chapter 9 solutions to serway's college physics

7/25/2019 Chapter 9 Solutions to Serway's College Physics

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Chapter 9

9

Solids and Fluids

CLICKER QUESTIONS

Question H1.01a

Description:Exploring and interrelating weight, force, pressure, and buoyancy.

Question

A block and a beaker of water are placed side by side on a scale (case A). The block is then placed into the beaker of water,

where it floats (case B). How do the two scale readings compare?

1. Scale A reads more than scale B.

2. Scale A reads the same as scale B.

3. Scale A reads less than scale B.

4. Not enough information

Commentary

Purpose:To explore and interrelate ideas about weight, force, pressure, and buoyancy.

Discussion:Consider the beaker, water, and block as one compound object or system. In both cases, that system has the

same mass and therefore the same weight. Since the system is not accelerating, the net force on it must be zero, so the force

the scale exerts on this system must be equal to the total weight. Scales measure force. The forces are the same, so the scale

readings are identical.

But, howdoes the weight of the floating block in case B affect the scale reading?

It is useful to look at the beaker. There are three forces on the beaker: (1) gravitation pulling down;

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Chapter 9

(2) water pushing down; and (3) scale pushing up. Since the beaker is at rest, the force of the scale pushing up must balance

the forces exerted down.

Force #1 is the same for both cases; it is the weight of the beaker.

Force #2 is different for the two cases. In case A, the force pushing down is the weight of the water. In case B, the force

pushing down is larger than the weight of the water, because the water level is higher in B than in A, so the pressure on the

bottom of the beaker is larger. The amount the force is larger in B is exactly equal to the weight of the block. That is, it is the

weight of the displaced water.

Therefore, to support the beaker in B, the scale must push up with a force equal to the total weight of the water, block, and

beaker, which is exactly the force exerted by the scale in A.

Key Points:

Some questions are easiest to answer by considering a set of objects as if they were one compound object or system.

A scale measures force. If the system is not accelerating and interactions with its environment are small (e.g., via the

buoyancy of air), then the scale reading is the weight of the system.

Newtons laws hold for a body of fluid as well as for a solid object.

When thinking about fluids and forces, the concept ofpressureis often useful.

For Instructors Only

This is the first of three related questions. It is very easy if approached the right way, but students can get themselves quite

confused. If students give the straightforward, correct answer, we recommend challenging them to explain how the floating

block can influence the scale reading. Resolving the confusion this generally causes will strengthen their understanding of

several related concepts.

For students already familiar with the concept of buoyancy and with Archimedes principle (perhaps from high school), this

question can be used to introduce, motivate, and provide context for the concept ofpressure.

Scales measure force, not weight. If three criteria are met, that force is the weight: (1) the system is not accelerating; (2)

buoyancy due to air is negligible; (3) the only external forces on the system are due to gravitation and the scale.

Question H1.01b

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Chapter 9

Description:Exploring and interrelating weight, force, pressure, and buoyancy.

Question

A block and a beaker of water are placed side by side on a scale (case A). The block is then placed into the beaker of water,

where it sinks (case B). How do the two scale readings compare?

1. Scale A reads more than scale B.

2. Scale A reads the same as scale B.

3. Scale A reads less than scale B.


Commentary

Purpose:To explore and interrelate ideas about weight, force, pressure, and buoyancy.

Discussion:As with the previous question, the scale readings must be the same because the total mass supported by the scale,

and therefore the total weight, are the same for both cases.

As with the previous question, we would like to understand howthe scale readings can be the same. The scale is only

sensitive to the normal force exerted upon it by the beaker contacting it. How does it know about the block?

It is useful to focus on the beaker and think about the forces on it. There are four: (1) gravitation pulling down; (2) block

pushing down; (3) water pushing down; and (4) scale pushing up. The force of the scale must balance the other three forces

exerted down.

Force #1 is the same in both cases; it is the weight of the beaker.

Force #2 is different for the two cases. In case A, it is not exerted on the beaker. (However, a force down due to the block is

exerted directly on the scale.) In case B, the block is pushing down on the beaker with a force smaller than its weight,

because part of its weight is supported by buoyancy. In case A, the full weight of the block is pushing down on the scale.

Force #3 is also different for the two cases. In case A, a force equal to the weight of the water is pushing down on the beaker,

but in case B, the force pushing down is larger than the weight of the water, because the water level is higher in B than it is in

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Chapter 9

A, so the pressure at the bottom is higher too. The net effect is that the force due to the water is increased by exactly the

amount that the force due to the block is decreased. The result is that the scale readings are the same!

Note that there must be a thin layer of water underneath the block, even though it is touching the bottom of the beaker.

Otherwise, there would be no buoyant force.

Key Points:

When an object is placed in water, whether it float or sinks, the water level rises, which causes the pressure at the bottom

of the water to increase.

An object in a fluid experiences a buoyant force, whether it float or sinks.

Scales measure force, not weight. If the system is not accelerating and interactions with its environment are small (e.g.,

via the buoyancy of air), then the scale reading is the weight of the system.

Old force ideas such as free-body diagrams are useful for understanding fluids. New ideas such as pressure and

buoyancy are also useful.


This is the second of three related questions. It is very similar to the first, but understanding howthe scale reading comes to

be the same (by analyzing the forces on the various bodies) introduces a new wrinkle: the fluid only supports part of the

weight of the block, and the pressure exerted on the inside bottom of the beaker by the water and block is not uniform.

Many students will think the scale readings are the same without fully appreciating what the fuss is all about. They might

have trouble understanding why some people are confused.

Students who think the scale readings are different might need a demonstration to be convinced of the predicted result.

Students might think that the force exerted by the water in case B is actually smaller than in case A, perhaps because the

effective area of water in contact with the beaker is smaller. They do not realize that there must be water beneath the block in

order for there to be a buoyant force. Otherwise, we are talking about a suction cup, for which there is no water on one side,and an enormous force due to the water on the other.

Question H1.01c

Description:Honing understanding of buoyancy.

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Chapter 9

Question

Two blocks, A and B, have the same size and shape. Block A floats in water, but block B sinks in water.

Which block has the larger buoyant force on it?

1. Block A has the larger buoyant force on it.

2. Block B has the larger buoyant force on it.

3. Neither; they have the same buoyant force on them.

4. Impossible to determine from the given information

Commentary

Purpose:To develop your understanding of buoyancy.

Discussion:According to Archimedess Principle, the buoyant force on an object is equal to the weight of the fluid displaced

by the object. Whether the block floats or sinks is irrelevant.

