s2)

Fnet = 39 N

(450 kg)[(1.1 m/s)2 − (0 m/s)2]

(2)(7.0 m)(cos 0°)

5. vi = 2.40 × 102 km/h

vf = 0 km/h

anet = 30.8 m/s2

m = 1.30 × 104 kg

q = 180°


mvf2 − vi

2

Wnet = Fnet d (cos q) = manet d(cos q)

manetd(cos q) = 12

m(vf2 − vi

2)

d = 2 a

v

n

f

e

2

t (

−c

v

oi

s

2

q) =

d = = 72.2 m(−5.76 × 104 km2/h2)(103 m/km)2 (1h/3600 s)2

−(2)(30.8 m/s2)

[(0 km/h)2 − (2.40 × 102 km/h)2] (103 m/km)2(1h/3600 s)2

(2)(30.8 m/s2)(cos 180°)


V

6. m = 50.0 kg

vi = 47.00 m/s

vf = 5.00 m/s


mvf2 − 1

2 mvi

2

Wnet = 12

m(vf2 − vi

2) = 12

(50.0 kg)[(5.00 m/s)2 − (47.00 m/s)2]

Wnet = 12

(50.0 kg)(25.0 m2/s2 − 2209 m2/s2) = 12

(50.0 kg)(−2184 m2/s2)

Wnet = −5.46 × 104 J

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7. m = 2.0 × 106 kg

d = 7.5 m

anet = 7.5 × 10−2 m/s2

KEi = 0 J

Wnet = ∆KE = KEf − KEi = KEf

Wnet = Fnet d (cos q) = manetd(cos q)

KEf = manet d (cos q) = (2.0 × 106 kg)(7.5 × 10−2 m/s2)(7.5 m)(cos 0°)

KEf = 1.1 × 106 J

8. Fapplied = 92 N

m = 18 kg

mk = 0.35

d = 7.6 m

g = 9.81 m/s2

q = 0°

KEi = 0 J

Wnet = ∆KE = KEf − KEi = KEf

Wnet = Fnet d (cos q)

Fnet = Fapplied − Fk = Fapplied − mk mg

KEf = (Fapplied − mk mg) d (cos q)

= [92 N − (0.35)(18 kg)(9.81 m/s2)](7.6 m)(cos 0°)

KEf = (92 N − 62 N)(7.6 m) = (3.0 × 101 N)(7.6 m)

KEf = 228 J

9. m = 2.00 × 102 kg

Fwind = 4.00 × 102 N

d = 0.90 km

vi = 0 m/s

q = 0°


mvf2 − 1

2 mvi

2

Wnet = Fnet d (cos q) = Fwind d(cos q)

12

mvf2 − 1

2 mvi

2 = Fwind d (cos q)

vf = 2Fwindm

d (cosq)+ vi

2 = + (0m/s)2vf = vf = 6.0 × 101 m/s

(2)(4.00 × 102 N)(9.0 × 102 m)

2.00 × 102 kg

(2)(4.00 × 102 N)(0.90 × 103 m)(cos 0°)

2.00 × 102 kg

10. m = 20.0 g

d = 2.5 m

Fforward = 7.3 × 10−2 N

mk = 0.20

vi = 0 m/s

g = 9.81 m/s2

q = 0°


mvf2 − 1

2 mvi

2

Wnet = Fnetd(cos q) = (Fforward − Fk) d(cos q) = (Fforward − mk mg)d(cos q)

12

mvf2 − 1

2 mvi

2 = (Fforward − mk mg)d(cos q)

vf = vf = vf = = vf = 2.9 m/s

(2)(3.4 × 10−2 N)(2.5 m)

2.00 × 10−2 kg

(2)(7.3 × 10−2 N − 3.9 × 10−2 N)(2.5 m)

20.0 × 10−3 kg

(2)[7.3 × 10−2 N − (0.20)(20.0 × 10−3 kg)(9.81 m/s2)](2.5 m)(cos 0°) + (0 m/s)2

20.0 × 10−3 kg

2[(Fforward − mk mg) d (cos q) + vi2]

m


V

1. x = 5.00 cm

KEcar = 1.09 × 10−4 J

Assuming all of the kinetic energy becomes stored elastic potential energy,

KEcar = PEelastic = 12

kx2

k = 2 PE

xe2lastic =

(

(

5

2

.

)

0

(

0

1.

×09

10

×−120

m

4

)

J2)

k = 8.72 × 106 N/m


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2. PEelastic = 5.78 × 107 J

x = 102 m for each spring

PEelastic = 2 12

kx2PEelastic = kx2

k = PE

xel

2astic =

5

(

.7

1

8

02

×m

10

)

7

2J

k = 5.56 × 103 N/m

3. m = 0.76 kg

x = 2.3 cm

g = 9.81 m/s2

elastic potential energy stored = decrease in gravitational potential energy

PEelastic = mgx

PEelastic = 12

kx2 = mgx

k = 2 m

x2gx

= 2 m

x

g =

k = 6.5 × 102 N/m

(2)(0.76 kg)(9.81 m/s2)

2.3 × 10−2 m

4. m = 5.0 kg

q = 25.0°

PEg = 2.4 × 102 J

PEg = mgh = mgd(sin q)

d = mg

P

(

E

si

g

n q) =

d = 12 m

2.4 × 102 J(5.0 kg)(9.81 m/s2)(sin 25.0°)

5. k = 1.5 × 104 N/m

PEelastic = 120 J

PEelastic = 12

kx2

x = ± 2PE

kelastic = ± 1.5

(2×) (

1

10

240

NJ)

/m

Spring is compressed, so negative root is selected.

x = −0.13 m = −13 cm

6. m = 1750 kg

PEg = 1.69 × 1010 J

g = 6.44 m/s2

PEg = mgh

h = P

m

E

g

g =

h = 1.50 × 106 m = 1.50 × 103 km

1.69 × 1010 J(1750 kg)(6.44 m/s2)

7. h = 7.0 m

PEg = 6.6 × 104 J

g = 9.81 m/s2

PEg = mgh

m = P

g

E

hg =

(9.81

6.

m

6

/

×s2

1

)

0

(

4

7.

J

0 m)

m = 9.6 × 102 kg


V

8. PEg = 3.36 × 109 J

h = 1.45 km

PEg = mgh

m = P

g

E

h

g =

m = 2.36 × 105 kg

3.36 × 109 J(9.81 m/s2)(1.45 × 103 m)

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9. k = 550 N/m

x = −1.2 cm

PEelastic = 12

kx2 = 12

(550 N/m)(–1.2 × 10−2 m)2 = 4.0 × 10−2 J

1. m = 0.500 g

h = 0.250 km

g = 9.81 m/s2

PEi = KEf

mgh = KEf

KEf = (0.500 × 10−3 kg)(9.81 m/s2)(0.250 × 103 m) = 1.23 J

2. d = 96.0 m

q = 18.4°

m = 70.0 kg

g = 9.81 m/s2

a. PE1 = mgh = mgd(sin q)

PE2 = 0 J

MEi = PE1 + PE2 = mgd(sin q)

MEi = (70.0 kg)(9.81 m/s2)(96.0 m)(sin 18.4°) =

b. PE1 = 0 J

PE2 = mgh = mgd(sin q)

MEf = PE1 + PE2 = mgd(sin q) = MEi

MEf =

c. MEi = MEf

PE1,i + PE2,i + KE1 + KE2 = PE1,f + PE2,f + KE1 + KE2

The kinetic energy of each passenger remains unchanged during the trip once thecars are in motion, so

PE1,i + PE2,i = PE1,f + PE2,f

mgh1,i + 0 J = mgh1,f + PE2,f

PE2,f = mgh1,f − mgh1,f = mgd(sin q) − mgh1,f

PE2,f = (70.0 kg)(9.81 m/s2)(96 m)(sin 18.4°) − (70.0 kg)(9.81 m/s2)(20.0 m)

PE2,f = 2.1 × 104 J − 1.37 × 104 J

PE2,f = 7.0 × 103 J

2.08 × 104 J

2.08 × 104 J

10. h = 5334 m

m = 64.0 kg

g = 9.81 m/s2

PEg = mgh = (64.0 kg)(9.81 m/s2)(5334 m) = 3.35 × 106 J


v1 = v2 = 1.0 m/s

h1,f = 20.0 m


V

3. hi = 75.0 m

vi = 1.2 m/s + 3.5 m/s = 4.7 m/s

vf = 0 m/s

g = 9.81 m/s2

m = 20.0 g

PEi + KEi = PEf + KEf

mghi + 12

mvi2 = mghf + 1

2mvf

2

hf =12

m(vi2 − vf

2)+ hi = + himg

hf = + 75.0 m = 1.1 m + 75.0 m = 76.1 m(4.7 m/s)2 − (0 m/s)2

(2)(9.81 m/s2)

vi2 − vf

2

2g

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4. m = 25.0 kg

v = 12.5 m/s

g = 9.81 m/s2

PEi = KEf

mgh = 12

mv2

h = 2

v2

g =

(2

(

)

1

(

2

9

.

.

5

81

m

m

/s

/

)

s

2

2) = 7.96 m

5. m = 50.0 g

vi = 3.00 × 102 m/s

vf = 89.0 m/s

MEi + ∆ME = MEf

MEi = KEi = 12

mvi2

MEf = KEf = 12

mvf2

∆ME = MEf − MEi = 12

m(vf2 − vi

2)

∆ME = 12

(50.0 × 10−3 kg)[(89.0 m/s)2 − (3.00 × 102 m/s)2]

∆ME = 12

(5.00 × 10−2 kg)(7.92 × 103 m2/s2 − 9.00 × 104 m2/s2)

∆ME = 12

(5.00 × 10−2 kg) (−8.21 × 104 m2/s2)

∆ME = −2.05 × 103 J

6. m = 50.0 g

vi = 3.00 × 102 m/s

vf = 89.0 m/s

For upward flight,

PE1,f − KE1,i = ∆ME1

where

∆ME1 = Wnet,1 = Fnet,1 h (cos 180°) = (mg + Fresist) h

For downward flight,

KE2,f − PE2,i = ∆ME2

Where

∆ME2 = Wnet,2 = Fnet,2 h (cos 0°) = (mg − Fresist) h

Solving for h,

∆ME2 − ∆ME1 = (mg − Fresist) h − [−(mg + Fresist)h] = 2 mgh

h = ∆ME

22 −

m

∆g

ME1 =

KE2,f = 12

mvf2

KE1,i = 12

mvi2

PE1,f = PE2,i = mgh

h = = vf

2

4

+g

vi2

− h

h = vf

2

8

+g

vi2

=(89.0 m/s)2 + (3.00 × 102 m/s)2

(8)(9.81 m/s2)

12

mvf2 − mgh − mgh + 1

2 mvi

2

2 mg

(KE2,f − PE2,i) − (PE1,f − KE1,i)2 mg


V

h = = 9

(

.7

8

9

)(

×9.

1

8

0

1

3

m

m

/s

2

2/

)

s2

h = 1.25 × 103 m = 1.25 km

7.92 × 103 m2/s2 + 9.00 × 104 m2/s2

(8)(9.81 m/s2)

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7. m = 50.0 kg

k = 3.4 × 104 N/m

x = 0.65 m

hf = 1.00 m − 0.65 m = 0.35 m

PEg,i = PEelastic,f + PEg,f

mghi = 12

kx2 + mghf

hi = hf + 2

kx

m

2

g = 0.35 m + = 0.35 m + 15 m

hi = 15 m

(3.4 × 104 N/m)(0.65 m)2

(2)(50.0 kg)(9.81 m/s2)

8. h = 3.0 m

g = 9.81 m/s2

PEi = KEf

mgh = 12

mvf2

vf =√

2gh =√

(2)(9.81 m/s2)(3.0 m)

vf = 7.7 m/s

10. mw = 546 kg

h = 5.64 m

g = 9.81 m/s2

mflyer = 273 kg

PEw = KEflyer

mw gh = 12

mflyer vflyer2

vflyer = 2

m

mf

w

ly

e

g

rh =

vflyer = 14.9 m/s

(2)(546 kg)(9.81 m/s2)(5.64 m)

273 kg

9. m = 100.0 g

x = 30.0 cm

k = 1250 N/m

PEelastic = KE

12

kx2 = 12

mv2

v = k

mx2

= v = 33.5 m/s

(1250 N/m)(30.0 × 10−2 m)2

100.0 × 10−3 kg


1. P = 380.3 kW

W = 4.5 × 106 J∆t =

W

P =

38

4

0

.

.

5

3

××

1

1

0

0

6

3J

W = 12 s

2. P = 331 W

h = 442 m

m = 55 kg

g = 9.81 m/s2

W = mgh

∆t = W

P =

m

P

gh =

(55 kg

3

)(

3

9

1

.8

W

1 m/s2) = 720 s = 12 min

3. F = 334 N

d = 50.0 m

q = 0°

P = 2100 W

W = Fd(cos q)

∆t = W

P =

Fd(c

P

os q) = = 7.95 s

(334 N)(50.0 m)(cos 0°)

2100 W


V

4. P = 13.0 MW

∆t = 15.0 minW = P ∆t = (13.0 × 106 W)(15.0 min)(60 s/min) = 1.17 × 1010 J

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5. P = 1 hp = 745.7 W

∆t = 0.55 sW = P ∆t = (745.7 W)(0.55 s) = 4.1 × 102 J

6. P = (4)(300.0 kW)

∆t = 25 sW = P ∆t = (4)(300.0 × 103 W)(25 s) = 3.0 × 107 J

7. ∆t = 39 s

P = 158 kWW = P ∆t = (158 × 103 W)(39 s) = 6.2 × 106 J

8. W = 1.4 × 1013 J

∆t = 8.5 minP =

∆W

t = = 2.7 × 1010 W = 27 GW

1.4 × 1013 J(8.5 min)(60 s/min)

9. W = 2.82 × 107 J

∆t = 30.0 minP =

∆W

t = =

P = (1.57 × 104 W)(1 hp/745.7 W) = 21.1 hp

1.57 × 104 W = 15.7 kW2.82 × 107 J

(30.0 min)(60 s/min)

10. W = 3.0 × 106 J

∆t = 5.0 minP =

∆W

t = = 1.0 × 104 W

3.0 × 106 J(5.0 min)(60 s/min)


Chapter 6Momentum and Collisions

V

1. m = 1.46 × 105 kg

p = 9.73 × 105 kg•m/s to thesouth

v = m

p =

v = 6.66 m/s to the south

9.73 × 105 kg•m/s

1.46 × 105 kg


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2. m = 25 kg

p = 6.8 × 104 kg•m/sv =

m

p = = 2.7 × 103 m/s

v = (2.7 km/s)(3600 s/h) = 9.7 × 103 km/h

v = 2.7 × 103 m/s = 9.7 × 103 km/h

6.8 × 104 kg•m/s

25 kg

3. m = 5.00 × 102 kg

p = 8.22 × 103 kg•m/s to thewest

v = m

p =

v = 16.4 m/s to the west

8.22 × 103 kg•m/s

5.00 × 102 kg

4. mc = 177.4 kg

md = 61.5 kg

p = 4.416 × 103 kg•m/s

v = m

p =

mc +p

md = =

v = 18.48 m/s = (18.48 m/s)(3600 s/h)(1 km/103 m) = 66.5 km/h

v = 18.48 m/s = 66.53 km/h)

4.416 × 103 kg•m/s

238.9 kg

4.416 × 103 kg•m/s177.4 kg + 61.5 kg

5. ∆x = 200.0 m

∆t = 19.32 s

m = 77 kg

vavg = ∆∆

x

t =

2

1

0

9

0

.3

.0

2

m

s = 10.35 m/s

p = mv

pavg = mvavg = (77 kg)(10.35 m/s) = 7.8 × 102 kg•m/s

6. ∆x = 274 m to the north

∆t = 8.65 s

m = 50.0 kg

vavg = ∆∆

x

t =

2

8

7

.6

4

5

m

s = 31.7 m/s to the north

p = mv

pavg = mvavg = (50.0 kg)(31.7 m/s) = 1.58 × 103 kg•m/s to the north

7. m = 7.10 × 105 kg

v = 270 km/h

p = mv = (7.10 × 105 kg)(270 km/h)(103 m/km)(1 h/3600 s)

p = 5.33 × 107 kg•m/s

1. F = 10.0 N to the right

m = 3.0 kg

∆t = 5.0 s

vi = 0 m/s

∆p = mvf − mvi = F∆t

vf = F∆t

m

+ mvi = = 17 m/s

vf = 17 m/s to the right

(10.0 N)(5.05) + (3.0 kg)(0 m/s)

3.0 kg


V

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8. v = 50.0 km/h

p = 0.278 kg•m/sm =

p

v =

m = 2.00 × 10−2 kg = 20.0 g

0.278 kg•m/s(50.0 km/h)(103 m/km)(1 h/3600 s)

9. vavg = 96 km/h to the southeast

pavg = 4.8 × 104 kg•m/s tothe southeast

m = p

v =

p

va

a

v

v

g

g =

m = 1.8 × 103 kg

4.8 × 104 kg•m/s(96 km/h)(103 m/km)(1 h/3600 s)

10. v = 255 km/s

p = 8.62 × 1036 kg•m/sm =

p

v = = 3.38 × 1031 kg

8.62 × 1036 kg•m/s

255 × 103 m/s


2. m = 60.0 g

F = −1.5 N

∆t = 0.25 s

vf = 0 m/s


vi = mvf

m

− F∆t = =

vi = 6.2 m/s

(1.5 N)(0.25 s)60.0 × 10−3 kg

(60.0 × 10−3 kg)(0 m/s) − (−1.5 N)(0.25 s)

60.0 × 10−3 kg

3. F = 75 N

m = 55 kg

∆t = 7.5 s

vi = 0 m/s


vf = F∆t

m

+ mvi =

vf = 1.0 × 101 m/s

(75 N)(7.5 s) + (55 kg)(0 m/s)

55 kg

4. m = 0.195 kg

vi = 0.850 m/s to the right = +0.850 m/s

F = 3.50 N to the left = −3.50 N

∆t = 0.0750 s


vf = F∆t

m

+ mvi =

vf = = = −0.49 m/s

vf = 0.49 m/s to the left

−0.096 kg•m/s

0.195 kg

−0.262 kg•m/s + 0.166 kg•m/s

0.195 kg

(−3.50 N)(0.0750 s) + (0.195 kg)(0.850 m/s)

0.195 kg

5. m = 5.00 g

vi = 255 m/s to the right

vf = 0 m/s

∆t = 1.45 s


F = mvf

∆−t

mvi = = −0.879 N

F = 0.879 N to the left

(5.00 × 10−3 kg)(0 m/s) − (5.00 × 10−3 kg)(255 m/s)

1.45 s


V

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6. m = 1.1 × 103 kg

vf = 9.7 m/s to the east

vi = 0 m/s

∆t = 19 s

F = ∆∆

p

t =

mvf

∆−t

mvi

F = = 560 N

F = 560 N to the east

(1.1 × 103 kg)(9.7 m/s) − (1.1 × 103 kg)(0 m/s)

19 s

7. m = 3.00 × 103 kg

vi = 0 m/s

vf = 8.9 m/s to the right

∆t = 5.5 s

F = ∆∆

p

t =

mvf

∆−t

mvi

F = =

F = 4.9 × 103 N to the right

(3.00 × 103 kg)(8.9 m/s)

5.5 s

(3.00 × 103 kg)(8.9 m/s) − (3.00 × 103 kg)(0 m/s)

5.5 s

8. m = 0.17 kg

∆v = −9.0 m/s

g = 9.81 m/s2

mk = 0.050

10. vf = 15.8 km/s

vi = 0 km/s

F = 12.0 N

m = 0.20 g

∆t = ∆F

p =

mvf −F

mvi

∆t =

∆t =

∆t = 0.26 s

(0.20 × 10−3 kg)(15.8 × 103 m/s)

12.0 N

(0.20 × 10−3 kg)(15.8 × 103 m/s) − (0.20 × 10−3 kg)(0 m/s)

12.0 N

1. vi = 382 km/h to the right

vf = 0 km/h

mc = 705 kg

md = 65 kg

∆t = 12.0 s

F = ∆∆

p

t =

F =

F = = − 6.81 × 103 N

F =

∆x = 12

(vi + vf)∆t = 12

(382 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(12.0 s)

∆x = 637 m to the right

6.81 × 103 N to the left

−(7.70 × 102 kg)(382 km/h)(103 m/km)(1 h/3600 s)

12.0 s

[(705 kg + 65 kg)(0 km/h) − (705 kg + 65 kg)(382 km/h)](103 m//km)(1 h/3600 s)

12.0 s

(mc + md)vf − (mc + md)vi∆t

F∆t = ∆p = m∆v

F = Fk = −mgmk

∆t = −m

m

∆gm

v

k =

−∆gm

v

k =

−(9.81

−9

m

.0

/s

m2)

/

(

s

0.050)

∆t = 18 s

9. m = 12.0 kg

Fapplied = 15.0 N

q = 20.0°

Ffriction = 11.0 N

vi = 0 m/s

vf = 4.50 m/s

F = Fapplied(cos q) − Ffriction

∆t = ∆F

p = =

∆t = =

∆t = 1.7 s

5.40 kg•m/s

3.1 N

5.40 kg•m/s − 0 kg•m/s

14.1 N − 11.0 N

(12.0 kg)(4.50 m/s) − (12.0 kg)(0 m/s)

(15.0 N)(cos 20.0°) − 11.0 N

mvf − mviFapplied(cos q) − Ffriction



V

2. vi = 7.82 × 103 m/s

vf = 0 m/s

m = 42 g

∆t = 1.0 × 10−6 s

F = = mvf

∆−t

mvi

F =

F =

F =

∆x = 12

(vi + vf)∆t = 12

(7.82 × 103 m/s + 0 m/s)(1.0 × 10−6 s)

∆x = 3.9 × 10−3 m = 3.9 mm

−3.3 × 108 N

−(42 × 10−3 kg)(7.82 × 103 m/s)

1.0 × 10−6 s

(42 × 10−3 kg)(0 m/s) − (42 × 10−3 kg)(7.82 × 103 m/s)

1.0 × 10−6 s

∆p∆t

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3. m = 63 kg

vi = 7.0 m/s to the right

vf = 0 m/s

∆t = 14.0 s

F = ∆∆

p

t =

mvf

∆−t

mvi

F = = = −32 N

F =

∆x = 12

(vi + vf) ∆t = 12

(7.0 m/s + 0 m/s)(14.0 s)


32 N to the left

−(63 kg)(7.0 m/s)

14.0 s

(63 kg)(0 m/s) − (63 kg)(7.0 m/s)

14.0 s

4. m = 455 kg

∆t = 12.2 s

mk = 0.071

g = 9.81 m/s2

vf = 0 m/s

∆p = F∆t

F = Fk = −mgmk

∆p = −mgmk∆t = −(455 kg)(9.81 m/s2)(0.071)(12.2 s) = −3.9 × 103 kg•ms

∆p =

vi = mvf

m

− ∆p =

vi = = 8.6 m/s

∆x = 12

(vi + vf)∆t = 12

(8.6 m/s + 0 m/s)(12.2 s)

∆x = 52 m

3.9 × 103 kg•ms

455 kg

(455 kg)(0 m/s) − (−3.9 × 103 kg•ms)

455 kg

3.9 × 103 kg•ms opposite the polar bear’s motion

5. m = 75.0 g

∆t = 1.2 s

g = 9.81 m/s2

vf = 0 m/s

∆p = F∆t

F = −mg

∆p = −mg∆t = −(75.0 × 10−3 kg)(9.81 m/s2)(1.2 s)

∆p = −0.88 kg•m/s

∆p =

vi = mvf

m

− ∆p =

vi = 7

0

5

.

.

8

0

8

×k

1

g

0

•−m

3/

k

s

g = 12 m/s upward

∆x = 12

(vi + vf)∆t = 12

(12 m/s + 0 m/s)(1.2 s)

∆x = 7.2 m upward

(75.0 × 10−3 kg)(0 m/s) − (−0.88 kg•m/s)

75.0 × 10−3 kg

0.88 kg•m/s downward


V

6. m = 4400 kg

F = 2200 N to the left = −2200 N

∆t = 8.0 s

vi = 6.5 m/s to the right = +6.5 m/s

∆p = F∆t = (−2200 N)(8.0 s) = −1.8 × 104 kg•m/s

∆p =

vf = ∆p +

m

mvi =

vf = = = 2.5 m/s to the right

∆x = 12

(vi + vf)∆t = 12

(6.5 m/s + 2.5 m/s)(8.0 s) = 12

(9.0 m/s)(8.0 s)


1.1 × 104 kg•m/s

4400 kg

−1.8 × 104 kg•m/s + 2.9 × 104 kg•m/s

4400 kg

−1.8 × 104 kg•m/s + (4400 kg)(6.5 m/s)

4400 kg

1.8 × 104 kg•m/s to the left

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7. F = 25.0 N

∆t = 7.00 s

m = 14.0 kg

vi = 0 m/s

∆p = F∆t = (25.0 N)(7.00 s) =

vf = ∆p +

m

mvi = = 12.5 m/s

∆x = 12

(vi + vf)∆t = 12

(0 m/s + 12.5 m/s)(7.00 s)

∆x = 43.8 m

175 kg•m/s + (14.0 kg)(0 m/s)

14.0 kg

175 kg•m/s

8. m = 2.30 × 103 kg

vi = 22.2 m/s

vf = 0 m/s

F = −1.26 × 104 N

∆t = ∆F

p =

mvf −F

mvi

∆t = =

∆t =

∆x = 12

(vi + vf)∆t = 12

(22.2 m/s + 0 m/s)(4.06 s)

∆x = 45.1 m

4.06 s

−5.11 × 104 kg•m/s

−1.26 × 104 N

(2.30 × 103 kg)(0 m/s) − (2.30 × 103 kg)(22.2 m/s)

−1.26 × 104 N

9. m = 1.35 × 104 kg

vi = 66.1 m/s to the west = −66.1 m/s

vf = 0 m/s

F = 4.00 × 105 N to the east= +4.00 × 105 N

∆t = ∆F

p =

mvf −F

mvi

∆t = =

∆t =

∆x = 12

(vi + vf)∆t = 12

(−66.1 m/s + 0 m/s)(2.23 s) = −73.7 m

∆x = 73.7 m to the west

2.23 s

8.92 × 105 kg•m/s

4.00 × 105 N

(1.35 × 104 kg)(0 m/s) − (1.35 × 104 kg)(−66.1 m/s)

4.00 × 105 N

10. vi = 14.5 m/s

vf = 0 m/s

m = 1.50 × 103 kg

mk = 0.065

a = −1.305 m/s2

∆t = ∆F

p =

mvf −F

mvi

F = Fk = mamk

∆t = v

a

f

m−

k

vi

∆t =

∆t =

∆x = 12

(vi + vf)∆t = 12

(14.5 m/s + 0 m/s)(1.7 × 102 s)

∆x = 1.2 × 103 m = 1.2 km

1.7 × 102 s

0 m/s − 14.5 m/s(− 1.305 m/s2)(0.065)


V

1. m1 = 68 kg

m2 = 68 kg

v2, i = 0 m/s

v1, f = 0.85 m/s to the west = −0.85 m/s

v2, f = 0.85 m/s to the west = −0.85 m/s

m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, i = =

v1, i = −0.85 m/s + (−0.85 m/s) = −1.7 m/s

v1, i = 1.7 m/s to the west

(68 kg)(−0.85 m/s) + (68 kg)(−0.85 m/s) − (68 kg)(0 m/s)

68 kg

m1v1, f + m2v2, f − m2v2, imi


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2. mi = 1.36 × 104 kg

m2 = 8.4 × 103 kg

v2, i = 0 m/s

v1, f = v2, f = 1.3 m/s

m1v1, i + m2v2, i = m1 v1, f + m2v2, f

v1, i =

v1, i =

v1, i = =

v1, i = 2.1 m/s

2.9 × 104 kg•m/s

1.36 × 104 kg

1.8 × 104 kg•m/s + 1.1 × 104 kg•m/s

1.36 × 104 kg

(1.36 × 104 kg)(1.3 m/s) + (8.4 × 103 kg)(1.3 m/s) − (8.4 × 103 kg)(0 m/s)

1.36 × 104 kg

m1v1, f + m2v2, f − m2v2, im1

3. v1, f = 2.2 m/s backwards= −2.2 m/s

v2, f = 5.5 m/s forward= +5.5 m/s

m1 = 38 kg

m2 = 68 kg

m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, i = v2, i, so

v1, i = m1v

m1,

1

f ++

m

m2

2

v2, f =

v1, i = = = 2.7 m/s

v1, i = 2.7 m/s forward

290 kg•m/s

106 kg

−84 kg•m/s + 370 kg•m/s

106 kg

(38 kg)(−2.2 m/s) + (68 kg)(5.5 m/s)

38 kg + 68 kg

4. m1 = 38 kg

v1, i = 1.6 m/s to the north

m2 = 142 kg

v1, f = 0.32 m/s to the north


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v2, i =

v2, i =

v2, i = = = −2.8 × 10−2 m/s

v2, i = 2.8 × 10−2 m/s to the south

−4.0 kg•m/s

142 kg

12 kg•m/s + 45 kg•m/s − 61 kg•m/s

142 kg

(38 kg)(0.32 m/s) + (142 kg)(0.32 m/s) − (38 kg)(1.6 m/s)

142 kg

m1v1, f + m2v2, f − m1v1, im2

5. m1 = 50.0 g

v1, i = 0 m/s

v1, f = 400.0 m/s forward

m2 = 3.00 kg

v2, i = 0 m/s

Because the initial velocities for both rifle and projectile are zero, the momentum con-servation equation takes the following form:

m1v1, f + m2v2, f = 0

v2, f = = = −6.67 m/s

v2, f = 6.67 m/s backward

−(50.0 × 10−3 kg)(400.0 m/s)

3.00 kg

−m1v1, fm2


V

6. mi = 1292 kg

vi = 88.0 km/h to the east

mf = 1255 kg

mivi = mfvf

vf = m

miv

f

i =

vf = 90.6 km/h to the east

(1292 kg)(88.0 km/h)

1255 kg

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7. m = 5.0 × 1014 kg

vi = 74.0 km forward

mi = m2 = 12

m

v1, f = 105 km/s at 15.0°above forward

v2, f is at an angle of −30.0°to the forward direction

mvi = m1v1, f + m2v2, f

To solve for velocity in two dimensions, the momentum conservation equation mustbe written as two equations, one for both the x and y directions.

In the x-direction:

mvi = m1v1, f (cos q1) + m2v2, f (cos q2)

vi = 12

v1, f (cos q1) + 12

v2, f (cos q2)

v2, f =

v2, f =

v2, f = =

v2, f = 54 km/s

In the y-direction (check):

0 = m1v1, f (sin q1) + m2v2, f (sin q2)

v1, f (sin q1) = −v2, f (sin q2)

(105 km/s)(sin 15.0°) = −(54 km/s)[sin(−30.0°)]27.2 km/s = 27 km/s

The slight difference arises from differences in the number of significant figures andfrom rounding.

v2,f = 54 km/s at 30.0° below the initial forward direction

47 km/scos(−30.0°)

148 km/s − 101 km/s

cos(−30.0°)

(2)(74.0 km/s) − (105 km/s)(cos 15.0°)

cos(−30.0°)

2vi − v1, f (cos q1)

cos q2

8. v1, i = 0 cm/s

v1, f = 1.2 cm/s forward = +1.2 cm/s

v2, i = 0 cm/s

v2, f = 0.40 cm/s backward = −0.40 cm/s

m1 = 2.5 g

m1v1, f + m2v2, f = 0

m2 = =

m2 = 7.5 g

−(2.5 g)(1.2 cm/s)

−0.40 cm/s

−m1v1, fv2, f


V

9. vi = 0 cm/s

m1 = 25.0 g

m2 = 25.0 g

v1 = 7.0 cm/s to the south= 7.0 cm/s at −90° fromeast

v2 = 7.0 cm/s to the west

= 7.0 cm/s at 180° from east

v3 = 3.3 m/s at 45° north ofeast

m1v1, f + m2v2, f + m3v3, f = 0

In the x-direction:

m1v1, f (cos q1) + m2v2, f (cos q2) + m3v3, f (cos q3) = 0

m3 =

m3 =

m3 = =

In the y-direction (check):

m1v1, f (sin q1) + m2v2, f (sin q2) + m3v3, f (sin q3) = 0

(25.0 g)(7.0 cm/s)[sin(−90°)] + (25.0 g)(7.0 cm/s)(sin 180°) + (75 g)(3.3 cm/s)(sin 45°) = 0

−180 g•cm/s + 0 g•cm/s + 180•g•cm/s = 0

75 g(25.0 g)(7.0 cm/s)(3.3 cm/s)(cos 45°)

− (25.0 g)(7.0 cm/s)(cos −90°) + (25.0 g)(7.0 cm/s)(cos 180°)

−(3.3 cm/s)(cos 45°)

m1v1, f (cos q1) + m2v2, f (cos q2)

−v3, f (cos q3)

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10. v1, i = 0 m/s




m1 = 63 kg

m1v1, i + m2v2, i = m1v1, f + m2v2, f

m2 = = =

m2 = 24 kg

(63 kg)(1.5 m/s)

3.9 m/s

(63 kg)(1.5 m/s) − (63 kg)(0 m/s)

5.4 m/s − 1.5 m/s

m1v1, f − m1v1, iv2, i − v2, f


1. m1 = 1550 kg

m2 = 770 kg

v2, i = 0 m/s

vf = 9.44 m/s forward

v1, i =

v1, i = =

v1, i = 14.0 m/s forward

(2320 kg)(9.44 m/s)

1550 kg

(1550 kg + 770 kg)(9.44 m/s) − (770 kg)(0 m/s)

1550 kg

(m1 + m2)vf − m2v2, im1

2. m1 = 0.17 kg

m2 = 0.75 kg

v2, i = 0.50 m/s to the left = −0.50 m/s

vf = 4.2 m/s to the right = +4.2 m/s

v1, i =

v1, i =

v1, i =

v1, i = =

v1, i = 25 m/s to the right

4.3 kg•m/s

0.17 kg

3.9 kg•m/s + 0.38 kg•m/s

0.17 kg

(0.92 kg)(4.2 m/s) + 0.38 kg•m/s

0.17 kg

(0.17 kg + 0.75 kg)(4.2 m/s) − (0.75 kg)(−0.50 m/s)

0.17 kg

(m1 + m2)vf − m2v2, im1


V

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3. m1 = 45 g

m2 = 75 g

v2, i = 0 m/s

h = 8.0 cm

g = 9.81 m/s2

Use the conservation of mechanical energy to calculate vf .

KE = PE

12

(m1 + m2)vf2 = (m1 + m2)gh

vf =√

2gh =√

(2)(9.81 m/s2)(8.0 × 10−2m)

vf = 1.3 m/s

v1, i = =

v1, i = =

The height that the first ball must have can be determined by using the conservationof mechanical energy.

PE = KE

m1gh1 = 12

m1v1, i2

h1 = v

21,

gi2

= = 0.62 m = 62 cm(3.5 m/s)

(2)(9.81 m/s2)

3.5 m/s(1.20 × 102 g)(1.3 m/s)

45 g

(45 g + 75 g)(1.3 m/s) − (75 g)(0 m/s)

45 g

(m1 + m2)vf − m2v2, im1

4. m1 = m2 = m3 = 5.00 × 102 kg

vf = 3.67 m/s

v3, i = 3.00 m/s

v2, i = 3.50 m/s

m1v1, i + m2v2, i + m3v3, i = (m1 + m2 + m3)vf

v1, i =

v1, i =

−

v1, i =

v1, i =

v1, i = = 4.50 m/s2250 kg•m/s5.00 × 102 kg

5.50 × 103 kg•m/s − 3250 kg•m/s

5.00 × 102 kg

(15.00 × 102 kg)(3.67 m/s) − 1750 kg•m/s − 1.50 × 103 kg•m/s

5.00 × 102 kg

(5.00 × 102 kg)(3.50 m/s) − (5.00 × 102 kg)(3.00 m/s)

5.00 × 102 kg

(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)(3.67 m/s)

5.00 × 102 kg

(m1 + m2 + m3)vf − m2v2, i − m3v3, im1

5. m1 = 8500 kg

v1, i = 4.5 m/s to the right= +4.5 m/s

m2 = 9800 kg

v2, i = 3.9 m/s to the left= −3.9 m/s

vf =

vf = =

vf = 0 m/s

3.8 × 104 kg•m/s − 3.8 × 104 kg•m/s

1.83 × 104 kg

(8500 kg)(4.5 m/s) + (9800 kg)(−3.9 m/s)

8500 kg + 9800 kg

m1v1, i + m2v2, im1 + m2


V

6. m1 = 1400 kg

v1, i = 45 km/h to the north

m2 = 2500 kg

v2, i = 33 km/h to the east

vf =

The component of vf in the x-direction is given by

vf, x = = =

vf, x = 21 km/h

The component of vf in the y-direction is given by

vf, y = = =

vf, y = 16 km/h

vf =√

vf, x2+ vf,y2 =

√(21km/h)2 + (16 km/h)2

vf =√

440km2/h2+ 260 km2/h2 =√

7.0× 102 km2/h2

vf = 26 km/h

q = tan−1 v

v

f

f

,

,

x

y = tan−1126

1

k

k

m

m

/

/

h

h = 37°

vf = 26 km/h at 37° north of east

(1400 kg)(45 km/h)

3900 kg

(1400 kg)(45 km/h)1400 kg + 2500 kg

m1v1, im1 + m2

(2500 kg)(33 km/h)

3900 kg

(2500 kg)(33 km/h)1400 kg + 2500 kg

m2v2, im1 + m2

m1v1, i + m2v2, im1 + m2

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7. m1 = 50.0 g

v1, i = 0.80 m/s to the west = −0.80 m/s

m2 = 60.0 g

v2, i = 2.50 m/s to the north= +2.50 m/s

m3 = 100.0 g

v3, i = 0.20 m/s to the east = +0.20 m/s

m4 = 40.0 g

v4, i = 0.50 m/s to the south= −0.50 m/s

The component of final velocity in the x-direction is given by

vf, x = =

vf, x = =

vf, x = −8.0 × 10−2 m/s

The component of final velocity in the y-direction is given by

vf, y = =

vf, y = =

vf, y = 0.520 m/s

vf =√

vf, x2+ vf,y2 =

√(−8.0× 10−2m/s)2 + (0.520m/s)2

vf =√

6.4× 10−3m2/s2 + 0.270 m2/s2 =√

0.276m2/s2

vf = 0.525 m/s

q = tan−1v

v

f

f

,

,

x

y = tan−1−8.

0

0

.5

×2

1

0

0

m−2

/s

m/s = −81°

q = 81° north of west, or 9° west of north

vf = 0.53 m/s at 9° west of north

1.30 × 102 g•m/s

250.0 g

1.50 × 102 g•m/s − 2.0 × 101 g•m/s

250.0 g

(60.0 g)(2.50 m/s) + (40.0 g)(−0.50 m/s)

50.0 g + 60.0 g + 100.0 g + 40.0 g

m2v2, i + m4v4, im1 + m2 + m3 + m4

−2.0 × 101 g•m/s

250.0 g

−4.0 × 101 g•m/s + 2.0 × 101 g•m/s

250.0 g

(50.0 g)(−0.80 m/s) + (100.0 g)(0.20 m/s)

50.0 g + 60.0 g + 100.0 g + 40.0 g

m1v1, i + m3v3, im1 + m2 + m3 + m4


V

8. ms = 25.0 kg

mc = 42.0 kg

v1, i = 3.50 m/s

v2, i = 0 m/s

vf = 2.90 m/s

m1 = mass of child and sled = ms + mc = 25.0 kg + 42.0 kg = 67.0 kg

m1v1, i + m2v2, i = (m1 + m2)vf

m2 = =

m2 = =

m2 = 13.8 kg

40.0 kg•m/s

2.90 m/s234 kg•m/s − 194 kg•m/s

2.90 m/s

(67.0 kg)(3.50 m/s) − (67.0 kg)(2.90 m/s)

2.90 m/s − 0 m/s

m1v1, i − m1vfvf − v2, i

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9. v1, i = 5.0 m/s to the right= +5.0 m/s

v2, i = 7.00 m/s to the left= −7.00 m/s

vf = 6.25 m/s to the left= −6.25 m/s

m2 = 150.0 kg

m1 = =

m1 = =

m1 = 9.8 kg

−110 kg•m/s

−11.2 m/s

−1050 kg•m/s + 938 kg•m/s

−11.2 m/s

(150.0 kg)(−7.00 m/s) − (150.0 kg)(−6.25 m/s)

−6.25 m/s − 5.0 m/s

m2v2, i − m2vfvf − v1, i

10. v1, i = 8.0 × 103 m/s to theright = +8.0 × 103 m/s

v2, i = 8.0 × 103 m/s to theleft = −8.0 × 103 m/s

vf = (0.900)v1, i

m2 = (52 000)(2.5 g) = 1.3 × 105 g

m1 = =

m1 = =

m1 = 2.4 × 103 kg

1.9 × 106 kg•m/s

8.0 × 102 m/s

9.4 × 105 kg•m/s + 1.0 × 106 kg•m/s

(8.0 × 103 m/s)(0.100)

(1.3 × 102 kg)(0.900)(8.0 × 103 m/s) − (1.3 × 102 kg)(−8.0 × 103 m/s)

(8.0 × 103 m/s)(1 − 0.900)

m2vf − m2v2, iv1, i − vf

1. m1 = 55 g

v1, i = 1.5 m/s

m2 = 55 g

v2, i = 0 m/s

vf = = =

vf = 0.75 m/s

percent decrease of KE = × 100 = × 100 = − 1 × 100

KEi = 12

m1v1, i2 + 1

2m2v2, i = 1

2(55 × 10−3 kg)(1.5 m/s)2 + 1

2(55 × 10−3 kg)(0 m/s)2

KEi = 6.2 × 10−2 J + 0 J = 6.2 × 10−2 J

KEf = 12

(m1 + m2)vf2 = 1

2(55 g + 55 g)(10−3 kg/g)(0.75 m/s)3

KEf = 3.1 × 10−2 J

percent decrease of KE = − 1 × 100 = (0.50 − 1) × 100 = (−0.50) × 100

percent decrease of KE = −5.0 × 101 percent

3.1 × 10−2 J6.2 × 10−2 J

KEfKEi

KEf − KEiKEi

∆KEKEi

(55 g)(1.5 m/s)

1.10 × 102 g

(55 g)(1.5 m/s) + (55 g)(0 m/s)

55 g + 55 g

m1v1, i + m2v2, im1 + m2



V

2. m1 = 4.5 kg

v1, i = 0 m/s

m2 = 1.3 kg

vf = 0.83 m/s

v2, i = =

v2, i = = 3.7 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 = 12

(4.5 kg)(0 m/s)2 + 12

(1.3 kg)(3.7 m/s)2

KEi = 0 J + 8.9 J = 8.9 J

KEf = 12

(m1 + m2)vf2 = 1

2(4.5 kg + 1.3 kg)(0.83 m/s)2 = 1

2(5.8 kg)(0.83 m/s)2

KEf = 2.0 J

∆KE = KEf − KEi = 2.0 J − 8.9 J = −6.9 J

(5.8 kg)(0.83 m/s)

1.3 kg

(4.5 kg + 1.3 kg)(0.83 m/s) − (4.5 kg)(0 m/s)

1.3 kg

(m1 + m2)vf − m1v1, im2

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3. m1 = 1.50 × 1013 kg

v1, i = 250 m/s

m2 = 6.5 × 1012 kg

v2, i = 420 m/s

vf = =

vf = = = 3.0 × 102 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 = 12

(1.50 × 1013 kg)(250 m/s)2 + 12

(6.5 × 1012 kg) (420 m/s)2

KEi = 4.7 × 1017 J + 5.7 × 1017 J = 10.4 × 1017 J = 1.04 × 1018

KEf = 12

(m1 + m2)vf2 = 1

2(1.50 × 1013 kg + 6.5 × 1012 kg)(3.0 × 102 m/s)2

KEf = 12

(2.15 × 1013 kg)(3.0 × 102 m/s)2

KEf = 9.7 × 1017 J

∆KE = KEf − KEi = 9.7 × 1017 J − 1.04 × 1018 J = −7.0 × 1016 J

6.5 × 1015 kg•m/s

2.15 × 1013 kg

3.8 × 1015 kg•m/s + 2.7 × 1015 kg•m/s

2.15 × 1013 kg

(1.50 × 1013 kg)(250 m/s) + (6.5 × 1012 kg)(420 m/s)

1.50 × 1013 kg + 6.5 × 1012 kg

m1v1, i + m2v2, im1 + m2

4. m1 = 0.650 kg

v1, i = 15.0 m/s to the right = +15.0 m/s

m2 = 0.950 kg

v2, i = 13.5 m/s to the left = −13.5 m/s

vf = =

vf = = = −1.91 m/s

vf = 1.91 m/s to the left

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 = 12

(0.650 kg)(15.0 m/s)2 + 12

(0.950 kg)(−13.5 m/s)2

KEi = 73.1 J + 86.6 J = 159.7 J

KEf = 12

(m1 + m2)vf2 = 1

2(0.650 kg + 0.950 kg)(1.91 m/s)2 = 1

2(1.600 kg)(1.91 m/s)2

KEf = 2.92 J

∆KE = KEf − KEi = 2.92 J − 159.7 J = −1.57 × 102 J

−3.0 kg•m/s

1.600 kg

9.75 kg•m/s − 12.8 kg•m/s

1.600 kg

(0.650 kg)(15.0 m/s) + (0.950 kg)(−13.5 m/s)

0.650 kg + 0.950 kg

m1v1, i + m2v2, im1 + m2


V

5. m1 = 75.0 kg

v1, i = 1.80 m/s downstream= 1.80 m/s

m2 = (8)(0.30 kg) = 2.4 kg

v2, i = 1.3 m/s upstream = −1.3 m/s

vf = =

vf = = = 1.71 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 = 12

(75.0 kg)(1.80 m/s)2 + 12

(2.4 kg)(−1.3 m/s)2

KEi = 122 J + 2.0 J = 124 J

KEf = 12

(m1 + m2)vf2 = 1

2(75.0 kg + 2.4 kg)(1.71 m/s)2 = 1

2(77.4 kg)(1.71 m/s)2 = 113 J

∆KE = KEf − KEi = 113 J − 124 J = −11 J

132 kg•m/s

77.4 kg

135 kg•m/s − 3.1 kg•m/s

77.4 kg

(75.0 kg)(1.80 m/s) + (2.4 hg)(−1.3 m/s)

75.0 kg + 2.4 kg

m1v1, i + m2v2, im1 + m2

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6. m1 = 8500 kg

v1, i = 4.5 m/s

m2 = 9800 kg

v2, i = −3.9 m/s

vf = 0 m/s

KEi = 12

m1v1, i2 + 1

2 m2v2, i

2 = 12

(8500 kg)(4.5 m/s)2 + 12

(9800 kg)(−3.9 m/s)2

KEi = 8.6 × 104 J + 7.5 × 104 J = 16.1 × 104 J = 1.61 × 105 J

KEf = 12

(m1 + m2)vf2 = 1

2(8500 kg + 9800 kg)(0 m/s)2 = 0 J

∆KE = KEf − KEi = 0 J − 1.61 × 105 J = −1.61 × 105 J

7. m1 = 45 g

v1, i = 3.5 m/s

m2 = 75 g

v2, i = 0 m/s

vf = 1.3 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 = 12

(45 × 10−3 kg)(3.5 m/s)2 + 12

(75 × 10−3 kg)(0 m/s)2

KEi = 0.28 J + 0 J = 0.28 J

KEf = 12

(m1 + m2)vf2 = 1

2(45 g + 75 g)(10−3 kg/g)(1.3 m/s)2 = 1

2(1.20 × 10−3 kg)(1.3 m/s)2

KEf = 1.0 × 10−3 J

∆KE = KEf − KEi = 1.0 × 10−3 J − 0.28 J =

The decrease in kinetic energy is almost total.

−0.28 J

8. m1 = 2.4 × 103 kg

v1, i = 8.0 × 103 m/s

m2 = 1.3 × 102 kg

v2, i = −8.0 × 103 m/s

vf = (0.900)(8.0 × 103 m/s)

KEi = 12

m1v1, i2 + 1

2m2v2, i

2

KEi = 12

(2.4 × 103 kg)(8.0 × 103 m/s)2 + 12

(1.3 × 102 kg)(−8.0 × 103 m/s)2

KEi = 7.7 × 1010 J + 4.2 × 109 J = 8.1 × 1010 J

KEf = 12

(m1 + m2)vf2 = 1

2(2.4 × 103 kg + 1.3 × 102 kg)[(0.900)(8.0 × 103 m/s)]2

KEf = 12

(2.5 × 103 kg)(7.2 × 103 m/s)2

KEf = 6.5 × 1010 J

∆KE = KEf − KEi = 6.5 × 1010 J − 8.1 × 1010 J = −1.6 × 1010 J

9. m1 = m2 = m3 = 5.00 × 102 kg

v1, i = 4.50 m/s

v2, i = 3.50 m/s

v3, i = 3.00 m/s

vf = 3.67 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 + 12

m3v3, i2

KEi = 12

(5.00 × 102 kg)(4.50 m/s)2 + 12

(5.00 × 102 kg)(3.50 m/s)2

+ 12

(5.00 × 102 kg)(3.00 m/s)2

KEi = 5.06 × 103 J + 3.06 × 103 J + 2.25 × 103 J = 10.37 × 103 J = 10.37 kJ

KEf = 12

(m1 + m2 + m3)vf2 = 1

2(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)

(3.67 m/s)2 = 12

(1.500 × 103 kg)(3.67 m/s)2

KEf = 1.01 × 104 J = 10.1 kJ

∆KE = KEf − KEi = 10.1 kJ − 10.37 kJ = −300 J


V

10. m1 = 50.0 × 10−3 kg

v1, i = −0.80 m/s

m2 = 60.0 × 10−3 kg

v2, i = 2.50 m/s

m3 = 0.1000 kg

v3, i = 0.20 m/s

m4 = 40.0 × 10−3 kg

v4, i = −0.50 m/s

vf = 0.53 m/s

KEi = 12

m1v1, i2 + 1

2m2v2, i

2 + 12

m3v3, i2 + 1

2m4v4, i

2

KEi = 12

(50.0 × 10−3 kg)(−0.80 m/s)2 + 12

(60.0 × 10−3 kg)(2.50 m/s)2

+ 12

(0.1000 kg)(0.20 m/s)2 + 12

(40.0 × 10−3 kg)(−0.50 m/s)2

KEi = 1.6 × 10−2 J + 18.8 × 10−2 J + 0.20 × 10−2 J + 0.50 × 10−2 J

KEi = 0.211 J

KEf = 12

(m1 +m2 + m3 + m4)vf2

KEf = 12

(50.0 × 10−3 kg + 60.0 × 10−3 kg + 0.1000 kg + 40.0 × 10−3 kg)(0.53 m/s)2

KEf = 12

(0.2500 kg)(0.53 m/s)2 = 3.5 × 10−2 J

∆KE = KEf − KEi = 3.5 × 10−2 J − 0.211 J = −0.18 J

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1. v1, i = 6.00 m/s to the right = +6.00 m/s

v2, i = 0 m/s

v1, f = 4.90 m/s to the left = −4.90 m/s

v2, f = 1.09 m/s to the right = +1.09 m/s

m2 = 1.25 kg

Momentum conservation

m1v1, i + m2v2, i = m1v1, f + m2v2, f

m1 = = =

m1 =

Conservation of kinetic energy (check)

12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(0.125 kg)(6.00 m/s)2 + 12

(1.25 kg)(0 m/s)2 = 12

(0.125 kg)(−4.90 m/s)2

+ 12

(1.25 kg)(1.09 m/s)2

2.25 J + 0 J = 1.50 J + 0.74 J

2.25 J = 2.24 J


0.125 kg

1.36 kg•m/s10.90 m/s

(1.25 kg)(1.09 m/s) − (1.25 kg)(0 m/s)

6.00 m/s − (−4.90 m/s)

m2v2, f − m2v2, iv1, i − v1, f

2. m1 = 2.0 kg

v1, i = 8.0 m/s

v2, i = 0 m/s

v1, f = 2.0 m/s

m1v1, i + m2v2, i = m1v1, f + m2v2, f

12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

m2 =

12

m1v1, i2 + 1

2 v2, i

2 = 12

m1v1, f2 + 1

2 v2, f

2

v1, i2(v2, i - v2, f) + (v1, f - v1, i )v2, i

2 = v1, f2(v2, i - v2, f) + (v1, f - v1, i )v2, f

2

(v1, i2v2, i + v1, f v2, i

2 - v1, iv2, i2 - v1, f

2v2, i + v2, f (v1, f2 - v1, i

2) = v2, f2(v1, f - v1, i )

Because v2, i = 0, the above equation simplifies to

v1, f2 − v1, i

2 = v2, f (v1, f − v1, i)

v2, f = v1, f + v1, i = 2.0 m/s + 8.0 m/s = 10.0 m/s

m2 = = =

m2 = 1.2 kg

−12 kg•m/s−10.0 m/s

4.0 kg•m/s − 16 kg•m/s

−10.0 m/s

(2.0 kg)(2.0 m/s) − (2.0 m/s)(8.0 m/s)

0 m/s − 10.0 m/s

m1v1, f − m1v1, iv2, i − v2, f

m1v1, f − m1v1, iv2, i − v2, f

m1v1, f − m1v1, iv2, i − v2, f

Additional Practice 6G


V

3. m1 = m2 = 45 g

v2, i = 0 m/s

v1, f = 0 m/s

v2, f = 3.0 m/s


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, i = v1, f + v2, f − v2, i = 0 m/s + 3.0 m/s − 0 m/s

v1, i =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

v1, i2 + v2, i

2 = v1, f2 + v2, f

2

(3.0 m/s)2 + (0 m/s)2 = (0 m/s)2 + (3.0 m/s)2

9.0 m2/s2 = 9.0 m2/s2

3.0 m/s

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4. m1 = 3.0 × 107 kg

m2 = 2.5 × 107 kg

v2, i = 4.0 km/h to the north= +4.0 km/h

v1, f = 3.1 km/h to the north= +3.1 km/h

v2, f = 6.9 km/h to the south= −6.9 km/h


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, i =

v1, i =

v1, i = = −6.0 km/h

v1, i =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(3.0 × 107 kg)[(−6.0 × 103 m/h)(1 h/3600 s)]2 + 12

(2.5 × 107 kg)[(4.0 × 103 m/h)(1 h/3600 s)]2

= 12

(3.0 × 107 kg)[(3.1 × 103 m/h)(1 h/3600 s)]2 + 12

(2.7 × 107 kg)[(−6.9 × 103 m/h)(1 h/3600 s)]2

4.2 × 107 J + 1.5 × 107 J = 1.1 × 107 J + 4.6 × 107 J

5.7 × 107 J = 5.7 × 107 J

6.0 km/h to the south

−1.8 × 108 kg•km/h

3.0 × 107 kg

9.3 × 107 kg•km/h − 1.7 × 108 kg•km/h − 1.0 × 108 kg•km/h

3.0 × 107 kg

(3.0 × 107 kg)(3.1 km/h) + (2.5 × 107 kg)(−6.9 km/h) − (2.5 × 107 kg)(4.0 km/h)

3.0 × 107 kg

m1v1, f + m2v2, f − m2v2, im1


V

5. m1 = m2


v1, f = 4.0 m/s to the west


v1, i = 3.0 m/s

q1, i = 90° counterclockwisefrom east

v1, f = 4.0 m/s

q1, f = 180° counterclock-wise from east

v2, f = 3.0 m/s

q2, f = 90° counterclockwisefrom east


In the x-direction:

m1 v1, i(cos q1, i) + m2 v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2 v2, f (cos q2, f )

v2, i (cos q2, i) = v1, f (cos q1, f ) + v2, f (cos q2, f ) − v1, i (cos q1, i ) = (4.0 m/s)(cos 180°)

+ (3.0 m/s)(cos 90°) − (3.0 m/s)(cos 90°)

v2, i (cos q2, i) = −4.0 m/s + 0 m/s + 0 m/s = −4.0 m/s

In the y-direction:

m1v1, i (sin q1, i) + m2v2, i(sin q2, i) = m1v1, f (sin q1, f ) + m2v2, f (sin q2, f )

v2, i (sin q2, i ) = v1, f (sin q1, f ) + v2, f (sin q2, f ) − v1, i (sin q1, i ) = (4.0 m/s)(sin 180°)

+ (3.0 m/s)(sin 90°) − (3.0 m/s)(sin 90°)

v2, i (sin q2, i) = 0 m/s + 3.0 m/s − 3.0 m/s = 0 m/s

This equation indicates that

sin q2, i = 0, or q2, i = 0° or 180°

v2, i (cos q2, i) = −4.0 m/s

v2, i = 4.0 m/s

q2, i = cos−1(−1.0) = 180°

v2, i =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

v1, i2 + v2, i

2 = v1, f2 + v2, f

2

(3.0 m/s)2 + (4.0 m/s)2 = (4.0 m/s)2 + (3.0 m/s)2

9.0 m2/s2 + 16 m2/s2 = 16 m2/s2 + 9.0 m2/s2

25 m2/s2 = 25 m2/s2

4.0 m/s to the west

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V

6. m1 = 0.75 kg

m2 = 0.50 kg

m3 = 0.50 kg

v2, i = 0 m/s

v3, i = 0 m/s

v1, f = 0.80 m/s to the east(at 0°)

v2, f = 3.4 m/s at 45° northof east (at 45°)

v3, f = 3.4 m/s at 45° southof east (at −45°)


In the x-direction:

m1v1, i (cos q1, i) + m2v2, i (cos q2, i) + m3v3, i (cos q3, i) = m1v1, f (cos q1, f) + m2v2, f (cos q2, f) + m3v3, f (cos q3, f)

m1v1, i (cos q1, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f ) + m3v3, f (cos q3, f ) − m2v2, i(cos q2, i) − m3v3, i(cos q3, i)

m1v1, i (cos q1, i) = (0.75 kg)(0.80 m/s)(cos 0°) + (0.50 kg)(3.4 m/s)(cos 45°) + (0.50 kg)(3.4 m/s)[cos (−45°)] − (0.50 kg)(0 m/s) − (0.50 kg)(0 m/s)

m1v1, i (cos q1, i) = 0.60 kg•m/s + 1.2 kg•m/s + 1.2 kg•m/s − 0 kg•m/s − 0 kg•m/s m1v1, i (cos q1, i) = 3.0 kg•m/s

v1, i (cos q1, i ) = = 4.0 m/s

In the y-direction:

m1v1, i (sin q1, i ) + m2v2, i (sin q2, i ) + m3v3, i (sin q3, i ) = m1v1, f (sin q1, f ) + m2v2, f (sin q2, f ) + m3v3, f (sin q3, f )

Because v2, i, v3, i, and sin q1, f equal 0, m1v1, i (sin q1, i ) = m2v2, f (sin q2, f ) + m3v3, f (sin q3, f ) = (0.50 kg)(3.4 m/s)(sin 45°) + (0.50 kg)(3.4 m/s)[sin(−45°)] = 1.2 kg•m/s − 1.2 kg•m/s = 0 kg•m/s

This result indicates that sin q1, i = 0, or q1, i = 0° or 180°

v1, i (cos q1, i) = 4.0 m/s

v1, i = 4.0 m/s

q1, i = cos−1(1.0) = 0°

v1, i =


12

m1v1, i2 + 1

2m2v2, i

2 + 12

m3v3, i2 = 1

2m1v1, f

2 + 12

m2v2, f2 + 1

2m3v3, f

2

12

(0.75 kg)(4.0 m/s)2 + 12

(0.50 kg)(0 m/s)2 + 12

(0.50 kg)(0 m/s)2

= 12

(0.75 kg)(0.80 m/s)2 + 12

(0.50 kg)(3.4 m/s)2 + 12

(0.50 kg)(3.4 m/s)2

6.0 J + 0 J + 0 J = 0.24 J + 2.9 J + 2.9 J

6.0 J = 6.0 J

4.0 m/s to the east

3.0 kg•m/s

0.75 kg

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V

7. v2, i = 2.000 m/s upward

v2, f = 1.980 m/s upward

∆yi = −20.4 m

a = −g = −9.81 m/s2

m1 = 0.150 kg

m2 = 325.0 kg

hi = 20.4 m

Velocity of ball dropped from rest is

v1, i = ±√

2a∆yi = ±√

(2)(−9.81 m/s2)(−20.4 m) = ± 20.0m/s

v1, i = −20.0 m/s = 20.0 m/s downward


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, f =

v1, f =

v1, f = = = 23.3 m/s

v1, f =

∆yf = = = 26.9 m

h = ∆yf − hi = 27 m − 20.4 m =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

12

(0.150 kg)(−20.0 m/s)2 + 12

(325.0 kg)(2.000 m/s)2 = 12

(0.150 kg)(23 m/s)2

+ 12

(325.0 kg)(1.980 m/s)2

30.0 J + 650.0 J = 4.0 × 101 J + 637.1 J

680.0 J = 677 J


7 m above the shaft

(23.3 m/s)2

(2)(9.81 m/s2)

v1, f2

2g

23.3 m/s upward

3.50 kg•m/s

0.150 kg

−3.00 kg•m/s + 650.0 kg•m/s − 643.5 kg•m/s

0.150 kg

(0.150 kg)(−20.0 m/s) + (325.0 kg)(2.000 m/s)−(325.0 kg)(1.980 m/s)

0.150 kg

m1v1, i + m2v2, i − m2v2, fm1

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8. v2, i = 2.000 m/s downward= −2.000 m/s

v2, f = 2.017 m/s downward= −2.017 m/s

v1, i = 20.0 m/s downward= −20.0 m/s

g = 9.81 m/s2

m1 = 0.150 kg

m2 = 325.0 kg

hi = 20.4 m


m1v1, i + m2v2, i = m1v1, f + m2v2, f

v1, f =

v1, f =

v1, f = = = 17 m/s

v1, f =

∆yf = = = 14.2 m

h = ∆yf − hi = 14.2 m − 20.4 m = −6.20 m

h = 6.20 m below the top of the shaft

(17 m/s)2

(2)(9.81 m/s2)

v1, f2

2g

16.7 m/s upward

2.5 kg•m/s0.150 kg

−3.00 kg•m/s − 650.0 kg•m/s + 655.5 kg•m/s

0.150 kg

(0.150 kg)(−20.0 m/s) + (325.0 kg)(−2.000 m/s)−(325.0 kg)(−2.017 m/s)

0.150 kg

m1v1, i + m2v2, i − m2v2, fm1


V


12

m1v1, i2 + 1

2m2v2, i

2 = 12

mi v1, f2 + 1

2m2v2, f

2

12

(0.150 kg)(−20.0 m/s)2 + 12

(325.0 kg)(−2.000 m/s)2 = 12

(0.150)(17 m/s)2

+ 12

(325.0 kg)(−2.017 m/s)2

30.0 J + 650.0 J = 22 J + 661.1 J

680.0 J = 683 J


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9. m1 = 0.500 kg

h = 40.0 cm

g = 9.81 m/s2

m2 = 2.50 kg

v2, i = 0 m/s

The velocity of steel ball at point of collision can be determined through conserva-tion of mechanical energy.

PEi = KEf

m1gh = 12

m1v1, i2

v1, i =√

2gh =√

(2)(9.81 m/s2)(0.400 m) = 2.80 m/s

m1v1, i + m2v2, i = m1v1, f + m2v2, f

v2, f =

12

m1v1, i2 + 1

2m2v2, i

2 = 12

m1v1, f2 + 1

2m2v2, f

2

vi, f2 = v1, i

2 + − 2

Because v2, i = 0 m/s, the equation simplifies to the following:

v1, f2 = v1, i

2 − + -

1 + v1, f2 − v1, f + − 1(v1, i)

2 = 0

1 + v1, f2 − v1, f + − 1(2.80 m/s)2 = 0

(1.20)v1, f2 − (1.12 m/s)v1, f − 6.27 m2/s2 = 0

Solving for v1, f by using the quadratic equation,

v1, f =

v1, f = =

v1, f =1.12 m/s ± 5.59 m/s

2.40

1.12 m/s ±√

31.3 m2/s22.40

1.12 m/s ±√

1.25 m2/s2 + 30.1m2/s22.40

1.12 m/s ±√

(−1.12 m/s)2 − (4)(1.20)(−6.27 m2/s2)(2)(1.20)

0.500 kg2.50 kg

(2)(0.500 kg)(2.80 m/s)

2.50 kg

0.500 kg2.50 kg

m1m2

2m1v1, im2

m1m2

m1v1, f2

m2

2m1v1, iv1, fm2

m1v1, i2

m2

m1v1, i + m2v2, i − m1v1, fm2

m2m1

m2v2, i2

m1

m1v1, i + m2v2, i − m1v1, fm2

10. m1 = 7.00 kg

v1, i = 2.00 m/s to the east(at 0°)

m2 = 7.00 kg

v1, i = 0 m/s

v1, f = 1.73 m/s at 30.0°north of east


In the x-direction:

m1v1, i (cos q1, i) + m2v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f )

v2, f (cos q2, f) = v1, i (cos q1, i) + v2, i (cos q2, i) − vi, f (cos q1, f)

v2, f (cos q2, f) = (2.00 m/s)(cos 0°) + 0 m/s − (1.73 m/s)(cos 30.0°)

v2, f = 2.00 m/s − 1.50 m/s = 0.50 m/s

In the y-direction:

m1v1, i (sin q1, i) + m2v2, i (sin q2, i) = m1v1, f (sin q1, i) m2v2, f (sin q2, f)

v2, f (sin q2, f ) = v1, i (sin q1, i) + v2, i (sin q2, i) − v2, f (sin q2, f)

v2, f (sin q2, f) = (2.00 m/s)(sin 0°) + 0 m/s − (1.73 m/s)(sin 30.0°) = −0.865 m/s

=

tan q2, f = −1.7

q2, f = tan−1(−1.7) = (−6.0 × 101)°

v2, f = = 1.0 m/s

v2, f =


12

m1v1, i2 + 1

2m2v2, i

2 = 12

miv1, f2 + 1

2m2v2, f

2

12

(7.00 kg)(2.00 m/s)2 + 12

(7.00 kg)(0 m/s)2 = 12

(7.00 kg)(1.73 m/s)2 + 12

(7.00 kg)(1.0m/s)2

14.0 J + 0 J = 10.5 J + 3.5 J

14.0 J = 14.0 J

1.0 m/s at (6.0 × 101)° south of east

0.50 m/scos(−6.0 × 101)°

− 0.865 m/s

0.50 m/s

v2, f (sin q2, f)v2, f (cos q2, f)


V

The ball’s speed must have a smaller magnitude after collision than before. The positiveroot gives a final, forward speed that is close to the ball’s initial speed. Therefore the neg-ative root gives a more realistic result.

v1, f = −4.

2

4

.

7

40

m/s = −1.86 m/s

v2, f =

v2, f =

v2, f = = = 0.932 m/s

v1, f =

v2, f = 0.932 m/s forward

1.86 m/s backwards

2.33 kg•m/s

2.50 kg

1.40 kg•m/s + 0 kg•m/s + 0.930 kg•m/s

2.50 kg

(0.500 kg)(2.80 m/s) + (2.50 kg)(0 m/s) − (0.500 kg)(−1.86 m/s)

2.50 kg

m1v1, i + m2v2, i − m1v1, fm2

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Chapter 7Rotational Motion and the Law of Gravity

V

1. ∆s = +24.0 m

r = 3.50 m∆θ =

∆r

s =

2

3

4

.5

.0

0

m

m = 6.86 rad


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2. r = 5.55 m

∆s = +31.3 m∆θ =

∆r

s =

3

5

1

.5

.3

5

m

m = 5.64 rad

3. area = πr2 = 2730 km2

∆s = −545 kmr =

arπea =

2730πkm2

= 29.5 km

∆θ = ∆r

s =

−2

5

9

4

.5

5

k

k

m

m = −18.5 rad

4. ∆s = 4.3 × 1011 m

∆θ = 0.39 radr =

∆∆

θs

= 4.3

0.

×39

10

ra

11

d

m = 1.1 × 1012 m

5. ∆s = 35.0 km

∆θ = 1.75 radr =

∆∆

θs

= 1

3

.

5

7

.

5

0

r

k

a

m

d = 20.0 km

6. ∆s = −36.6 µm

∆θ = −6

π rad

r = ∆∆

θs

= = 69.9 µm−36.6 µm

−6

π rad

7. r = 10.0 m

∆θ = +5.7 rad∆s = r∆θ = (10.0 m)(5.7 rad) = 57 m

8. r = 1.08 × 108 km

∆θ = + π3

rad∆s = r∆θ = (1.08 × 108 km)

π3

rad = 1.13 × 108 km

9. r = 4.48 × 109 km

∆θ = + π3

rad∆s = r∆θ = (4.48 × 109 km)

π3

rad = 4.69 × 109 km

10. r = 28.1 m

∆θ = –7.50 rad∆s = r∆θ = (28.1 m)(–7.50 rad)= 2.11 × 102 m

4. ωavg = 0.75 rad/s

∆θ = 3.3 rad∆t =

ω∆

a

θ

vg =

0.

3

7

.

5

3

r

r

a

a

d

d

/s = 4.4 s


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2. ∆θ = +2π rad

∆t = 4.56 minωavg =

∆∆

θt

= = 2.30 × 10−2 rad/s2π rad

(4.56 min)(60 s/min)

3. v = 280 m/s

∆x = 2.0 m

∆θ = +0.54 rad

∆t = ∆v

x

ωavg = ∆∆

θt

= = v

∆∆x

θ = = 12 rev/s

(280 m/s)(0.54 rad)(1 rev/2π rad)

2.0 m

∆θ

∆v

x

5. ωavg = 8.6 × 10−3 rad/s

∆θ = (6)(2π rad)∆t =

ω∆

a

θ

vg =

8.6

(6

×)(

1

2

0

π−3

ra

r

d

a

)

d/s = 4.4 × 103 s = 1.2 h

6. ωavg = 2.75 rad/s

∆θ = (3)(2π rad)∆t =

ω∆

a

θ

vg =

(

2

3

.

)

7

(

5

2πra

r

d

a

/

d

s

) = 6.85 s

7. ∆t = 4.2 h

ωavg = +2π rad/day∆θ = ωavg∆t = (2π rad/day)(1 day/24 h)(4.2 h) = 1.1 rad

8. ωavg = 1 rev/212 × 106 year

∆t = 4.50 × 109 year

∆θ = ωavg∆t = (1 rev/212 × 106 year)(4.50 × 109 year)(2π rad/rev) = 133 rad

9. ωavg = −1 rev/243 day

∆tE = 365.25 day

∆tV = 224.7 day

ωavg = (−1 rev/243 day)(2π rad/rev) =

∆θ = ωavg∆tE = (−2.59 × 10−2 rad/day)(365.25 day) =

∆θ = ωavg∆tV = (−2.59 × 10−2 rad/day)(224.7 day) = −5.82 rad

−9.46 rad

−2.59 × 10−2 rad/day

10. ∆x = 6.0 m

v = 5.0 m/s

ωavg = 6.00 rad/s

∆t = ∆v

x

∆θ = ωavg∆t = ωavg∆v

x =

(6.00 r

5

a

.0

d/

m

s)

/

(

s

6.0 m) = 7.2 rad

1. αavg = 1.5 rad/s2

ω1 = 3.0 rad/s

∆t = 4.0 s

ω2 = ω1 + αavg∆t = 3.0 rad/s + (1.5 rad/s2)(4.0 s) = 3.0 rad/s + 6.0 rad/s

ω2 = 9.0 rad/s


1. ∆θ = −106 rad

∆t = 7.5 sωavg =

∆∆

θt

= −1

7

0

.

6

5

r

s

ad = −14.1 rad/s



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2. ω1 = 9.5 rad/s

αavg = −5.4 × 10−3 rad/s2

∆t = 22 min

ω2 = ω1 + αavg∆t = 9.5 rad/s + (−5.4 × 10−3 rad/s2)(22 min)(60 s/min)

ω2 = 9.5 rad/s − 7.1 rad/s = 2.4 rad/s

3. αavg = 32 rad/s2

∆t = 1.5 s

ω1 = 0 rad/s

ω2 = ω1 + αavg∆t = 0 rad/s + (32 rad/s2)(1.5 s) = 48 rad/s

4. ω2 = 76 rad/s

ω1 = 0 rad/s

αavg = 9.5 rad/s2

6. ω1 = 5.14 × 10−2 rad/s

ω2 = 3.09 × 10−2 rad/s

αavg = −1.75 × 10−3 rad/s2

∆t = ω2

α−

av

ω

g

1 = =

∆t = 11.7 s

−2.05 × 10−2 rad/s−1.75 × 10−3 rad/s2

3.09 × 10−2 rad/s − 5.14 × 10−2 rad/s

−1.75 × 10−3 rad/s2

9. ω1 = 2.07 rad/s

ω2 = 1.30 rad/s

∆t = 2.2 s

αavg = ω2

∆−t

ω1 = = −0.7

2

7

.2

r

s

ad/s = −0.35 rad/s21.30 rad/s − 2.07 rad/s

2.2 s

∆t = ω2

α−

av

ω

g

1 = 76 r

9

ad

.5

/s

ra

−d

0

/s

r2ad/s

= 8.0 s

5. αavg = 3.91 rad/s2

ω2 = 7.70 rad/s

ω1 = 2.50 rad/s

∆t = ω2

α−

av

ω

g

1 = = 1.33 s7.70 rad/s − 2.50 rad/s

3.91 rad/s2

7. ω1 = 4.0 rad/s

ω2 = 5.0 rad/s

∆t = 7.5 s

αavg = ω2

∆−t

ω1 = 5.0 rad/

7

s

.

−5

4

s

.0 rad/s =

1.0

7.

r

5

ad

s

/s = 0.13 rad/s2

8. ω1 = 7.14 rad/s

ω2 = 2.38 rad/s

∆t = 9.00 s

αavg = ω2

∆−t

ω1 = = 9

−.

4

0

.

0

76

s = −0.529 rad/s22.38 rad/s − 7.14 rad/s

9.0 s

10. ω1 = 2π rad/23.66 h

ω2 = 2π rad/24.00 h

∆t = 70.0 × 106 year

αavg = ω2

∆−t

ω1 =

αavg =

αavg = = −6.2 × 10−15 rad/h2−3.8 × 10−3 rad/h(70.0 × 106 year)(365.25 day/year)(24 h/day)

0.2618 rad/h − 0.2656 rad/h(70.0 × 106 year)(365.25 day/year)(24 h/day)

(2 π rad/24.00 h) − (2π rad/23.66 h)(70.0 × 106 year)(365.25 day/year)(24 h/day)


V

1. ωi = 0 rad/s

ωf = 3.33 rad/s

α = 0.183 rad/s2

∆t = ωf

α− ωi =

3.33

0

r

.1

a

8

d

3

/s

r

−ad

0

/s

r2ad/s

= 18.2 s

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2. ωi = 0 rad/s

α = 0.13 rad/s2

∆θ = 1.6 rad

Because ωi = 0 rad/s

∆θ = 12

α∆t2

∆t = 2∆αθ =

(

0

2.