Since block B sinks, it displaces its entire volume, whereas block A displaces only part of its volume.

Since the two blocks have the same total volume, block B displaces the larger volume of water, so it also has the larger

buoyant force on it.

If block B experiences a larger buoyant force but sinks, it must have a larger mass, and therefore a larger density. This is the

only way they can have the same size and shape yet behave as they do.

Key Points:

The buoyant force on an object is equal to the weight of the fluid displaced by the object. It does not depend on other

factors, such as whether the object floats or sinks.


This is the third of three related questions. It uses a different situation, but makes a good follow-up to the first two in that it

focuses attention on one specific difference between the first two questions, helping to resolve lingering confusion and

solidify students understanding. It can also be used effectively as a stand-alone question, if desired.

Many students will be overly focused on the state of an object: in this case, whether the block sinks or floats. Many will think

that the buoyant force must be smaller on the sinking block, and that is why it sinks. Encourage them to consider other

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Chapter 9

reasons why one might sink.

Students often assume that the masses of the two blocks are the same.

A demonstration can be useful, if only to let students see that the masses of the objects are definitely not the same.

Question H1.02a

Description:Developing understanding of buoyancy.

Question

A metal block sits on top of a floating wooden block. If the metal block is placed on the bottom of the beaker, what happens

to the level of waterin the beaker?

1. The level decreases.

2. The level stays the same.

3. The level increases.


Commentary

Purpose:To explore buoyancy and Archimedes principle.

Discussion:Archimedes principlestates that the buoyant force on a floating object is equal in magnitude to the weight of

displaced fluid. The buoyant force on the two-block object in the first case must equal the weight of the two blocks (so the

net force on them is zero), so the amount of water displaced must have that same weight. The displaced water must go

somewhere, so the water level in the beaker rises.

In the second case, the floating wooden block will displace an amount of water with weight equal to the wooden block alone.

The metal block at the bottom will displace a volumeof water equal to its volume, but no more: the normal force due to the

beakers bottom helps support it. Since the density of water is less than the density of the metal (or it wouldnt sink), this

means the volume of water displaced will weigh

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Chapter 9

less than the metal block. So, in the second case the two blocks will displace less total water than in the first case, and the

water level rises less in the second case.

Key Points:

Archimedes principlestates that the buoyant force on a floating object is equal in magnitude to the weight of displaced

fluid.

A floating object displaces a weight of fluid equal to its own weight.

A submerged object displaces a volume of fluid equal to its own volume.

It is sometimes helpful to think of a combination of objects as a single object.


This is the first of three related questions that help students develop a robust understanding of buoyancy, weight, floating,

sinking, and fluid displacement. The situations lend themselves to a live demonstration, using a predict, observe, and

reconcile pattern.

One possible source of confusion with this question is how the weight of the metal block in the first case can displace any

water, when the block is not in the water. Having students draw free-body diagrams for each block can be helpful for

resolving this.

Question H1.02b


Question

A metal block sits on top of a floating wooden block. If the metal block is suspended from the bottom of the wooden block,

what happens to the volume of the wooden blockthat is submerged in the water?

1. The volume decreases.

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Chapter 9

2. The volume stays the same.

3. The volume increases.


Commentary



displaced fluid. In both cases, the water is supporting the same weightthe combined weight of the two blocksso the

volume of water displaced must be the same.

However, in the first case, all the water is displaced by the wooden block, while in the second some of the water is displaced

by the hanging metal block and the rest by the wooden block. So for the second case, more of the wooden block will be

above the waters surface, and the submerged volume of the wooden block has decreased.

Key Points:


fluid.




This is the second of three related questions that help students develop a robust understanding of buoyancy, weight, floating,


reconcile pattern.

Students might wonder what the difference between the two cases is: in one the small block pushes down on the big, and in

the second it pulls down. Having students draw a free body diagram for each block should help them realize that the

magnitude of that pull is less.

Question H1.02c


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Chapter 9

Question

A metal block sits on top of a floating wooden block. If the metal block is suspended from the bottom of the wooden block,

what happens to the level of waterin the beaker?

1. The level decreases.

2. The level stays the same.

3. The level increases.


Commentary



displaced fluid. In both cases, the water is supporting the same weightthe combined weight of the two blocksso the total

volume of water displaced must be the same. Thus, the water level in the beaker must be the same as well.

Key Points:


fluid.




This is the third of three related questions that help students develop a robust understanding of buoyancy, weight, floating,


reconcile pattern.

This question should be rather easy for students who have grasped the ideas raised in the previous two questions; it serves

primarily to confirm their understanding.

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Chapter 9

QUICK QUIZZES

1. (c). The mass that you have of each element is as follows:

( ) ( )3 3 3 3gold gold gold 19.3 10 kg/m 1 m 19.3 10 kgm V= = =

( ) ( )3 3 3 3silver silver silver 10.5 10 kg/m 2 m 21.0 10 kgm V= = =

( ) ( )3 3 3 3aluminum aluminum aluminum 2.70 10 kg/m 6 m 16.2 10 kgm V= = =

2. (a). At a fixed depth, the pressure in a fluid is directly proportional to the density of the fluid. Since ethyl alcohol is

less dense than water, the pressure is smaller than Pwhen the glass is filled with alcohol.

3. (c). For a fixed pressure, the height of the fluid in a barometer is inversely proportional to the density of the fluid.

Of the fluids listed in the selection, ethyl alcohol is the least dense.

4. (b). The blood pressure measured at the calf would be larger than that measured at the arm. If we imagine the

vascular system of the body to be a vessel containing a liquid (blood), the pressure in the liquid will increase with

depth. The blood at the calf is deeper in the liquid than that at the arm and is at a higher pressure.

Blood pressures are normally taken at the arm because that is approximately the same height as the heart. If blood

pressures at the calf were used as a standard, adjustments would need to be made for the height of the person, and

the blood pressure would be different if the person were lying down.

5. (c). The level of floating of a ship is unaffected by the atmospheric pressure. The buoyant force results from the

pressure differential in the fluid. On a high-pressure day, the pressure at all points in the water is higher than on a

low-pressure day. Because water is almost incompressible, however, the rate of change of pressure with depth is

the same, resulting in no change in the buoyant force.

6. (b). Since both lead and iron are denser than water, both objects will be fully submerged and

(since they have the same dimensions) will displace equal volumes of water. Hence, the buoyant forces acting on

the two objects will be equal.