)

1(

3

1 .

r

6adra

/s

d2)

= 5.0 s

3. ωi = 5.2 rad/s

ωf = 20.9 rad/s

∆θ = 216 rad

∆t = ω

2

i

∆+

θωf

= = = 16.6 s(2)(216 rad)

26.1 rad/s

(2)(216 rad)5.2 rad/s + 20.9 rad/s

4. ωi = 0.111 rad/s

ωf = 0.178 rad/s

α = 1.1 × 10−2 rad/s2

∆θ = ωf

2

2

−a

ωi2

=

∆θ = = = 0.87 rad1.94 × 10−2 rad2/s2

2.2 × 10−2 rad/s2

3.17 × 10−2 rad2/s2 − 1.23 × 10−2 rad2/s2

2.2 × 10−2 rad/s2

(0.178 rad/s)2 − (0.111 rad/s)2

(2)(1.1 × 10−2 rad/s2)


5. ωi = 78.0 rev/min

∆t = 30.0 s

α = −0.272 rad/s2

∆θ = ωi∆t + 12

α∆t2

∆θ = (78.0 rev/min)(2π rad/rev)(1 min/60 s)(30.0 s) + 12

(−0.272 rad/s2)(30.0 s)2

∆θ = 245 rad − 122 rad = 123 rad

∆θ = (123 rad)(1 rev/2π rad) = 19.6 rev = 123 rad = 19.6 rev

6. ωi = 298 rad/s

α = −44.0 rad/s2

∆θ = 276 rad

ωf2 = ωi

2 + 2α∆θ

ωf =√

ωi2 + 2α∆θ =

√(298 rad/s)2 + (2)(−44.0 rad/s2)(276 rad)

ωf =√

8.88 × 104 rad2/s2 − 2.43× 104 rad2/s2 =√

6.45 × 104 rad2/s2 f = 254 rad/s

7. ωi = 0 rad/s

∆t = 13.0 s

∆θ = 10.0 rev

ωf = 2

∆∆

t

θ − ωi = − 0 rad/s f = 9.67 rad/s

(2)(10.0 rev)(2π rad/rev)

13.0 s

8. ωi = 1200 rev/min

ωf = 3600 rev/min

∆t = 12 s

α = ωf

∆−t

ωi =

α = = 21 rad/s2(2400 rev/min)(2π rad/rev)(1 min/60 s)

12 s

(3600 rev/min − 1200 rev/min)(2π rad/rev)(1 min/60 s)

12 s


V

9. ∆θ = 158 rad

ωi = 0 rad/s

ωf = 70.0 rad/s

α = ωf

2

2∆−

θωi

2

= = 15.5 rad/s2(70.0 rad/s)2 − (0 rad/s)2

(2)(158 rad)

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10. ωi = 3.29 rad/s

∆t = 2.50 s

∆θ = 12.3 rad

α = 2(∆θ

∆−t2ωi∆t) =

α = = (2)

(

(

2

4

.5

.1

s)

r2ad)

α = 1.3 rad/s2

(2)(12.3 rad − 8.2 rad)

(2.5 s)2

(2)[(12.3 rad) − (3.29 rad/s)(2.5 s)]

(2.5 s)2

1. ω = 2.07 × 10−3 rad/s

r = 1.5 km

vt = rω = (1.5 × 103 m)(2.07 × 10−3 rad/s) = 3.1 m/s

2. ω = 188.5 rad/s

r = 3.73 cmvt = rω = (3.73 × 10−2 m)(188.5 rad/s) = 7.03 m/s

4. r = 0.30 m

vt = 4.5 m/sω =

v

rt =

4

0

.

.

5

30

m

m

/s = 15 rad/s


3. r = 15.2 m

ω = 6.28 rad/svt = rω = (15.2 m)(6.28 rad/s) = 95.5 m/s

5. r = 2.00 m

vt = 94.2 m/sω =

v

rt =

9

2

4

.

.

0

2

0

m

m

/s = 47.1 rad/s

6. vt = 0.63 m/s

r = 1.5 mω =

v

rt =

0.

1

6

.

3

5

m

m

/s = 0.42 rad/s

7. ω = 3.14 × 10−2 rad/s

vt = 0.45 m/sr =

v

wt =

3.14

0

×.4

1

5

0

m−2

/

r

s

ad/s = 14 m

8. ω = 10.0 rad/s

vt = 4.60 m/sr =

ωvt =

1

4

0

.

.

6

0

0

r

m

ad

/

/

s

s = 0.460 m = 46.0 cm

9. ω = 11 rad/s

vt = 4.0 cm/sr =

ωvt =

4

1

.

1

0

r

c

a

m

d/

/

s

s = 0.36 cm = 3.6 mm

10. vt = 1.5 m/s

ω = 0.33 rad/sr =

ωvt =

0.

1

3

.

3

5

r

m

ad

/s

/s = 4.5 m


V

1. r = 6.0 cm

α = 35.2 rad/s2at = rα = (6.0 10−2 m)(35.2 rad/s2) = 2.1 m/s2


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2. α = 105 rad/s2

r = 1.75 cm

at = rα = (1.75 cm)(105 rad/s2) = 184 cm/s2 = 1.84 m/s2

3. r = 5.87 m

α = 1.40 × 10−2 rad/s2at = rα = (5.87 m)(1.40 × 10−2 rad/s2) = 8.22 × 10−2 m/s2

4. ∆ω = 1.23 × 10−2 rad/s

∆t = 10.0 s

at = 7.50 × 10−2 m/s2

α = ∆∆ωt =

1.23 ×1

1

0

0

.0

−2

s

rad/s = 1.23 × 10−3 rad/s2

r = a

αt =

1

7

.2

.5

3

0

××

1

1

0

0−

−

3

2

r

m

ad

/

/

s

s

2

2 = 61.0 m

5. α = 42 rad/s2

at = 64 m/s2r =

a

rt =

4

6

2

4

r

m

ad

/

/

s

s

2

2 = 1.5 m

6. α = 6.25 × 10−2 rad/s2

at = 0.75 m/s2r =

a

rt =

6.25

0

×.7

1

5

0

m−2

/

r

s2

ad/s2 = 12 m

7. at = 0.157 m/s2

r = 0.90 mα =

a

rt =

0.1

0

5

.9

7

0

m

m

/s2

= 0.17 rad/s2

8. r = 1.75 m

at = 0.83 m/s2α =

a

rt =

0.

1

8

.

3

75

m

m

/s2

= 0.47 rad/s2

9. r = 0.50 m

∆v = 5.0 m/s

∆t = 8.5 s

at = ∆∆

v

t =

5.

8

0

.5

m

s

/s = 0.59 m/s2

α = a

rt =

0.

0

5

.

9

50

m

m

/s2

= 1.2 rad/s2

10. r = 16 cm

at = 0.59 m/s2α =

a

rt =

16

0.

×59

1

m

0−/2s2

m = 3.7 rad/s2

1. r = 3.81 m

vt = 124 m/sac =

v

rt2

= (1

3

2

.

4

81

m

m

/s)2

= 4.04 × 103 m/s2

Additional Practice 7G


V

2. r = 6.50 cm

ω = 30.0 rad/sac = rω2 = (6.50 × 10−2 m)(30.0 rad/s)2 = 58.5 m/s2

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3. r = 11 m

vt = 1.92 × 10−2 m/sac =

v

rt2

= (1.92 ×

1

1

1

0

m

−2 m/s)2

= 3.4 × 10−5 m/s2

4. r = 8.9 m

ac = (20.0) g

g = 9.81 m/s2

vt =√

rac =√

(8.9 m)(20.0)(9.81 m/s2) = 42 m/s

5. r = 4.2 m

ac = 2.13 m/s2vt =

√rac =

√(4.2 m)(2.13 m/s2) = 3.0 m/s

6. ac = g = 9.81 m/s2

r = 150 m

vt =√

rac =√

(150 m)(9.81 m/s2) = 38 m/s

7. vt = 75.0 m/s

ac = 22.0 m/s2r =

v

at

c

2

= (7

2

5

2

.

.

0

0

m

m

/

/

s

s

)2

2

= 256 m

8. ω = 3.5 rad/s

ac = 2.0 m/s2r =

ωac

2 =

(3

2

.5

.0

r

m

ad

/

/

s

s

2

)2 = 0.16 m = 16 cm

9. vt = 0.35 m/s

ac = 0.29 m/s2r =

v

at

c

2

= (0

0

.

.

3

2

5

9

m

m

/

/

s

s

)2

2

= 0.42 m = 42 cm

10. ac = 9.81 m/s2

vt = 15.7 m/sr =

v

at

c

2

= (1

9

5

.8

.7

1

m

m

/

/

s

s

)2

2

= 25.1 m

1. m = 40.0 kg

ω = 0.50 rad/s

r = 18.0 m

Fc = mrω2 = (40.0 kg)(18.0 m)(0.50 rad/s)2 = 180 N

Additional Practice 7H

2. r = 0.25 m

vt = 5.6 m/s

m = 0.20 kg

Fc = mv

rt2

= (0.20 kg)(5

0

.6

.2

m

5 m

/s)2

= 25 N


V

3. vt = 48.0 km/h

r = 35.0 m

mk = 0.500

m = 1250 kg

g = 9.81 m/s2

Fc = mv

rt2

= (1250 kg)

Fc =

Ff = µkFn = µkmg = (0.500)(1250 kg)(9.81 m/s2)

Ff =

The available frictional force is not large enough to maintain the automobile’scircular motion.

6130 N

6350 N

[(48.0 km/h)(103 m/km)(1 h/3600 s)]2

35.0 m

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4. m = 1250 kg

r = 35.0 m

θ = 9.50°

g = 9.81 m/s2

µk = 0.500

F = Ff + mg(sin θ) = µkFn + mg(sinθ) = µkmg(cos θ) + mg(sin θ)

F = (0.500)(1250 kg)(9.81 m/s2)(cos 9.50°) + (1250 kg)(9.81 m/s2)(sin 9.50°)

F = 6.05 × 103 N + 2.02 × 103 N

F =

Fc = F = 8.07 × 103 N

vt = F

mcr = vt = 15.0 m/s = 54.0 km/h

(8.07 × 103 N)(35.0 m)

1250 kg

8.07 × 103 N

5. m = 2.05 × 108 kg

r = 7378 km

Fc = 3.00 × 109 N

vt = F

mcr = vt = 1.04 × 104 m/s = 10.4 km/s

(3.00 × 109 N)(7378 × 103 m)

2.05 × 108 kg

6. m = 55 kg

ω = 2.0 rad/s

Fc = 135 N

r = m

F

ωc

2 = (55 kg)

1

(

3

2

5

.0

N

rad/s)2 = 0.61 m = 61 cm

7. m = 7.55 × 1013 kg

vt = 0.173 km/s

Fc = 505 N

r = m

F

v

c

t2

= = 4.47 × 1015 m(7.55 × 1013 kg)(0.173 × 103 m/s)2

505 N

8. ω = 36.7 rad/s

r = 0.10 m

Fc = 670 N

m = r

F

ωc2 = = 5.0 kg

670 N(0.10 m)(36.7 rad/s)2

9. r = 35.0 cm

vt = 2.21 m/s

Fc = 0.158 N

m = F

vt

c2r

= = 1.13 × 10−2 kg = 11.3 g(0.158 N)(35.0 × 10−2 m)

(2.21 m/s)2

10. Fc = 8.00 × 102 N

r = 0.40 m

vt = 6.0 m/s

m = F

vt

c2r

= = 8.9 kg(8.00 × 102 N)(0.40 m)

(6.0 m/s)2


V

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.

1. m1 = 2.04 × 104 kg

m2 = 1.81 × 105 kg

r = 1.5 m

G = 6.673 × 10−11 N•m2

kg2

Fg = G m

r1m

22 = 6.673 × 10−11 = 0.11 N

(2.04 × 104 kg)(1.81 × 105 kg)

(1.5 m)2

N•m2

kg2

2. m1 = 1.4 × 1021 kg

m2 = 5.98 × 1024 kg

r = 3.84 × 108 m

G = 6.67 × 10−11 N•m2

kg2

Fg = G m

r1m

22 = 6.673 × 10−11 = 3.8 × 1018 N

(1.4 × 1021 kg)(5.98 × 1024 kg)

(3.84 × 108 m)2N•m2

kg2

3. m1 = 0.500 kg

m2 = 2.50 × 1012 kg

r = 10.0 km

G = 6.673 × 10−11 N•m2

kg2

Fg = G m

r1m

22 = 6.673 × 10−11 = 8.34 × 10−7 N

(0.500 kg)(2.50 × 1012 kg)

(10.0 × 103 m)2N•m2

kg2

Additional Practice 7I

4. Fg = 2.77 × 10−3 N

r = 2.50 × 10−2 m

m1 = 157 kg

G = 6.673 × 10−11 N•m2

kg2

m2 = = = 165 kg(2.77 × 10−3 N)(2.50 × 10−2 m)2

6.673 × 10−11 (157 kg)

Fgr2

Gm1

5. Fg = 1.636 × 1022 N

m1 = 1.90 × 1027 kg

r = 1.071 × 106 km

G = 6.673 × 10−11 N•m2

kg2

m2 = = = 1.48 × 1023 kg(1.636 × 1022 N)(1.071 × 109 m)2

6.673 × 10−11 (1.90 × 1027 kg)

Fgr2

Gm1

6. Fg = 1.17 × 1018 N

m1 = 1.99 × 1030 kg

r = 4.12 × 1011 m

G = 6.673 × 10−11 N•m2

kg2

m2 = = = 1.50 × 1021 kg(1.17 × 1018 N)(4.12 × 1011 m)2

6.673 × 10−11 (1.99 × 1030 kg)

Fgr2

Gm1

7. m1 = m2 = 9.95 × 1041 kg

Fg = 1.83 × 1029 N

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2 = = 1.90 × 1022 m6.673 × 10−11

N

k

•

g

m2

2

(9.95 × 1041 kg)2

1.83 × 1029 N

N•m2

kg2

N•m2

kg2

N•m2

kg2


V

8. m1 = 1.00 kg

m2 = 1.99 × 1030 kg

Fg = 274 N

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2 = = 6.96 × 108 m

6.673 × 10−11 N

k

•

g

m2

2

(1.00 kg)(1.99 × 1030 kg)

274 N

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9. m1 = 1.00 kg

m2 = 3.98 × 1031 kg

Fg = 2.19 × 10−3 N

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2 = = 1.10 × 1012 m

6.673 × 10−11 N

k

•

g

m2

2

(1.00 kg)(3.98 × 1031 kg)

2.19 × 10−3 N

10. Fg = 125 N

m1 = 4.5 × 1013 kg

m2 = 1.2 × 1014 kg

G = 6.673 × 10−11 N•m2

kg2

r = Gm

F1

g

m2 = = 5.4 × 107 m

6.673 × 10−11 N

k

•

g

m2

2

(4.5 × 1013 kg)(1.2 × 1014 kg)

125 N


Chapter 8Rotational Equilibrium and Dynamics

V

1. d = 1.60 m

t = 4.00 × 102 N • m

q = 80.0°

F = d(si

tn q) =

F = 254 N

4.00 × 102 N • m(1.60 m)(sin 80.0°)


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2. tnet = 14.0 N • m

d ′ = 0.200 m

q ′ = 80.0°

t = 4.00 × 102 N • m

tnet = t − t ′

t ′ = Fgd ′(sin q ′) = t − tnet

Fg = d

t′(

−sin

tn

qet

′) =

Fg = = 1.96 × 103 N386 N • m

(0.200 m)(sin 80.0°)

4.00 × 102 N • m − 14. N • m

(0.200 m)(sin 80.0°)

3. d = 2.44 m

t = 50.0 N • m

q = 90°

F = =

F = 20.5 N

50.0 N • m(2.44 m)(sin 90°)

td(sin q)

4. t = 1.4 N • m

d = 0.40 m

q = 60.0°

F = d(si

tn q) =

F =

tmax is produced when q = 90°, or

tmax = Fd = (4.0 N)(0.40 m) = 1.6 N • m

4.0 N

1.4 N • m(0.40 m)(sin 60.0°)

5. Fmax = 2.27 × 105 N • m

r = 0.660 m

d = 12

r

tmax = Fmaxd = Fm

2axr

tmax = = 7.49 × 104 N • m(2.27 × 105 N • m)(0.660 m)

2

6. m = 1.6 kg

l = 43 cm

x = 15 cm

q = 90°

g = 9.81 m/s2

t = F d(sin q) = mg (l − x)(sin q)

t = (1.6 kg)(9.81 m/s2)(0.43 m − 0.15 m)(sin 90°) = (1.6 kg)(9.81 m/s2)(0.28 m)

t = 4.4 N • m

7. t = 0.46 N • m

F = 0.53 N

q = 90°

d = F(si

tn q) =

d = 0.87 m

0.46 N • m(0.53 N)(sin 90°)

9. m = 28 kg

g = 9.81 m/s2

q = 89°

t = 1.84 × 104 N • m

d = F(si

tn q) =

mg(s

tin q) =

d = 67 m

1.84 × 104 N • m(28 kg)(9.81 m/s2)(sin 89°)


V

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8. t = 8.25 × 103 N • m

F = 587 N

q = 65.0°

d = F(si

tn q) =

(5

8

8

.2

7

5

N

×)

1

(s

0

i

3

n

N

65

•

.0

m

°)

d = 15.5 m

10. Fb = 1.200 × 103 N

qb = 90.0°

m = 60.0 kg

g = 9.81 m/s2

qg = 87.7°

tnet = −2985 N • m

tnet = tg − tb = Fgd(sin qg) − Fbd(sin qb)

d = =

d =

d = 588

−−2

1

9

.

8

2

5

00

N

×•

1

m

03 N

d = −29

−8

6

5

12

N

N

• m = 4.88 m

−2985 N • m(60.0 kg)(9.81 m/s2)(sin 87.7°) − (1.200 × 103 N)(sin 90.0°)

tnetmg(sin qg) − Fb(sin qb)

tnetFg(sin qg) − Fb(sin qb)

1. m1 = 2.3 kg

m2 = 0.40 kg

l = 1.00 m

g = 9.81 m/s2

Apply the second condition of equilibrium.

tnet = t1 − t2 = 0

t1 = Fg,1d1 = m1gx

t2 = Fg,2d2 = m2g(l − x)

m1gx = m2g(l − x)

(m1 + m2)x = m2l

x = m1

m

+2l

m2 =

(0

2

.

.

4

3

0

k

k

g

g

+)(

0

1

.

.

4

0

0

0

k

m

g

) =

(0.40 k

2

g

.7

)(

k

1

g

.00 m)

x = 0.15 m from the ostrich egg

2. mms = 139 g

dms = 49.7 cm

mw = 50.0 g

dw = 10.0 cm

g = 9.81 m/s2

Apply the second condition of equilibrium.

tnet = tms − tw = 0

Let x be the distance from the fulcrum to the zero mark.

tms = mmsg(dms − x)

tw = mwg(x − dw)

mmsg(dms − x) = mwg(x − dw)

mmsdms + mwdw = (mw + mms)x

x = mm

msd

m

m

s

s

++

m

m

w

wdw =

x = =

x = 39.2 cm from the zero mark

7.41 × 103 g • cm

189 g

6.91 × 103 g • cm + 5.00 × 102 g • cm

189 g

(139)(49.7 cm) + (50.0 g)(10.0 cm)

139 g + 50.0 g



V

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3. mA = 4.64 × 107 kg

dA = 3.00 × 102 m

Fcs = 3.22 × 107 N

Fs = 7.55 × 108 N

g = 9.81 m/s2

Apply the first condition of equilibrium to solve for the weight of the cantilever arm, mcg.

Fs − mAg − mCg − Fcs = 0

mCg = Fs − mAg − Fcs = 7.55 × 108 N − (4.64 × 107 kg)(9.81 m/s2) − 3.22 × 107 N

mCg = 7.55 × 108 N − 4.55 × 108 N − 3.22 × 107 N = 2.68 × 108 N

To solve for dC, apply the second condition of equilibrium using the support for the can-tilever as the pivot point.

mAgd

2A − mCg

d

2C − FcsdC = 0

m2Cg + FcsdC =

mA

2

gdA

dC = m

mA

C

gd

gA + 2Fcs =

dC = =

dC = 411 m

(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)

3.32 × 108 N

(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)

2.68 × 108 N + 6.44 × 107 N

(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)

2.68 × 108 N + (2)(3.22 × 107 N)

4. mf = 70.0 kg

g = 9.81 m/s2

q = 10.0°

t = 7.08 × 103 N • m

Fu = 3.14 × 103 N

Apply the first condition of equilibrium to find the weight of the ladder, ml g.

Fu − mfg − mlg = 0

mlg = Fu − mf g

mlg = 3.14 × 103 N − (70.0 kg)(9.81 m/s2) 3.14 × 103 N − 687 N = 2.45 × 103 N

To solve for the length of the ladder (d), apply the second condition of an equilib-rium, using the base of the ladder as the pivot point.

t − mlgd

2(sin q) − mf gd(sin q) = 0

m

2l g + mf g(sin q)d = t

d = =

d = = = 21.3 m(2)(7.08 × 103 N • m)(3.82 × 103 N)(sin 10.0°)

(2)(7.08 × 103 N • m)(2.45 × 103 N + 1.37 × 103 N)(sin 10.0°)

(2)(7.08 × 103 N • m)[2.45 × 103 N + (2)(70.0 kg)(9.81 m/s2)](sin 10.0°)

2t(mlg + 2 mf g)(sin q)

5. dc = 32.0 m

q = 60.0°

FT,x = 1.233 × 104 N

FT,y = 1.233 × 104 N

g = 9.81 m/s2

Apply the first condition of equilibrium in the x direction.

Fx = Rx,base − FI,x = 0

Rx,base = R(cos q) = FT,x

R = c

F

oT

s,x

q

Apply the first condition of equilibrium in the y direction; solve for the flagpole’s weight.

Fy = Ry,base − mg − FT,y = 0

Ry,base = R(sin q) − FT,y = mg

cFoT

s,x

q(sin q) − FT,y = mg

mg = FT,x(tan q) − FT,y = (1.233 × 104 N)(tan 60.0°) − 1.233 × 104 N

mg = (1.233 × 104 N)[(tan 60.0°) − 1.00] = (1.233 × 104 N)(1.73 − 1.00)

mg = (1.233 × 104 N)(0.73) = 9.0 × 103 N


V

To solve for the length of the flagpole (l ), apply the second condition of equilibrium,using the base of the flagpole as the pivot point.

FT,ydc(sin q) − FT,ydc(cos q) − mg (cos q) = 0

l =

l =

l = =

l = 64 m

1.45 × 105 N • m12

(9.0 × 103 N)(cos 60.0°)

3.42 × 105 N • m − 1.97 × 105 N • m

12

(9.0 × 103 N)(cos 60.0°)

(1.233 × 104 N)(32.0 m)(sin 60.0°) − (1.233 × 104 N)(32.0 m)(cos 60.0°)

12

(9.0 × 103 N)(cos 60.0°)

FT,xdc(sin q) − FT,ydc(cos q)

12

mg(cos q)

l2

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6. mE = 5.98 × 1024 kg

g = 9.81 m/s2

dE = 1.00 m

dapplied = 3.8 × 1016 m

Apply the first condition of equilibrium.

Ffulcrum − Fapplied − mEg = c

Ffulcrum = Fapplied + mEg

To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum asthe pivot point.

Fapplieddapplied − mEgdE = 0

Fapplied = d

m

ap

E

p

g

l

d

ie

E

d =

Fapplied =

Substitute the value for Fapplied into the first-condition equation and solve forFfulcrum.

Ffulcrum = 1.5 × 109 N + (5.98 × 1024 kg)(9.81 m/s2) = 1.5 × 109 N + 5.87 × 1025 N

Ffulcrum = 5.87 × 1025 N

1.5 × 109 N

(5.98 × 1024 kg)(9.81 m/s2)(1.00 m)

3.8 × 1016 m

7. d = 2.00 m

q = 30.0°

t = 1.47 × 103 N • m

Apply the first condition of equilibrium in the x and y directions.

Fx = Frod − FT(cos q) = 0

Frod = FT(cos q)

Fy = FT(sin θ) − Fg = 0

Fg = FT(sin q)

Apply the second condition of equilibrium, using the end of the rod anchored in thewall as the pivot point.

t − Fgd = 0

Fg = d

t =

Fg =

Substitute the value for Fy into the second first-condition equation to solve for FT.Substitute the value of FT into the first first-condition equation to solve for Frod.

FT = si

F

ng

q =

si

7

n

3

3

5

0

N

.0° = 1470 N

Frod = FT(cos q) = (1470 N)(cos 30.0°)

Frod = 1270 N

735 N

1.47 × 103 N • m

2.00 m


V

8. Fg = 7.10 × 102 N

ms = 0.75

dx = 1.22 m

dy = 1.00 m

Apply the first condition of equilibrium to determine the largest force that will notovercome static friction.

Fapplied − Fs = 0

Fapplied = Fs = msFn = msFg

Fapplied = (0.75)(7.10 × 102 N) = 532 N

Apply the second condition of equilibrium to determine the largest force that will notlift the door from the rail. Choose one of the wheels for the axis of rotation.

Fapplieddy − Fgd

2x = 0

Fapplied = F

2g

d

d

y

x =

Fapplied = 433 N

The largest force that will not upset either equilibrium condition is the smaller of thetwo forces.

Fapplied = 433 N

(7.10 × 102 N)(1.22 m)

(2)(1.00 m)

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9. m = 307 kg

h = 2.44 m

q = 70.0°

d = 1.22 m

g = 9.81 m/s2

Apply the first condition of equilibrium in the x direction.

Fx = Fapplied(sin q) − R(cos q) = 0

R = Fapplied(tan q)

To solve for Fapplied, either use the first condition of equilibrium in the y direction orthe second condition of equilibrium.

Fy = Fapplied(cos q) + R(sin q) − mg = 0

Fapplied(cos q) + [Fapplied(tan q)](sin q) = mg

Fapplied = =

Fapplied = =

Fapplied =

Alternatively,

Fappliedd − mgh

2(cos q) = 0

Fapplied = mgh(

2

c

d

os q) =

Fapplied = 1030 N

Substitute the value of Fapplied into the first-condition equation in the x direction tosolve for R.

R = Fapplied(tan q) = (1030 N)(tan 70.0°) = 2830 N

(307 kg)(9.81 m/s2)(2.44 m)(cos 70.0°)

(2)(1.22 m)

1030 N

(307 kg)(9.81 m/s2)

2.92

(307 kg)(9.81 m/s2)

0.342 + 2.58

(307 kg)(9.81 m/s2)(cos 70.0°) + (tan 70.0°)(sin 70.0°0)

mg(cos q) + (tan q)(sin q)


V

10. Fg = 1.96 × 103 N

d1 = 0.250 m

d2 = 1.50 m

Apply the first condition of equilibrium to the x and y directions.

Fx = F − Rx = 0

Rx = F

Fy = Ry − Fg = 0

Ry = Fg = 1.96 × 103 N

To solve for F, apply the second condition of equilibrium, using the hinge as the pivotpoint.

Fgd1 − Fd2 = 0

F = F

dgd

2

1 =

F =

Substitute the value for F in the first-condition equation for the x direction, thensolve for R using the Pythagorean theorem.

Rx = 327 N

R =√

Rx2 + Ry

2 =√

(327 N)2 + (1.96 × 103 N)2R =

√1.07 × 105 N2+ 3.84× 106 N2 =

√3.95 × 106 N2

R = 1.99 × 103 N

327 N

(1.96 × 103 N)(0.250 m)

1.50 m

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1. t = 2.98 N • m

wi = 0 rad/s

wf = 55 rad/s

Dt = 0.75 s

R = 20.0 cm

I = at

= =

I =

For a hoop spinning on an axis along its diameter,

I = 12

MR2

M = R

2I2 =

M = 2.0 kg

(2)(0.041 kg • m2)

(0.200 m)2

0.041 kg • m2

2.98 N • m

55 rad

0

/

.

s

7

−5

0

s

rad/s

t

wf

∆−twi

2. t = 1.7 N • m

a = 5.5 rad/s2I =

at

= = 0.3 kg • m21.7 N • m5.5 rad/s2


3. t = 0.750 N • m

a = 499 rad/s2I =

at

= = 1.50 × 10−3 kg • m30.750 N • m499 rad/s2

4. M = 7.91 × 103 kg

R = 1.83 m

a = 6.13 rad/s2

I = 25

MR2 = 25

(7.91 × 103 kg)(1.83 m)2 = 1.06 × 104 kg • m2

t = Ia = (1.06 × 104 kg • m2)(6.13 rad/s2) = 6.50 × 104 N • m


V

5. a = −6.53 × 10−27 rad/s2

M = 5.98 × 1024 kg

R = 6.37 × 106 m

I = 0.331 MR2 = (0.331)(5.98 × 1024 kg)(6.37 × 106 m)2 = 8.03 × 1037 kg • m2

t = Ia = (8.03 × 1037 kg • m2)(−6.53 × 10−22 rad/s2) = −5.24 × 1016 N • m

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6. a = 1.05 rad/s2

I = 8.14 × 104 kg • m2

t = Ia = (8.14 × 104 kg • m2)(1.05 rad/s2) = 8.55 × 104 N • m

8. t = 1.01 N • m

I = 3.85 × 10−5 kg • m2a =

tI

= = 2.62 × 104 rad/s21.01 N • m3.85 × 10−5 kg • m2

10. M = 15 kg

R = 0.25 m

wi = 9.5 rad/s

wf = 0 rad/s

t = −0.80 N • m

I = 12

MR2 = 12

(15 kg)(0.25 m)2 = 0.47 kg • m2

a = tI

= =

∆t = wf

a− wi =

0 ra

−d

1

/s

.7

−r

9

ad

.5

/s

r2ad/s

= −−1

9

.

.

7

5

r

r

a

a

d

d

/

/

s

s2

∆t = 5.6 s

−1.7 rad/s2−0.80 N • m0.47 kg • m2


7. t = 108 N • m

I = 5.40 kg • m2a =

tI

= = 20.0 rad/s2108 N • m5.40 kg • m2

9. m = 0.15 kg

r = 0.35 m

t = 1.5 N • m

Dt = 0.26 s

a = tI

= m

tr2

a = =

w = a∆t = (82 rad/s2)(0.26 s) = 21 rad/s

82 rad/s21.5 N • m(0.15 kg)(0.35 m)2

1. r = 0.120 m

m = 22.0 g

wi = 50.00 rad/s

wf = 50.24 rad/s

Li = Lf

Iiwi = Ifwf

(I + 25 mr2)wi = Iwf

I(wf − wi) = 25 mr2 wf

I = 25

wm

f −r2

ww

i

f =

I = 1.7 kg • m2

(25)(22.0 × 10−3 kg)(0.120 m)2 (50.24 rad/s)

50.24 rad/s − 50.00 rad/s

2. M = 755 kg

li = 1.75 m × 2

= 3.50 m

wi = 1.25 rad/s

wf = 1.70 × 10−2 rad/s

Li = Lf

Iiwi = If wf

1

1

2Mli

2wi = Ifwf

If = =

If = 5.67 × 104 kg • m2

(755 kg)(3.50 m)2 (1.25 rad/s)

(12)(1.70 × 10−2 rad/s)

Mli2wi

12wf


V

3. ri = 3.00 m

rf = 0.20 m

m = 55.0 kg

wi = 2.00 rad/s

wf = 2.35 rad/s

Li = Lf

Iiwi = If wf

(I + 4 mri2)wi = (I + 4 mrf

2)wf

I(wf − wi) = 4 m(ri2wi − rf

2wf)

I = 4 m(r

wi2

f

w−i −

wr

i

f2wf) =

I = =

I = 1.1 × 104 kg • m2

(4)(55.0 kg)(17.9 m2/s)

0.35 rad/s

(4)(55.0 kg)(18.0 m2/s − 0.094 m2/s)

0.35 rad/s

(4)(55.0 kg)[(3.00 m)2 (2.00 rad/s) − (0.20 m)2 (2.35 rad/s)]

2.35 rad/s − 2.00 rad/s

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4. wi = 2.1 rad/s

ri = 1.2 m

rf = 0.50 m

Li = Lf

Iiwi = Ifwf

mri2wi = mrf

2wf

wf = ri

r

2

f

w2

i = (1.2

(

m

0

)

.5

2

0

(2

m

.1

)

r2ad/s)

wf = 12 rad/s

5. v1 = 43.5 km/s

r1 = 7.00 × 107 km

r2 = 1.49 × 108 km

L1 = L2

I1w1 = I2w2

mr12w1 = mr2

2w2

r1v1 = r2v2

v2 = =

v2 = 20.4 km/s

(7.00 × 107 km)(43.5 km/s)

1.49 × 108 km

r1v1r2

6. wi = 57.7 rad/s

Mw = 3.81 kg

Rw = 0.350 m

Itot = 2.09 kg • m2

Li = Lf

Iwwi = Itotwf

MwRw2wi = Itotwf

wf = Mw

I

R

to

w

t

2wi =

wf = 12.9 rad/s

(3.81 kg)(0.350 m)2 (57.7 rad/s)

2.09 kg • m2

7. wi = 1.50 rad/s

wf = 2.04 × 10−2 rad/s

M = 7.55 kg

li = 3.50 m

Li = Lf

Iiwi = If wf

1

1

2Mli

2wi = 1

1

2Mlf

2wf

lf = = lf = 30.0 m

(3.50 m)2 (1.50 rad/s)

2.04 × 10−2 rad/s

l i2wiwf


V

8. v1 = 3.68 km/s

r1 = 7.35 × 109 km

v2 = 6.14 km/s

L1 = L2

I1w1 = I2w2

mr12w1 = mr2

2w2

r1vi = r2v2

r2 = r1

v

v

2

1 =

r2 = 4.41 × 109 km

(7.35 × 109 km)(3.68 km/s)

6.14 km/s

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9. v1 = 3403 m/s

r1 = 3593 km

h1 = 2.00 × 102 km

v2 = 3603 m/s

Li = Lf

I1w1 = I2w2

mr12w1 = mr2

2w2

r1v1 = r2v2

r2 = r1

v

v

2

1 =

r2 =

r = r2 − h2 = r1 − h1

h2 = r2 − r1 + h1 = 3394 km − 3593 km + 2.00 × 102 km

h2 = 1 km

3394 km

(3593 km)(3403 m/s)

3603 m/s

10. m1 = m2 = 55.0 kg

ri = 5.0

2

0 m = 2.50 m

vi = 5.00 m/s

vf = 15.0 m/s

Li = Lf

Iiwi = Ifwf

(m1ri2 + m2ri

2)wi = (mirf2 + m2rf

2)wf

ri2wi = rf

2wf

rivi = rf vf

rf = r

viv

f

i = (2.50

1

m

5.

)

0

(5

m

.0

/

0

s

m/s)

rf = 0.833 m

distance between skaters = 2rf2 = 1.67 m

1. k = 1.05 × 104 N/m

x = 4.0 × 10−2 m

KErot = 2.8 J

MEi = MEf

PEelastic = KEtrans + KErot

12

kx2 = KEtrans + KErot

KEtrans = 12

kx2 − KErot = 12

(1.05 × 104 N/m)(4.0 × 10−2 m)2 − 2.8 J

KEtrans = 8.4 J − 2.8 J − 5.6 J



V

2. vi = 2.2 m/s

m = 55 g

MEi = MEf

KEi = KEtrans − KErot

12

mvi2 = 1

2mvf

2 + 12

Iwf2 = 1

2mvf

2 + 12

23

mr2v

rf

2

12

mvi2 = 1

2mvf

2+23

= 12

mvf25

3 = 5

3KEtrans

KEtrans = 1

3

0mvi

2

KErot = KEi − KEtrans = 12

mvi2 −

1

3

0mvi

2 = 1

2

0mvi

2 = 15

mvi2

KErot = 15

(55 × 10−3 kg)(2.2 m/s)2 = 5.3 × 10−2 J

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3. h = 46.0 m

m = 25.0 kg

g = 9.81 m/s2

MEi = Ef

PEg = KEtrans + KErot

mgh = 12

mvf2 + 1

2Iwf

2 = 12

mvf2 + 1

2(mr2)

v

rf

2

= 12

mvf2(1 + 1) = mvf

2

KEtrans = 12

mgh = 12

(25.0 kg)(9.81 m/s2)(46.0 m)

KEtrans = 5.64 × 103 J

4. k = 1.05 × 104 N/m

x = 4.0 cm

wf = 43.5 rad/s

MEi = MEf


12

kx2 = 12

mvf2 + 1

2Iwf

2 = 12

mr2v

rf

2

+ 12

Iwf2 = 1

2Iwf

2 + 12

Iwf2 = Iwf

2

I = 2

k

wx

f

2

2 =

I = 4.4 × 10−3 kg • m2

(1.05 × 104 N/m)(4.0 × 10−2 m)2

(20943.5 rad/s)2

5. KEi = 45 J

wf = 27 rad/s

r = 0.11 m

MEi = MEf

KEi = KEtrans + KErot = 12

mvf2 + 1

2Iwf

2

KEi = 12

mr2v

rf

2

+ 12

Iwf2 = 1

25

2Iwf

2 + 12

Iwf2 = 1

2Iwf

2(52

+ 1) = 74

Iwf2

I = 4

7

K

wE

f2i =

(7)

(

(

4

2

)

7

(4

r

5

ad

J

/

)

s)2 =

m = = = 7.2 kg(5)(3.5 × 10−2 kg • m2)

(2)(0.11 m)2

52

I

r2

3.5 × 10−2 kg • m2


V

6. h = 0.60 m

g = 9.81 m/s2

MEi = MEf

KEtrans + KErot = PEg

12

mvi2 + 1

2Iwi

2 = mgh

12

mvi2 + 1

22

5mr2

v

ri

2

= 12

mvi 1 + 25

= 1

7

0mvi

2 = mgh

vi = = vi = 2.9 m/s

(10)(9.81 m/s2+)(0.60 m)

7

10gh

7

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7. k = 150 N/m

x = 6.0 cm

m = 67 g

MEi = Ef


12

kx2 = 12

mvf2 + 1

2Iwf

2 = 12

mvf2 + 1

22

5mr2

v

rf

2

= mvf21

2 + 1

5 =

1

7

0mvf

2

vf = 5

7

k m

x2

= vf = 2.4 m/s

(5)(150 N/m)(6.0 × 10−2 m)2

(7)(67 × 10−3 kg)

8. vi = 2.4 m/s

q = 3.5°

g = 9.81 m/s2

MEi = MEf


12

mvi2 + 1

2Iwf

2 = mgh

12

mvi2 + 1

22

5mr2

v

ri

2

= mgd(sin q)

mvi21

2 + 1

5 =

1

7

0mvi

2 = mgd(sin q)

d = 10g

7

(

v

sii

n

2

q) =

d =

The ball has more than enough kinetic energy to reach the back of the pinball machine.

6.7 m

(7)(2.4 m/s)2

(10)(9.81 m/s2)(sin 3.5°)

9. vi = 4.6 m/s

g = 9.81 m/s2

MEi = MEf


12

mvi2 + 1

2Iwi

2 = mgh

12

mvi2 + 1

22

5mr2

v

ri

2

= mvi21

2 + 1

5 =

1

7

0mvi

2 = mgh

h = 7

1

v

0i

g

2

= (1

(7

0

)

)

(

(

4

9

.

.

6

81

m

m

/s

/

)

s

2

2)

h = 1.5 m


V

10. r = 15 cm

wf = 102 rad/s

g = 9.81 m/s2

MEi = MEf

PEg = KEtrans + KErot

mgh = 12

mvf2 + 1

2Iwf

2

mgh = 12

(mr2)v

rf

2

+ 12

Iwf2 = 1

23

2Iwf

2 + 12

Iwf2 = 1

2Iwf

232

+ 1 = 54

Iwf2

mgh = gh = 54

Iwf2

h = 5r

6

2wg

f2

=

h = 2.0 × 101 m

(5)(0.15 m)2(102 rad/s)2

(6)(9.81 m/s2)

32

Ir2

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Chapter 9Fluid Mechanics

V

1. rgasoline = 675 kg/m3

Vs = 1.00 m3

g = 9.81 m/s2

FB = Fg rga

rso

s

line = m

rs

s

g rgasoline = Vsg rgasoline

FB = (1.00 m3)(9.81 m/s2)(675 kg/m3) = 6.62 × 103 N


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2. rr = 2.053 × 104kg/m3

Vr = (10.0 cm)3

g = 9.81 m/s2

apparent weight = 192 N

FB = Fg − apparent weight

FB = mrg − apparent weight = rrVrg − apparent weight

FB = (2.053 × 104 kg/m3)(10.0 cm)3(10−2 m/cm)3(9.81 m/s2) − 192 N = 201 N − 192 N

FB = 9 N

3. mh = 1.47 × 106 kg

Ah = 2.50 × 103 m2

rsw = 1.025 × 103 kg/m3

g = 9.81 m/s2

FB = Fg = mhg

FB = (1.47 × 106 kg)(9.81 m/s2) =

volume of hull submerged = Vsw = m

rs

s

w

w = rm

sw

h

h = V

As

h

w = Ah

m

rh

sw

h = = 0.574 m1.47 × 106 kg

(2.50 × 103 m2)(1.025 × 103 kg/m3)

1.44 × 107 N

4. msh = 1.47 × 106 kg

rsteel = 7.86 × 103 kg/m3

rgold = 1.93 × 104 kg/m3

Ah = 2.50 × 103 m2

rsw = 1.025 × 103 kg/m3

g = 9.81 m/s2

FB = Fg = mghg = rgoldVhg = rgold rm

st

s

e

h

el g

FB = =

h = V

As

h

w = A

m

hrgh

sw =

Ah

F

rB

swg

h = = 1.41 m3.54 × 107 N

(2.50 × 103 m2)(1.025 × 103 kg/m3)(9.81 m/s2)

3.54 × 107 N(1.93 × 104 kg/m3)(1.47 × 106 kg)(9.81 m/s2)

7.86 × 103 kg/m3

5. V = 166 cm3

apparent weight = 35.0 N

rw = 1.00 × 103 kg/m3

g = 9.81 m/s2

Fg = FB + apparent weight

rosmiumVg = rwVg + apparent weight

rosmium = rw + appare

V

nt

g

weight

rosmium = 1.00 × 103 kg/m3+

rosmium = 1.00 × 103 kg/m3+ 2.15 × 104 kg/m3

rosmium = 2.25 × 104 kg/m3

35.0 N(166 cm3)(10−6 m3/cm3)(9.81 m/s2)


V

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6. V = 2.5 × 10−3 m3


rw = 1.0 × 103 kg/m3

g = 9.81 m/s2


rebonyVg = rwVg + apparent weight

rebony = rw + appare

V

nt

g

weight

rebony = 1.0 × 103 kg/m3+

= 1.0 × 103 kg/m3+ 3.0 × 102 kg/m3

rebony = 1.3 × 103 kg/m3

7.4 N(2.5 × 10−3 m3)(9.81 m/s2)

7. Vsg = 7.52 cm3

msg = 45.0 g

Vlg = 7.38 × 10−6 m3

g = 9.81 m/s2

rsg = m

Vs

s

g

g =

r of solid gallium =

Fg = FB

msgg = rlgVlgg

rlg = m

Vl

s

g

g = 7

4

.

5

3

.

8

0

××

1

1

0

0−

−

6

3

m

kg3

r of liquid gallium = 6.10 × 103 kg/m3

5.91 × 103 kg/m3

45.0 × 10−3 kg(7.62 cm3)(10−6 m3/cm3)

8. rplatinum = 21.5 g/cm3

rw = 1.00 g/cm3


g = 9.81 m/s2


mg = rwVg + apparent weight = rwrpla

m

tinumg + apparent weight

mg 1 − rpla

r

t

w

inum = apparent weight

m = =

m = =

m = 4.30 kg

40.2 N(9.81 m/s2)(0.953)

40.2 N(9.81 m/s2)(1 − 0.047)

40.2 N

(9.81 m/s2)1 − 1

2

.

1

0

.

0

5

g

g

/

/

c

c

m

m

3

3apparent weight

g 1 − rpla

r

t

w

inum

9. rlithium = 534 kg/m3

rgasoline = 675 kg/m3

Vgasoline = 5.93 × 10−4 m3

Fg = FB

mlithiumg = rgasolineVgasolineg

mlithium = rgasolineVgasoline = (675 kg/m3)(5.93 × 10−4 m3)

mlithium =

Vlithium = m

rl

l

i

i

t

t

h

h

i

i

u

u

m

m = 5

0

3

.

4

40

k

0

g/

k

m

g3 = 7.49 × 10−4 m3

0.400 kg


V

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10. h1 = 5.00 cm

h2 = 4.30 cm

A = 3.4 m2

rw = 1.0 × 103kg/m3

Before the mass is submerged,

Fg = FB

(mp + mb + mm)g = rwV1g = rwAh1g

After the mass is submerged,

(mp + mb)g = rwV2g = rwAh2g

Substituting the second equation into the first,

rwAh2g + mmg = rwAh1g

mm = rwA(h1 − h2)

mm = (1.0 × 103 kg/m3)(3.4 m2)(5.00 cm − 4.3 cm)(10−2 m/cm)

mm = (1.0 × 103kg/m3)(3.4 m2)(0.70 × 10−2 m)

mm = 24 kg

2. P = 1.01 × 105 Pa

F = 2.86 × 108 N

A = P

F =

1

2

.

.

0

8

1

6

××

1

1

0

05

8

P

N

a =

A = 4pr2

r = 4

A

p = 2.83 ×4

1

p03

m2

= 15.0 m

2.83 × 103 m2


1. P = 1.50 × 106 Pa

F = 1.22 × 104 N

A = P

F =

1

1

.

.

5

2

0

2

××

1

1

0

06

4

P

N

a = 8.13 × 10−3 m2

3. F = 5.0 N

P = 9.6 × 103 PaA =

P

F =

9.6

5

×.0

10

N3 Pa

= 5.2 × 10−4 m2

4. m = 1.40 × 103 kg

h = 0.076 m

rice = 917 kg/m3

P1 = P2

A

F1

1 =

A

F2

2

m

A1

g =

m

Aic

2

eg = m

Vic

i

e

c

h

e

g = ricehg

A1 = r

m

iceh = = 2.0 × 101 m21.40 × 103 kg

(917 kg/m3)(0.076 m)


V

5. A = 1.54 m2

P1 = (1 + 1.00 × 10−2)P2

P2 = 1.013 × 105 Pa

g = 9.81 m/s2

F = PnetA = (P1 − P2)A = (1 + 1.00 × 10−2 − 1)P2A = (1.00 × 10−2)P2A

F = (1.00 × 10−2)(1.013 × 105 Pa)(1.54 m2) =

m = F

g =

1.

9

5

.

6

81

×m

10

/

3

s2N

= 159 kg

1.56 × 103 N

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6. r = 6.0 km

P = 1.2 × 1016 Pa

F = PA = P(4pr2) = (1.2 × 1016 Pa)(4p)(6.0 × 103 m)2

F = 5.4 × 1024 N

7. F1 = 4.45 × 104 N

h1 = 448 m

h2 = 8.00 m

P1 = P2

A

F1

1 =

A

F2

2

A

F1

1

h

h1

1 =

F1

V

h1 = A

F2

2

h

h2

2 =

F2

V

h2

F2 = F

h1h

2

1 =

F2 = 2.49 × 106 N

(4.45 × 104 N)(448 m)

8.00 m

8. h = 760 mm

r = 13.6 × 103 kg/m3

g = 9.81 m/s2

P = A

F =

m

A

g =

m

A

g

h

h =

m

V

gh = rgh

P = (13.6 × 103 kg/m3)(9.81 m/s2)(760 × 10−3 m) = 1.0 × 105 Pa

9. m = 2.4 × 1013 kg

A = 3.14 km2

g = 9.81 m/s2

P = A

F =

m

A

g =

P = 7.5 × 107 Pa

(2.4 × 1013 kg)(9.81 m/s2)(3.14 km2)(106 m2/km2)

10. F = 4.4 × 103 N

A = 2.9 × 10−2 m2

Po = 1.0 × 105 Pa

P = A

F =

2

4

.9

.4

××1

1

0

0−2

3

m

N2 =

Pgauge = P − Po = 1.5 × 105 Pa − 1.0 × 105 Pa = 5 × 104 Pa

1.5 × 105 Pa

1. P = 6.9 × 104 Pa

r = 0.55 kg/m3

g = 9.81 m/s2

Po = 1.01 × 105 Pa

P = Po + pgh

h = P

r−

g

Po = =

h = −5.9 × 103 m = 5.9 km above sea level

−3.2 × 104 Pa(0.55 kg/m3)(9.81 m/s2)

6.9 × 104 Pa − 1.01 × 105 Pa

(0.55 kg/m3)(9.81 m/s2)

2. P − Po = 1.47 × 106 Pa

r = 1.00 × 103 kg/m3

g = 9.81 m/s2

P = Po + rgh

h = P

r−

g

Po =

h = 1.50 × 102 m

1.47 × 106 Pa(1.00 × 103 kg/m3)(9.81 m/s2)



V

3. Po = 9.0 × 104 Pa

P = 5.00 × 106 Pa

r = 1.0 × 103 kg/m3

g = 9.81 m/s2

P = Po + rgh

h = P

r−

g

Po = =

h = 5.0 × 102 m

4.91 × 106 Pa(1.0 × 103 kg/m3)(9.81 m/s2)

5.00 × 106 Pa − 9.0 × 108 Pa(1.0 × 103 kg/m3)(9.81 m/s2)

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4. P = 4.03 × 105 Pa

Po = 1.01 × 105 Pa

r = 1.025 × 103 kg/m3

g = 9.81 m/s2

P = Po − rgh

h = P

r−

g

Po = =

h = 30.0 m

3.02 × 105 Pa(1.025 × 103 kg/m3 (9.81 m/s2)

4.03 × 105 Pa − 1.01 × 105 Pa(1.025 × 103 kg/m3)(9.81 m/s2)

5. h = 9.1 m

r = 1.0 × 103 kg/m3

g = 9.81 m/s2

Pgauge = P − Po = rgh

Pgauge = (1.0 × 103 kg/m3)(9.81 m/s2)(9.1 m) = 8.9 × 104 Pa

6. h = 86 m

r = 1.29 kg/m3

Po = 1.01 × 105 Pa

g = 9.81 m/s2

P = Po + rgh = 1.01 × 105 Pa + (1.29 kg/m3)(9.81 m/s2)(86 m) = 1.01 × 105 Pa + 1.1 × 103 Pa

P = 1.02 × 105 Pa

7. r = 13.6 × 103 kg/m3

Po = 1.01 × 105 Pa

h = 1.50 m

g = 9.81 m/s2

P = Po + rgh = 1.01 × 105 Pa + (13.6 × 103 kg/m3)(9.81 m/s2)(1.50 m)

= 1.01 × 105 Pa + 2.00 × 105 Pa

P = 3.01 × 105 Pa

8. Po = 9.10 × 106 Pa

h = −1.00 km

P = 8.60 × 106 Pa

gv = 8.87 m/s2

P = Po + rgVh

r = P

g

−

Vh

Po = =

r = 56 kg/m3

−5.0 × 105 Pa(8.87 m/s2)(−1.00 × 103 m)

8.60 × 106 Pa − 9.10 × 106 Pa(8.87 m/s2)(−1.00 × 103 m)

9. Po = 1.01 × 105 Pa

P = 2.23 × 105 Pa

h = 3.99 m

g = 9.81 m/s2

P = Po + rgh

r = P

g

−h

Po = =

r = 3.12 × 103 kg/m3

1.22 × 105 Pa(9.81 m/s2)(3.99 m)

2.23 × 105 Pa − 1.01 × 105 Pa

(9.81 m/s2)(3.99 m)

10. Po = 1.01 × 105 Pa

P = 1.29 × 105 Pa

h = 3.99 m

g = 9.81 m/s2

P = Po + rgh

r = P

g

−h

Po = =

r = 7.2 × 102 kg/m3

2.8 × 104 Pa(9.81 m/s2)(3.99 m)

1.29 × 105 Pa −1.01 × 105 Pa

(9.81 m/s2)(3.99 m)


V

1. ∆P = P1 − P2 = 1.5 × 103 Pa

v2 = 17.0 m/s

r = 1.00 × 103 kg/m3

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

Assume v1 ≈ 0 m/s, so the above equation simplifies to

∆P = rg(h2 − h1) + 12

rv22

h2 − h1 = ∆r

P

g −

v

22

g

2

= − (2

(

)

1

(

7

9

.

.

0

81

m

m

/s

/

)

s

2

2)

h = h2 − h1 = 0.15 m − 14.7 m = −14.6 m

h = h2 − h1 = −14.6 m = 14.6 m below the surface

1.5 × 103 Pa(1.00 × 103 kg/m3)(9.81 m/s2)

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2. ∆x = 120 m

∆y = − 69 m

g = 9.81 m/s2

To calculate v2, use the equations for a horizontally-launched projectile.

v2 = vx = ∆∆

x

t

∆y = vy,i∆t − 12

g∆t2

vy,i = 0 m/s, so

∆y = − 12

g∆t2

∆t = −2

g

∆y

v2 =

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

Because P1 = P2, and assuming v1 ≈ 0 m/s, Bernoulli’s equation simplifies to

rg(hi − h2) = 12

rv22 = 1

2r

2

h1 − h2 = = −∆4

x

∆

2

y

h = h1 − h2 = (−4

(

)

1

(

2

−0

6

m

9.0

)2

m) = 52 m

∆x2

2g−2

g

∆t

∆x

−2

g

∆y

∆x

−2

g

∆y

3. v1 = 2.00 m/s

v2 = 7.93 m/s

g = 9.81 m/s2

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 +rgh2

P1 = P2, so

rg(h1 − h2) = 12

r(v22 + v1

2)

h1 − h2 = v2

2

2

−g

v12

=

h1 − h2 = = (2

5

)(

8

9

.9

.8

m

1 m

2/s

/

2

s2)

h = h1 − h2 = 3.00 m

62.9 m2/s2 − 4.00 m2/s2

(2)(9.81 m/s2)

(7.93 m/s)2 − (2.00 m/s)2

(2)(9.81 m/s2)



V

4. v1 = 24.45 m/s

v2 = 0.55 m/s

g = 9.81 m/s2

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + pgh2

P1 = P2, so

rg(h2 − h1) = 12

r(v12 − v2

2)

h2 − h1 = v1

2

2

−g

v22

=

h2 − h1 = = (2

5

)

9

(

7

9

.

.

5

81

m

m

2/

/

s

s

2

2)

h = h2 − h1 = 30.5 m

597.8 m2/s2 − 0.30 m2/s2

(2)(9.81 m/s2)

(24.45 m/s)2 − (0.55 m/s)2

(2)(9.81 m/s2)

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5. v1 = 2.50 m/s

h1 − h2 = 3.00 m

g = 9.81 m/s2

P1 + 12

rv12 + rgh1 = r2 + 1

2rv2

2 + rgh2

P1 = P2, so

12

rv12 + rg(h1 − h2) = 1

2rv2

2

v2 =√

v12+ 2g(h1− h2) =√

(2.50m/s)2 + (2)(9.81m/s2)(3.00m)

v2 =√

6.25 m2/s2 + 58.9m2/s2 =√

65.2 m2/s2

v2 = 8.07 m/s

6. v1 = 0.90 m/s

∆P = P2 − P1 = 311 Pa

P = 1.025 × 103 kg/m3

P1 = 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

Because h1 = h2, so the above equation simplifies to

12

rv12 − ∆P = 1

2rv2

2

v2 = v12− 2∆

PP = (0.90m/s)2 −

1.02

(2

5)

×(3

1

10

13 P

ka

g

)/m3v2 =

√0.81 m2/s2 − 0.607 m2s2 =

√0.20 m2/s2

v2 = 0.45 m/s

7. rair = 1.3 kg/m3

rmercury = 1.36 × 104 kg/m3

h2 − h1 = 3.5 cm

g = 9.81 m/s2

For the static mercury columns, v1 = v2.

P1 + 12

rmercuryv12 + rmercurygh1 = P2 + 1

2rmercuryv2

2 + rmercurygh2

P1 − P2 = rmercuryg(h2 − h1)

For the flowing air, v1 = 0 m/s, because the tube is fixed to the wing, and h1 = h2.

P1 + 12

rairv12 + rairgh1 = P2 + 1

2rairv2

2 + rairgh2

P1 − P2 = 12

rairv22

Substituting the first equation into the second,

rmercury g (h2 − h1) = 12

rairv22

v2 = 2rmercur

ryg

ai

(r

h2− h1) =

v2 = 85 m/s

(2)(1.36 × 104 kg/m3)(9.81 m/s2)(3.5 × 10−2 m)

1.3 kg/m3


V

8. r = 950 kg/m3

v1 = 10.00 m/s

v2 = 9.90 m/s

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

h1 = h2, so

P2 − P1 = 12

r(v12 − v2

2) = 12

(950 kg/m3)[(10.00 m/s)2 − (9.90 m/s)2]

P2 − P1 = 12

(950 kg/m3)(100.0 m2/s2 − 98.0 m2/s2) = 12

(950 kg/m3)(2.0 m2/s2)

∆P = P2 − P1 = 950 Pa

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9. A1 = 7.8 × 10−3 m2

v1 = 2.0 m/s

A2 = 3.1 × 10−4 m2

r = 1.00 × 103 kg/m3

h2 − h1 = 10.5 m

g = 9.81 m/s2

From the continuity equation, an expression for v2 can be derived.

A1v1 = A2v2

v2 = A

A1

2v1

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

P1 − P2 = 12

r(v22 − v1

2) + rg(h2 − h1) = 12

rv12

A

A1

2

2

2 − 1 + rg(h2 − h1)

P1 − P2 = 12

(1.00 × 103 kg/m3)(2.0 m/s)2((7

3

.

.

8

1

××

1

1

0

0

−

−

3

4m

m

2

2)

)

2

2 − 1+ (1.00 × 103 kg/m3)(9.81 m/s2)(10.5 m)

P1 − P2 = 12

(1.00 × 103 kg/m3)(2.0 m/s)2(6.3 × 102 − 1) + 1.03 × 105 Pa

P1 − P2 = 12

(1.00 × 103 kg/m3)(2.0 m/s)2 (6.3 × 102) + 1.03 × 105 Pa

P1 − P2 = 1.3 × 106 Pa + 1.03 × 105 Pa

∆P = P1 − P2 = 1.4 × 106 Pa

10. A1 = 9.2 × 10−2 m2

v1 = 8.3 m/s

A2 = 4.6 × 10−2 m2

r = 1.0 × 103 kg/m3

From the continuity equation, an expression for v2 can be derived.

A1v1 = A2v2

v2 = A

A1

2v1

P1 + 12

rv12 + rgh1 = P2 + 1

2rv2

2 + rgh2

h1 = h2, so

P1 − P2 = 12

r(v22 − v1

2) = 12

rv12

A

A

2

12

2 − 1P1 − P2 = 1

2(1.0 × 103 kg/m3)(8.3 m/s)((9

4

.

.

2

6

××

1

1

0

0

−

−

2

2m

m

2

2)

)

2

2 − 1P1 − P2 = 1

2(1.0 × 103 kg/m3)(8.3 m/s)2(4 − 1) = 1

2(1.0 × 103 kg/m3)(8.3 m/s)2(3)

∆P = P1 − P2 = 1.0 × 105 Pa

1. T1 = 184

T2 = 331 K

V1 = 3.70 m3

At constant pressure:

V

T1

1 = T

V

2

2

V2 = V

T1T

1

2 = (3.70 m

18

3

4

)(

K

331 K) = 6.66 m3



V

2. V1 = 0.455 m3

P1 = 9.1 × 106 Pa

P2 = 5.0 × 104 Pa

At constant temperature:

P1V1 = P2V2

V2 = P

P1V

2

1 = = 83 m3(9.1 × 106 Pa)(0.455 m3)

5.0 × 104 Pa

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3. V1 = 65.4 m3

P1 = 1.03 × 105 Pa

P2 = 5.84 × 104 Pa


P1V1 = P2V2

V2 = P

P1V

2

1 = = 115 m3(1.03 × 105 Pa)(65.4 m3)

5.84 × 104 Pa

4. T1 = 295 K

P1 = 1.01 × 105 Pa

T2 = 225 K

P2 = 5.10 × 104 Pa

N = 5.55 × 1022 particles

kB = 1.38 × 10−23 J/K

PV = NkBT

V1 = Nk

PB

1

T1 =

V1 =

V2 = Nk

PB

2

T2 =

V2 = 3.38 × 10−3 m3

(5.55 × 1022)(1.38 × 10−23 J/K)(225 K)

5.10 × 104 Pa

2.24 × 10−3 m3

(5.55 × 1022)(1.38 × 10−23 J/K)(295 K)

1.01 × 105 Pa

5. P1 = 7.5 × 104 Pa

T1 = 250 K

P2 = 2.0 × 106 Pa

At constant volume:

T

P1

1 =

T

P

2

2

T2 = P

P2T

1

1 = = 6.7 × 103 K(2.0 × 106 Pa)(250 K)

7.5 × 104 Pa


V

6. V1 = 1.00 m3

T1 = 295 K

V2 = 65.4 m3

At constant pressure:

V

T1

1 = T

V

2

2

T2 = V

V2T

1

1 = (65.4

1

m

.0

3

0

)

m

(2395 K)

= 1.93 × 104 K

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7. N = 2.1 × 1057 particles

P = 2.1 × 1016 Pa

V = 2.1 × 1025 m3

KB = 1.38 × 10−23 J/K

PV = NkBI

T = N

P

k

V

B = = 1.5 × 107 K

(2.1 × 1016 Pa)(2.1 × 1025 m3)(2.1 × 1057)(1.38 × 10−23 J/K)

8. T1 = 295 K

P1 = 2.50 × 105 Pa

T2 = 506 K

At constant volume:

T

P1

1 =

T

P

2

2

P2 = P

T1T

1

2 = = 4.29 × 105 Pa(2.50 × 105 Pa)(506 K)

295 K

9. T = 1.0 × 102 K

V = 3.3 × 1043 m2

r = 10.0 atoms/cm3

kB = 1.38 × 10−23 J/K

N = rV

PV = NkBT = rVkBT

P = rkBT = (10.0 atoms/cm3)(106 cm3/m3)(1.38 × 10−23 J/K)(1.0 × 102 K)

P = 1.38 × 10−14 Pa

10. P1 = 1.42 × 105 Pa

V1 = 1.00 m3

V2 = 1.83 m3


P1V1 = P2V2

P2 = P

V1V

2

1 = = 7.76 × 104 Pa(1.42 × 105 Pa)(1.00 m3)

1.83 m3


Chapter 10Heat

V

1. T1 = 463 K

T2 = 93 K

TC,1 = (T − 273)°C = (463 − 273)°C =

TF,1 = 9

5 TC,1 + 32 =

9

5 (1.90 × 102)°F + 32°F = 342°F + 32°F =

TC,2 = (T − 273)°C = (93 − 273)°C =

TF,2 = 9

5 TC,2 + 32 =

9

5 (−1.80 × 102)°F + 32°F = −324°F + 32°F = −292°F

−180 × 102 °C

374°F

1.90 × 102 °C


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2. T = 330.0 KTC = (T − 273)°C = (330.2 − 273.2)°C =

TF = 9

5 TC + 32 =

9

5 (56.8)°F + 32°F = 102°F + 32°F = 134°F

56.8°C

3. Ti = 237 K

Tf = 283 K

∆TC = (Tf − 273)°C − (Ti − 273)°C = Tf − Ti

∆TC = (283 − 237)°C =

∆TF = 9

5TC, f + 32°F −

99

5TC, i + 32°F =

9

5 ∆TC°F

∆TF = 9

5(46)°F = 83°F

46° C

4. TF,i = − 5 °F

TF,f = + 37°FTC,i =

5

9(TF, i − 32)°C =

5

9(−5 − 32)°C =

5

9(− 37)°C = −21°C

TC,f = 5

9(TF, f − 32)°C =

5

9(37 − 32)°C =

5

9(5)°C = 3°C

∆T = (TC,f + 273 K) − (TC,i + 273 K) = TC,f − TC,i

∆T = [3 − (−21)] K = 24 K

6. TC,1 = 47°C

TC, 2 = 42°CT1 = (TC,1 + 273)K = (47 + 273) K =

T2 = (TC,2 + 273)K = (42 + 273) K = 315 K

3.20 × 102 K

5. TF = 78°FT = TC + 273 =

5

9(TF − 32) + 273

T = 5

9 (78 − 32) K + 273 K =

5

9 (46) K + 273 K = 26 K + 273 K

T = 299 K

9. TF = 188.6°FTC =

5

9(TF − 32.0)°C =

5

9(188.6 − 32.0)°C =

5

9(156.6)°C

TC = 87.00°C


V

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8. TF = 2192°FTC =

5

9(TF − 32)°C =

5

9(2192 − 32)°C =

5

9(2.160 × 103)°C

TC = 1.200 × 103°C

10. T = 2.70 K TC = (T − 273)°C = (2.70 − 273)°C = −270°C

1. m = 5.25 g

vi = 3.27 m/s

∆Upenny = 12

∆U

k = 2.03 J/1.00° C

∆PE + ∆KE + ∆U = 0

The change in potential energy from before to after impact is zero, as is KEf.

∆PE + ∆KE + ∆U = 0 + KEf − KEi + ∆V = − KEi + ∆U = 0

∆U = KEi = 12

mvi2

∆Upenny = 12

∆U = 12

(12

mvi2) =

m

4

vi2

∆T = ∆Up

kenny =

∆T = = 6.91 × 10−3 °C(5.25 × 10−3 kg)(3.27 m/s)2

(4)(2.03 J/1.00°C)

mvi2

4k

2. h = 9.5 m

g = 9.81 m/s2

∆Uacorn = (0.85)∆U

k/m = 12

1

0

.0

0

°J

C

/kg

∆PE + ∆KE + ∆U = 0

The change in kinetic energy from before the acorn is dropped to after it has landedis zero, as is PEf.

∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0

∆U = PEi = mgh

∆Uacorn = (0.85) ∆U = (0.85)mgh

∆T = ∆U

kacorn =

(0.85

k

)mgh =

(0

(

.

k

8

/

5

m

)g

)

h

∆T = = 6.6 × 10−2 °C(0.85)(9.81 m/s2)(9.5 m)

12

1

0

.

0

0°J

C

/kg


7. TF = −67°FTC =

5

9(TF − 32) + 273K

T = 5

9(− 67 − 32) + 273K =

55

9(− 97) + 273K = (− 55 + 273)K

T = 218 K


V

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4. vi = 47.5 m/s

h = 151 m

m = 7.32 g

∆Utwig = (0.100)∆U

k = 8.5 J/1.0°C

g = 9.81 m/s2

∆PE + ∆KE + ∆U = 0

∆U = −∆PE − ∆KE = PEi − PEf + KEi − KEf

Both PEf and KEf equal zero.

∆U = PEi + KEi = mgh + 12

mvi2

∆Utwig = (0.100)(mgh + 12

mvi2)

∆T = ∆U

ktwig =

k

∆T =

∆T = = (0.1

8

0

.

0

5

)

J

(

/

1

°9

C

.1 J) = 0.22°C

(0.100)(10.8 J + 8.26 J)

8.5 J/°C

(0.100)[(7.32 × 10–3 kg)(9.81 m/s2)(151 m) + 12

(7.32 × 10–3 kg)(47.5 m/s)2]

8.5 J/1.0°C

3. PEi = 6.2 J

∆Ucone = (0.100)∆U

k = 180 J/1.0°C

∆PE + ∆KE + ∆U = 0

The change in kinetic energy from before the ice cream cone is dropped to after it haslanded is zero, as is PEf.

∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0

∆U = PEi

∆Ucone = (0.100)∆U = (0.100)PEi

∆T = ∆U

kcone =

(0.10

k

0)PEi

∆T = (0

1

.

8

1

0

00

J/

)

1

(

.

6

0

.

°2

C

J) = 3.4 × 10−3 °C

5. h = 561.7 m

∆U = 105 J

g = 9.81 m/s2

∆PE = ∆KE + ∆U = 0

When the stone lands, its kinetic energy is transferred to the internal energy of thestone and the ground. Therefore, overall, ∆KE = 0 J

∆PE = PEf − PEe = 0 − mgh = − ∆U

m = ∆g

U

h = = 1.91 × 10−2 kg = 19.1 g

105 J(9.81 m/s2)(561.7 m)

6. h = 5.5 m

g = 9.81 m/s2

∆U = 2.77 × 103 J

∆PE + ∆KE + ∆U = 0

When the prey lands, all of the kinetic energy is transferred to the ground. Therefore∆KE = 0 J

∆PE = PEf − PEi = 0 − mgh = −∆U

m = ∆g

U

h = = 51 kg

2.77 × 103 J(9.81 m/s2)(5.5 m)

7. vf = 0 m/s

vi = 13.4 m/s

∆U = 5836 J

∆PE + ∆KE + ∆U = 0

The bicyclist remains on the bicycle, which does not change elevation, so ∆PE = 0 J.

∆KE = KEf − KEi = 0 − 12

mvi2 = −∆U

m = 2

v

∆

i2U =

(

(

1

2

3

)(

.4

58

m

36

/s)

J)2 = 65.0 kg

(0.100)(mgh + 12

mvi2)


V

10. ∆KE = KEf − KEi= −7320 J

∆Uhands = (1 − 0.300)∆U

∆PE + ∆KE + ∆U = 0

The hands don’t change height, so ∆PE = 0 J.

∆U = −∆KE = −(−7320 J) = 7320 J

∆Uhands = (1− 0.300)(7320 J) = (0.700)(7320 J) = 5120 J

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1. mw = 15 g = 0.015 kg

mhp = 15 g = 0.015 kg

∆Tw = 1.0°C

∆Thp = 1.6°C

cp,w = 4186 J/kg • °C

cp,hpmhp∆Thp = cp,wmw∆Tw

cp,hp = cp

m,w

h

m

p∆w

T

∆

h

T

p

w =

cp,hp = 2.6 × 103 J/kg • °C

(4186 J)(kg • °C)(0.015 kg)(1.0°C)

(0.015 kg)(1.6°C)

2. mv = 0.340 kg

mw = 1.00 kg

Tv,i = 21.0°C

Tw,i = 90.0°C

Tf = 73.7°C

cp,w = 4186 J/kg • °C

cp,vmv∆Tv = cp,wmw∆Tw

∆Tw = Tw,i − Tf = 90.0°C − 73.7°C = 16.3°C

∆Tv = Tf − Tv,i = 73.7°C − 21.0°C = 52.7°C

cp,v = cp,

mwm

v∆w

T

∆

v

Tw =

cp,v = 3.81 × 103 J/kg •°C

(4186 J/kg • °C)(1.00 kg)(16.3°C)

(0.340 kg)(52.7°C)

3. ma = 0.250 kg

mw = 1.00 kg

∆Tw = 1.00° C

∆Ta = 17.5° C

cp,w = 4186 J/kg • °C

cp,ama∆Ta = cp,wmw∆Tw

cpa = =

cp,a = 957 J/kg • °C

(4186 J/kg • °C)(1.00 kg)(1.00°C)

(0.250 kg)(17.5° C)

cp,wmw∆Twma∆Ta

4. mi = 3.0 kg

mw = 5.0 kg

∆Tw = 2.25° C

∆Ti = 29.6° C

cp,w = 4186 J/kg • °C

cp,imi∆Ti = cp,wmw∆Tw

cp,i = =

cp,i = 530 J/kg • °C

(4186 J/kg • °C)(5.0 kg)(2.25°C)

(3.0 kg)(29.6°C)

cp,wmw∆Twmi∆Ti


8. ∆KE = KEf − KEi

= −2.15 × 104 J

∆PE + ∆KE + ∆U = 0

There is no change in the height of the sticks, so ∆PE = 0 J.

∆U = −∆KE = −(−2.15 × 104 J) = 2.15 × 104 J

9. vi = 20.5 m/s

vf = 0 m/s

m = 61.4 kg

∆PE + ∆E + ∆U = 0

The height of the skater does not change, so ∆PE = 0 J.

∆KE = KEf − KEi = 0 − 12

mvi2

∆U = −∆KF = −(− 12

mvi2) = 1

2(61.4 kg)(20.5 m/s)2 = 1.29 × 104 J


V

7. Q = 45 × 106 J

∆Ta = 55° C

cp,a = 1.0 × 103 J/kg • °C

cp,a = ma

Q

∆Ta

ma = cp,a

Q

∆Ta = a = 818 kg

45 × 106 J(1.0 × 103 J/kg • °C)(55°C)

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8. mc = 0.190 kg

Q = 6.62 × 104 J

cp,c = 387 J/kg • °C

cp,c = mc

Q

∆Tc

∆Tc = mc

Q

cp,c = = 9.00 × 102°C

6.62 × 104 J(0.190 kg)(387 J/kg • °C)

9. mt = 0.225 kg

cp,t = 2.2 × 103 J/kg • °C

Q = −3.9 × 104 J

cp,t = mt

Q

∆Tt

∆Tt = m

Q

tcp,t = = −79°C

−3.9 × 104 J(0.225 kg)(2.2 × 103 J/kg • °C)

10. cp,t = 140 J/kg • °C

mt = 0.23 kg

Q = −3.0 × 104 J

cp,t = mt

Q

∆Tt

∆Tt = m

Q

tcp,t = = −930°C

−3.0 × 104 J(0.23 kg)(140 J/kg • °C)

5. Q = −1.09 × 1010 J

cp,w = 4186 J/kg • °C

∆Tw = −5.0°C

cp,w = mw

Q

∆Tw

mw = cp,w

Q

∆Tw = = 5.2 × 105 kg

−1.09 × 1010 J(4186 J/kg • °C)(−5.0°C)

6. cp,b = 121 J/kg • °C

Q = 25 J

∆Tb = 5.0°C

cp,b = mb

Q

∆Tb

mb = cp,b

Q

∆Tb = = 4.1 × 10−2 kg

25 J(121 J/kg • °C)(5.0°C)


1. Q = 1.10 × 106 J

m = 5.33 kg

Q = mLf

Lf = m

Q =

1.1

5

0

.3

×3

1

k

0

g

6 J = 2.06 × 105 J/kg

2. Q = 9.6 × 105 J

m = 5.33 kg

Lf = m

Q =

9.

5

6

.3

×3

1

k

0

g

5 J = 1.8 × 105 J/kg

3. Q = 3.72 × 105 J

m = 0.65 kg

Lsubl = m

Q =

3.7

0

2

.6

×5

1

k

0

g

5 J = 5.7 × 105 J/kg

4. Q = 8.5 × 104 J

Lv = 2.26 × 106 J/kg

Q = mLv

m = L

Q

v =

2.2

8

6

.5

××1

1

0

06

4

J/

J

kg = 3.8 × 10−2 kg


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6. m = 250 kg

Qtot = 1.380 × 108 J

cp,i = 605 J/kg •°C

Lf = 2.47 × 105 J/kg

Qtot = mcp,i ∆T + mLf

mcp,i ∆T = Qtot − mLf

∆T = m

Q

ct

p

ot

,i −

c

L

p

f

,i

∆T = −

∆T = (910 − 408)°C = 5.0 × 102°C

2.47 × 105 J/kg605 J/kg •°C

1.380 × 108 J(250 kg)(605 J/kg •°C)

Givens Solutions

7. m = 1.45 kg

Tf = 330°C

Qtot = 4.46 × 104 J

cp,i = 120 J/kg •°C

Lf = 2.45 × 104 J/kg

Qtot = mcp,i ∆T + mLf

∆T = m

Q

ct

p

ot

,i −

c

L

p

f

,i = −

∆T = (2.6 × 102 − 2.0 × 102)°C = 60°C

∆T = Tf − Ti = 330°C − Ti = 60°C

Ti = 330°C − 60°C = 270°C

2.45 × 104 J/kg120 J/kg •°C

4.46 × 104 J(1.45 kg)(120 J/kg •°C)

8. m = 0.75 g

Ti = 100.0°C

Qtot = 2.0 × 103 J

cp,w = 4200 J/kg •°C

Lv = 2.26 × 106 J/kg

Qtot = mcp,w ∆T + mLv

∆T = m

Q

ct

p

o

,

t

w −

c

L

p,

v

w = −

∆T = (6.3 × 102 − 5.4 × 102)°C = 90°C

∆T = Ti − Tf = 100.0°C − Tf = 90°C

Tf = 100.0°C − 90°C = 10°C

2.26 × 106 J/kg4200 J/kg •°C

2.0 × 103 J(0.75 × 10−3 kg)(4200 J/kg •°C)

9. m = 0.35 kg

Lf = 8.02 × 104 J/kg

Q = mLf = (0.35 kg)(8.02 × 104 J/kg)

Q = 2.8 × 104 J

10. m = 55.0 g

Ti = 20.0°C

Tf = 357°C

cp,m = 130 J/kg •°C

Lv = 2.95 × 105 J/kg

Qtot = mcp,m ∆T + mLv = mcp,m (Tf − Ti) + mLv

Qtot = (55.0 × 10−3 kg)(130 J/kg •°C)(357°C − 20.0°C)

+ (55.0 × 10−3 kg)(2.95 × 105 J/kg)

Qtot = (55.0 × 10−3 kg)(130 J/kg •°C)(337°C)

+ (55.0 × 10−3 kg)(2.95 × 105 J/kg) = 2.4 × 103 J + 1.62 × 104 J

Qtot = 1.86 × 104 J

5. Q = 2.11 × 106 J

Lv = 8.45 × 105 J/kg

m = L

Q

v =

8.