7. (a). When there is a moving air stream in the region between the balloons, the pressure in this region will be less

than on the opposite sides of the balloons where the air is not moving. The pressure differential will cause the

balloons to move toward each other. This is demonstration of Bernoullis principle in action.

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Chapter 9

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. ( ) ( ) ( ) ( )3 3 2 2 2gold 19.3 10 kg m 4.50 10 m 11.0 10 m 26.0 10 m 24.8 kgm V = = = and

choice (a) is the correct response.

2. On average, the support force each nail exerts on the body is

( )( )21

66.0 kg 9.80 m s0.535 N

1 208 1 208

mgF = = =

so the average pressure exerted on the body by each nail is

1 5av 6 2

nail

end

0.535 N5.35 10 Pa

1.00 10 m

FP

A = = =

and (d) is the correct choice.

3. From Pascals principle, 1 1 2 2F A F A= , so if the output force is to be3

2 1.2 10 NF = , the required input force

is

( )2

1 31 2 2

2

0.050 m1.2 10 N 86 N

0.70 m

AF F

A

= = =

making (c) the correct answer.

4. According to Archimedess principle, the buoyant force exerted on the bullet by the mercury is equal to the weight

of a volume of mercury that is the same as the submerged volume of the bullet. If the bullet is to float, this buoyant

force must equal the total weight of the bullet. Thus, for a floating bullet,

mercury submerged lead bulletV g V g =and

3 3submerged lead

3 3bullet mercury

11.3 10 kg m0.831

13.6 10 kg m

V

V

= = =

so the correct response is (d).

5. The absolute pressure at depth hbelow the surface of a liquid with density , and with pressure 0P at its surface, is

0P P gh= + . Thus, at a depth of 754 ft in the waters of Loch Ness,

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Chapter 9

( ) ( ) ( )5 3 3 2 61 m

1.013 10 Pa 1.00 10 kg m 9.80 m s 754 ft 2.35 10 Pa3.281 ft

P = + =

and (c) is the correct response.

6. We assume that the air inside the well-sealed house has essentially zero speed and the thickness of the roof is

negligible so the air just above the roof and that just below the roof is at the same altitude. Then, Bernoullis

equation gives the difference in pressure just below and just above the roof (with the pressure below being the

greatest) as

( ) ( )2 21 2 air 2 1 air 2 11

2P P g y y = + v v

or

( ) ( )2

3 31 1

1.29 kg m 95 mi h 0 0 1.2 10 Pa2 2.237 mi h

P = + =

and the correct choice is (a).

7. From the equation of continuity, 1 1 2 2A A=v v, the speed of the water in the smaller pipe is

( )

( )

( )2

12 1 2

2

0.250 m1.00 m s 6.25 m s

0.100 m

A

A

= = =

v v

so (d) is the correct answer.

8. All of these phenomena are the result of a difference in pressure on opposite sides of an object due to a fluid

moving at different speeds on the two sides. Thus, the correct response to this question is choice (e). Bernoullis

equation can be used in the discussion of each of these phenomena.

9. The boat, even after it sinks, experiences a buoyant force,B, equal to the weight of whatever water it is displacing.

This force will support part of the weight, w, of the boat. The normal force exerted on the boat by the bottom of the

lake will be n w B w= < will support the balance of the boats weight. The correct response is (c).

10. The absolute pressure at depth hbelow the surface of a fluid having density is, 0P P gh= + where 0P is the

pressure at the upper surface of that fluid. The fluid in each of the three vessels has density water = , the top of

each vessel is open to the atmosphere so that 0 atmoP P= in each case, and the bottom is at the same depth hbelow

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Chapter 9

the upper surface for the three vessels. Thus, the pressure Pat the bottom of each vessel is the same and (c) is the

correct choice.

11. Since the pipe is horizontal, each part of it is at the same vertical level or has the samey-coordinate. Thus, from

Bernoullis equation1 22( )P gy constant + + =v , we see that the sum of the pressure and the kinetic energy

per unit volume 1 22

( )P + v must also be constant throughout the pipe, making (e) the correct choice.

12. Once the water droplets leave the nozzle, they are projectiles with initial speed 0y i=v vand having speed

0f y= =v v at their maximum altitude, h. From the kinematics equation 2 20 2 ( )y y ya y= + v v the maximum

height reached is 2 2ih g= v . Thus, if we want to quadruple the maximum height ( 4 )h h= we need to double

the speed of the water leaving the nozzle ( 2 )i i=v v . Using the equation of continuity, i iA A= v v, it is seen that

if I= 2i, it is necessary to have ( ) 2i i iA A A= = v v This says that the area needs to be decreased by a factor

of 2, and the correct choice is (d).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. We approximate the thickness of the atmosphere by using 0P P gh= + with 0 0P = at the top of the

atmosphere and 1 atmP = at sea level. This gives an approximation of

( ) ( )

50 4

3 1 2

10 Pa 0~ 10 m

1 kg m 10 m s

P Ph

g

= = or

~ 10 kmh

Because both the density of the air, , and the acceleration of gravity, g, decrease with altitude, the actual

thickness of the atmosphere will be greater than our estimate.

4. Both must have the same strength. The force on the back of each dam is the average pressure of the water times the

area of the dam. If both reservoirs are equally deep, the force is the same.

6. The external pressure exerted on the chest by the water makes it difficult to expand the chest cavity and take a

breath while under water. Thus, a snorkel will not work in deep water.

8. A fan driven by the motor removes air and hence decreases the pressure inside the cleaner. The greater air pressure

outside the cleaner pushes air in through the nozzle toward this region of lower pressure. This inward rush of air

pushes or carries the dirt along with it.

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Chapter 9

10. The water level on the side of the glass stays the same. The floating ice cube displaces its own weight of liquid

water, and so does the liquid water into which it melts.

12. The higher the density of a fluid, the higher an object will float in it. Thus, an object will float lower in low density

alcohol.

14. A breeze from any direction speeds up to go over the mound, and the air pressure drops at this opening. Air then

flows through the burrow from the lower to the upper entrance.