2

4

.

5

11

××10

150

J

6

/

J

kg = 2.50 kg

Chapter 11Thermodynamics

V

1. W = 3.29 × 106 J

∆V = 2190 m3P =

∆W

V =

3.

2

2

1

9

9

×0

1

m

03

6 J = 1.50 × 103 Pa = 1.50 kPa


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2. W = 1.06 × 106 J

Vi = 2.80 × 103 m3

Vf = 8.50 × 105 m3

∆V = Vf − Vi = 8.50 × 105 m3 − 2.80 × 103 m3 = 8.47 × 105 m3

P = ∆W

V =

8

1

.4

.0

7

6

××1

1

0

05

6

m

J3 = 1.25 Pa

3. W = 1.3 J

∆V = 5.4 × 10−4 m3P =

∆W

V =

5.4 ×1

1

.3

0−J4 m3 = 2.4 × 103 Pa = 2.4 kPa

4. W = 472.5 J

P = 25.0 kPa = 2.50 × 104 Pa∆V =

W

P =

2.50

47

×2

1

.5

04J

Pa = 1.89 × 10−2 m3

5. W = 393 J

P = 655 kPa = 6.55 × 105 Pa∆V =

W

P =

6.55

3

×93

10

J5 Pa

= 6.00 × 10−4 m3

6. W = 0.20 J

P = 39 Pa∆V =

W

P =

0

3

.

9

20

Pa

J = 5.1 × 10−3 m3

7. W = 3.2 × 102 J

P = 4.0 × 105 Pa∆V =

W

P =

4

3

.0

.2

××1

1

0

05

2

P

J

a = 8.0 × 10−4 m3

8. P = 2.07 × 107 Pa

∆V = 0.227 m3

W = P ∆V = (2.07 × 107 Pa)(0.227 m3) = 4.70 × 106 J

9. Vi = 3.375 × 10−6 m3

Vf = 5.694 × 10−6 m3

P = 101.33 kPa

= 1.0133 × 105 Pa

∆V = Vf − Vi = 5.694 × 10−6 m3 − 3.375 × 10−6 m3 = 2.319 × 10−6 m3

W = P ∆V = (1.0133 × 105 Pa)(2.319 × 10−6 m3) = 0.2350 J

10. Vi = 2.0 × 10−3 m3

Vf = 5.0 × 10−3 m3

P = 0.90 kPa = 900 Pa

∆V = Vf − Vi = 5.0 × 10−3 m3 − 2.0 × 10−3 m3 = 3.0 × 10−3 m3

W = P ∆V = (900 Pa)(3.0 × 10−3 m3) = 2.7 J

Holt Physics Solutions ManualV Ch. 11–2V Ch. 11–2

2. Ui = 39 J

Uf = 163 J

Q = 114 J

∆U = Uf − Ui = Q − W

W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui

W = 114 J − 163 J + 39 J = −10 J

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1. mi = 2.0 kg

mf = 15 kg

Ti = 28°C

Tf = 52°C

d = 2.3 mm

W = Fd = m2gd = (15 kg)(9.81 m/s2)(2.3 × 10−3 m) = 0.34 J

3. Ui = 8093 J

Uf = 2.092 × 104 J

Q = 6932 J

∆U = Uf − Ui = Q − W

W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui

W = 6932 J − 2.092 × 104 J + 8093 J = −5895 J

4. Q = 4.50 × 108 J

W = 3.21 × 108 J

∆U = Q − W = 4.50 × 108 J − 3.21 × 108 J = 1.29 × 108 J

5. Q = 632 kJ

W = 102 kJ∆U = Q − W = 632 kJ − 102 kJ = 5.30 × 102 kJ = 5.30 × 105 J

6. Q = 867 J

W = 623 J∆U = Q − W = 867 J − 623 J = 244 J

7. W = 192 kJ

∆U = 786 kJ

∆U = Q − W

Q = ∆U + W = 786 kJ + 192 kJ = 978 kJ

8. W = 602 kJ

∆U = 1.09 × 105 J

∆U = Q − W

Q = ∆U + W = 602 kJ + 109 kJ = 711 kJ


9. ∆U = 873 J ∆V = 0, so W =

∆U = Q − W

Q = ∆U + W = 873 J + 0 J = 873 J

0 J

10. ∆U = 986 J ∆V = 0, so W =

∆U = Q − W

Q = ∆U + W = 986 J + 0 J = 986 J

0 J

Section Five—Problem Bank V Ch. 11–3V Ch. 11–3

V

Givens SolutionsC

opyr

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©by

Hol

t, R

ineh

art a

nd W

inst

on.A

ll rig

hts

rese

rved

.

1. eff = 0.17

Qh = 5.5 × 109 JWnet = eff Qh = (0.17)(5.5 × 109 J) = 9.4 × 108 J

2. eff = 0.35

Qh = 7.37 × 108 J

4. eff = 0.29

Qh = 693 JWnet = eff Qh = (0.29)(693 J) = 2.0 × 102 J

7. Wnet = 544 J

eff = 0.2225Qh =

W

efn

fet =

0

5

.2

4

2

4

2

J

5 = 2.44 × 103 J

Wnet = eff Qh = (0.35)(7.37 × 108 J) = 2.6 × 108 J

3. eff = 0.15

Qh = 9.36 × 108 JWnet = eff Qh = (0.15)(9.36 × 108 J) = 1.4 × 108 J


5. eff = 0.11

Wnet = 1150 JQh =

W

efn

fet =

1

0

1

.

5

1

0

1

J = 1.0 × 104 J

6. eff = 0.19

Wnet = 998 JQh =

W

efn

fet =

9

0

9

.1

8

9

J = 5.3 × 103 J

8. Qh = 365 J

Qc = 223 Jeff = 1 −

Q

Q

h

c = 1 − 2

3

2

6

3

5

J

J = 1 − 0.611 = 0.389

9. Qh = 571 J

Qc = 463 Jeff = 1 −

Q

Q

h

c = 1 − 4

5

6

7

3

1

J

J = 1 − 0.811 = 0.189

10. Wnet = 128 J

Qh = 581 Jeff =

W

Qn

h

et = 1

5

2

8

8

1

J

J = 0.220


Chapter 12Vibrations and Waves

V

1. k = 420 N/m

x = 4.3 × 10−2 m

Felastic = −kx = −(420 N/m)(4.3 × 10−2 m) = −18 N


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2. k = 65 N/m

x = –1.5 × 10−1 mFelastic = −kx = −(65 N/m)(–1.5 × 10−1 m) = 9.8 N

3. k = 49 N/m

x = –1.2 × 10−1 mFelastic = −kx = −(49 N/m)(–1.2 × 10−1 m) = −5.9 N

4. k = 26 N/m

x = –5.0 × 10−2 mFelastic = −kx = −(26 N/m)(–5.0 × 10−2 m) = 1.3 N

5. Fg = –669 N

x = –6.5 × 10−2 m

Fnet = 0 = Felastic + Fg = −kx + Fg

Fg = kx

k = F

x

g = –6.5

–6

×6

1

9

0

N−2 m

= 1.0 × 104 N/m

6. Fg = –550 N

x = –15 m


Fg = kx

k = F

x

g = –

–

5

1

5

5

0

m

N = 37 N/m

7. Fg = 620 N

x = 7.2 × 10−2 m


Fg = kx

k = F

x

g = 7.2

6

×20

10

N−2 m

= 8.6 × 103 N/m

8. Felastic = 12 N

k = 180 N/m

Felastic = −kx

x = − Fel

kastic = −

18

1

0

2

N

N

/m = –0.067 m = −6.7 cm

9. Felastic = 52 N

k = 490 N/m

Felastic = −kx

x = − Fel

kastic = −

49

5

0

2

N

N

/m = − 0.11 m = −11 cm


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10. m = 3.0 kg

g = 9.81 m/s2

k = 36 N/m

Fnet = 0 = Felastic + Fg = −kx −mg

mg = −kx

x = − m

k

g = −

(3.0 k

3

g)

6

(

N

9.8

/m

1 m/s2) = –0.82 m = −82 cm

1. L = 3.0 × 10−1 m

T = 1.16 sT = 2p

L

g

T2 = 4p

g

2 L

g = 4p

T

2

2L

= = 8.8 m/s2(4p 2)(3.0 × 10−1 m)

(1.16 s)2

2. L = 0.650 m

T = 2.62 sT = 2p

L

g

T2 = 4p

g

2 L

g = 4p

T

2

2L

= (4p

(

2

2

)

.

(

6

0

2

.6

s

5

)

02

m) = 3.74 m/s2


3. L = 1.14 m

T = 3.55 sT = 2p

L

g

T2 = 4p

g

2 L

g = 4p

T

2

2L

= (4p

(

2

3

)

.5

(1

5

.1

s)

42m)

= 3.57 m/s2

4. L = 5.00 × 10−1 m

T = 2.99 sT = 2p

L

g

T2 = 4p

g

2 L

g = 4p

T

2

2L

= = 2.21 m/s2(4p 2)(5.00 × 10−1 m)

(2.99 s)2

5. f = 1.0 Hz

g = 9.81 m/s2 T = 2p L

g =

1

f

f

12 =

4pg

2 L

L = 4p

g2f 2 =

(4p9

2.8

)

1

(1

m

.0

/

s

s−

2

1)2 = 0.25 m = 25 cm


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8. L = 6.200 m

g = 9.819 m/s2 T = 2p L

g = 2p

9.6

8

.1

2

900mm

/s2 =

f = T

1 =

4.9

1

93 s = 0.2003 Hz

4.993 s

9. L = 2.500 m

g = 9.780 m/s2 T = 2p L

g = 2p

9.2

7

.8

5

000mm

/s2 =

f = T

1 =

3.1

1

77 s = 0.3148 Hz

3.177 s

10. L = 3.120 m

g = 9.793 m/s2 T = 2p L

g = 2p

9.3

7

.9

1

320mm

/s2 =

f = T

1 =

3.5

1

46 s = 0.2820 Hz

3.546 s

6. f = 0.50 Hz

g = 9.81 m/s2 T = 2p L

g =

1

f

f

12 =

4pg

2 L

L = 4p

g2f 2 =

(4p92.

)

8

(

1

0.

m

50

/s

s

2

−1)2 = 0.99 m = 99 cm

7. f = 2.5 Hz

g = 9.81 m/s2 T = 2p L

g =

1

f

f

12 =

4pg

2 L

L = 4p

g2f 2 =

(4p9

2.8

)(

1

2

m

.5

/

s

s−

2

1)2 = 4.0 × 10−2 m

1. f = 3.00 × 102 Hz

k = 8.65 × 104 N/m T = 2p m

k =

1

f

f

12 =

4pk

2 m

m = 4p

k2f 2 =

(

8

4

.6

p52)

×(3

1

0

0

0

4 N

s−/1m

)2 = 2.43 × 10−2 kg



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5. F = 66 N

T = 2.9 s

g = 9.81 m/s2

f = mg

m = F

g

T = 2p m

k

T 2 = 4p

k

2 m

k = 4p

T

2

2m

= 4

g

pT

2

2F

= (9.8

(

1

4pm

2

/

)

s

(26

)

6

(2

N

.9

)

s)2 = 32 N/m

6. f = 87 N

g = 9.81 m/s2

T = 0.64 s

f = mg

m = F

g =

9.8

8

1

7

m

N

/s2 = 8.9 kg

T = 2pm

k

T 2 = 4p2

k

m

k = 4p

T

2

2m

= 4

g

pT

2

2F

= (9.8

(

1

4pm

2)

/s

(28

)

7

(0

N

.6

)

4 s) = 850 N/m

3. k = 2.03 × 103 N/m

f = 0.79 HzT = 2p

m

k =

1

f

f

12 =

4pk

2 m

m = 4p

k2f 2 =

(

2

4

.

p03

2)

×(0

1

.

0

7

3

9

N

H

/

z

m

)2 = 82 kg

4. F = 32 N

T = 0.42 s

g = 9.81 m/s2

F = mg

m = F

g =

9.8

3

1

2

m

N

/s2 = 3.3 kg

T = 2p m

k

T 2 = 4p

k

2 m

k = 4p

T

2

2m

= 4

g

pT

2

2F

= = 730 N/m4p2 (32 N)

(9.81 m/s2)(0.42 s)2

2. T = 0.079 s

k = 63 N/mT = 2p

m

k

T 2 = 4p

k

2 m

m = 4

kT

p

2

2 = (63 N/m

4

)

p(20.079 s)2

= 9.96 × 10−3 kg


V

4. f = 288 Hz

l = 5.00 mv = fl = (288 Hz)(5.00 m) = 1.44 × 103 m/s

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7. f = 20.0 Hz

v = 331 m/s

v = fl

l = v

f =

3

2

3

0

1

.0

m

H

/

z

s = 16.6 m

7. k = 364 N/m

m = 24 kgT = 2p

m

k = 2p

362

4

4 N

kg

/m = 1.6 s

8. m = 55 kg

k = (36)(458 N/m)T = 2p

m

k = 2p(36)(

5

45

58

kg

N/m) = 0.36 s


9. m = 8.2 kg

k = 221 N/mT = 2p

m

k = 2p

228

1

.2

Nk

/

gm = 1.2 s

10. m = 3(24 g) = 72 g = 7.2 × 10−2 kg

k = 99 N/m

T = 2pm

k = 2p

7.2

9×9 1

N0

/

−

m

2

kg = 0.17 s

1. l = 2.3 × 104 m

f = 0.065 Hz

v = fl = (0.065 Hz)(2.3 × 104 m) = 1.5 × 103 m/s

2. f = 2.8 × 105 Hz

l = 0.51 cm = 5.1 × 10−3 m

v = fl = (2.8 × 105 Hz)(5.1 × 10−3 m) = 1.4 × 103 m/s

3. l = 6.0 m

T = 2.0 sv = fl =

T

l =

6

2

.

.

0

0

m

s = 3.0 m/s

5. f = 1.6 Hz

l = 2.5 mv = fl = (1.6 Hz)(2.5 m) = 4.0 m/s

6. f = 2.5 × 104 Hz

v = 331 m/s

v = fl

l = v

f =

2.5

33

×1

1

m

04/s

Hz = 1.3 × 10−2 m

8. l = 14 m

v = 7.0 m/s

v = fl

f = lv

= 7

1

.0

4

m

m

/s = 0.50 Hz


V

9. l = 10.6 m

v = 331 m/s

v = fl

f = lv

= 3

1

3

0

1

.6

m

m

/s = 31.2 Hz

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10. l = 1.1 m

v = 2.42 × 104 m/s

v = fl

f = lv

= 2.42

1

×.1

10

m

4 m/s = 2.2 × 104 Hz

Chapter 13Sound

V

1. P = 5.88 × 10−5 W

Intensity = 3.9 × 10−6 W/m2Intensity =

4pP

r2

r = (4p)(InP

tensity) =

r = 1.1 m

5.88 × 10−5 W(4p) (3.9 × 10−6 W/m2)


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2. P = 1.57 × 10−3 W


4pP

r2

r = (4p)(InP

tensity) =

r = 0.155 m

1.57 × 10−3 W(4p)(5.20 × 10−3 W/m2)

3. P = 4.80 W


4pP

r2

r = (4p)(InP

tensity) = = 2.3 m

4.80 W(4p)(7.2 × 10−2 W/m2)

4. P = 151 kW = 1.51 × 105 W

Intensity = .025 W/m2Intensity =

4pP

r2

r = (4p)(InP

tensity) = (41

p.)

5

(1

.0×25

1 0

W5

/

W

m2) = 693 m

5. P = 402 W

r = 32 mIntensity =

4pP

r2 = (4p

4

)

0

(

2

32

W

m)2 = 3.1 × 10−2 W/m2

6. P = 3.5 W

r = 0.50 mIntensity =

4pP

r2 = (4p)

3

(

.

0

5

.5

W

0 m)2 = 1.11 W/m2

7. P = 2.76 × 10−2 W

r = 5.0 × 10−2 mIntensity =

4pP

r2 = (4p

2

)

.

(

7

5

6

.0

××10

1

−

0

2

−2W

m)2 = 0.88 W/m2

8. Intensity = 9.3 × 10−8 W/m2


4pP

r2

P = 4pr2 Intensity = (4p)(0.21 m)2 (9.3 × 10−3 W/m2)

P = 5.2 × 10−3 W


10. Intensity = 1.0 × 104 W/m2


4pP

r2

P = 4pr2 Intensity = (4p) (1.0 m)2 (1.0 × 104 W/m2)

P = 1.3 × 105 W

V

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9. Intensity = 4.5 × 10−4 W/m2


4pP

r2

P = 4pr2 Intensity = (4p) (1.5 m)2 (4.5 × 10−4 W/m2)

P = 1.3 × 10−2 W

1. n = 2

f2 = 466.2 Hz

v = 331 m/s

fn = 4

nv

L

L = 4

n

f

v

n =

(

(

4

2

)

)

(

(

4

3

6

3

6

1

.2

m

H

/s

z

)

) = 0.355 m = 35.5 cm

2. n = 3

f3 = 370 Hz

v = 331 m/s

fn = 4

nv

L , n = 1, 3, 5, . . .

L = 4

n

f

v

n =

(

(

3

4

)

)

(

(

3

3

3

7

1

0

m

H

/

z

s

)

) = 0.671 m = 67.1 cm

3. n = 1

f1 = 392.0 Hz

v = 331 m/s

fn = 4

nv

L

L = 4

n

f

v

n =

(

(4

1

)

)

(

(

3

3

9

3

2

1

.0

m

H

/s

z

)

) = 0.211 m = 21.1 cm

4. n = 1

f1 = 370.0 Hz

v = 331 m/s

fn = 2

nv

L

L = 2

n

f

v

n =

(

(

2

1

)

)

(

(

3

3

7

3

0

1

.0

m

H

/s

z

)

) = 0.447 m = 44.7 cm

5. n = 1

L = 35.0 cm = 3.50 × 10−1 m

v = 346 m/s

fn = 2

nv

L

f1 = (2)

(

(

1

3

)

.5

(3

0

4

×6

1

m

0−/s

1)

m) = 494 Hz

6. n = 1

L = 4.20 × 10−1 m

v = 329 m/s

fn = 2

nv

L

f1 = (2)

(

(

1

4

)

.2

(3

0

2

×9

1

m

0−/s

1)

m) = 392 Hz



V

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7. n = 1

v = 499 m/s

L = 0.85 m

fn = 2

nv

L

f1 = (

(

1

2

)

)

(

(

4

0

9

.8

9

5

m

m

/s

)

) = 294 Hz

8. n = 1

f1 = 277.2 Hz

L = 0.75 m

fn = 2

nv

L

v = 2

n

Lfn =

v = 420 m/s

(2)(0.75 m)(277.2 Hz)

1

9. n = 7

f1 = 466.2 Hz

L = 1.53 m

fn = 4

nv

L

v = 4

n

Lfn = = 408 m/s(4)(1.53 m)(466.2 Hz)

7

10. n = 1

f1 = 125 Hz

L = 1.32 m

fn = 2

nv

L

v = 2L

n

fn = = 330 m/s(2)(1.32 m)(125 Hz)

1


Chapter 14Light and Reflection

V

1. f = 7.6270 × 108 Hz

l = 3.9296 × 10−1 m

c = fl = (7.6270 × 108 s−1)(3.9296 × 10−1 m) =

The radio wave travels through Earth’s atmosphere.

2.9971 × 108 m/s


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2. f = 1.17306 × 1011 Hz

l = 2.5556 × 10−3 m

c = fl = (1.17306 × 1011 s−1)(2.5556 × 10−3 m) =

The microwave travels through space.

2.9979 × 108 m/s

3. l = 3.2 × 10−9 m f = lc

= 3

3

.0

.2

0

××

1

1

0

0−

8

9m

m

/s = 9.4 × 1016 Hz

4. l = 5.291 770 × 10−11 ma. f =

lc

= =

b. You cannot see atoms because light in the visible part of the spectrum has wavelengthsthat are much larger than atoms.

5.67 × 1018 Hz3.00 × 108 m/s

5.291 770 × 10−11 m

5. UVA: l1 = 3.2 × 10−7 m

l2 = 4.0 × 10−7 m

UVB: l1 = 2.8 × 10−7 m

l2 = 3.2 × 10−7 m

for UVA waves:

f1 = lc

1 =

3

3

.0

.2

0

××

1

1

0

0−

8

7m

m

/s =

f2 = lc

2 =

3

4

.0

.0

0

××

1

1

0

0−

8

7m

m

/s =

for UVB waves:

f1 = lc

1 =

3

2

.0

.8

0

××

1

1

0

0−

8

7m

m

/s =

f2 = lc

2 =

3

3

.0

.2

0

××

1

1

0

0−

8

7m

m

/s = 9.4 × 1014 Hz

1.1 × 1015 Hz

7.5 × 1014 Hz

9.4 × 1014 Hz

6. l = 1.67 × 10−10 m f = lc

= 1

3

.

.

6

0

7

0

××

1

1

0

0−

8

1m0 m

/s = 1.80 × 1018 Hz

7. f = 9.5 × 1014 Hz l = c

f =

3

9

.

.

0

5

0

××

1

1

0

01

8

4m

s−/1s

= 3.2 × 10−7 m = 320 nm

8. f = 2.85 × 109 Hz l = c

f =

3

2

.

.

0

8

0

5

××

1

1

0

0

8

9m

s−/1s

= 0.105 m = 10.5 cm

Holt Physics Solutions ManualV Ch. 14–2

10. f = 2.5 × 1010 Hz a. l = c

f =

3

2

.

.

0

5

0

××

1

1

0

01

8

0m

s−/1s

= 1.2 × 10−2 m =

b. 1.2 cm > 1. 2 mm

The microwave’s wavelength is larger than the holes, so it cannot pass through.This is analogous to a ball being trapped by a net.

c. 400 nm < 1.2 mm

700 nm < 1.2 mm

Yes, visible light can pass through the holes in the microwave oven’s door. This iswhy we can look through the holes and see the food as it cooks.

1.2 cm

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9. f1 = 1800 MHz = 1.8 × 109 Hz

f2 = 2000 MHz = 2.0 × 109 Hz

l1 = f

c

1 =

3

1

.0

.8

0

××

1

1

0

09

8

s

m−1

/s = 0.17 m =

l2 = f

c

2 =

3

2

.0

.0

0

××

1

1

0

09

8

s

m−1

/s = 0.15 m = 15 cm

17 cm

1. f = 32.0 cm

h = 75 cm

a. You want to appear to be shaking hands with yourself, so the image must appearto be where your hand is. So

p = q 1

f =

p

1 +

1

q =

p

2

p = 2f = (2)(32.0 cm) =

q = p =

b. M = h

h

′ = −

p

q

h′ = − q

p

h = −

(64.0

6

c

4

m

.0

)(

c

7

m

.5 cm) = −7.5 cm

64.0 cm

64.0 cm

2. f = 9.5 cm

q = 15.5 cm

h = 3.0 cm

a. 1

f =

p

1 +

q

1

p

1 =

1

f −

1

q =

9.5

1

cm −

15.5

1

cm

p

1 =

0

1

.1

c

0

m

5 −

0

1

.0

c

6

m

5 =

0

1

.0

c

4

m

0

p =

b. h′ = − q

p

h = −

(15.5 c

2

m

5

)

c

(

m

3.0 cm) = −1.9 cm

25 cm


3. f = 17 cm

q = 23 cm

h = 2.7 cm

a. 1

f =

p

1 +

q

1

p

1 =

1

f −

1

q =

17

1

cm −

23

1

cm

p

1 =

0

1

.0

c

5

m

9 −

0

1

.0

c

4

m

35 =

0

1

.0

c

1

m

6

p =

b. h′ = − q

p

h = −

(23 cm

62

)(

c

2

m

.7 cm) = −1.0 cm

62 cm


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4. p = 5.0 cm A car’s beam has rays that are parallel, so q = ∞.

1

q =

∞1

= 0 1

f =

p

1 +

1

q =

p

1 + 0 =

p

1

f = p =

R

2 =

p

1 +

1

q =

1

f

R = 2 f = (2)(5.0 cm) = 1.0 × 101 cm

5.0 cm

7. p = 3.00 cm = 3.00 × 102 cm

f = 30.0 cm

h = 15 cm

a. 1

q =

1

f −

p

1 =

30.0

1

cm −

3.00 ×1

102 cm

1

q =

0

1

.0

c

3

m

33 −

0.

1

00

c

3

m

33 =

0

1

.0

c

3

m

00

q =

b. M = h

h

′ = −

p

q

h′ = − q

p

h = −

(3

3

3

.

.

0

3

0

c

×m

1

)(

0

125

c

c

m

m) = 1.7 cm

33.3 cm

5. p = 19 cm

q = 14 cm1

f =

p

1 +

1

q =

19

1

cm +

14

1

cm

1

f =

0

1

.0

c

5

m

3 +

0

1

.0

c

7

m

1 =

1

0.

c

1

m

2

f =

R

2 =

p

1 +

1

q =

1

f

R = 2f = (2)(8.3 cm) = 17 cm

8.3 cm

6. p = 35 cm

q = 42 cm1

f =

p

1 +

1

q =

R

2 =

35

1

cm +

42

1

cm

1

f =

0

1

.0

c

2

m

9 +

0

1

.0

c

2

m

4 =

0

1

.0

c

5

m

3

f =

R

2 =

1

f

R = 2f = (2)(19 cm) = 38 cm

19 cm

8. f = 17.5 cm

p = 15.0 cma.

1

q =

1

f −

p

1 =

17.5

1

cm −

15.0

1

cm

1

q =

0

1

.0

c

5

m

71 −

0

1

.0

c

6

m

67 =

0.

1

00

c

9

m

60

q =

b. M = − p

q = −

−1

1

5

0

.0

4

c

c

m

m = 6.93

−104 cm


V

9. f = 60.0 cm

p = 35.0 cm

a. 1

q =

1

f −

p

1 =

60.0

1

cm −

35.0

1

cm =

0

1

.0

c

1

m

67 −

0

1

.0

c

2

m

86 =

−0

1

.0

cm

119

q =

b. M = − p

q = −

−3

8

5

4

.0

.0

c

c

m

m = 2.4

−84.0 cm

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10. f = 23.0 cm

p = 3.00 cm

p = 33.0 cm

a. 1

q =

1

f −

p

1 =

23.0

1

cm −

3.00

1

cm =

0

1

.0

c

4

m

35 −

0

1

.3

c

3

m

3 =

−1

0.

c

2

m

90

q =

M = − p

q = −

3

3

.

.

4

0

5

0

c

c

m

m =

The image is real, and upright, so you can read the writing.

b. 1

q =

1

f −

p

1 =

23.0

1

cm −

33.0

1

cm =

0

1

.0

c

4

m

35 −

0

1

.0

c

3

m

03 =

0

1

.0

c

1

m

32

q =

M = − p

q = −

7

3

5

3

.

.

8

0

c

c

m

m =

The image is inverted and virtual, so you cannot read the writing (unless you canread upside-down).

−2.30

75.8 cm

1.15

−3.45 cm

1. h′ = 9.0 cm

h = 1.5 m = 150 cm

p = 3 m = 300 cm

a. M = h

h

′ =

1

9

5

.0

0

c

c

m

m =

b. M = − p

q

q = −Mp = −(0.060)(300 cm) =

1

f =

p

1 +

1

q =

3

1

m +

−0.1

1

8 m =

0

1

.3

m

3 −

−1

5

m

.6 =

−1

5

m

.3

f = −0.19 m =

c. R = 2f = 2 (19 cm) = −38 cm

−19 cm

−18 cm

0.060



V

3. f = −6.3 cm

q = −5.1 cmp

1 =

1

f −

1

q =

−6.3

1

cm −

−5.1

1

cm =

0

1

.1

c

5

m

9 −

0

1

.1

c

9

m

6 =

0

1

.0

c

3

m

7

p =

M = −p

q =

−(−2

5

7

.1

cm

cm) = 0.19

27 cm

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4. f = −33 cm

q = −16.1 cmp

1 =

1

f −

1

q =

−33

1

cm −

−16.

1

1 cm =

−1

0.

c

0

m

30 −

0

1

.0

c

6

m

2 =

0

1

.0

c

3

m

2

p =

M = − q

f = −

(−1

3

6

1

.1

cm

cm) = 0.52

31 cm

6. h = 1.75 m

M = 0.11

q = −42 cm = −0.42 m

a. M = h

h

′ = −

p

q

h′ = Mh = (0.11)(1.75 m) =

b. p = − M

q = −

0.

0

4

.

2

11

m = 3.8 m

0.19 m

5. f = −12 cm

q = −9.0 cmp

1 =

1

f −

1

q =

−12

1

cm −

−9.0

1

cm =

0

1

.0

c

8

m

3 +

0

1

.1

c

1

m

1 =

0

1

.0

c

2

m

8

p =

M = − p

q = −

9

3

.

6

0

c

c

m

m = 0.25

36 cm

7. f = −27 cm

p = 43 cm1

q =

1

f −

p

1 =

−27

1

cm −

43

1

cm =

−1

0.

c

0

m

37 −

0

1

.0

c

2

m

3 =

−1

0.

c

0

m

60

q =

M = − p

q = −

−4

1

3

7

c

c

m

m = 0.40

−17 cm

2. q = −5.2 cm

p = 17 cm

h = 3.2 cm

a. 1

f =

p

1 +

q

1 =

17

1

cm +

−5.2

1

cm =

0

1

.0

c

5

m

9 −

1

0.

c

1

m

9 =

−1

0

c

.

m

13

f =

b. 1

f =

R

2 =

p

1 +

q

1

R = 2f = (2)(−7.7 cm) = −15 cm

−7.7 cm

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10. M = 0.24

p = 12 cm

M = − p

q

q = −Mp = −(0.24)(12 cm) = −2.9 cm

9. f = −39 cm

p = 16 cm

h = 6.0 cm

a. 1

q =

1

f −

p

1 =

−39

1

cm −

16

1

cm =

−1

0.

c

0

m

26 −

0

1

.0

c

6

m

2 =

−1

0.

c

0

m

88

q =

b. M = h

h

′ = −

p

q

h′ = − q

p

h = −

(−11 c

1

m

6

)

c

(

m

6.0 cm) = 4.1 cm

−11 cm

8. f = −8.2 cm

p = 18 cm1

q =

1

f −

p

1 =

−8.2

1

cm −

18

1

cm =

−1

0.

c

1

m

22 −

0

1

.0

c

5

m

6 =

−1

0

c

.

m

18

q =

M = − p

q = −

−1

5

8

.6

c

c

m

m = 0.31

−5.6 cm

Chapter 15Refraction

V

1. qr = 35°

nr = 1.553

ni = 1.000

qi = sin−1nr(s

n

in

i

qr) = sin−1(1.553

1

)

.0

(s

0

i

0

n 35°) = 63°


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2. qr = 41°

nr = 1.486

ni = 1.00

qi = sin−1nr(s

n

in

i

qr) = sin−1(1.486

1

)

.0

(s

0

i

0

n 41°) = 77°

3. qr = 33°

nr = 1.555

ni = 1.000

qi = sin−1nr(s

n

in

i

qr) = sin−1(1.555

1

)

.0

(s

0

i

0

n 33°) = 58°

4. c = 3.00 × 108 m/s

v = 2.07 × 108 m/sn =

v

c =

3

2

.

.

0

0

0

7

××

1

1

0

0

8

8m

m

/

/

s

s = 1.45

5. c = 3.00 × 108 m/s

v = 1.97 × 108 m/sn =

v

c =

3

1

.

.

0

9

0

7

××

1

1

0

0

8

8m

m

/

/

s

s = 1.52

6. c = 3.00 × 108 m/s

v = 1.95 × 108 m/sn =

v

c =

3

1

.

.

0

9

0

5

××

1

1

0

0

8

8m

m

/

/

s

s = 1.54

7. qi = 59.2°

nr = 1.61

ni = 1.00

qr = sin−1ni(s

n

in

r

qi) = sin−1(1.00)(

1

s

.

i

6

n

1

59.2°) = 32.2°

8. qi = 35.2°

ni = 1.00

nr,1 = 1.91

nr,2 = 1.66

qr,1 = sin−1ni(s

n

i

r

n

,1

qi) = sin−1(1.00)(

1

s

.

i

9

n

1

35.2°) =

qr,2 = sin−1ni(s

n

i

r

n

,2

qi) = sin−1(1.00)(

1

s

.

i

6

n

6

35.2°) = 20.3°

17.6°


10. qi = 22°

ni = 1.000

nr = 1.544

qi = 14°

ni = 1.544

nr = 1.000

1st surface:

qr = sin−1ni(s

n

in

r

qi) = sin−1(1.000

1

)

.5

(s

4

i

4

n 22°) =

2nd surface:

qr = sin−1(1.544

1

)

.0

(s

0

i

0

n 14°) = 22°

14°

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9. qi = 17°

ni = 1.00

nr = 1.54

qi = 11°

ni = 1.54

nr = 1.00

1st surface:

qr = sin−1ni(s

n

in

r

qi) = sin−1(1.00)

1

(

.

s

5

i

4

n 17°) =

2nd surface:

qr = sin−1(1.54)

1

(

.

s

0

i

0

n 11°) = 17°

11°

1. p = 13 cm

q = 19 cm

h′ = 3.0 cm

1

f =

p

1 +

1

q =

13

1

cm +

19

1

cm =

0

1

.0

c

7

m

7 +

0

1

.0

c

5

m

3 =

1

0.

c

1

m

3

f =

M = h

h

′ = −

p

q

h = − p

q

h′ = −

(13 cm

19

)(

c

3

m

.0 cm) = 2.1 cm

7.7 cm

2. p = 44 cm

q = −14 cm

h = 15 cm

1

f =

p

1 +

1

q =

44

1

cm +

−14

1

cm =

0

1

.0

c

2

m

3 +

0

1

.0

c

7

m

1 =

−1

0

c

.0

m

48

f =

M = h

h

′ = −

p

q

h′ = − q

p

h = −

(−14 c

4

m

4

)

c

(

m

15 cm) = 4.8 cm

−21 cm

3. f = −13.0 cm

M = 5.00M = −

p

q

q = −Mp = −(5.00)p

1

f =

p

1 +

1

q =

p

1 +

−(5.

1

00)p =

−5

−.0

(5

0

.0

+0

1

)p

.00 =

−(

−5

4

.0

.0

0

0

)p =

(5

4

.0

.0

0

0

)p

M = h

h

′ = −

p

q

p = (4

5

.

.

0

0

0

0

)f =

(4.00)(

5

−.0

1

0

3.0 cm) = −10.4 cm



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4. h = 18 cm

h′ = −9.0 cm

f = 6.0 cm

a. M = h

h

′ =

−1

9

8

.0

c

c

m

m =

b. M = − p

q

q = −Mp = −(−0.50)p = (0.50)p

1

f =

p

1 +

1

q =

p

1 +

(0.5

1

0)p =

(0

0

.5

.5

0

0

)p +

(0.5

1

0)p =

(0

1

.5

.5

0

0

)p =

3

p

.0

p = (3.0) f = (3.0)(6.0 cm) =

c. q = −Mp = (0.50)(18 cm) = 9.0 cm

18 cm

−0.50 cm

8. q = −12 cm

f = −44 cmp

1 =

1

f −

1

q =

−44

1

cm −

−12

1

cm = −1

0

c

.0

m

23 − −1

0.

c

0

m

83 =

0

1

.0

c

6

m

0

p = 17 cm

5. p = 4 m

f = 4 m1

f =

p

1 +

1

q, but f = p, so

1

q = 0. That means q =

M = − p

q = −

4

•m =

The rays are parallel, and the light can be seen from very far away.

•

•

6. p = 0.5 m

f = 0.5 m1

f =

p

1 +

1

q

p = f, so 1

q = 0, and q =

M = −p

q =

−0.

•5 m =

The rays are parallel, and the light can be seen from far away.

−•

•

7. f = 3.6 cm

q = 15.2 cmp

1 =

1

f −

1

q =

3.6

1

cm −

15.2

1

cm =

1

0.

c

2

m

8 −

0

1

.0

c

6

m

6 =

1

0.

c

2

m

1

p = 4.8 cm

9. f = 9.0 cm = 0.09 m

q = 18 mp

1 =

1

f −

1

q =

0.09

1

0 m −

18

1

m =

1

1

m

1 −

0

1

.0

m

56 =

1

1

m

1

p = 0.091 m = 9.1 cm

10. f = 5.5 m

q = 5.5 cm = 0.055 mp

1 =

1

f −

1

q =

5.5

1

m −

−0.0

1

55 m =

0

1

.1

m

8 −

1

−1

m

8 =

1

1

m

8

p = 5.5 × 10−2 m = 5.5 cm


V

1. qc = 37.8°

nr = 1.00sin qc =

n

nr

i

ni = sin

nr

qc =

sin

1.

3

0

7

0

.8° = 1.63

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2. qc = 39.18°

nr = 1.000sin qc =

n

nr

i

ni = sin

nr

qc =

sin

1.