PROBLEM SOLUTIONS

9.1 The elastic limit is the maximum stress, F A where Fis the tension in the wire, that the wire can withstand and

still return to its original length when released. Thus, if the wire is to experience a tension equal to the weight of the

performer without exceeding the elastic limit, the minimum cross-sectional area is

2min

min4

D F mgA

elastic limit elastic limit

= = =

and the minimum acceptable diameter is

( )

( )( )( )

2

3min 8

4 70 kg 9.8 m s41.3 10 m 1.3 mm

5.0 10 Pa

mgD

elastic limit

= = = =

9.2 (a) In order to punch a hole in the steel plate, the

superhero must punch out a plug with cross-

sectional area, csA , equal to that of his fist and

a height tequal to the thickness of the steel plate. The area shearA of the face that is sheared as the plug is

removed is the cylindrical surface with radius rand height tas shown in the sketch. SinceAcs= r2, then

csr A = and

( ) ( )2 2

2shear

1.00 10 cm2 2 2 2.00 cm 70.9 cmcs

AA r t t

= = = =

If the ultimate shear strength of steel (i.e., the maximum shear stress it can withstand before shearing) is,

8 8 22.50 10 Pa 2.50 10 N m = the minimum force required to punch out this plug is

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Chapter 9

22

shear

1 m70.9 cmF A stress= = 8

4 2 2

N2.50 10

10 cm m

61.77 10 N

=

(b) By Newtons third law, the wall would exert a force of equal magnitude in the opposite direction on the

superhero, who would be thrown backward at a very high recoil speed.

9.3 Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank,

separated by distance 2.00 mh = and each with area, ( ) ( )22.00 cm 15.0 cm 30.0 cmA = = move a distance

25.00 10 mx = parallel to each other. The force causing this shearing effect in the plank is the weight of the

man F mg= applied perpendicular to the length of the plank at its outer end. Since the shear modulus Sis given

by

( )shear stress F A FhSshear strain x h x A

= = =

we have

( )( )( )

( ) ( ) ( )

2

7

2 2 2 4 2

80.0 kg 9.80 m s 2.00 m1.05 10 Pa

5.00 10 m 30.0 cm 1 m 10 cmS

= =

9.4 As a liquid, the water occupied some volume lV . As ice, the water would occupy volume 1.090Vlif it were not

compressed and forced to occupy the original volume. Consider the pressure change required to squeeze ice back

into volume lV . Then, 0 1.09 and 0.090l lV V V V = = , so

9 82

0

0.090N2.00 10 1.65 10 Pa 1600 atm

m 1.09

l

l

VVP B

V V

= = =

9.5 Using 0 ( )Y F L A L= with 2 4A d= and F mg= , we get

( )( ) ( )

( ) ( )

2

82

2

4 90 kg 9.80 m s 50 m3.5 10 Pa

1.0 10 m 1.6 mY

= =

9.6 From 0 ( )Y F L A L= the tension needed to stretch the wire by 0.10 mm is

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Chapter 9

( ) ( )( )

( ) ( ) ( )( )

2

0 0

210 3 3

2

4

18 10 Pa 0.22 10 m 0.10 10 m22 N

4 3.1 10 m

Y d LY A LF

L L

= =

= =

The tension in the wire exerts a force of magnitude Fon the tooth in each direction along the length of the wire as show

in the above sketch. The resultant force exerted on the tooth has anx-component of

cos 30 cos 30 0x xR F F F= = + = , and ay-component of sin 30 sin 30 22 Ny yR F F F F= = = =

Thus, the resultant force is

22 N directed down the page in the diagram=R .

9.7 From 0 0( ) ( ) ( ) ( )Y F A L L stress L L= = , the maximum compression the femur can withstand is

( )( ) ( )( )60 39

160 10 Pa 0.50 m4.4 10 m 4.4 mm

18 10 Pa

stress LL

Y

= = = =

9.8 (a) When at rest, the tension in the cable equals the weight of the 800-kg object, 37.84 10 N .

Thus, from 0 ( )Y F L A L= , the initial elongation of the cable is

( )( )

( ) ( )

30 3

4 2 10

7.84 10 N 25.0 m2.45 10 m = 2.5 mm

4.00 10 m 20 10 Pa

F LL

A Y

= = =

(b) When the load is accelerating upward, Newtons second law gives

yF mg ma =

( )yF m g a= + [1]

If 2800 kg and 3.0 m sym a= = + , the elongation of the cable will be

( ) ( ) ( )

( ) ( )

20 3

4 2 10

800 kg 9.80 3.0 m s 25.0 m3.2 10 m 3.2 mm

4.00 10 m 20 10 Pa

F LL

A Y

+ = = = =

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Chapter 9

Thus, the increase in the elongation has been

( ) ( )initial

3.20 mm 2.45 mm 0.75 mmincrease L L= = =

(c) From the definition of the tensile stress, stress F A= , the maximum tension the cable can withstand is

( ) ( ) ( )4 2 8 4max max 4.00 10 m 2 .2 10 Pa 8.8 10 NF A stress = = =

Then, Equation [1] above gives the mass of the maximum load as

( )

4max 3

max 2

8.8 10 N6.9 10 kg

9.8 3.0 m s

Fm

g a

= = =

+ +

9.9 From the defining equation for the shear modulus, we find the displacement, x , as

( ) ( )( )

( ) ( )

3 4 2

26 2

5

5.0 10 m 20 N 10 cm

1 m3.0 10 Pa 14 cm

2.4 10 m 0.024 mm

h F A h Fx

S S A

= = =

= =

9.10 The shear modulus is given by

( )shear stress stress

Sshear strain x h

= =

Hence, the stress is

( )10 63

5.0 m1.5 10 Pa 7.5 10 Pa

10 10 m

xstress S

h

= = =

9.11 The tension and cross-sectional area are constant through the entire length of the rod, and the total elongation is the

sum of that of the aluminum section and that of the copper section.

( ) ( ) ( ) ( )0 00 0Cu CuAl Alrod Al Cu

Al Cu Al Cu

F L LF L LFL L L

AY AY A Y Y

= + = + = +

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Chapter 9

where 2A r= with 30.20 cm 2.0 10 mr = = . Thus,

( )

( )

3

2rod 2 10 10

3

5.8 10 N 1.3 m 2.6 m1.9 10 m 1.9 cm

7.0 10 Pa 11 10 Pa2.0 10 mL

= + = =

9.12 The acceleration of the forearm has magnitude

3

3 23

km 10 m 1 h80

h 1 km 3 600 s4.4 10 m s

5.0 10 sa

t

= = = v

The compression force exerted on the arm is F ma= and the compressional stress on the bone material is

( )( )( )

3 2

7

2 4 2 2

3.0 kg 4.4 10 m s5.5 10 Pa

2.4 cm 10 m 1 cm

FStress

A

= = =

Since the stress is less than the allowed maximum, the arm should survive.

9.13 The average density of either of the two original worlds was

0 3 334 3 4M M MV R R

= = =

The average density of the combined world is

( )

( )

2total

3 32 3

4 22 32

934 3

3 4

MM M M

V RRR

= = = =

so

3

30

32 4 1284.74

9 3 27

M R

R M

= = =

or04.74 =

9.14 (a) The mass of gold in the coin is

Page 9.18


19/55

Chapter 9

( )( )total 3 3Au total

# karats 22 117.988 10 kg 7.322 10 kg

24 24 12

mm m = = = =

and the mass of copper is

( )3 4Cu total1 1

7.988 10 kg 6.657 10 kg12 12

m m = = =

(b) The volume of the gold present is

3Au 7 3

Au 3 3Au

7.322 10 kg3.79 10 m

19.3 10 kg m

mV

= = =

and the volume of the copper is

4Cu 8 3

Cu 3 3Cu

6.657 10 kg7.46 10 m

8.92 10 kg m

mV

= = =

(c) The average density of the British sovereign coin is

3total total 4 3

av 7 3 8 3total Au Cu

7.988 10 kg1.76 10 kg m

3.79 10 m 7.46 10 m

m m

V V V

= = = =

+ +

9.15 (a) The total normal force exerted on the bottom acrobats shoes by the floor equals the total weight of the acro

bats in the tower. That is

( ) ( )2 3total 75.0 68.0 62.0 55.0 kg 9.80 m s 2.55 10 Nn m g = = + + + =

(b) ( )

34

2 2 4 2total shoesole

2.55 10 N3.00 10 Pa

2 2 425 cm 1 m 10 cm

n nP

A A

= = = =

(c) If the acrobats are rearranged so different ones are at the bottom of the tower, the total weight supported, and

hence the total normal force n, will be unchanged. However, the total area total shoe sole2A A= , and hence the

pressure, will change unless all the acrobats wear the same size shoes.

Page 9.19


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Chapter 9

9.16 We shall assume that each chair leg supports one-fourth of the total weight so the normal force each leg exerts on

the floor is 4n mg= . The pressure of each leg on the floor is then

( )( )

( )

2

6

leg 22 2leg

95.0 kg 9.80 m s4

2.96 10 Pa4 0.500 10 m

n mg

P A r = = = =

9.17 (a) If the particles in the nucleus are closely packed with negligible space between them, the average nuclear

density should be approximately that of a proton or neutron. That is

( )

( ): :

27proton proton 17 3

nucleus 3315proton

3 1.67 10 kg 4 10 kg m

4 3 4 1 10 m

m m

V r

=

(b) The density of iron is3 3

Fe 7.86 10 kg m = kg/m3 and the densities of other solids and liquids are on

the order of3 310 kg m . Thus, the nuclear density is about 1410 times greater than that of common solids

and liquids, which suggests that atoms must be mostly empty space. Solids and liquids, as well as gases, are

mostly empty space.

9.18 Let the weight of the car be W. Then, each tire supports 4W , and the gauge pressure is

4

F WP

A A= =

Thus, ( ) ( )2 5 44 4 0.024 m 2.0 10 Pa 1.9 10 NW A P= = = .

9.19 The volume of concrete in a pillar of height hand cross-sectional areaAis V Ah= , and its weight is

( )( )4 35.0 10 N mgF Ah= . The pressure at the base of the pillar is then

( )( )( )

4 3

4 35.0 10 N m

5.0 10 N mg AhF

P hA A

= = =

Thus, if the maximum acceptable pressure on the base is,7

max 1.7 10 PaP = , the maximum allowable height is

Page 9.20


21/55

Chapter 9

7max 2

max 4 3 4 3

1.7 10 Pa3.4 10 m

5.0 10 N m 5.0 10 N m

Ph

= = =

9.20 Assuming the spring obeys Hookes law, the increase in force on the piston required to compress the spring an

additional amount x is ( ) ( )0 0F F F P P A k x = = = . The gauge pressure at depth hbeneath the

surface of a fluid is 0P P gh = , so we have, ( )ghA k x = or the required depth is ( )h k x gA= , If

1 250 N mk = , 2 2with 1.20 10 mA r r = = , with ( )3 31.00 10 kg m= , and the fluid is water (

= 1.00 103 kg/m3), the depth required to compress the spring an additional is 30.750 cm 7.50 10 m= is

( )( )

( ) ( ) ( )

3

23 3 2 2

1 250 N m 7.50 10 m2.11 m

1.00 10 kg m 9.80 m s 1.20 10 m

h

= =

9.21 (a) ( ) ( )( )3 3 3 2 50 101.3 10 Pa 1.00 10 kg m 9.80 m s 27.5 m 3.71 10 PaP P gh= + = + =

(b) The inward force the water will exert on the window is

( ) ( )2

22 5 4

35.0 10 m3.71 10 Pa 3.57 10 N

2F PA P r

= = = =

9.22 The gauge pressure in a fluid at a level distance hbelow where gauge 0P = is gaugeP gh= with hbeingpositive

when measured in the downward direction. The difference in gauge pressures at two levels is

gauge 2 gauge 1( ) ( ) ( )P P g h = or (Pgauge)2= (Pgauge)1+ g(h) with h being positive if, in going from level 1 to

level 2, one is moving lower in the fluid.

(a) In moving from the heart to the head, one is moving higher in the blood column so 0h < and we find

( ) ( ) ( ) ( ) ( )( )3 3 2gauge gaugehead heart

13.3 10 Pa 1 060 kg m 9.80 m s 0.500 mP P g h= + = +

or

( ) 3gaugehead

8.11 10 Pa 8.11 kPaP = =

Page 9.21


22/55

Chapter 9

(b) In going from heart to feet, one moves deeper in the blood column, so 0h < and

( ) ( ) ( ) ( ) ( )( )3 3 2gauge gaugefeet heart

13.3 10 Pa 1 060 kg m 9.80 m s 1.30 mP P g h= + = + +

or

( ) 3gaugefeet

26.8 10 Pa 26.8 kPaP = =

9.23 The density of the solution is3 3

water1.02 1.02 10 kg m = = . The gauge pressure of the fluid at the level

of the needle must equal the gauge pressure in the vein, so 3gauge 1.33 10 PaP gh= = , and

( ) ( )

3gauge

3 3 2

1.33 10 Pa0.133 m

1.02 10 kg m 9.80 m s

P

hg = = =

9.24 (a) From the definition of bulk modulus, ( )0B P V V= , the change in volume of the31.00 m of seawater

will be

( ) ( ) ( )3 8 50 310

water

1.00 m 1.13 10 Pa 1.013 10 Pa0.053 8 m

0.210 10 Pa

V PV

B

= = =

(b) The quantity of seawater that had volume3

0 1.00 mV = at the surface has a mass of 1 030 kg. Thus, the

density of this water at the ocean floor is

( )3 3

30

1 030 kg1.09 10 kg m

1.00 0.053 8 m

m m

V V V= = = =

+

(c) Considering the small fractional change in volume (about 5%) and enormous change in pressure

generated, we conclude that it is a good approximation to think of water as incompressible.