3

0

9

0

.

0

18 = 1.583

3. qc,1 = 35.3°

qc,2 = 33.1°

nr = 1.00

sin qc = n

nr

i

ni,1 = sin

n

qr

c,1 =

sin

1.

3

0

5

0

.3° =

ni,2 = sin

n

qr

c,2 =

sin

1.

3

0

3

0

.1° = 1.83

1.73

4. ni = 1.670

qc = 62.85°sin qc =

n

nr

i

nr = ni(sin qc) = (1.670)(sin 62.85°) = 1.486


5. ni = 1.80

qc = 57.0°sin qc =

n

nr

i

nr = ni(sin qc) = (1.80)(sin 57.0°) = 1.51

6. ni = 1.64

qc = 69.9°sin qc =

n

nr

i

nr = ni(sin qc) = (1.64)(sin 69.9°) = 1.54

7. ni- = 1.766

nr = 1.000

sin qc = n

nr

i

qc = sin−1n

nr

i = sin−111.

.

0

7

0

6

0

6 = 34.49°

8. ni = 1.774

nr = 1.000

sin qc = n

nr

i

qc = sin−1n

nr

i = sin−111.

.

0

7

0

7

0

4 = 34.31°

9. ni = 1.61

nr = 1.00sin qc =

n

nr

i

qc = sin−1n

nr

i = sin−1

1

1

.

.

0

6

0

1 = 38.4°

10. ni = 1.576

nr = 1.000sin qc =

n

nr

i

qc = sin−1n

nr

i = sin−111.

.

0

5

0

7

0

6 = 39.38°


Chapter 16Interference and Diffraction

V

1. 14 450 lines/cm

l = 6.250 × 10–7 m

q < 90°

m = 1: q1 = sin–1(ml/d)

q1 = sin–1[(1)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]

q1 = 64.99°

m = 2: q2 = sin–1[(2)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]

q2 = ∞



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2. 12 260 lines/cm

l = 5.896 × 10–7 m

q < 90°

m = 1: q1 = sin–1(ml/d)

q1 = sin–1[(1)(5.896 × 10–7 m) ÷ (1 226 000 lines/m)–1]

q1 = 46.29°

m = 2: q2 = sin–1[(2)(5.896 × 10–7 m) ÷ (1 450 000 lines/m)–1]

q2 = ∞


3. l= 5.461 × 10–7 m

m = 1

q = 75.76°

d = s

m

in

lq

= = 5.634 × 10–7 m

d = 5.634 × 10–5 cm

# lines/cm = (5.634 × 10–5 cm)–1 = 1.775 × 104 lines/cm

(1)(5.461 × 10–7 m)

sin (75.76°)

4. l = 4.471 × 10–7 m

m = 1

q = 40.25°

d = s

m

in

lq

= = 6.920 × 10–7 m

d = 6.920 × 10–7 cm

# lines/cm = (6.920 × 10–7 cm)–1 = 1.445 × 104 lines/cm

(1)(4.471 × 10–7 m)

sin (40.25°)

5. 1950 lines/cm

l = 4.973 × 10–7 m

m = 1: q1 = sin–1(ml /d)

q1 = sin–1[(1)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]

q1 =

m = 2: q2 = sin–1[(2)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]

q2 = 11.18°

5.565°

10. 2400 lines/cm

m = 2

q = 26.54 °

l = d(s

m

in q) = = 9.296 × 10–7 m

(240 000 m)–1sin (26.54°)

2

1. l = 5.875 × 10–7 m

m = 2

q = 0.130°

d = s

m

in

lq

= = 5.18 × 10–4 m

d = 0.518 mm

2(5.875 × 10–7 m)

sin (0.130°)

2. l = 6.563 × 10–7 m

m = 4

q = 0.626°

d = s

m

in

lq

= = 2.40 × 10–4 m

d = 0.240 mm

4(6.563 × 10–7 m)

sin (0.626°)


V

6. 7500 lines/cm

l1 = 5.770 × 10–7 m

l2 = 5.790 × 10–7 m

m = 2

q1 = sin–1(ml1/d)

q1 = sin–1[(2)(5.770 × 10–7 m) ÷ (750 000 lines/m)–1]

q1 = 59.94°

q2 = sin–1(ml2/d)

q2 = sin–1[(2)(5.790 × 10–7 m) ÷ (750 000 lines/m)–1]

q2 = 60.28°

q = q2 – q1 = 59.94° – 60.28° = 0.3400°

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7. 3600 lines/cm

m = 3

q = 76.54°

l = d(s

m

in q) = = 9.000 × 10–7 m

(360 000 m)–1sin (76.54°)

3

8. 9550 lines/cm

m = 2

q = 54.58°

l = d(s

m

in q) = = 4.267 × 10–7 m

(955 000 m)–1sin (54.58°)

2

9. 12 510 lines/cm

m = 1

q = 38.77°

l = d(s

m

in q) = = 5.006 × 10–7 m

(1 251 000 m)–1sin (38.77°)

1



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3. l = 6.93 × 10–7 m

m = 3

q = 0.578°

d = s

m

in

lq

= = 2.06 × 10–4 m

d = 0.206 mm

3(6.93 × 10–7 m)

sin (0.578°)

5. d = 4.43 × 10–6 m

m = 3

q = 21.7 °

10. d = 2.67 × 10–4 m

l = 6.87 × 10–7 m

m = 1

q = sin–1(ml /d)

q = sin–1[(1)(6.87 × 10–7 m) ÷ (2.67 × 10–4 m)] = 0.147°

4. d = 8.04 × 10–6 m

m = 3

q = 13.1 °

l = d(s

m

in q) = = 6.07 × 10–7 m

l = 607 nm

(8.04 × 10–6 m) sin (13.1°)

3

l = d(s

m

in q) = = 5.46 × 10–7 m

l = 546 nm

(4.43 × 10–6 m) sin (21.7°)

3

6. d = 3.92 × 10–6 m

m = 2

q = 13.1 °

l = d(s

m

in q) = = 4.44 × 10–7 m

l = 444 nm

(3.92 × 10–6 m) sin (13.1°)

2

7. d = 2.20 × 10–4 m

l= 5.27 × 10–7 m

m = 1

q = sin–1(ml /d)

q = sin–1[(1)(5.27 × 10–7 m) ÷ (2.20 × 10–4 m)] = 0.137°

8. d = 1.63 × 10–4 m

l = 4.308 × 10–7 m

m = 1

q = sin–1(ml /d)

q = sin–1[(1)(4.308 × 10–7 m) ÷ (1.63 × 10–4 m)] = 0.151°

9. d = 3.29 × 10–4 m

l = 5.83 × 10–7 m

m = 1

q = sin–1(ml /d)

q = sin–1[(1)(5.83 × 10–7 m) ÷ (3.29 × 10–4 m)] = 0.102°


Chapter 17Electric Forces and Fields

V

1. q1 = − 1.30 × 10−5 C

q2 = −1.60 × 10−5 C

Felectric = 12.5 N

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

r = k

FCel

qec

1t

q

r

i

2

c =

r = 0.387 m = 38.7 cm

(8.99 × 109 N •m2/C2)(−1.30 × 10−5 C)(−1.60 × 10−5 C)

12.5 N


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2. q1 = 9.99 × 10−5 C

q2 = 3.33 × 10−5 C

Felectric = 87.3 N

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

r = k

FCel

qec

1t

q

r

i

2

c =

r = 0.585 m = 58.5 cm

(8.99 × 109 N •m2/C2)(9.99 × 10−5 C)(3.33 × 10−5 C)1

87.3 N

3. q1 = −4.32 × 10−5 C

q2 = 2.24 × 10−5 C

Felectric = −6.5 N

KC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

r = k

FCel

qec

1t

q

r

i

2

c =

r = 1.15 m

(8.99 × 109 N •m2/C2)(−4.32 × 10−5 C)(2.24 × 10−5 C)

−6.5 N

4. q1 = −5.33 × 10−6 C

q2 = +5.3 × 1−6 C

r = 4.2 × 10−2 m

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

Felectric =

Felectric = −143 N

(8.99 × 109 N •m2/C2)(−5.3 × 10−6 C)(5.3 × 10−6 C)

(4.2 × 10−2 m)2

5. q1 = −1.40 × 10−8 C

q2 = +1.40 × 10−8 C

r = 7.1 × 10−2 m

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

Felectric =

Felectric = 3.50 × 10−4 N

(8.99 × 109 N •m2/C2)(−1.40 × 10−8 C)(1.4 × 10−8 C)

(7.1 × 10−2 m)2

6. q1 = −8.0 × 10−9 C

q2 = +8.0 × 10−9 C

r = 2.0 × 10−2 m

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2

Felectric =


(8.99 × 109 N •m2/C2)(−8.0 × 10−9 C)(8.0 × 10−9 C)

(2.0 × 10−2 m)2


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7. r = 8.3 × 10−10 m

Felectric = 3.34 × 10−10 N

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2 =

kC

r2q2

q = Fele

kct

C

ricr2

= q = 1.60 × 10−19 C

(3.34 × 10−10 N)(8.3 × 10−10 N)2

8.99 × 109 N •m2/C2

8. r = 6.4 × 10−8 m

Felectric = 5.62 × 10−14 N

kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2 =

kC

r2q2

q = Fele

kct

C

ricr2

= q = 1.60 × 10−19 C

(5.62 × 10−14 N)(6.4 × 10−8 m)2

8.99 × 109 N •m2C2

9. r = 9.30 × 10−11 m


kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2 =

kC

r2q2

q = Fele

kct

C

ricr2

= q = 1.60 × 10−19 C

(2.66 × 10−8 N)(9.30 × 10−11 m)2

8.99 × 109 N •m2/C2

10. r = 6.5 × 10−11 m


kC = 8.99 × 109 N •m2/C2

Felectric = kC

r

q21q2 =

kC

r2q2

q = Fele

kct

C

ricr2

= q = 2.2 × 10−17 C

(9.92 × 10−4 N)(6.5 × 10−11 m)2

8.99 × 109 N •m2/C2


1. qp = 1.60 × 10−19 C

r4,1 = r2,1 = 1.52 × 10−9 m

kC = 8.99 × 109 N •m2/C2

qp = q1 = q2 = q3 = q4

r3,2 =√

(1.52× 10−9m)2 + (1.52 × 10−9m)2 = 2.15 × 10−9 m

F2,1 = k

rC

2

q

,1

p2

2

= = 9.96 × 10−11 N

F3,1 = k

rC

3

q

,1

p2

2

= = 4.98 × 10−11 N

F4,1 = k

rC

4

q

,1

2

2p = = 9.96 × 10−11 N

j = tan−1 11.

.

5

5

2

2

××

1

1

0

0

−

−

9

9m

m = 45°

F2,1: Fx = 0 N

Fy = 9.96 × 10−11 N

F3,1: Fx = F3,1 cos 45° = (4.98 × 10−11 N)(cos 45°) = 3.52 × 10−11 N

Fy = F3,1 sin 45° = (4.98 × 10−11 N)(sin 45°) = 3.52 × 10−11 N

F4,1: Fx = 9.96 × 10−11 N

Fy = 0 N

(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2

(1.52 × 10−9 m)2

(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2

(2.15 × 10−9 m)2

(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2

(1.52 × 10−9 m)2


V

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Fx,tot = 0 N + 3.52 × 10−11 N + 9.96 × 10−11 N = 1.3 × 10−10 N

Fy,tot = 9.96 × 10−10 N + 3.52 × 10−11 N + 0 N = 1.3 × 10−10 N

Ftot =√

(Fx,tot)2+ (Fy,tot)2 =√

(1.3 × 10−10 N)2 + (1.3× 10−10 N)2

Ftot =

j = tan−111.

.

3

3

××

1

1

0

0

−

−

1

1

0

0N

N = 45°

1.8 × 10−10 N

2. h = 3.50 m

q1 = q2 = 4.50 C

q3 = 6.30 C

r1,2 = 5.00 m

kC = 8.99 × 109 N • m2/C2

Set the origin at the midpoint.

r1,3 = r2,3 =√

(2.50m)2 + (3.50 m)2 = 4.30 m

F1,3 = kC

r1

q

,3

12q3 = = 1.38 × 1010 N

F2,3 = kC

r2

q

,3

22q3 = = 1.38 × 1010 N

q = tan−1 32.

.

5

5

0

0

m

m = 54.5°

F1,3: Fx = F1,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N

Fy = F1,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N

F2,3: Fx = F2,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N

The electrical force on q2 points in the −x direction, so Fx must be negative.

Fx = −8.01 × 109 N

Fy = F2,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N

Fx,tot = 8.01 × 109 N − 8.01 × 109 N = 0 N

Fy,tot = 1.12 × 1010 N + 1.12 × 1010 N = 2.24 × 1010 N

There is no x-component of the resultant force, so

Ftot = Fy,tot = 2.24 × 1010 N pointing upward along the y-axis

(8.99 × 109 N •m2/C2)(4.50 C)(6.30 C)

(4.30 m)2

(8.99 × 109 N •m2/C2)(4.50 C)(6.30 C)

(4.30 m)2

3. q1 = −9.00 × 10−9 C

q2 = −8.00 × 10−9 C

q3 = 7.00 × 10−9 C

r1,2 = 2.00 m

r1,3 = 3.00 m

kC = 8.99 × 109 N •m2/C2

F1,2 = kC

r1

q

,2

12q2 = = 1.62 × 10−7 N

F1,3 = kC

r1

q

,3

12q3 = = −6.29 × 10−8 N

F1,2: Fx = 4.05 × 10−8 N

Fy = 0 N

F1,3: Fx = 0 N

Fy = −6.29 × 10−8 N

Ftot =√

(1.62× 10−7N)2 + (−6.29 × 10−8N)2 =

Ftot is negative because the larger, y-component of the force is negative.

j = tan−1 −16.6.229××

1

1

0

0−

−

7

8

N

N = −21.22°

1.74 × 10−7 N

(8.99 × 109 N •m2/C2)(−9.00 × 10−9 C)(−7.00 × 10−9 C)

(3.00 m)2

(8.99 × 109 N •m2/C2)(−9.00 × 10−9 C)(−8.00 × 10−9 C)

(2.00 m)2


V

4. q1 = −2.34 × 10−8 C

q2 = 4.65 × 10−9 C

q3 = 2.99 × 10−10 C

r1,2 = 0.500 m

r1,3 = 1.00 m

kC = 8.99 × 109 N •m2/C2

F1,2 = kC

r1

q

,2

12q2 =

F1,2 = Fy = −3.91 × 10−6 N

F1,3 = kC

r1

q

,3

12q3 =

F1,3 = Fy = − 6.29 × 10−8 N

Fy,tot = −3.91 × 10−6 N + −6.29 × 10−8 N = 3.97 × 10−6 N

There are no x-components of the electrical force, so the magnitude of the electricalforce is

√(Fy,tot)2.

Ftot = 3.97 × 10−6 N directed along the −y-axis

(8.99 × 109 N •m2/C2)(−2.34 × 10−8 C)(2.99 × 10−10 C)

(1.00 m)2

(8.99 × 109 N •m2/C2)(−2.34 × 10−8 C)(4.65 × 10−9 C)

(0.500 m)2

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5. qe = −1.60 × 10−19 C

r2,3 = r4,3 = 3.02 × 10−5 m

r1,3 =√

2(3.02 × 10−5m)2= 4.27 × 10−5 m

kC = 8.99 × 109 N •m2/C2

qe = q1 = q2 = q3 = q4

F3,2 = Fx = k

rC

3

q

,2

22e

= = 2.52 × 10−19 N

F3,4 = Fy = k

rC

3

q

,4

e2

2

= = 2.52 × 10−19 N

F3,1 = k

rC

3,

q

1

e2

2

= = 1.26 × 10−19 N

j = tan−133.

.

0

0

2

2

××

1

1

0

0

−

−

5

5m

m = 45°

F3,1: Fx = F3,1 cos 45° = (1.26 × 10−19 N) cos 45° = 8.91 × 10−20 N

Fy = F3,1 sin 45° = (1.26 × 10−19 N) sin 45° = 8.91 × 10−20 N

Fx,tot = 8.91 × 10−20 N + 2.52 × 10−19 N + 0 N = 3.41 × 10−19 N

Fy,tot = 8.91 × 10−20 N + 0 N + 2.52 × 10−19 N = 3.41 × 10−19 N

Ftot =√


(3.41× 10−19 N)2 (3.41 × 10−19 N)2

Ftot =

j = tan−1FFx

y,

,

t

t

o

o

t

t = tan−133.

.

4

4

1

1

××

1

1

0

0

−

−

1

1

9

9N

N = 45°

4.82 × 10−19 N

(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2

(4.27 × 10−5 m)2

(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2

(3.02 × 10−5 m)2

(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2

(3.02 × 10−5 m)2

6. q1 = 2.22 × 10−10 C

q2 = 3.33 × 10−9 C

q3 = 4.44 × 10−8 C

r1,2 = 1.00 m

h = 0.250 m

kC = 8.99 × 109 N •m2/C2

r3,1 = r3,2

=√

(0.500 m)2 + (0.250m)2= 0.559 m

F3,1 = kC

r3

q

,1

12q3 = = 2.84 × 10−7 N

F3,2 = kC

r3

q

,2

32q2 = = 4.25 × 10−6 N

j = tan−100.

.

2

5

5

0

0

0

m

m = 26.6°

F3,1: Fx = F3,1 cos q = (2.84 × 10−7 N) cos (−26.6°) = 2.54 × 10−7 N

Fy = F3,1 sin q = (2.84 × 10−7 N) sin (−26.6°) = 1.27 × 10−7 N

F3,2: Fx = F3,2 cos q = (4.25 × 10−6 N) cos (−26.6°) = 3.80 × 10−6 N

Fy = F3,2 sin q = (4.25 × 10−6 N) sin (−26.6°) = 1.90 × 10−6 N

(8.99 × 109 N •m2/C2)(4.44 × 10−8 C)(3.33 × 10−9 C)

(0.559 m)2

(8.99 × 109 N •m2/C2)(2.22 × 10−10 C)(4.44 × 10−8 C)

(0.559 m)2


V

Fx,tot = 2.54 × 10−7 N + 3.80 × 10−6 N = 4.05 × 10−6 N

Fy,tot = 1.27 × 10−7 N + 1.90 × 10−6 N = 2.03 × 10−6 N

Ftot =√

(Fx,tot2 + (Fy,tot)2 =

√(4.05× 10−6N)2 + (2.03 × 10−6N)

Ftot =

j = tan−1FFx

y,

,

t

t

o

o

t

t = tan−124.

.

0

0

3

5

××

1

1

0

0

−

−

6

6N

N = 26.6°

4.53 × 10−6 N

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7. q1 = q2 = q3 = 2.0 × 10−9 C

r1,2 = 1.0 m

r1,3 =√

(1.0 m)2 + (2.0m)2= 2.24 m

kC = 8.99 × 109 N •m2/C2

F1,2 = kC

r1

q

,2

12q2 = = 3.60 × 10−8 N

F1,3 = kC

r1

q

,3

12q3 = = 7.17 × 10−9 N

F1,2 = Fx = 3.60 × 10−8 N

Fy = 0 N

j = tan−121.

.

0

0

m

m = 63.4°

F1,3: Fx = F1,3 cos q = (7.17 × 10−9 N) cos (63.4°) = 3.21 × 10-−9 N

Fy = F1,3 sin q = (7.17 × 10−9 N) sin (63.4°) = 6.41 × 10−9 N

Fx,tot = 3.60 × 10−8 N + 3.21 × 10−9 N = 3.92 × 10−8 N

Fy,tot = 0 N + 6.41 × 10−9 N = 6.41 × 10−9 N

Ftot =√


(3.92× 10−8N)2 + (6.41 × 10−9N)2Ftot = 3.97 × 10−8 N

j = tan−1FFx

y,

,

t

t

o

o

t

t = tan−163.

.

4

9

1

2

××

1

1

0

0

−

−

9

8N

N = 9.29°

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)2

(−2.24 m)2

(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)2

(1.0 m)2

8. q1 = −4.0 × 10−3 C

q2 = −8.0 × 10−3 C

q3 = 2.0 × 10−3 C

kC = 8.99 × 109 N •m2/C2

r1,2 = 2.0 m

r1,3 = 2.0 m

F1,2 = kC

r1

q

,

1

2

q2

2 =

F1,2 = Fx,tot =

F1,3 = kC

r1

q

,

1

3

q2

3 =

F1,2 = Fy,tot =

Ftot =√


(7.2 × 104 N)2 + (−1.8× 104 N)2

Ftot =

j = tan−1FFx

y,

,

t

t

o

o

t

t = tan−1–71.2.8×

×1

1

0

04

4

N

N = –14°

7.4 × 104 N

−1.8 × 104 N

(8.99 × 109 N •m2/C2)(−4.0 × 10−3 C)(2.0 × 10−3 C)

(2.0 m)2

7.2 × 104 N

(8.99 × 109 N •m2/C2)(−4.0 × 10−3 C)(−8.0 × 10−3 C)

(2.0 m)2


V

9. q1 = 9.00 × 10−3 C

q2 = 6.00 × 10−3 C

q3 = 3.00 × 10−3 C

kC = 8.99 × 109 N •m2/C2

x1,2 = 1.00 m

y1,2 = 1.00 m

x1,3 = 1.00 m

y1,3 = 1.00 m

r1,2 = r1,3 =√

x1,22+ y1,22 =

√x1,32+ y1,3

2 =√

(1.00m)2 + (1.00 m)2 = 1.41 m

r2,3 =√

r1,22+ r1,32 =

√(1.41m)2 + (1.41 m)2 = 1.99 m

F1,2 = kC

r1

q

,2

12q2 = = 2.44 × 105 N

F1,3 = kC

r1

q

,3

12q3 = = 1.22 × 105 N

j = tan−1 = 45°

F1,2: Fx = F1,2 cos q = (2.44 × 105 N) cos 45° = 1.73 × 105 N

Fy = F1,2 sin q = (2.44 × 105 N) sin 45° = 1.73 × 105 N

F1,3: Fx = F1,3 cos q = (1.22 × 105 N) cos 45° = 8.63 × 104 N

Fy = F1,3 sin q = (1.22 × 105 N) sin 45° = 8.63 × 104 N

Fx,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N

Fy,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N

Ftot =√


(2.59× 105 N)2 + (2.59 × 105 N)2

Ftot =

j = tan−122.

.

5

5

9

9

××

1

1

0

0

5

5N

N = 45°

3.66 × 105 N

1.00 m1.00 m

(8.99 × 109 N •m2/C2)(9.00 × 10−3 C)(3.00 × 10−3 C)

(1.41 m)2

(8.99 × 109 N •m2/C2)(9.00 × 10−3 C)(6.00 × 10−3 C)

(1.41 m)2

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10. q1 = q2 = q3 = 4.00 × 10−9 C

kC = 8.99 × 109 N •m2/C2

r2,1 = r2,3 = 4.00 m

All forces are along the x-axis, so there are no y-components.

F2,1 = F2,3 = k

r2

C

,

q

12

2

= = 8.99 × 10−9 N

Fx,tot = 2(8.99 × 10−9 N) = 1.80 × 10−8 N

Fy,tot = 0 N

Ftot =√

(Fx,tot)2+ (Fy,tot)2 = 1.80 × 10−8 N along the x-axis

(8.99 × 109 N •m2/C2)(4.00 × 10−9 C)2

(4.00 m)2

1. q1 = 9.0 mC

q2 = −19 mC

q3 = 9.0 mC

kC = 8.99 × 109 N •m2/C2

r2,1 = 3.0 m

The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because theforces point in the same direction. Because q2 is larger than q1, q3 will be close to q1,and opposite q2.

F3,1 = −F3,2 = (y −

kC

3

q

.03q

m1

)2 = − kC

y

q23q2

q1y2 = −q2(y − 3.0 m)2 = −q2y2 + 6q2y − (9.0 m2)q2

(q1 + q2)y2 − 6q2y + (9.0 m2)q2 = 0

y =

y =

y = 9.6 m = r2,3

F3,2 = kC

r3

q

,2

32q2 =

F3,2 = −1.7 × 10−2 N

(8.99 × 109 N •m2/C2)(−1.9 × 10−5 C)(9.0 × 10−6 C)

(9.6 m)2

6(−19 mC) ±√

(6.0 m)2(−19 mC)2 − 4(9.0 mC − 19mC)(9.0m2)(−19 mC)2(9.0 mC − 19 mC)

6q2 ±√

(6q2)2 − 4(q1+ q2)(9.0 m2)(q2)2(q1 + q2)



V

2. q1 = 25 mC

q2 = −5.0 mC

q3 = −35 mC

kC = 8.99 × 109 N •m2/C2

r1,2 = 0.25 m

The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because the forces point in the same direction. Because q1 is larger than q2, q3 will be closest to q2on the side opposite of q1.

F3,2 = −F3,1 = k

(C

−q

x3

)

q22 =

(x +−k

0C

.2

q

53q

m1

)2

(q2 + q1)x2 − (0.50 m)q2x + (0.625 m2)q2 = 0

x =

x =

x = −0.20 m = r3,2

F3,2 = kC

r

q

3,

3

2

q1 = = 39 N(8.99 × 109 N •m2/C2)(−3.5 × 10−5 C)(−5.0 × 10−6 C)

(−0.20 m)2

(0.50 m)(−5.0 mC) ±√

(0.50m)2(−5.0mC)2 − 4(−5.0mC + 25mC)(0.625m2)(−5.0mC)2(−5.0 mC + 25 mC)

(0.50 m)q2 ±√

(0.50m)2q22− 4(q2+ q1)(0.625 m2)q22(q2 + q1)

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3. q1 = 6.0 mC

q2 = −12.0 mC

q3 = 6.0 mC

kC = 8.99 × 109 N •m2/C2

r1,0 = 5.0 × 10−2 m

F2,3 = −F1,2 = −k

rC

1,

q

2

12q2 =

F2,3 = 259 N

−(8.99 × 109 N •m 2/C2)(6.0 × 10−6 C)(−12.0 × 10−6 C)

(5.0 × 10−2 m)2

4. q1 = 7.2 nC

q2 = 6.7 nC

q3 = −3.0 nC

kC = 8.99 × 109 N •m2/C2

r1,2 = 3.2 × 10−1 m = 0.32 m

The charge, q3, must be between the charges to achieve electrostatic equilibrium.

F1,3 + F1,2 = kC

x

q21q3 −

(x −kC

0.

q

32

2

q3

m)2 = 0

(q1 − q2)x2 − (0.64 m)q1x + (0.32 m)2q1x = 0

x =

x = 16 cm

(0.64 m)(7.2 nC) ±√

(0.64m)2 (7.2nC)2 − 4(7.2 nC − 6.7 nC)(0.32 m)2(7.2 nC)2(7.2 nC − 6.7 nC)

5. q1 = 5.5 nC

q2 = 11 nC

q3 = −22 nC

kC = 8.99 × 109 N •m2/C2

r1,2 = 88 cm


F1,3 + F1,2 = kC

x

q21q3 −

(x −kC

8

q

82q

cm3

)2 = 0

(q1 − q2)x2 − (176 cm)q1x + (88 cm)2q1x = 0

x =

x = 36 cm

(176 cm)(5.5 nC) ±√

(176 cm)2(5.5 nC)2 − 4(5.5 nC − 11nC)(88 cm)2(5.5 nC)2(5.5 nC − 11 nC)


V

6. q1 = −2.5 nC

q2 = − 7.5 nC

q3 = 5.0 nC

kC = 8.99 × 109 N •m2/C2

r1,2 = 20.0 cm


F1,3 + F1,2 = kC

x

q21q3 −

(x −k

2C

0

q

.2

0

q

c3

m)2 = 0

(q1 − q2)x2 − (40.0 cm)q1x + (20.0 cm)2q1x = 0

x =

x = 7.3 cm

(40.0 cm)(−2.5 nC) ±√

(40.0cm)2(−2.5nC)2 − 4(−2.5nC + 7.5 nC)(20.0 cm)2(−2.5nC)2(−2.5 nC + 7.5 nC)

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7. q1 = −2.3 C

q3 = −4.6 C

r1,2 = r3,1 = 2.0 m

r3,2 = 4.0 m

kC = 8.99 × 109 N •m2/C2

F3,1 + F3,2 = −k

rC

3,

q

1

32q1 −

kC

r

q

3,

3

2

q2 = 0

q2 = −q

r1

3

r

,

3

1

,22

2 = −(−2.

(

3

2.

C

0

)

m

(4

)

.20 m)2

= 9.2 C

8. q1 = 8.0 C

q3 = −4.0 C

r1,2 = 1.0 m

r3,1 = 1.0 m

r3,2 = 2.0 m

F3,1 + F3,2 = −k

rC

3,

q

1

32q1 −

kC

r3

q

,2

32q2 = 0

q2 = −q

r1

3

r

,1

32,2

2

= −(8.0

(1

C

.0

)(

m

2.

)

02

m)2

= −32 C

9. q1 = 49 C

q3 = −7.0 C

r1,2 = 7.0 m

r3,1 = −18.0 m

r3,2 = 25.0 m

F3,1 + F3,2 = − k

rC

3,

q

123q1 −

kC

r3

q

,2

32q2 = 0

q2 = −q

r1

3

r

,1

32,2

2

= −(4

(

9

−C

18

)(

.0

25

m

.0

)2m)2

= −94.5 C

10. q1 = 72 C

q3 = −8.0 C

r1,2 = 15 mm = 1.5 × 10−2 m

r3,1 = −9.0 mm = −9.0 × 10−3 m

r3,2 = 2.4 × 10−2 m

F3,1 + F3,2 = − k

rC

3,

q

123q1 −

kC

r3

q

,2

32q2 = 0

q2 = −q

r1

3

r

,1

32,2

2

= = −512 C−(72 C)(2.4 × 10−2 m)2

(−9.0 × 10−3 m)2


1. Ex = 9.0 N/C

q = −6.0 CEx =

Fele

qctric

Felectric = Exq = (9.0 N/C)(−6.0 C)

Felectric = −54 N in the −x direction

2. Ey = 1500 N/C

q = 5.0 × 10−9 CEy =

Fele

qctric

Felectric = Eyq = (1500 N/C)(5.0 × 10−9 C)

Felectric = 7.5 × 10−6 N in the +y direction


V

3. m = 3.35 × 10−15 kg

q = 1.60 × 10−19 C

g = 9.81 m/s2

a. Felectric = −Fgravity = −mg = −(3.35 × 10−15 kg)(9.81 m/s2)

Felectric = −3.29 × 10−14 N upward

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4. qq = 3.00 × 10−6 C

q2 = 3.00 × 10−6 C

kC = 8.99 × 109 N •m2/C2

r1 = 0.250 m

r2 =√

(2.00m)2 + (2.00 m)2 = 2.02 m

E1 = Ey = Ey = k

rC

1

q21 =

E1 = Ey,1 = 4.32 × 105 N/C

E2 = k

rC

2

q22 = = 6.61 × 103 N/C

j = tan−1x

y = tan−102.2.0500m

m = 7.12°

Ex,2 = E2 cos 7.12° = (6.61 × 103 N/C)(cos 7.12°) = 6.56 × 103 N/C

Ey,2 = E2 sin 7.12° = (6.61 × 103 N/C)(sin 7.12°) = 8.19 × 103 N/C

Ex,tot = 0 N/C + 6.56 × 103 N/C = 6.56 × 103 N/C

Ey,tot = 4.32 × 105 N/C + 8.19 × 103 N/C = 4.40 × 105 N/C

Etot =√

(Ex,tot2 + (Ey,tot)2

Etot =√

(6.56× 103 N/C)2 + (4.40 × 105 N/C)2

Etot =

tan j = E

E

x

y,

,

t

t

o

o

t

t =

4

6

.

.

4

5

0

6

××

1

1

0

0

5

3N

N

/

/

C

C

j = 89.1°

4.40 × 105 N/C

(8.99 × 109 N •m2/C2)(3.00 × 10−6 C)

(2.02 m)2

(8.99 × 109 N •m2/C2)(3.00 × 10−6 C)

(0.250 m)2

5. q1 = 1.50 × 10−5 C

q2 = 5.00 × 10−6 C

kC = 8.99 × 109 N •m2/C2

r1 = 1.00 m

r2 = 0.500 m

E1 = Ey,1 = k

rC

1

q21 = = 1.35 × 105 N/C

E2 = Ey,2 = k

rC

2

q22 = = 1.80 × 105 N/C

Ey,tot = Etot = 1.35 × 105 N/C + 1.80 × 105 N/C =

The electric field points along the y-axis.

3.15 × 105 N/C

(8.99 × 109 N •m2/C2)(5.00 × 10−6 C)

(0.500 m)2

(8.99 × 109 N •m2/C2)(1.50 × 10−5 C)

(1.00 m)2


V

6. E = 1663 N/C

Felectric = 8.4 × 10−9 NE =

Fele

qctric

q = = 8.4

1

2

66

×3

1

N

0−

/C

9 N = 5.06 × 10−12 C

FelectricE

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7. E = 4.0 × 103 N/C

Felectric = 6.43 × 10−9 NE =

Fele

qctric

q = = 4

6

.

.

0

43

××10

130−

N

9

/

N

C = 1.61 × 10−12 C

FelectricE

8. q1 = −1.60 × 10−19 C

kC = 8.99 × 109 N •m2/C2

q2 = −1.60 × 10−19 C

q3 = 1.60 × 10−19 C

r1 = 3.00 × 10−10 m

r2 = 2.00 × 10−10 m

Fx,tot = kC

r2q1 +

k

rC

2

q22 +

kC

x2q3 = 0

q1 = q2 = −q3

kCq1r

1

12 +

r

1

22 −

x

12 = 0

kCq1r

1

12 +

r

1

22 =

kC

x2q1

x2 = =

x = 1.66 × 10−10 m

1(3.00 × 1

1

0−10 m)2 + (2.00 × 1

1

0−10 m)

1r

1

12 +

r

1

22

9. q1 = −9.00 C

q2 = 6.00 C

q3 = 3.00 C

r1 = 1.5 mm

r2 = 1.5 mm

kC = 8.99 × 109 N • m2/C2

The electric field of both q2 and q1 point toward q1. To balance the electric field, q3must be placed opposite of q2. Also, because the electric field of q2 points in the −xdirection, E2,x is negative.

Ex,tot = Ex − E2,x + E3,x

Ex,tot = k

rC

1

q21 −

k

rC

2

q22 +

kC

x2q3 = 0

r1 = r2

kC(q

r1

12− q2) +

kC

x2q3 = 0

(q1

r

−

12q2)

= −x

q23

x2 = (q

−

1

q

−3r1

q

2

2) = = 0.45 mm2

x = ±0.67 mm Since q3 must be opposite q2, and the position of q3 is positive, q2 must

be on the negative x-axis, so x = −0.67 mm

−(3.00 C)(1.5 mm)2

(−9.00 C − 6.00 C)


V

10. q1 = 5.50 × 10−8 C

q2 = 1.10 × 10−8 C

q3 = 5.00 × 10−9 C

r1 = −5.00 × 10−7 m

r2 = 5.00 × 10−7 m

kC = 8.99 × 109 N •m2/C2

The electric field of both q2 and q1 point toward q1. To balance the electric field, q3must be on the side opposite q2. Also, because q2 points in the −x direction, E2,x isnegative.

Ex,tot = E1,x − E2,x + E3,x

Ex,tot = k

rC

1

q21 +

k

rC

2

q21 +

kC

x2q3 = 0

r1 = r2, so r12 = r2

2

k

rC

1

q21 −

k

rC

2

q22 =

−k

xC2q3

q1

r

−

12q2 =

−x

q23

x2 = −(q

q

1

3

q

r

2

1

)

2

= = 1.89 × 10−14 m2

x = −1.37 × 10−7 m

−(5.00 × 10−9 C)(−5.00 × 10−7 m)2

(−5.50 × 10−8 C − 1.10 × 10−8 C)

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Chapter 18Electrical Energy and Capacitance

V

1. q = 1.45 × 10−8 C

E = 105 N/C

d = 290 m

PEelectric = −qEd = −(1.45 × 10−8 C)(−105 N/C)(290 m)

PEelectric = 4.42 × 10−4 J


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2. q1 = 6.4 × 10−8 C

q2 = −6.4 × 10−8 C

r = 0.95 m

PEelectric = kcq

r1q2

PEelectric =

PEelectric = −3.9 × 10−5 J

(8.99 × 109 N•m2/C2)(6.4 × 10−8 C)(−6.4 × 10−8 C)

0.95 m

3. q = 1.60 × 10−19 C

E = 3.0 × 106 N/C

d = 7.3 × 10−7 m

PEelectric = −qEd

PEelectric = −(1.60 × 10−19 C)(3.0 × 106 N/C)(7.3 × 10−7 m)


4. q1 = −4.2 × 10−8 C

q2 = 6.3 × 10−8 C


r = P

k

Ec

e

q

le

1

c

q

tr

2

ic

r =

r = 3.4 × 10−2 m = 0.34 cm

(8.99 × 109 N•m2/C2)(−4.2 × 10−8 C)(6.3 × 10−8 C)

−6.92 × 10−4 J

5. q1 = 1.6 × 10−8 C

q2 = 1.4 × 10−8 C


r = P

k

Ec

e

q

le

1

c

q

tr

2

ic

r =

r = 9.5 × 10−1 m = 96 cm

(8.99 × 109 N•m2/C2)(1.6 × 10−8 C)(1.4 × 10−8 C)

2.1 × 10−6 J

6. q1 = −5.5 × 10−8 C

q2 = 7.7 × 10−8 C


r = P

k

Ec

e

q

le

1

c

q

tr

2

ic

r =

r = 2.9 × 10−3 m = 2.9 mm

(8.99 × 109N•m2/C2)(−5.5 × 10−8 C)(7.7 × 10−8 C)

−1.3 × 10−2 J

10. r = 1.25 × 10−6 m


q1 = q2

q2 = rPE

kel

c

ectric

q = rPEk

el

cectric =

q = 2.9 × 10−11 C

a. q1 =

b. q2 = −2.9 × 10−11 C

2.9 × 10−11 C

(1.25 × 10−6 m)(6.3 × 10−6 J)

(8.99 × 109 N•m2/C2)


1. ∆V = 114.0 V

r = 6.695 × 106 mq =

r∆kc

V = = 8.49 × 10−2 C

(6.695 × 106 m)(114.0 V)

8.99 × 109 N•m2/C2


V

7. PEelectric = −1.39 × 1011 J

E = 3.4 × 105 N/C

d = 7300 m

q = −PE

Ee

dlectric =

q = 56 C

−(−1.39 × 1011 J)(3.4 × 105 N/C)(7300 m)

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8. r = 1.25 × 10−3 m


q1 = q2

q2 = rPE

kel

c

ectric

q = rPEk

el

cectric =

q = 4.17 nC

a. q1 =

b. q2 =

c. It is unlikely that the hat would stay on a friend’s head only by charge becausethe charge on the friend would most likely be different than the hat.