9.25 We first find the absolute pressure at the interface between oil and water.

( ) ( )( )

1 0 oil oil

5 3 2 51.013 10 Pa 700 kg m 9.80 m s 0.300 m 1.03 10 Pa

P P gh= +

= + =

Page 9.22


23/55

Chapter 9

This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use, P2 =P1 + waterghwater,

or

( ) ( )( )5 3 3 2 52 1.03 10 Pa 10 kg m 9.80 m s 0.200 m 1.05 10 PaP = + =

9.26 If we assume a vacuum exists inside the tube above the wine column, the pressure at the base of the tube (that is, at

the level of the wine in the open container) is Patmo= 0 + gh= gh. Thus,

( ) ( )

5atmo

3 2

1.013 10 Pa10.5 m

984 kg m 9.80 m s

Ph

g

= = =

Some alcohol and water will evaporate, degrading the vacuum above the column.

9.27 Pascals principle, 1 1 2 2F A F A= , or pedal Master brake brakecylinder cylinderF A F A= , gives

( )2

brake cylinder

brake pedal 2master cylinder

6.4 cm44 N 156 N

1.8 cm

AF F

A

= = =

This is the normal force exerted on the brake shoe. The frictional force is

( )0.50 156 N 78 Nkf n= = =

and the torque is ( ) ( )drum 78 N 0.34 m 27 N mf r = = = .

9.28 First, use Pascals principle,F1/A1=F2/A2, to find the force

piston 1 will exert on the handle when a 500-lb force

pushes downward on piston 2.

( )

( )( )

2 2

1 1 11 2 2 22 2

2 2 2

2

2

44

0.25 in500 lb 14 lb

1.5 in

A d dF F F F A d d

= = =

= =

Now, consider an axis perpendicular to the page, passing through the left end of the jack handle. = 0 yields

Page 9.23


24/55

Chapter 9

( ) ( ) ( )14 lb 2 .0 in 12 in 0F+ = , 2.3 lbF =

9.29 When held underwater, the ball will have three forces acting on it: a downward gravitational force, mg; an upward

buoyant force,B =waterV = 4waterr3/3; and an applied force, F. If the ball is to be in equilibrium, we have (taking

upward as positive) 0yF F B mg = + = ,or

3 3

water water

4 4

3 3

r rF mg B mg g m g

= = =

giving

( ) ( )3

3 3 24 0.250 m0.540 kg 1.00 10 kg m 9.80 m s 74.9 N3 2

F = =

so the required applied force is .ur

74.9 N directed downward=F

9.30 (a) To float, the buoyant force acting on the person must equal the weight of that person, or the weight of the

water displaced by the person must equal the persons own weight. Thus,

sea submerged body totalB mg gV gV = =or

submerged body

total sea

V

V

=

After inhaling,

3submerged

3total

945 kg m0.768 76.8%

1 230 kg m

V

V= = =

leaving .23.2% above surface.

After exhaling,

3submerged

3total

1 020 kg m0.829 82.9%

1 230 kg m

V

V= = =

Page 9.24


25/55

Chapter 9

leaving .17.1% above surface.

(b) In general, sinkers would be expected to be thinner with heavier bones, whereas floaters would have

lighter bones and more fat.

9.31 The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus,

( )

( ) ( )( )

truck water

3 23 2

kg m10 4.00 m 6.00 m 4.00 10 m 9.80

m s

w V g

=

=

or

3truck 9.41 10 N 9.41 kNw = =

9.32 (a)

(b) Since the system is in equilibrium, 0 .y rF B w w = =

(c)( )

( ) ( )( )( )submerged

3 2 21 025 kg m 9.80 m s 0.024 0 m 4.00 m 964 N

w wB gV g d A = =

= =

(d) From 0rB w w = ,

( )( )2964 N 62.0 kg 9.80 m s 356 Nr sw B w B m g= = = =

(e)

( )( )

( )

3foam 2 2

356 N101 kg m

9.80 m s 0.090 m 4.00 m

r r

r

m w g

V t A = = = =

(f)( )

( ) ( )( )( )max

3 2 2 31 025 kg m 9.80 m s 0.090 0 m 4.00 m 3.62 10 N

w r wB gV g t A = =

= =

(g) The maximum weight of survivors the raft can support is max max max rw m g B w= = so

Page 9.25


26/55

Chapter 9

3max

max 2

3.62 10 N 356 N333 kg

9.80 m s

rB wmg

= = =

9.33 (a) While the system floats, total block steelB w w w= = + .or wg submerged bV g= steelbV m g+ .

When steel 0.310 kgm = ,4 3

submerged 5.24 10 mbV V = = giving

steel steel 3 3 34 3

0.310 kg1.00 10 kg m 408 kg m

5.24 10 m

w bb w

b b

V m m

V V

= = = =

(b) If the total weight of the block + steel system is reduced, by having msteel


27/55

Chapter 9

( ) ( ) ( )3 2 3 3air 1.29 kg m 9.80 m s 325 m 4.11 10 NbB gV= = =

(c)( )

( ) ( ) ( )He He He

3 3 3 2 34.11 10 N 226 kg 0.179 kg m 325 m 9.80 m s 1.33 10 N

y b b b bF B w w B m g gV B m V g = = = +

= + = +

Since Fy = may>0, aywill be positive (upward), and the balloon rises.

(d) If the balloon and load are in equilibrium, ( )He load 0y bF B w w w = = and

( ) 3load He 1.33 10 Nbw B w w= = . Thus, the mass of the load is

3load

load 2

1.33 10 N136 kg

9.80 m s

wm

g

= = =

(e) If mload


28/55

Chapter 9

1wya g

=

(c) ( )

3 32 2 2

3

1.00 10 m kg1 9.80 m s 0.467 m s 0.467 m s downward

1 050 m kgya

= = =

(d) From2

0 2y yy t a t = +v with 0y = 0, we find

( ) ( )2

2 2 8.00 m5.85 s

0.467 m sy

yt

a

= = =

( )

( ) ( ) ( )

3total single air balloon air

balloon

33 2 3

4 (a) 600 600 600

34

600 1.29 kg m 9.80 m s 0.50 m 4.0 10 N 4.0 kN3

B B gV g r

= = =

= = =

9.37

(b) ( )( )3 2 3total total 4.0 10 N 600 0.30 kg 9.8 m s 2.2 10 N 2.2 kNyF B m g = = = =

(c) Atmospheric pressure at this high altitude is much lower than at Earths surface, so the balloons expanded

and eventually burst.

9.38 Note:We deliberately violate the rules of significant figures in this problem to illustrate a point.

(a) The absolute pressure at the level of the top of the block is

( )

top 0 water top

5 3 23 2

5

kg m1.0130 10 Pa 10 9.80 5.00 10 m

m s

1.0179 10 Pa

P P gh

= +

= +

=

and that at the level of the bottom of the block is

( )

bottom 0 water bottom

5 3 23 2

5

kg m1.0130 10 Pa 10 9.80 17.0 10 m

m s

1.0297 10 Pa

P P gh

= +

= +

=

Page 9.28


29/55

Chapter 9

Thus, the downward force exerted on the top by the water is

( )( )25

top top 1.0179 10 Pa 0.100 m 1017.9 NF P A= = =

and the upward force the water exerts on the bottom of the block is

( )( )25

bot bot 1.0297 10 Pa 0.100 m 1029.7 NF P A= = =

(b) The scale reading equals the tension, T, in the cord supporting the block. Since the block is in equilibrium,

bot top 0yF T F F mg = + = , or

( )( ) ( )210.0 kg 9.80 m s 1029.7 1017.9 N 86.2 NT = =

(c) From Archimedess principle, the buoyant force on the block equals the weight of the displaced water. Thus,

( )

( ) ( ) ( ) ( )

water block

23 3 210 kg m 0.100 m 0.120 m 9.80 m s 11.8 N

B V g=

= =

From part (a), ( )bot top 1 029.7 1 017.9 N 11.8 NF F = = , which is the same as the buoyant force found

above.

9.39 Constant velocity means that the submersible is in equilibrium under the gravitational force, the upward buoyant

force, and the upward resistance force:

0y yF ma = =

( )4 sea water1.20 10 kg 1 100 N 0m g gV + + + =

where mis the mass of the added sea water and Vis the spheres volume.

Thus,

( )33 4

3 2

kg 4 1 100 N1.03 10 1.50 m 1.20 10 kg

m 3 9.80 m sm

= +

or

Page 9.29


30/55

Chapter 9

32.67 10 kgm =

9.40 At equilibrium, spring 0yF B F mg = = , so the spring force is

( )spring water block F B mg V m g = =

where

3 3block 3

wood

5.00 kg7.69 10 m

650 kg m

mV

= = =

Thus, ( ) ( ) ( )

3 3 3 3 2

spring

10 kg m 7.69 10 m 5.00 kg 9.80 m s 26.4 N.F = =

The elongation of the spring is then

spring 26.4 N0.165 m 16.5 cm

160 N m

Fx

k = = = =

9.41 (a) The buoyant force is the difference between the weight in air and the apparent weight when immersed in the

alcohol, orB= 300 N 200 N = 100 N. But, from Archimedess principle, this is also the weight of thedisplaced alcohol, soB= (alcoholV)g. Since the sample is fully submerged, the volume of the displaced

alcohol is the same as the volume of the sample. This volume is

( ) ( )2 3

3 2alcohol

100 N1.46 10 m

700 kg m 9.80 m s

BV

g= = =

(b) The mass of the sample is

2

300 N30.6 kg

9.80 m s

weight in airm

g= = =

and its density is

Page 9.30


31/55

Chapter 9

3 32 3

30.6 kg2.10 10 kg m

1.46 10 m

m

V

= = =

9.42 The difference between the weight in air and the apparent weight when immersed is the buoyant force exerted on

the object by the fluid.

(a) The mass of the object is

2

300 N30.6 kg

9.80 m s

weight in airm

g= = =

The buoyant force when immersed in water is the weight of a volume of water equal to the volume of the

object, orBw= (wV)g. Thus, the volume of the object is

( ) ( )3 3

3 3 2

300 N 265 N3.57 10 m

10 kg m 9.80 m s

w

w

BV

g= = =

and its density is

3 3object 3 3

30.6 kg8.57 10 kg m

3.57 10 m

m

V

= = =

(b) The buoyant force when immersed in oil is equal to the weight of a volume V= 3.57 10.3m3of oil. Hence

Boil= (oilV)g, or the density of the oil is

( ) ( )oil 3

oil 3 3 2

300 N 275 N714 kg m

3.57 10 m 9.80 m s

B

Vg

= = =

9.43 The volume of the iron block is

iron 4 33 3

iron

2 .00 kg 2 .54 10 m7.86 10 kg m

mV

= = =

and the buoyant force exerted on the iron by the oil is

( ) ( ) ( ) ( )3 4 3 2oil 916 kg m 2 .54 10 m 9.80 m s 2 .28 NB V g = = =

Page 9.31


32/55

Chapter 9

Applying Fy= 0 to the iron block gives the support force exerted by the upper scale (and hence the reading on that

scale) as

upper iron 19.6 N 2.28 N 17.3 NF m g B= = =

From Newtons third law, the iron exerts forceBdownward on the oil (and hence the beaker). Applying Fy= 0 to

the system consisting of the beaker and the oil gives

( )lower oil beaker 0F B m m g + =

The support force exerted by the lower scale (and the lower scale reading) is then

( ) ( ) ( )2

lower oil beaker 2.28 N 2.00 1.00 kg 9.80 m s 31.7 NF B m m g = + + = + + =

9.44 (a) The cross-sectional area of the hose is ( )22 2 4 2.74 cm 4A r d = = = , and the volume flow rate

(volume per unit time) isA=25.0 L/1.50 min. Thus,

25.0 L 1.50 min 25.0 L

A= =v

1.50 min ( )2 2

4 1 min

2.74 cm

3 310 cm

60 s 1 L

( )2

1 m47.1 cm s 0.471 m s

10 cm

= =

(b)

2 222 2 2

21 1 1

4 1 1

4 3 9

A d d

A d d

= = = = or 1

29

AA =

Then from the equation of continuity, 2 2 1 1A A=v v, we find

( )12 12

9 0.471 m s 4.24 m sA

A

= = =

v v

9.45 (a) The volume flow rate isA, and the mass flow rate is

( ) ( )( )3 21.0 g cm 2.0 cm 40 cm s 80 g sA = =v

Page 9.32


33/55

Chapter 9

(b) From the equation of continuity, the speed in the capillaries is

( )2

aortacapiliaries aorta 3 2

capillaries

2 .0 cm40 cm s

3.0 10 cm

A

A

= =

v v

or2

capiliaries 2.7 10 cm s 0.27 mm s .= =v

9.46 (a) From the equation of continuity, the flow speed in the second pipe is

( )2

12 1 2

2

10.0 cm2.75 m s 11.0 m s

2.50 cm

A

A

= = =

v v

(b) Using Bernoullis equation and choosing y = 0 along the centerline of the pipes gives

( ) ( ) ( ) ( )2 22 2 5 3 3

2 1 1 2

1 11.20 10 Pa 1.65 10 kg m 2.75 m s 11.0 m s

2 2P P = + = +

v v

or4

2 2.64 10 PaP = .