−4.2 nC

4.2 nC

(1.25 × 10−3 m)(1.25 × 10–4 J)

(8.99 × 109N•m2/C2

9. r = 9.4 × 10−4 m


q1 = q2

q2 = rPE

kel

c

ectric

q =rPEk

el

cectric =

q = 9.6 × 10−12 C

a. q1 =

b. q2 = −9.6 × 10−12 C

9.6 × 10−12 C

(9.4 × 10−4 m)(8.89 × 10–10 J)

(8.99 × 109 N•m2/C2)

2. ∆V = 18600 V

r = 1991 mq = = = 4.12 × 10−3 C

(1991 m)(18600 V)8.99 × 109 N•m2/C2

r∆Vkc


V

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4. V = 9.0 V

q = 9.4 × 10−8 Cr =

k

Vcq =

r = 93.4 m

(8.99 × 109 N•m2/C2)(9.4 × 10−8 C)

9.0 V

6. E = 3.0 × 106 N/C

∆d = 6.25 × 10−4 m

10. q1 = 1.60 × 10−19 C

q2 = −1.60 × 10−19 C

q3 = 1.60 × 10−19 C

q4 = −1.60 × 10−19 C

d = 2.82 × 10−10 m

r = r1 + r2 + r3 + r4 = (0.707)d = (0.707)(2.82 × 10−10 m)

r = 1.99 × 10−10 m

V = k

rc [q1 + q2 + q3 + q4] =

k

rc [2q1 + 2q2]

V = 8.9

1

9

.9

×9

1

×09

10

N−•1m0 m

2/C2

[2(1.60 × 10−19 C) + 2(−1.60 × 10−19 C)]

V = 0 V

5. q = 1.28 × 10−18 C

r = 3.95 × 10−2 m∆V =

kc

r

q =

∆V = 2.91 × 10−7 V

(8.99 × 109 N•m2/C2)(1.28 × 10−18 C)

3.95 × 10−2 m

∆V = –E∆d = −(3.0 × 106 N/C)(6.25 × 10−4 m)

∆V = −1.9 × 103 V

7. E = 95 N/C

∆d = 3.0 × 102 m

∆V = −E∆d = −(95 N/C)(3.0 × 102 m)

∆V = −2.8 × 104 V

8. q1 = 8(1.60 × 10−19 C) = 1.28 × 10−18 C

q2 = 1.60 × 1−19 C

d = 9.58 × 10−11 m

q = 105°

f = q2

= = 52.5°

r1 = dcosf = (9.58 × 10−11 m) cos (52.5°) = 5.83 × 10−11 m r2 = dsinf = (9.58 × 10−11 m) sin (52.5°) = 7.60 × 10−11 m

V = kc

r

q

1

1 + 2k

rc

2

q2 = kcq

r1

1 + 2

r

q

2

2V = (8.99 × 109 N•m2/C2) + V = 197 V + 37.9 V = 235 V

2(1.60 × 10−19 C)

7.60 × 10−11 m

1.28 × 10−18 C5.83 × 10−11 m

105°

2

9. q1 = 3.04 × 10−18 C

q2 = 5.60 × 10−18 C

r1 = 1.89 × 10−10 m

r2 = −9.30 × 10−11 m

V = kc

r

q

1

1 + k

rc

2

q2 = kcq

r1

1 + q

r2

2V = (8.99 × 109 N•m2/C2) 13.

.

8

0

9

4

××

1

1

0

0−

−

1

1

0

8

m

C +

−5

9

.

.

6

3

0

0

××

1

1

0

0

−

−

1

1

8

1C

m

V = 145 V − 541 V = −396 V

3. V = 1.0 V

q = –1.60 × 10–19 Cr =

k

Vcq =

r = 1.4 × 10−9 m

(8.99 × 109 N•m2/C2)(−1.60 × 10−19 C)

1.0 V


V

1. A = 4.8 × 10−3 m2

C = 1.8 × 10−5 F

k = 1

d = ke

C0A =

d = 2.4 × 10−9 m

(1)(8.85 × 10−12 C2/N•m2)(4.8 × 10−3 m2)

1.8 × 10−5 F


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2. A = 6.4 × 10−3 m2

C = 4.55 × 10−9 Fd =

eC0A =

d = 1.2 × 10−5 m

(8.85 × 10−12 C2/N•m2)(6.4 × 10−3 m2)

4.55 × 10−9 F

3. R = 6.4 × 106 m Csphere = k

R

c =

8.99

6

×.4

1

×09

1

N

06

•m

m2/C2 = 7.1 × 10−4 F

8. C = 1.0 × 10−6 F

Q = 3.0 × 10−2 C

9. C = 2.0 × 10−6 F

Q = 4.0 × 10−4 C

∆V = Q

C =

3

1

.

.

0

0

××

1

1

0

0

−

−

2

6C

F = 3.0 × 104 V = 30 kV

∆V = Q

C =

4

2

.

.

0

0

××

1

1

0

0

−

−

4

6C

F = 2.0 × 102 V

10. C = 5.0 × 10−5 F

Q = 6.0 × 10−4 C

∆V = Q

C =

6

5

.

.

0

0

××

1

1

0

0

−

−

4

5C

F = 12 V

4. R = 0.10 mCsphere =

k

R

c =

8.99 ×0

1

.

0

190

N

m

•m2/C2 = 1.1 × 10−11 F

5. C = 1.4 × 10−5 F

∆V = 1.5 × 104 V

Q = C∆V = (1.4 × 10−5 F)(1.5 104 V) =

PEelectric = 2

Q

C

2

= 2(1

(

.

0

4

.2

×1

1

C

0−)2

5 F) = 1.6 × 103 J

0.21 C

6. C = 1.0 × 10−9 F

∆V = 600 V

Q = C∆V = (1.0 × 10−9 F)(600 V) =

PEelectric = Q

2C

2

= (

2

6

(

.

1

0

.0

××10

1

−

0

7

−9C

F

)

)

2

= 1.8 × 10−4 J

6.0 × 10−7 C

7. C = 5.0 × 10−13 F

∆V = 1.5 V

Q = C∆V = (5 × 10−13 F)(1.5 V) = 7.5 × 10−13 C


19ChapterCurrent and Resistance

V

1. ∆Q = 76 C

∆t = 19 sI =

∆∆Q

t =

7

1

6

9

C

s = 4.0 A


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2. ∆Q = 1.14 × 10−4 C

∆t = 0.36 sI =

∆∆Q

t =

1.14

0

×.3

1

6

0

s

−4 C = 0.32 mA

3. ∆Q = 2.9 × 10−2 C

∆t = 11 sI =

∆∆Q

t =

2.9 ×11

10

s

−2 C = 2.6 mA

4. ∆Q = 98 C

I = 1.4 A∆t =

∆I

Q =

1

9

.

8

4

C

A = 70 s

1. R = 1.0 × 105 ΩI1 = 1.0 × 10−3 A

I2 = 1.5 × 10−2 A

∆V1 = I1R = (1.0 × 10−3 A) (1.0 × 105 Ω) =

∆V2 = I2R = (1.5 × 10−2 A) (1.0 × 105 Ω) = 1.5 × 103 V

1.0 × 102 V

5. ∆Q = 30.9 C

I = 9.65 A∆t =

∆I

Q =

9

3

.

0

6

.

5

9 C

A = 3.20 s

6. ∆Q = 56 C

I = 7.8 A∆t =

∆I

Q =

7

5

.

6

8

C

A = 7.2 s

7. ∆t = 15 s

I = 9.3 A∆Q = I∆t = (9.3 A) (15 s) = 1.4 × 102 C

8. ∆t = 2.0 min = 120 s

I = 3.0 A∆Q = I∆t = (3.0 A) (120 s) = 3.6 × 102 C

9. ∆t = 2.0 s

I = 0.70 A∆Q = I∆t = (0.70 A) (2.0 s) = 1.4 C


10. ∆t = 4.3 s

I = 5.6 A∆Q = I∆t = (5.6 A) (4.3 s) = 24 C


V

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3. I = 4.66 A

R = 25.0 Ω∆V = IR = (4.66 A) (25.0 Ω) = 116 V

4. ∆V = 120 V

R = 12.2 ΩI =

∆R

V =

1

1

2

2

.

0

2

V

Ω = 9.84 A

5. ∆V = 650 V

R = 1.0 × 102 ΩI =

∆R

V =

1.0

6

×50

10

V2 Ω

= 6.5 A

6. R = 40.0 Ω∆V1 = 120 V

V2 = 240 V

I1 = ∆

R

V1 = 4

1

0

2

.

0

0

V

Ω =

I2 = ∆

R

V2 = 4

2

0

4

.

0

0

V

Ω = 6.00 A

3.00 A


7. I = 0.75 A

∆V = 120 VR =

∆I

V =

0

1

.

2

7

0

5

V

A = 1.6 × 102 Ω

8. I = 0.89 A

∆V = 5.00 × 102 VR =

∆I

V =

5.0

0

0

.8

×9

1

A

02 V = 5.6 × 102 Ω

9. I = 0.545 A

∆V = 120 VR =

∆I

V =

0

1

.5

2

4

0

5

V

A = 220 Ω

10. I = 0.65 A

∆V = 117 VR =

∆I

V =

0

1

.

1

6

7

5

V

A = 1.8 × 102 Ω

2. I = 0.75 A

R = 6.4 Ω∆V = IR = (0.75 A) (6.4 Ω) = 4.8 V

1. ∆V = 2.5 × 104 V

I = 20.0 A

P = I∆V = (20.0 A) (2.5 × 104 V) = 5.0 × 105 W

2. ∆V = 720 V

R = 0.30 ΩP =

(∆V

R

)2

= (7

0

2

.3

0

0

V

Ω)2

= 1.7 × 106 W

3. ∆V = 120 V

R1 = 144 ΩR2 = 240 Ω

P1 = (∆

R

V

1

)2

= (1

1

2

4

0

4

V

Ω)2

=

P2 = (∆

R

V

2

)2

= (1

2

2

4

0

0

V

Ω)2

=

The brighter bulb is the 60.0-W bulb, which has the 240−Ω resistance.

60.0 W

100 W

4. ∆V = 120 V

P = 1750 WR =

(∆V

P

)2

= (

1

1

7

2

5

0

0

V

W

)2

= 8.22 Ω


V

5. ∆V = 120 V

P = 650 WR =

(∆V

P

)2

= (1

6

2

5

0

0

V

W

)2

= 22.2 Ω

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6. ∆V = 120 V

P = 370 W R =

(∆V

P

)2

= (1

3

2

7

0

0

V

W

)2

= 38.9 Ω

7. P = 350 W

R = 75 ΩI2 =

R

P I =

R

P =

3

7

55

0ΩW

= 2.16 A

8. P = 230 W

R = 91 ΩI2 =

R

P I =

R

P =

2

9

31

0ΩW

= 1.59 A

9. I = 8.0 × 106 A

P = 6.0 × 1013 W ∆V =

P

I =

6

8

.0

.0

××1

1

0

0

1

6

3

A

W = 7.5 × 106 V

10. I = 16.3 A

P = 1.06 × 104 W ∆V =

P

I =

1.06

16

×.3

10

A

4 W = 6.50 × 102 V

1. ∆t = 1.0 h

Energy = 2.7 × 108 J P = En

∆er

t

gy = = 75 kW

2.7 × 108 J(1.0 h) (3.6 × 106 J/kW•h)

2. ∆t = 3.0 h

Energy = 4.86 × 108 J P = En

∆er

t

gy = = 45 kW

4.86 × 108 J(3.0 h) (3.6 × 106 J/kW•h)

3. P = 1200 W = 1.200 kW

Energy = 1.512 × 108 J ∆t = En

P

ergy = = 35.00 h

1.512 × 108 J(1.200 kW) (3.6 × 106 J/kW•h)


4. P = 600 W = 0.600 kW

Energy = 8.64 × 109 J ∆t = En

P

ergy = = 4.00 × 103 h

8.64 × 109 J(0.600 kW) (3.6 × 106 J/kW•h)

5. Cost of electricity = $0.0650/kW•h

Energy = 200.0 kW•h

Cost = (Energy) ($0.065/kW•h)

Cost = (200.0 kW•h) ($0.0650/kW•h) = $13.00

6. Cost of electricity = $0.078/kW•h

Final Meter Reading = 24422 kW•h

Previous Meter Reading = 24204 kW•h

Energy = Final Meter Reading – Previous Meter Reading

Energy = 24422 kW•h – 24204 kW•h = 218 kW•h

Cost = (Energy) ($0.078/kW•h)

Cost = (218 kW•h) ($0.078/kW•h) = $17.00


V

7. ∆t = 8.0 h

P = 0.125 kWEnergy = P∆t = (0.125 kW) (8.0 h) = 1.0 kW•h

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8. ∆t = 24.0 h

P = 0.75 kWEnergy = P∆t = (0.75 kW) (24.0 h) (3.6 × 106 J/kW•h) = 6.5 × 107 J

9. ∆t = 10.0 min = 0.167 h

P = 0.55 kWEnergy = P∆t = (0.55 kW) (0.167 h) (3.6 × 106 J/kW•h) = 3.3 × 105 J

10. ∆t = 3.0 min

= 5.0 × 10–2 h

P = 0.85 kW

Energy = P∆t = (0.85 kW) (5.0 × 10–2h) (3.6 × 106 J/kW•h)

Energy = 1.5 × 105 J


Chapter 20Circuits and Circuit Elements

V

1. ∆V = 12 V

R1 = 16 Ω

I = 0.42 A

R2 = ∆

I

V − R1 =

0

1

.4

2

2

V

A − 16 Ω = 29 Ω − 16 Ω = 13 Ω


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2. ∆V = 3.0 V

R1 = 24Ω

I = 0.062 A

R2 = ∆

I

V − R1 =

0

3

.0

.0

62

V

A − 24 Ω = 48 Ω − 24 Ω = 24 Ω

3. ∆V = 9.0 V

R1 = 9.1 Ω

I = 0.33 A

R2 = ∆

I

V − R1 =

0

9

.

.

3

0

3

V

A − 9.1 Ω = 27 Ω − 9 Ω = 18 Ω

4. 73 bulbs

Reach bulb = 3.0 Ω

Req = ΣReach bulb All bulbs have equal resistance.

Req = (73) (3.0 Ω) = 219 Ω

5. 25 speakers

Reach speaker = 12.0 Ω

Req = ΣReach speaker All speakers have equal resistance.

Req = (25)(12.0 Ω) = 3.0 × 102 Ω

6. 57 lights

Reach light = 2.0 Ω

Req = ΣReach light All lights have equal resistance.

Req = (57)(2.0 Ω) = 114 Ω

7. 4 speakers

Reach speaker = 4.1 Ω

∆V = 12 V

Req = ΩReach speaker All speakers have equal resistance.

Req = (4)(4.1 Ω) = 16.4 Ω

I = R

∆

e

V

q =

1

1

6

2

.4

V

Ω = 7.3 × 10−1 A

8. 10 bulbs

Reach bulb = 10 Ω

∆V = 115 V

Req = ΣReach bulb All bulbs have equal resistance.

Req = (10)(10 Ω) = 100 Ω

I = R

∆

e

V

q =

1

1

0

0

0

0

ΩV

= 1 A

9. R1 = 96 Ω

R2 = 48 Ω

R3 = 29 Ω

∆V = 115 V

Req = ΣR = R1 + R2 + R3 = 96 Ω + 48 Ω + 29 Ω = 173 Ω

I = R

∆

e

V

q =

1

1

7

1

3

5

ΩV

= 665 mA

1. ∆V = 3.0 V

R1 = 3.3 Ω

I = 1.41 A

∆V = IReq

I = R

∆

e

V

q =

∆R

V

1 +

∆R

V

2

∆R

V

2 = I −

∆R

V

1

R2 = = = [1.41 A

3.0

−V

0.91 A] = 6.0 Ω

3.0 V

1.41 A − 3

3

.

.

3

0

ΩV

∆V

I − ∆R

V

1


V

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10. R1 = 56 Ω

R2 = 82 Ω

R3 = 24 Ω

∆V = 9.0 V

Req = ΣR = R1 + R2 + R3 = 56Ω + 82Ω + 24Ω = 162Ω

I = R

∆

e

V

q =

1

9

6

.0

2

V

Ω = 55.6 mA

2. ∆V = 12 V

R1 = 56 Ω

I = 3.21 A

∆V = IReq

I = R

∆

e

V

q =

∆R

V

1 +

∆R

V

2

∆R

V

2 = I −

∆R

V

1

R2 = = = = 4.0 Ω12 V

[3.21 A − 0.21 A]

12 V

3.21 A − 5

1

6

2

ΩV

∆V

I − ∆R

V

1

3. ∆V = 1.5 V

R1 = 18 Ω

I = 0.103 A

∆V = IReq

I = R

∆

e

V

q =

∆R

V

1 +

∆R

V

2

∆R

V

2 = I −

∆R

V

1

R2 = = = = 75 Ω1.5 V

[0.103 A − 0.083 A]

1.5 V

0.103 A − 1

1

.

8

5

ΩV

∆V

I − ∆R

V

1


4. R1 = 39 Ω

R2 = 82 Ω

R3 = 12 Ω

R4 = 22 Ω

∆V = 3.0 V

R

1

eq =

R

1

1 +

R

1

2 +

R

1

3 +

R

1

4 =

39

1

Ω +

82

1

Ω +

12

1

Ω +

22

1

Ω

R

1

eq =

0

1

.0

Ω26 +

0

1

.0

Ω12 +

0

1

.0

Ω83 +

0

1

.0

Ω45 =

01.1Ω7

Req = 6.0 Ω


V

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5. R1 = 10.0 Ω

R2 = 12 Ω

R3 = 15 Ω

R4 = 18 Ω

∆V = 12 V

R

1

eq =

R

1

1 +

R

1

2 +

R

1

3 +

R

1

4 =

10.

1

0 Ω +

12

1

Ω +

15

1

Ω +

18

1

Ω

R

1

eq =

0

1

.1

Ω0

+ 0

1

.0

Ω83 +

0

1

.0

Ω67 +

0

1

.0

Ω56

Req = 3.3 Ω

6. R1 = 33 Ω

R2 = 39 Ω

R3 = 47 Ω

R4 = 68 Ω

V = 1.5 V

R

1

eq =

R

1

1 +

R

1

2 +

R

1

3 +

R

1

4 =

33

1

Ω +

39

1

Ω +

47

1

Ω +

68

1

Ω

R

1

eq =

0

1

.0

Ω30 +

0

1

.0

Ω26 +

0

1

.0

Ω21 +

0

1

.0

Ω15

Req = 11 Ω

7. ∆V = 120 V

R1 = 75 Ω

R2 = 91 Ω

I1 = R

V

1 I2 =

R

V

2

I1 = 1

7

2

5

0

ΩV

=

I2 = 1

9

2

1

0

ΩV

= 1.3 A

1.6 A

8. ∆V = 120 V

R1 = 82 Ω

R2 = 24 Ω

I1 = R

V

1 I2 =

R

V

2

I1 = 1

8

2

2

0

ΩV

=

I2 = 1

2

2

4

0

ΩV

= 5.0 A

1.5 A

9. ∆V = 120 V

R1 = 11 Ω

R2 = 36 Ω

I1 = R

V

1 I2 =

R

V

2

I1 = 1

1

2

1

0

ΩV

=

I2 = 1

3

2

6

0

ΩV

= 3.3 A

11 A

10. ∆V = 1.5 V

R1 = 3.3 Ω

R2 = 4.3 Ω

I1 = R

V

1 I2 =

R

V

2

I1 = 3

1

.

.

3

5

ΩV

=

I2 = 4

1

.

.

3

5

ΩV

= 0.35 A

0.45 A


V

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2. I = 0.375 A

V = 9.00 V

R1 = 18.0 Ω

R2 = 3.0 Ω

R3 = 4.0 Ω

R4 = 8.00 Ω

R5 = 20.0 Ω

R6 = 22.0 Ω

R7 = 6.0 Ω

Series:

Group (a): Req,a = R2 + R3 = 3.0Ωeq + 4.0Ωeq = 7.0Ω

Group (b): Req,b = R5 + R6 = 20.0 Ω + 22.0 Ω = 42.0 Ω

Parallel:

Group (c): R

1

eq =

18.

1

0 Ω +

7.0

1

Ω =

0.

1

05

Ω56

+ 0

1

.1

Ω43

Req = 5.03 Ω

Group (d): R

1

eq =

42.

1

0 Ω +

6.0

1

Ω =

0.

1

02

Ω38

+ 0

1

.1

Ω67

Req = 5.24 Ω

Series:

Group (e): Req,e = Req,c + R4 + Req,d + R

= 5.03 Ω + 8.00 Ω + 5.24 Ω + R = 18.27 Ω + R

R = ∆

I

V − 18.27 Ω =

0

9

.

.

3

0

7

0

5

V

A − 18.27 Ω = 5.7 Ω

1. I = 0.680 A

V = 15.0 V

R1 = 15 Ω

R2 = 11 Ω

R3 = 6.0 Ω

R4 = 7.0 Ω

R5 = 12.0 Ω

Series:

Group (a): Req,a = R1 + R2 = 15 Ω + 11 Ω = 26 Ω


Parallel:

Groups (a) + (b): R

1

eq =

26

1

Ω +

13.

1

0 Ω =

0

1

.0

Ω38 +

0

1

.0

Ω77 =

0

1

.1

Ω2

Req = 8.3 Ω

Series:

Req = 12.0 Ω + 8.3 Ω + R = 20.3 Ω + R

∆V = IReq = I(20.3 Ω + R)

R = ∆

I

V − 20.3 Ω =

0

1

.

5

6

.

8

0

0

V

A − 20.3 Ω = 1.8 Ω



V

3. I = 0.185 A

V = 12.0 V

R1 = 2.0 Ω

R2 = 3.0 Ω

R3 = 3.0 Ω

R4 = 4.0 Ω

R5 = 8.0 Ω

R6 = 2.0 Ω

R7 = 3.0 Ω

R8 = 4.0 Ω

R9 = 4.0 Ω

R10 = 5.0 Ω

R11 = 6.0 Ω

Series:

Group (a): Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω


Group (c): Req,c = R6 + R7 + R8 = 2.0 Ω + 3.0 Ω + 4.0 Ω = 9.0 Ω

Group (d): Req,d = R9 + R10 + R11 = 4.0 Ω + 5.0 Ω + 6.0 Ω = 15.0 Ω

Parallel:

Group (e): Re

1

q,e =

Re

1

q,a +

Re

1

q,b =

5.0

1

Ω +

7.0

1

Ω =

0

1

.2

Ω0

+ 0

1

.1

Ω4

Req,e = 2.9 Ω

Group (f): R

1

eq,f =

Re

1

q,c +

Re

1

q,d =

9.0

1

Ω +

15.

1

0 Ω =

0

1

.1

Ω11 +

0

1

.0

Ω67

Req,f = 5.6 Ω

Series:

Req = 4R5 + 2Req,e + Req,f + R = 4(8.0 Ω) + 2(2.9 Ω) + 5.6 Ω + R

Req = 32.0 Ω + 5.8 Ω + 5.6 Ω + R = 43.4 Ω + R

R = ∆

I

V − 43.4 Ω =

0

1

.

2

1

.

8

0

5

V

A − 43.4 Ω = 21.5 Ω

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4. each resistor = 10.0 Ω Parallel:

Group (a): Re

1

q,a =

10.

1

0 Ω +

10.

1

0 Ω =

0

1

.1

Ω00 +

0

1

.1

Ω00

Req,a = 5.00 Ω

Group (b): Re

1

q,b =

10.

1

0 Ω +

10.

1

0 Ω +

10.

1

0 Ω +

10.

1

0 Ω

Re

1

q,b =

0

1

.1

Ω00 +

0

1

.1

Ω00 +

0

1

.1

Ω00 +

0

1

.1

Ω00

Req,b = 2.50 Ω

Group (c): Req,c = 10.0 Ω + 10.0 Ω + 5.00 Ω + 2.50 Ω + 5.00 Ω + 2.50 Ω + 10.0 Ω + 10.0 Ω

Req,c = 55.0 Ω

5. each resistor = 2.0 Ω Series:

Group (a): Req,a = 2.0 Ω + 2.0 Ω = 4.0 Ω

Group (b): Req,b = 2.0 Ω + 2.0 Ω + 2.0 Ω = 6.0 Ω

Group (c): Req,c = 2.0 Ω + 2.0 Ω + 2.0 Ω = 2.0 Ω = 8.0 Ω

Parallel:

Group (d): Re

1

q,d =

2.0

1

Ω +

4.0

1

Ω +

6.0

1

Ω +

8.0

1

Ω =

0

1

.5

Ω0

+ 0

1

.2

Ω5

+ 0

1

.1

Ω7

+ 0

1

.1

Ω3

Req,d = 0.95 Ω

Series:

Req = 2.0 Ω + 2.0 Ω + 0.95 Ω = 5.0 Ω


V

6. each resistor = 10.0 Ω Parallel:

Group (a): Re

1

q,a =

10.

1

0 Ω +

10.

1

0 Ω +

10.

1

0 Ω =

0

1

.1

Ω00 +

0

1

.1

Ω00 +

0

1

.1

Ω00

Req,a = 3.33 Ω

Series:

Req = 4(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 3(10.0 Ω)

Req = 1.20 × 102 Ω

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7. ∆Vtot = 12.0 V

R1 = 3.0 Ω

R2 = 5.0 Ω

R3 = 2.0 Ω

R4 = 4.0 Ω

R5 = 5.0 Ω

R6 = 6.0 Ω

R7 = 1.5 Ω

Parallel:

Group (a): Re

1

q,a =

R

1

2 +

R

1

3 =

5.0

1

Ω +

2.0

1

Ω =

0

1

.2

Ω0

+ 0

1

.5

Ω0

Req,a = 1.4 Ω

Group (b): Re

1

q,b =

R

1

5 +

R

1

6 =

5.0

1

Ω +

6.0

1

Ω =

0

1

.2

Ω0

+ 0

1

.1

Ω7

Req,b = 2.7 Ω

Series:

Group (c): Req,c = R1 + Req,a + R4 + Req,b + R7

Req,c = 3.0 Ω + 1.4 Ω + 4.0 Ω + 2.7 Ω + 1.5 Ω = 13 Ω

I = ∆R

V

eq

t

,

o

c

t = 1

1

2

3

.0

ΩV

= 0.92 A

8. ∆Vtot = 15.0 V

R1 = 5.0 Ω

R2 = 5.0 Ω

R3 = 5.0 Ω

R4 = 3.0 Ω

Parallel:

Group (a): Re

1

q,a =

R

1

1 +

R

1

2 +

R

1

3 =

5.0

1

Ω +

5.0

1

Ω +

5.0

1

Ω =

0

1

.2

Ω0

3Req,a = 1.7 Ω

Series:

Req = 2R4 + 2Req,a = (2)(3.0 Ω) + (2)(1.7 Ω) = 9.4 Ω

I = ∆R

V

e

t

q

ot = 1

9

5

.4

.0

ΩV

= 1.6 A


V

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9. ∆Vtot = 24.0 V

R1 = 4.0 Ω

R2 = 4.0 Ω

R3 = 4.0 Ω

R4 = 3.0 Ω

R5 = 1.0 Ω

R6 = 2.0 Ω

R7 = 4.0 Ω

R8 = 2.0 Ω

R9 = 4.0 Ω

R10 = 2.0 Ω

R11 = 2.0 Ω

Series:

Group (a): Req,a = R6 + R7 + R8 = 2.0 Ω + 4.0 Ω + 2.0 Ω = 8.0 Ω


Parrallel:

Group (c): Re

1

q,c =

R

1

1 +

R

1

2 =

4.0

1

Ω +

4.0

1

Ω =

0

1

.2

Ω5

+ 0

1

.2

Ω5

Req,c = 2.0 Ω

Group (d): Re

1

q,d =

Re

1

q,a +

Re

1

q,b =

8.0

1

Ω +

6.0

1

Ω =

0

1

.1

Ω3

+ 0

1

.1

Ω7

Req,d = 3.4 Ω

Series:

Group (e): Req,e = Req,c + R3 + R4 + R5 + Req,d + R11

Req,e = 2.0 Ω + 4.0 Ω + 3.0 Ω + 1.0 Ω + 3.4 Ω + 2.0 Ω = 15 Ω

I = ∆R

V

eq

t

,

o

e

t = 2

1

4

5

.0

ΩV

= 1.6 A

10. ∆V = 24.0 V

R1 = 4.0 Ω

R2 = 8.0 Ω

R3 = 2.0 Ω

R4 = 5.0 Ω

R5 = 3.0 Ω

R6 = 2.0 Ω

R7 = 3.0 Ω

Parallel:

Group (a): Re

1

q,a =

R

1

1 +

R

1

2 =

4.0

1

Ω +

8.0

1

Ω =

0

1

.2

Ω5

+ 0

1

.1

Ω3

Req,a = 2.6 Ω

Group (b): Re

1

q,b =

R

1

6 +

R

1

7 =

2.0

1

Ω +

3.0

1

Ω =

0

1

.5

Ω0

+ 0

1

.3

Ω3

Req,b = 1.2 Ω

Series:

Req = 2Req,a + R3 + R4 + R5 + Req,b = (2)(2.7 Ω) + 2.0 Ω + 5.0 Ω + 1.2 Ω + 3.0 Ω

Req = 17 Ω

I = ∆R

V

e

t

q

ot = 2

1

4

7

.0

ΩV

= 1.4 A

Givens Solutions

1. ∆Vtot = 12.0 V

R1 = 3.0 Ω

R2 = 10.0 Ω

R3 = 10.0 Ω

R4 = 10.0 Ω

R5 = 4.0 Ω

Parallel:

Re

1

q,a =

R

1

2 +

R

1

3 +

R

1

4 =

10.0

1

Ω +

10.

1

0 Ω +

10.

1

0 Ω =

0

1

.1

Ω00 +

0

1

.1

Ω00 +

0

1

.1

Ω00

Req,a = 3.3 Ω

Series:

Req,b = R1 + Req,a + R5 = 3.0 Ω + 3.3 Ω + 4.0 Ω = 10 Ω

I = ∆R

V

eq

t

,

o

b

t = 1

1

2

0

.0

ΩV

=

∆V = IR5

∆V = (1.2 A)(4.0 Ω) = 4.8 V

1.2 A


3. ∆Vtot = 9.0 V

R1 = 5.0 Ω

R2 = 4.0 Ω

R3 = 7.0 Ω

R4 = 6.0 Ω

R5 = 3.0 Ω

R6 = 2.0 Ω

Parallel:

Re

1

q,a =

R

1

3 +

R

1

4 =

7.0

1

Ω +

6.0

1

Ω =

0

1

.1

Ω4

+ 0

1

.1

Ω7

Req,a = 3.2 Ω

Series:

Req = R1 + R2 + Req,a + R5 + R6 = 5.0 Ω + 4.0 Ω + 3.2 Ω + 3.0 Ω + 2.0 Ω = 17 Ω

I = ∆R

V

e

t

q

ot = 9

1

.

7

0

ΩV

= 0.52 A

∆V = I Req,a = (0.52 A)(3.2 Ω) =

I = ∆R

V

4 =

6

1

.

.

0

7

ΩV

= 0.28 A

1.7 V

Givens Solutions

4. ∆Vtot = 9.0 V

R1 = 7.0 Ω

R2 = 5.0 Ω

R3 = 4.0 Ω

R4 = 3.0 Ω

R5 = 2.0 Ω

R6 = 10.0 Ω

R7 = 6.0 Ω

Series:

Req,a = R2 + R3 + R4 = 5.0 Ω + 4.0 Ω + 3.0 Ω = 12.0 Ω

Req,b = R5 + R6 + R7 = 2.0 Ω + 10.0 Ω + 6.0 Ω = 18.0 Ω

Parallel:

Re

1

q,b =

Re

1

q,a +

Re

1

q,b =

12.

1

0 Ω +

18.

1

0 Ω =

0

1

.0

Ω83 +

0

1

.0

Ω56

Req,b = 7.2 Ω

Series:

Req,d = R1 + Req,c = 7.0 Ω + 7.2 Ω = 14 Ω

I = ∆R

V

e

t

q

ot = 9

1

.

4

0

ΩV

= 0.63 A

∆V = IReq,c = (0.63 A)(7.2 Ω) = 4.5 V

I = ∆R

V

6 =

1

4

8

.

.

5

0

V

Ω =

∆V = IR6 = (0.25 A)(10.0 Ω) = 2.5 V

0.25 A


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2. ∆Vtot = 1.5 V

R1 = 6.0 Ω

R2 = 2.0 Ω

R3 = 4.0 Ω

R4 = 8.0 Ω

R5 = 9.0 Ω

Parallel:

Re

1

q,a =

R

1

4 +

R

1

5 =

8.0

1

Ω +

9.0

1

Ω =

0

1

.1

Ω25 +

0.1

Ω11

Req,a = 4.2 Ω

Series:

Req = R1 + R2 + R3 + Req,a = 6.0 Ω + 2.0 Ω + 4.0 Ω + 4.2 Ω = 16 Ω

I = ∆R

V

e

t

q

ot = 1

1

.

6

5

ΩV

= 0.094 A

∆V = IReq,a = (0.094 A)(4.2 Ω) =

I = ∆R

V

5 =

0

9

.

.

3

0

9

ΩV

= 0.043 A

0.39 V


V

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5. ∆Vtot = 3.0 V

R1 = 2.0 Ω

R2 = 4.0 Ω

R3 = 6.0 Ω

R4 = 1.0 Ω

R5 = 3.0 Ω

R6 = 5.0 Ω

Series:

Req,a = R1 + R2 + R3 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω

Req,b = R4 + R5 = 1.0 Ω + 3.0 Ω = 4.0 Ω

Parallel:

Re

1

q,c =

Re

1

q,a +

Re

1

q,b +

R

1

eq =

12

1

Ω +

4.0

1

Ω +

5.0

1

Ω =

0

1

.0

Ω83 +

0

1

.2

Ω5

+ 0

1

.2

Ω0

Req,c = 1.9 Ω

I = ∆R

V

eq

t

,

o

c

t = 1

3

.

.

9

0

ΩV

= 1.6 A

I = R

∆

e

V

q,a =

3

1

.

2

0

ΩV

=

∆V = IR2 = (0.25 A)(4.0 Ω) = 1.0 V

0.25 A

Givens Solutions

6. ∆Vtot = 3.0 V

R1 = 2.0 Ω

R2 = 3.0 Ω

R3 = 2.0 Ω

R4 = 4.0 Ω

R5 = 2.0 Ω

R6 = 2.0 Ω

Series:

Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω

Req,b = R3 + R4 = 2.0 Ω + 4.0 Ω = 6.0 Ω

Parallel:

Re

1

q,c =

Re

1

q,a +

Re

1

q,b =

5.0

1

Ω +

6.0

1

Ω =

0

1

.2

Ω0

+ 0

1

.1

Ω7

Req,c = 2.7 Ω

Re

1

q,d =

R

1

5 +

R

1

6 =

2.0

1

Ω +

2.0

1

Ω

Req,d = 1.0 Ω

Series:

Req = Req,c + Req,d = 2.7 Ω + 1.0 Ω = 3.7 Ω

I = ∆R

V

e

t

q

ot = 3

3

.

.

7

0

ΩV

= 0.81 A

∆V = IReq,c = (0.81 A)(2.7 Ω) = 2.2 V

I = R

∆

e

V

q,a =

5

2

.

.

0

2

ΩV

=

∆V = IR2 = (0.44 A)(3.0 Ω) = 1.3 V

0.44 A

8. ∆Vtot = 12.0 V

R1 = 4.0 Ω

R2 = 5.0 Ω

R3 = 2.0 Ω

R4 = 3.0 Ω

R5 = 7.0 Ω

R6 = 3.0 Ω

Parallel:

Re

1

q,a =

R

1

3 +

R

1

4 +

R

1

5 =

2.0

1

Ω +

3.0

1

Ω +

7.0

1

Ω =

0

1

.5

Ω0

+ 0

1

.3

Ω3

+ 0

1

.1

Ω4

Req,a = 1.0 Ω

Re

1

q,b =

R

1

1 +

R

1

2 =

4.0

1

Ω +

5.0

1

Ω =

0

1

.2

Ω5

+ 0

1

.2

Ω0

Req,b = 2.2 Ω

Series:

Req,c = Req,a + Req,b + R4 = 1.0 Ω + 2.2 Ω + 3.0 Ω = 6.2 Ω

I = ∆R

V

eq

t

,

o

c

t = 1

6

2

.2

.0

ΩV

= 1.9 A

∆V = IReq,a = (1.9 A)(1.0 Ω) = 1.9 V

Givens Solutions


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7. ∆Vtot = 12.0 V

R1 = 5.0 Ω

R2 = 5.0 Ω

R3 = 5.0 Ω

R4 = 2.0 Ω

R5 = 5.0 Ω

R6 = 5.0 Ω

R7 = 5.0 Ω

Parallel:

Re

1

q,a =

R

1

1 +

R

1

2 =

5.0

1

Ω +

5.0

1

Ω =

0

1

.2

Ω0

+ 0

1

.2

Ω0

Req,a = 2.5 Ω

Series:

Req,b = R4 + R5 = 2.0 Ω + 5.0 Ω = 7.0 Ω

Req,c = R6 + R7 = 5.0 Ω + 5.0 Ω = 10 Ω

Parallel:

Re

1

q,d =

Re

1

q,b +

Re

1

q,c =

7.0

1

Ω +

10

1

Ω =

0

1

.1

Ω4

+ 0

1

.1

Ω0

Req,d = 4.2 Ω

Series:

Req = Req,a + R3 + Req,d = 2.5 Ω + 5.0 Ω + 4.2 Ω = 12 Ω

I = ∆R

V

e

t

q

ot = 1

1

2

2

.0

ΩV

= 1.0 A

∆V = IReq,d = (1.0 A)(4.2 Ω) =

I = R

∆

e

V

q,b =

7

4

.

.

0

2

ΩV

=

∆V = IR4 = (0.6 A)(2.0 Ω) = 1.2 V

0.6 A

4.2 V

I = ∆R

V

5 =

7

1

.

.

0

9

ΩV

= 0.27 A


V

Givens Solutions

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9. ∆Vtot = 1.5 V

R1 = 4.0 Ω

R2 = 5.0 Ω

R3 = 6.0 Ω

R4 = 4.0 Ω

R5 = 6.0 Ω

R6 = 12.0 Ω

R7 = 6.0 Ω

R8 = 3.0 Ω

R9 = 3.0 Ω

R10 = 3.0 Ω

Series:

Req,a = R3 + R4 = 6.0 Ω + 4.0 Ω = 10 Ω

Req,b = R5 + R6 + R7 = 6.0 Ω + 12.0 Ω + 6.0 Ω = 24 Ω

Req,c = R8 + R9 = 3.0 Ω + 3.0 Ω = 6.0 Ω

Parallel:

Re

1

q,d =

Re

1

q,a +

Re

1

q,b +

Re

1

q,c =

0

1

.1

Ω0

+ 0

1

.0

Ω42 +

0

1

.1

Ω7

Req,d = 3.2 Ω

Series:

Req = R1 + R2 + Req,d + R10 = 4.0 Ω + 5.0 Ω + 3.2 Ω + 3.0 Ω = 15 Ω

I = ∆R

V

e

t

q

ot = 1

1

.

5

5

ΩV

= 0.10 A

∆V = IReq,d = (0.10 A)(3.2 Ω) = 0.32 V

I = R

∆

e

V

q,b =

0

2

.3

4

2

ΩV

=

∆V = IR6 = (0.013 A)(12 Ω) = 0.16 V

0.013 A

10. ∆Vtot = 12.0 V

R1 = 5.0 Ω

R2 = 6.0 Ω

R3 = 15.0 Ω

R4 = 30.0 Ω

Parallel:

Re

1

q,a =

R

1

3 +

R

1

4 =

15.

1

0 Ω +

30.

1

0 Ω =

0

1

.0

Ω67 +

0

1

.0

Ω33

Req,a = 10 Ω

Series:

Req = R1 + R2 + Req,a = 5.0 Ω + 6.0 Ω + 10.0 Ω = 21 Ω

I = ∆R

V

e

t

q

ot = 1

2

2

1

.0

ΩV

= 0.57 A

∆V = IReq,a = (0.57 A)(10 Ω) = 5.7 V

I = ∆R

V

3 =

1

5

5

.

.

7

0

V

Ω = 0.38 A


21ChapterMagnetism

V

1. q = 1.60 × 10−19 C

v = 1.2 × 106 m/s

Fmagnetic = 1.2 × 10−17 N

B = Fma

q

g

v

netic = = 60.3 × 10−5 T1.2 × 10−17 N

(1.60 × 10−19 C)(1.2 × 106 m/s)


Givens Solutions

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2. q = 1.60 × 10−19 C

v = 3.9 × 106 m/s


B = Fma

q

g

v

netic = = 3.0 × 10−10 T1.9 × 10−22 N

(1.60 × 10−19 C)(3.9 × 106 m/s)

3. q = 1.60 × 10−19 C

v = 7.8 × 106 m/s


B = Fma

q

g

v

netic = = 03.0 T3.7 × 10−13 N

(1.60 × 10−19 C)(7.8 × 106 m/s)

4. q = 1.60 × 10−19 C

r = 1.0 km = 1.0 × 103 m

m = 1.67 × 10−27 kg

B = 3.3 T

m

r

v2

= qv B

v = q

m

r B = = 3.2 × 1011 m/s

(1.60 × 10−19 C)(1.0 × 103 m)(3.3 T)

1.67 × 10−27 kg

5. q = 1.60 × 10−19 C

B = 5.0 × 10−5 T


v = Fm

q

ag

B

netic = = 7.6 × 106 m/s6.1 × 10−17 N

(1.60 × 10−19 C)(5.0 × 10−5 T)

6. B = 1 × 10−8 T

q = 1.60 × 10−19 C


v = Fm

q

ag

B

netic = = 2 × 105 m/s3.2 × 10−22 N

(1.60 × 10−19 C)(1 × 10−8 T)

7. q = 1.60 × 10−19 C

v = 6 × 106 m/s to the right

q = 45°

B = 3 × 10−4 T upward

Fmagnetic = qvBsin q = (1.60 × 10−19 C)(6 × 106 m/s)(3 × 10−4 T)sin 45°

Fmagnetic = 2 × 10−16 N

8. q = 1.60 × 10−19 C

B = 0.8 T

v = 3.0 × 107 m/s

Fmagnetic = qvB = (1.60 × 10−19 C)(3.0 × 107 m/s)(0.8 T) = 4 × 10−12 N

9. q = 1.60 × 10−19 C

v = 2.2 × 106 m/s

B = 1.1 × 10−2 T

Fmagnetic = qvB = (1.60 × 10−19 C)(2.2 × 106 m/s)(1.1 × 10−2 T) = 3.9 × 10−15 N


V

Fmagnetic = qvB = (1.60 × 10−19 C)(9.3 × 105 m/s)(4.1 × 10−10 T) = 6.1 × 10−23 N

Givens Solutions

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10. q = 1.60 × 10−19 C

v = 9.3 × 105 m/s

B = 4.1 × 10−10 T

1. B = 4.6 × 10−4 T


T = 10.0 A

l = Fma

Bg

Inetic = = 0.63 m

2.9 × 10−3 N(4.6 × 10−4 T)(10.0 A)

2. I = 1 A

B = 2.8 × 10−5 T


l = Fma

Bg

Inetic =

(2.8

5

×.6

1

×0−

150−

T

5

)(

N

1 A) = 2 m

3. l = 12 m

I = 12 A


B = Fma

I

g

l

netic = (

7

1

.

2

3

A

×)

1

(

0

1

−

2

2

m

N

) = 5.1 × 10−4 T

4. Fmagnetic = 7.8 × 105 N

l = 12 m = 1.2 × 104 m

I = 1.8 × 104 A

B = Fma

I

g

l

netic = = 3.6 × 10−3 T7.8 × 105 N

(1.8 × 104 A)(1.2 × 104 m)

5. I = 14.32 A

l = 15.0 cm = 0.150 m


B = Fma

I

g

l

netic = (14

6

.3

.6

2

2

A

×)(

1

0

0

.

−

1

4

50

N

m) = 3.08 × 10−4 T


6. l = 10 m

m = 75 kg

B = 4.8 × 10−4 T

g = 9.81 m/s2

mg = BIl

I = mBl

g = = 1.5 × 105 A(75 kg)(9.81 m/s2)(4.8 × 10−4 T)(10 m)

7. l = 1.0 mFmagnetic = 9.1 × 10−5 N

B = 1.3 × 10−4 T

I = Fma

B

g

l

netic = = 0.7 A9.1 × 10−5 N

(1.3 × 10−4 T)(1.0 m)

8. I = 1.5 × 103 A

l = 15 km = 1.4 × 104 m

q = 45°

B = 2.3 × 10−5 T

Fmagnetic = Bcos qI l = (2.3 × 10−5 T)cos 45°(1.5 × 103 A)(1.5 × 104 m)

Fmagnetic = 3.7 × 102 N

9. I = 14 A

l = 2 m

B = 3.6 × 10−4 T

Fmagnetic = BI l = (3.6 × 10−4 T)(14 A)(2 m) = 1 × 10−2 N

10. I = 0.5 A

l = 5 cm = 5 × 10−2 m

B = 1.3 × 10−4 T

Fmagnetic = BI l = (1.3 × 10−4 T)(0.5 A)(5 × 10−2 m) = 3 × 10−6 N


22ChapterInduction and Alternating Current

V

1. N = 540 turns

A = 0.016 m2

qi = 0°

qf = 90.0°

∆t = 0.05 s

emf = 3.0 V

B = −N

em

A∆f

c

∆o

t

sq =

B =

B = 1.7 × 10−2 T

(3.0 V)(0.05 s)−(540)(0.016 m2)[cos 90.0° − cos 0°]

emf ∆t−NA[cosqf − cosqi]


Givens Solutions

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2. N = 320 turns

A = 0.068 m2

qi = 0°

qf = 90°

∆t = 0.25 s

emf = 4.0 V

B = −N

em

A∆f

c

∆o

t

sq =

B =

B = 4.6 × 10−2 T

(4.0 V)(0.25 s)−(320)(0.068 m2)(cos 90.0° − cos 0°)

emf ∆t−NA(cosqf − cosqi)

3. N = 628 turns

A = 0.93 m2

qi = 0°

∆t = 0.30 s

qf = 30.0°

emf = 62 V

B = =

B =

B = 0.24 T

(62 V)(0.30 s)−(628)(0.93 m2)(cos 30.0° − cos 0°)

emf ∆t−NA(cosqf − cosqi)

emf ∆t−NA∆cosq

4. N = 550 turns

A = 5.0 × 10−5 m2

∆B = 2.5 × 10−4 T

∆t = 2.1 × 10−5 s

q = 0°

emf = −NA∆

∆B

t

cos q

emf =

emf = 0.33 V

−(550)(5.0 × 10−5 m2)(2.5 × 10−4 T)(cos 0°)

2.1 × 10−5 s

5. N = 220 turns

A = 6.0 × 10−6 m2

∆B = 9.7 × 10−4 T

∆t = 1.7 × 10−6 s

q = 0°

emf = −NA∆

∆B

t

cos q

emf =

emf = 0.75 V

−(220)(6.0 × 10−6 m2)(9.7 × 10−4 T)(cos 0°)

1.7 × 10−6 s


V

emf = −NA∆

∆B

t

cos q

emf =

emf = 0.11 V

−(148)(1.25 × 10−8 m2)(5.2 × 10−4 T)(cos 0°)

8.5 × 10−9 s

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6. N = 148 turns

A = 1.25 × 10−8 m2

q = 0°

∆B = 5.2 × 10−4 T

∆t = 8.5 × 10−9 s

7. emf = 220 V

R = 120 ΩI =

em

R

f =

1

2

2

2

0

0

ΩV

= 1.8 A

8. N = 180 turns

A = 5.0 × 10−5 m2

∆B = 5.2 × 10−4 T

q = 0°

∆t = 1.9 × 10−5 s

R = 1.0 × 102Ω

emf = −NA∆

∆B

t

cos q

emf =

emf =

I = em

R

f =

1.0

0.

×25

10

V2Ω

= 2.5 × 10−3 A = 25 mA

0.25 V

−(180)(5.0 × 10−5 m2)(5.2 × 10−4 T)(cos 0°)

1.9 × 10−5 s

9. N = 246 turns

A = 0.40 m2

q = 0°

Bi = 0.237 T

Bf = 0.320 T

∆t = 0.9 s

emf = 9.1 V

∆t = −NA∆

em

B

f

cos q =

−NA[Bf

e

−m

B

fi] cos q

∆t =

∆t = 0.90 s

−(246)(0.40 m2)[0.320 T − 0.237 T](cos 0°)

9.1 V

1. N = 220 turns

A = 0.080 m2

B = 4.8 × 10−3 T

maximum emf = 12 V

w = maxi

N

m

A

um

B

emf =

w = 1.4 × 102 rad/s

12 V(220)(0.080 m2)(4.8 × 10−3 T)

2. N = 140 turns

D = 0.33 m

B = 9.3 × 10−2 T

maximum emf = 150 V

w = maxi

N

m

A

um

B

emf

A = πD

22

= π0.3

2

3 m2

= 8.6 × 10−2 m2


10. N = 785 turns

A = 7.3 × 10−2 m2

∆B = 6.9 × 10−3 T

emf = 2.8 V

q = 0°

∆t = −NA∆

em

B

f

cos q =

∆t =

∆tearthquake = (0.14 s/oscillation)(120 oscillations)

∆t =

The earthquake lasted for 16.8 s.

16.8 s

0.14 s/oscillation

−(785)(7.3 × 10−2 m2)(6.9 × 10−3 T)(cos 0°)

2.8 V


w = = 130 rad/s150 V

(140)(8.6 × 10−2 m2)(9.3 × 10−2 T)

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VV

3. N = 195 turns

A = 0.052 m2

B = 3.2 × 10−3 T

Imax = 1.2 A

R = 16 Ω

maximum emf = Imax R = (1.2 A)(16 Ω) =

w = = 590 rad/s19 V

(195)(0.052 m2)(3.2 × 10−3 T)

19 V

4. N = 385 turns

A = 0.38 m2

B = 9.4 × 10−3 T

f = 45 Hz

w = 2 π fmaximum emf = NABw = NAB2πf

maximum emf = (385)(0.38 m2)(9.4 × 10−3 T)(2π)(45 Hz)

maximum emf = 390 V

5. Imax = 14 A

R = 5 Ωmaximum emf = Imax R = (14 A)(5 Ω) = 70 V

6. N = 119 turns

A = 4.9 × 10−2m2

B = 9.4 × 10−3 T

w = 345 rad/s

maximum emf = NABwmaximum emf = (119)(4.9 × 10−2 m2)(9.4 × 10−3 T)(345 rad/s)

maximum emf = 19 V

7. maximum emf = 40 VR = 8 Ω Imax =

maxim

R

um emf =

4

8

0

ΩV

= 5 A

8. N = 425 turns

A = 2.16 × 10−2m2

B = 3.9 × 10−2 T

f = 33 Hz

R = 25 Ω

w = 2 π fmaximum emf = NABw = NAB2πf

maximum emf = (425)(2.16 × 10−2 m2)(3.9 × 10−2 T)(2π)(33 Hz)

maximum emf =

Imax = maxim

R

um emf =

2

7

5

4

ΩV

= 3.0 × 101 A

74 V

9. A = 1.20 × 10−2m2

B = 6.0 × 10−2 T

w = 393 rad/s

maximum emf = 213 V

N = maxim

AB

u

wm emf =

N = 750 turns

213 V(1.2 × 10−2 m2)(6.0 × 10−2 T)(393 rad/s)

10. A = 0.60 m2

B = 0.012 T

f = 44 Hz

maximum emf = 320 V

N = maxim

AB

u

wm emf =

max

A

im

B

u

2

m

πf

emf

N =

N = 160 turns

320 V(0.60 m2)(0.012 T)(2π)(44 Hz)

Section One—Pupil’s Edition Solutions V Ch. 22–4

1. ∆Vrms = 320 V

R = 100 Ω∆Vmax =

∆0.

V

7r

0m

7s =

3

0

2

.7

0

0

V

7 =

Irms = ∆V

Rrms =

1

3

0

2

0

0

ΩV

=

Imax = 0

I

.r

7m

0s

7 =

0

3

.7

A

07 = 4 A

3 A

450 V

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Givens Solutions

2. Irms = 1.3 AImax =

0

I

.r

7m

0s

7 =

0

1

.

.

7

3

0

A

7 = 1.8 A

3. Irms = 2.5 A

∆Vrms = 2.2 × 104 VImax =

0

I

.r

7m

0s

7 =

0

2

.

.

7

5

0

A

7 =

R = ∆

I

V

rm

rm

s

s = 2.2

2

×.5

1

A

04 V = 8.8 × 104Ω = 88 kΩ

3.5 A

4. ∆Vrms = 220 V

Irms = 1.7 A∆Vmax =

∆0.

V

7r

0m

7s =

2

0

2

.7

0

0

V

7 =

Imax = 0

I

.r

7m

0s

7 =

0

1

.

.

7

7

0

A

7 = 2.4 A

311 V

5. Imax = 1.2 A

∆Vmax = 211 VIrms = 0.707 Imax = 0.707(1.2 A) =

∆Vrms = 0.707 Vmax = 0.707(211 V) =

R = ∆

I

V

rm

rm

s

s = 0

1

.

4

8

9

5

V

A = 175 Ω

149 V

0.85 A

6. Vmax = 170 V ∆Vrms = 0.707 Vmax = 0.707(170 V) = 120 V

7. ∆Vrms = 115 V

R = 50.0 ΩIrms =

∆V

Rrms =

5

1

0

1

.

5

0

V

Ω =

Imax = 0

I

.r

7m

0s

7 =

2

0

.

.

3

7

0

07

A = 3.25 A

2.30 A

8. Irms = 2.1 A

R = 16 k Ω = 1.6 × 104 ΩImax =

0

I

.r

7m

0s

7 =

0

2

.

.

7

1

0

A

7 =

P = (Irms)2R = (2.1 A)2(1.6 × 104 W) = 7 × 104 Ω P = 70 kW

3.0 A

9. ∆Vrms = 1.56 × 104 V

Irms = 1.3 A

R = 1.2 × 104 Ω

∆Vmax = ∆0

V

.7

r

0

m

7

s =

1.56

0.

×70

1

7

04 V =

Imax = 0

I

.r

7m

0s

7 =

0

1

.

.

7

3

0

A

7 =

P = (Irms)2R = (1.3 A)2(1.2 × 104 Ω) = 2.0 × 104 W

1.8 A

2.2 × 104 V = 22kV


10. Irms = 2.2 × 1010 A

R = 6.1 × 10−10 ΩImax =

0

I

.r

7m

0s

7 =

2.2

0

×.7

1

0

0

7

10 A =

P = (Irms)2 R = (2.2 × 1010 A)2(6.1 × 10−10 Ω) = 2.9 × 1011 W

3.1 × 1010 A

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V

1. ∆V2 = 6.9 × 103 V

N1 = 1400 turns

N2 = 140 turns

∆V1 = ∆V2 N

N1

2 = (6.9 × 103 V)

1

1

4

4

0

0

0 = 6.9 × 104 V


2. ∆V2 = 3.4 × 103 V

N1 = 90 turns

N2 = 2250 turns

∆V1 = ∆V2 N

N1

2 = (3.4 × 103 V)

2

9

2

0

50 = 1.4 × 102 V

3. ∆V1 = 4.6 × 104 V

N1 = 1250 turns

N2 = 250 turns

∆V2 = ∆V1 N

N2

1 = (4.6 × 104 V)

1

2

2

5

5

0

0 = 9.2 × 103 V

4. ∆V1 = 5600 V

N1 = 140 turns

N2 = 840 turns

∆V2 = ∆V1 N

N2

1 = (5600 V)

8

1

4

4

0

0 = 3.36 × 104 V

5. ∆V1 = 9200 V

N1 = 120 turns

N2 = 1200 turns

∆V2 = ∆V1 N

N2

1 = (9200 V)

1

1

2

2

0

0

0 =

6. ∆V1 = 3.6 × 104 V

9.20 × 104 V

∆V2 = 7.2 × 103 V

N1 = 55 turns

N2 = N1 ∆∆

V

V2

1 = (55)73.

.

2

6

××

1

1

0

0

3

4V

V = 11 turns

7. ∆V1 = 240 V

∆V2 = 5.0 VN

N1

2 =

∆∆

V

V1

2 =

2

5

4

.0

0

V

V = 48:1

8. ∆V1 = 1800 V

∆V2 = 3600 V

N1 = 58 turns

N2 = N1 ∆∆

V

V2

1 = (58)316

8

0

0

0

0

V

V = 1.2 × 102 turns

9. ∆V1 = 4900 V

∆V2 = 4.9 × 104 V

N2 = 480 turns

N1 = N2 ∆∆

V

V1

2 = (480)4.9

49

×0

1

0

0

V4 V

= 48 turns

10. P = 1.38 × 106 W

∆V2 = 3.4 × 103 V

N1 = 340 turns

N2 = 17 turns

∆V1 = ∆V2 N

N1

2 = (3.4 × 103 V)

3

1

4

7

0 =

P = ∆V1I1

I1 = ∆

P

V1 =

1

6

.3

.8

8

××

1

1

0

04

6

V

W = 2.0 × 101 A

6.8 × 104 V


23ChapterAtomic Physics

V

1. λ = 527 nm = 5.27 × 10−7 mE = hf =

h

λc = = 3.77 × 10−19 J

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

5.27 × 10−7 m


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2. λ = 430.8 nm

= 4.308 × 10−7 m E = hf = h

λc = = 4.62 × 10−22 J

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

4.308 × 10−7 m

3. E = 20.7 eVf =

E

h = = 5.00 × 1015 Hz

(20.7 eV)(1.60 × 10−19 J/eV)

6.63 × 10−34 J•s

4. E = 1.24 × 10−3 eVf =

E

h = = 2.99 × 1011 Hz

(1.24 × 10−3 eV)(1.60 × 10−19 J/eV)

6.63 × 10−34 J•s

5. E = 1.78 eVf =

E

h = = 4.30 × 1014 Hz

(1.78 eV)(1.60 × 10−19 J/eV)

6.63 × 10−34 J•s

6. E = 12.4 MeV

= 1.24 × 107 eV λ = h

E

c = = 1.00 × 10−13 m

(6.63 × 10−34 J•s)(3.00 × 108 m/s)(1.24 × 107 eV)(1.60 × 10−19 J/eV)

7. E = 939.57 MeV

= 9.3957 × 108 eV λ = h

E

c = =

1.32 × 10−15 m = 1.32 × 10−6 nm

If a photon were to have this wavelength, it would not lie within the visible part ofthe spectrum.

1.32 × 10−15 m(6.63 × 10−34 J•s)(3.00 × 108 m/s)(9.3957 × 108 eV)(1.60 × 10 −19 J/eV)

8. E = 3.1 × 10−6 eV λ = h

E

c = = 0.401 m

(6.63 × 10−34 J•s)(3.00 × 108 m/s)(3.1 × 10−6 eV)(1.60 × 10−19 J/eV)


1. λ = 240 nm = 2.4 × 10−7 m

hft = 2.3 eV

KEmax = h

λc − hft

KEmax = − 2.3 eV

KEmax = 5.2 eV − 2.3 eV = 2.9 eV

(6.63 × 10−34 J•s)(3.00 × 108 m/s)2.4 × 10−7 m)(1.60 × 10−19 J/eV)


V

KEmax = h

λc − hft

KEmax = − 2.16 eV

KEmax = 2.40 eV − 2.16 eV = 0.24 eV

(6.63 × 10−34 J•s)(3.00 × 108 m/s)(5.19 × 10−7 m)(1.60 × 10−19 J/eV)

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2. λ = 519 nm = 5.19 × 10−7 m

hft = 2.16 eV

3. f = 6.5 × 1014 Hz

KEmax = 0.20 eVft =

hf − K

h

Emax

ft =

ft = 6.0 × 1014 Hz

[(6.63 × 10−34 J•s)(6.5 × 1014 Hz) − (0.20 eV)(1.60 × 10−19 J/eV)]

6.63 × 10−34 J•s

4. f = 9.89 × 1014 Hz

KEmax = 0.90 eVft =

hf − K

h

Emax

ft =

ft = 7.72 × 1014 Hz

(6.63 × 10−34 J•s)(9.89 × 1014 Hz) − (0.90 eV)(1.60 × 10−19 J/eV)

6.63 × 10−34 J•s

5. ft = 1.36 × 1015 Hzhft = = 5.64 eV

(6.63 × 10−34 J•s)(1.36 × 1015 Hz)

1.60 × 10−19 J/eV

6. ft = 1.1 × 1015 Hzhft = = 4.6 eV

(6.63 × 10−34 J•s)(1.1 × 1015 Hz)

1.60 × 10−19 J/eV

7. hft = 4.1 eVλ =

h

E

c = = 3.0 × 10−7 m = 300 nm

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

(4.1 eV)(1.60 × 10−19 eV)

8. hft = 5.0 eVλ =

h

E

c = = 2.5 × 10−7 m = 250 nm

(6.63 × 10−34 J•s)(3.00 × 108 m/s)

(5.0 eV)(1.60 × 10−19 eV)

9. KEmax = 0.62 V

me = 9.109 × 10−31 kgKEmax = hf − hft = 1

2mev

2

v = 2Km

Em

eax =

v = 4.7 × 105 m/s

2(0.62 eV)(1.60 × 10−19 J/eV)

9.109 × 10−31 kg

10. KEmax = 1.2 eV

me = 9.109 × 10−31 kgKEmax = hf − hft = 1

2 mev

2

v = 2Km

Em

eax =

v = 6.5 × 105 m/s

2(1.2 eV)(1.60 × 10−19 eV)

9.109 × 10−31 kg


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VV

1. v = 28 m/s

λ = 8.97 × 10−37 mm =

λh

v = = 26 kg

6.63 × 10−34 J•s(8.97 × 10−37 m)(28 m/s)

2. v = 7.1 × 102 m/s

λ = 5.8 × 10−42 mm =

λh

v = = 1.6 × 105 kg

6.63 × 10−34 J•s(5.8 × 10−42 m)(7.1 × 102 m/s)

3. v = 5.6 × 10−6 m/s

λ = 2.96 × 10−8 mm =

λh

v = = 4.0 × 10−21 kg

6.63 × 10−34 J•s(2.96 × 10−8 m)(5.6 × 10−6 m/s)

4. v = 12 m/s

λ = 2.6 × 10−29 mm =

λh

v = = 2.1 × 10−6 kg

6.63 × 10−34 J•s(2.6 × 10−29 m)(12 m/s)

5. me = 9.109 × 10−31 kg

v = 2.19 × 106 m/sλ =

m

h

v = = 3.3 × 10−10 m

6.63 × 10−34 J•s(9.109 × 10−31 kg)(2.19 × 106 m/s)

6. m = 7.6 × 107 kg

v = 35 m/sλ =

m

h

v = = 2.5 × 10−43 m

6.63 × 10−34 J•s(7.6 × 107 kg)(35 m/s)

7. m = 5.94 × 1024 kg

v = 3.0 × 104 m/sλ =

m

h

v = = 3.7 × 10−63 m

6.63 × 10−34 J•s(5.94 × 1024 kg)(3.0 × 104 m/s)

8. m = 4.0 × 1041 kg

v = 1.7 × 104 m/sλ =

m

h

v = = 9.7 × 10−80 m

6.63 × 10−34 J•s(4.0 × 1041 kg)(1.7 × 104 m/s)


Givens Solutions

9. me = 9.109 × 10−31 kg

λ = 9.87 × 10−14 mv =

m

h

λ = = 7.37 × 109 m/s

6.63 × 10−34 J•s(9.109 × 10−31 kg)(9.87 × 10−14 m)

10. mn = 1.675 × 10−27 kg

λ = 5.6 × 10−14 mv =

m

h

λ = = 7.1 × 106 m/s

6.63 × 10−34 J•s(1.675 × 10−27 kg)(5.6 × 10−14 m)


25ChapterSubatomic Physics

V

1. Z = 19

A = 39

atomic mass of K-39 = 38.963 708 u

atomic mass of H = 1.007 825 u

mn = 1.008 665 u

N = A − Z = 39 − 19 = 20

∆m = Z (atomic mass of H) + Nmn − atomic mass of K-39

∆m = 19(1.007 825) + 20 (1.008 665 u)–38.963 708 u

∆m = 19.156 75 u + 20.1733 u – 38.963 708 u

∆m = 0.3653 u

Ebind = (0.3653 u)(931.50 MeV/u) = 340.3 MeV

Problem Bank Solutions Chapter 25A

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d.

2. Z = 50

A = 120

atomic mass of Sn-120 = 119.902 197


mn = 1.008 665 u

N = A − Z = 120 − 50 = 70

∆m = Z(atomic mass of H) + Nmn − atomic mass of Sn-120

∆m = 50(1.007 825 u) + 70(1.008 665 u) − 119.902 197 u

∆m = 50.391 25 u + 70.606 55 u − 119.902 197 u

∆m = 1.095 60 u

Ebind = (1.095 60 u)(931.50 MeV/u) = 1020.55 MeV

3. For 10747 Ag:

Z = 47

A = 107

atomic mass of Ag-107 = 106.905 091 u


mn = 1.008 665 u

For 6329 Cu:

Z = 29

A = 63

atomic mass of Cu-63 = 62.929 599 u

N = A − Z = 107 − 47 = 60

∆m = Z(atomic mass of H) + Nmn − atomic mass of Ag-107

∆m = 47(1.007 825 u) + 60(1.008 665 u) − 106.905 091 u

∆m = 47.367 775 u + 60.5199 u − 106.905 091 u

∆m = 0.9826 u

Ebind = (0.9826 u)(931.50 MeV/u) = 915.29 MeV

N = A − Z = 63 − 29 = 34

∆m = Z(atomic mass of H) + Nmn − atomic mass of Cu-63

∆m = 29(1.007 825 u) + 34(1.008 665 u) − 62.929 599 u

∆m = 29.226 925 u + 34.2946 u − 62.929 599 u

∆m = 0.5919 u

Ebind = (0.5919 u)(931.50 MeV/u) = 551.4 MeV

The difference in binding energy is 915.29 MeV − 551.4 MeV = 363.9 MeV


V

N = A − Z = 12 − 6 = 6

∆m = Z (atomic mass of H) + Nmn − atomic mass of C-12

∆m = 6(1.007 825 u) + 6(1.008 665u) − 12.000 000 u

∆m = 6.046 95 u + 6.051 99 u − 12.000 000 u

∆m = 0.098 94 u

Ebind = (0.098 94 u)(931.50 MeV/u) = 92.163 MeV

N = A − Z = 16 − 8 = 8

∆m = Z (atomic mass of H) + Nmn − atomic mass of O-16

∆m = 8(1.007 825 u) + 8(1.008 665u) − 15.994 915 u

∆m = 8.0626 u + 8.06932 u − 15.994 915 u

∆m = 0.1370 u

Ebind = (0.1370 u)(931.50 MeV/u) = 127.62 MeV

The difference in binding energy is

127.62 MeV − 92.163 MeV = 35.46 MeV

Givens Solutions

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d.4. For 12

6 C:

Z = 6

A = 12

atomic mass of C-12 = 12.000 000 u

atomic mass of H= 1.007 825 u

mn = 1.008 665 u

For 168 O:

Z = 8

A = 16

atomic mass of O-16 = 15.994 915

5. Z = 17

A = 35

atomic mass of Cl-35 = 34.968 853 u


mn = 1.008 665 u

N = A − Z = 35 − 17 = 18

∆m = Z (atomic mass of H) + Nmn − atomic mass of CI-35

∆m = 17(1.007 825 u) + 18(1.008 665 u) − 34.968 853 u

∆m = 17.133 025 u + 18.155 97 u − 34.968 853 u

∆m = 0.320 14 u

Ebind = (0.320 14 u)(931.50 MeV/u) = 298.21 MeV

7. A = 58

Z = 28

atomic mass of Ni-58 = 57.935 345 u


mn = 1.008 665 u

N = A − Z = 58 − 28 = 30

∆m = Z(atomic mass of H) + Nmn − atomic mass of Ni-58

∆m = 28(1.007 825 u) + 30(1.008 665 u) − 57.935 345 u

∆m = 28.2191 u + 30.259 95 u − 57.935 345 u

∆m = 0.5437 u

6. Z = 1

A = 2

atomic mass of H-2 = 2.014 102 u


mn = 1.008 665 u

N = A − Z = 2 − 1 = 1

∆m = Z(atomic mass of H) + Nmn − atomic mass of H-2

∆m = 1(1.007 825 u) + 1(1.008 665 u) − 2.014 102 u

∆m = 2.388 × 10−3 u

Ebind = (2.388 × 10−3 u)(931.50 MeV/u) = 2.2244 MeV


9. A = 90

Z = 40

atomic mass of Zr-90 = 89.904 702 u


mn = 1.008 665 u

N = A − Z = 90 − 40 = 50

∆m = Z(atomic mass of H) + Nmn − atomic mass of Zr-90

∆m = 40(1.007 825 u) + 50(1.008 665 u) − 89.904 702 u

∆m = 40.313 u + 50.433 25 u − 89.940 702 u

∆m = 0.842 u

Ebind = (0.842 u)(931.50 MeV/u) = 784 MeV

Givens Solutions

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d.

VV

10. A = 32

Z = 16

atomic mass of S-32 = 31.972 071 u


mn = 1.008 665 u

N = A − Z = 32 − 16 = 16

∆m = Z(atomic mass of H) + Nmn − atomic mass of S-32

∆m = 16(1.007 825 u) = 16(1.008 665 u) − 31.972 071 u

∆m = 16.1252 u + 16.1386 u − 31.972 071 u

∆m = 0.2918 u

Problem Bank Solutions Chapter 25B

Givens Solutions

1. 21084 Po → ? + 4

2He A = 210 − 4 = 206

Z = 84 − 2 = 82

? = 20682 Pb

2. 167 N → ? + 0

−1e + v A = 16 − 0 = 16

Z = 7 − (−1) = 8

? = 168 O

3. 14762 Sm → 143

60 Nd + ? + v A = 147 − 143 = 4

Z = 62 − 60 = 2

? = 42 He

4. 1910 Ne → ? + 0

1e + v A = 19 − 0 = 19

Z = 10 − 1 = 9

? = 199 F

8. A = 64

Z = 30

atomic mass of Zn-64 = 63.929 144 u

atomic mass of H= 1.007 825 u

mn = 1.008 665 u

N = A − Z = 64 − 30 = 34

∆m = Z(atomic mass of H) + Nmn − atomic mass of Zn-64

∆m = 30(1.007 825 u) + 34(1.008 665 u) − 63.929 144 u

∆m = 30.234 75 u + 34.2946 u − 63.929 144 u

∆m = 0.6002 u


7. 16074 W → 156

72 Hf + ? A = 160 − 156 = 4

Z = 74 − 72 = 2

? = 42 He

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d.

V


Givens Solutions

8. ? → 10752 Te + 4

2He A = 107 + 4 = 111

Z = 52 + 2 = 54

? = 11154 Xe

9. 15772 Hf → 153

70 Yb + ? A = 157 − 153 = 4

Z = 72 − 70 = 2

? = 42 He

10. 14158 Ce → ? + 0

−1e + v A = 141 − 0 = 141

Z = 58 − (−1) = 59

? = 14159 Pr

Problem Bank Solutions Chapter 25C

Givens Solutions

1. mi = 5.25 × 10−3 g

mf = 3.28 × 10−4 g

∆t = 12 h

m

m

i

f = 3

5

.

.

2

2

8

5

××

1

1

0

0

−

−

4

3g

g =

1

1

6

If 1

1

6 of the sample remains after 12 h, then 1

8 of the sample must have remained after

6.0 h, 14

of the sample must have remained after 3.0 h, and 12

of the sample must have

remained after 1.5 h. So T1/2 = 1.5 h

2. mi = 3.29 × 10−3 g

mf = 8.22 × 10−4 g

∆t = 30.0 s

m

m

i

f = 8

3

.

.

2

2

2

9

××

1

1

0

0

−

−

4

3g

g =

1

4

If 14

of the sample remains after 30.0 s, then 12

of the sample must have remained after

15.0 s, so T1/2 = .15.0 s

3. mi = 4.14 × 10−4 g

mf = 2.07 × 10−4 g

∆t = 1.25 days

m

m

i

f = 2

4

.

.

0

1

7

4

××

1

1

0

0

−

−

4

4g

g =

1

2

If 12

of the sample remains after 1.25 days, then T1/2 = 1.25 days

5. ? → 13154 Xe + 0

−1e + v A = 131 + 0 = 131

Z = 54 + (−1) = 53

? = 13153 I

6. ? → 9039 Y + 0

−1e + v A = 90 + 0 = 90

Z = 39 + (−1) = 38

? = 9038 Sr


6. T1/2 = 2.7 y

N = 3.2 × 109λ =

0

T

.6

1

9

/2

3 = = 8.1 × 10−9 s−10.693

(2.7 y)(3.156 × 107 s/y)

Givens Solutions

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d.

V

7. activity = 765.3 mCi

T1/2 = 22 hλ =

0

T

.6

1

9

/2

3 =

(22.0 h)

0

(

.

3

6

6

9

0

3

0 s/h) = 8.75 × 10−6 s−1

N = = = 3.2 × 1015 nuclei(0.7653 Ci)(3.7 × 1010 s−1/Ci)

8.75 × 10−6 s−1activity

λ

8. T1/2 = 21.6 h

N = 6.5 × 106

λ = 0

T

.6

1

9

/2

3 =

(21.6 h

0

)

.

(

6

3

9

6

3

00 s/h) =

activity = N λ = = 1.5 × 10−9 Ci(8.9 × 10−6 s−1)(6.5 × 106)

3.7 × 1010 s−1/Ci

8.90 × 10−6 s−1

9. T1/2 = 12.33 y

N = 4.8 × 108λ =

0

T

.6

1

9

/2

3 = = 1.78 × 10–9 s–10.693

(12.33 y)(3.56 × 107 s/y)

10. activity = 0.3600 Ci

T1/2 = 17.2 s

λ = 0

T

.6

1

9

/2

3 =

1

0

7

.6

.2

93

s =

N = acti

λvity = = 3.3 × 104 nuclei

(0.3600 Ci)(3.7 × 1010 s−1/Ci)

(4.03 × 10−2 s−1)

4.03 × 10−2 s−1

4. T1/2 = 10.64 h For the sample to reach 12

its original strength, it takes 10.64 h. For the sample to

reach 14

its original strength, it takes 2(10.64 h) = 21.28 h. For the sample to reach 8

1 its

original strength, it takes 3(10.64 h) = 31.92 h

5. T1/2 = 462 days For the sample to reach 12

its original strength, it takes 462 days. For the sample to

reach 14

its original strength, it takes 2(462 days) = 924 days

blue book all questions[1]

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