9.47 From Bernoullis equation, choosing aty= 0 the level of the syringe and needle, so the flow speed in the needle,

1 12 22 2 1 12 2

P P + = +v v is

( )1 222 1 2 P P= +v v

In this situation,

( ) 41 2 1 atmo 1 gauge 5 21

2 .00 N8.00 10 Pa

2 .50 10 m

FP P P P P

A = = = = =

Thus, assuming 0,

( )42 3 3

2 8.00 10 Pa0 12.6 m s

1.00 10 kg m

= + =

v

9.48 We apply Bernoullis equation, ignoring the very small change in vertical position, to obtain

Page 9.33


34/55

Chapter 9

( ) ( )2 31 12 2 2 2

1 2 2 1 1 1 12 2 22P P = = =

v v v v v, or

( ) ( )2

3 2 23

1.29 kg m 15 10 m s 4.4 10 Pa2

P = =

9.49 (a) Assuming the airplane is in level flight, the net lift (the difference in the upward and downward forces

exerted on the wings by the air flowing over them) must equal the weight of the plane, or

lower upper wingssurface surface

( )P P A mg =. This yields

( ) ( )4 23

lower upper 2surface surface wings

8.66 10 kg 9.80 m s9.43 10 Pa

90.0 m

mgP P

A

= = =

(b) Neglecting the small difference in altitude between the upper and lower surfaces of the wings and applying

Bernoullis equation yields

2 2lower air lower upper air upper

1 1

2 2P P + = +v v

so

( ) ( ) ( )3

lower upper 22upper lower 3

air

2 2 9.43 10 Pa225 m s 255 m s1.29 kg m

P P

= + = + =v v

(c) The density of air decreases with increasing height, resulting in a smaller pressure difference. Beyond the

maximum operational altitude, the pressure difference can no longer support the aircraft.

9.50 For level flight, the net lift (difference between the upward and downward forces exerted on the wing surfaces by

air flowing over them) must equal the weight of the aircraft, or lower uppersurface surface

( )P P A Mg =. This gives the air

pressure at the upper surface as

upper lowersurface surface

MgP P

A=

9.51 (a) Since the pistol is fired horizontally, the emerging water stream has initial velocity components of (0x=

Page 9.34


35/55

Chapter 9

nozzle, 0y= 0) Then, y= 0yt+ ayt2/2, with ay= g, gives the time of flight as

( ) ( )2

2 2 1.50 m0.553 s

9.80 m sy

yt

a

= = =

(b) With ax= 0, and 0x= nozzle, the horizontal range of the emergent stream is x= nozzletwhere tis the time of

flight from above. Thus, the speed of the water emerging from the nozzle is

nozzle

8.00 m14.5 m s

0.553 s

x

t

= = =v

(c) From the equation of continuity,A11=A22, the speed of the water in the larger cylinder is 1 =(A2/A1)2=

(A2/A1)nozzle, or

( )2 22

2 21 nozzle nozzle2

1 1

1.00 mm14.5 m s 0.145 m s

10.0 mm

r r

r r

= = = = v v v

(d) The pressure at the nozzle is atmospheric pressure, or P2= 1.013 105m/s .

(e) With the two cylinders horizontal, y1= y2and gravity terms from Bernoullis equation can be neglected,

leaving2 2

1 1 2 22 2w wP P + = +v v so the needed pressure in the larger cylinder is

( ) ( ) ( )3 3

2 22 2 51 2 2 1

1.00 10 kg m1.013 10 Pa 14.5 m s 0.145 m s

2 2

wP P = + = +

v v

or

51 2.06 10 PaP =

(f) To create an overpressure of P= 2.06 105Pa = 1.05 105Pa in the larger cylinder, the force that must be

exerted on the piston is

( ) ( )( ) ( ) ( )2

2 5 21 1 1 1.05 10 Pa 1.00 10 m 33.0 NF P A P r

= = = =

9.52 (a) From Bernoullis equation,

Page 9.35


36/55

Chapter 9

2 21 2

1 1 2 22 2

w ww wP gy P gy

+ + = + +

v v

or

Figure 9.29

( )1 22 22 1 2 12w

P Pg y y

=

v v

and using the given data values, we obtain

( )( )5 5

2 2 22 1 3 3

1.75 10 Pa 1.20 10 Pa2 9.80 m s 2.50 m

1.00 10 kg m

=

v v

and [1]

2 2 2 22 1 61.0 m s =v v

From the equation of continuity,

22

1 1 12 1 1 12

2 2 2

A r r

A r r

= = =

v v v v

2

2 1

3.00 cm

1.50 cm

= v v or 2= 41 [2]

Substituting Equation [2] into [1] gives ( ) 2 2 2116 1 61.0 m s =v , or

2 2

1

61.0 m s2.02 m s

15= =v

(b) Equation [2] above now yields 2= 4(2.02 m/s) = 8.08 m/s.

(c) The volume flow rate through the pipe is: flow rate =A11=A22.

Looking at the lower point:

Page 9.36


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Chapter 9

( ) ( ) ( )2

2 2 3 31 1 3.00 10 m 2.02 m s 5.71 10 m sflow rate r

= = = v

9.53 First, consider the path from the viewpoint of projectile motion to find the speed at which the

water emerges from the tank. From1 2

0 2y yy t a t = +v with 0y= 0, we find the time of flight as

( ) ( )2

2 2 1.00 m0.452 s

9.80 m sy

yt

a

= = =

From the horizontal motion, the speed of the water coming out of the hole is

2 0

0.600 m1.33 m s

0.452 sx

x

t

= = = =v v

We now use Bernoullis equation, with point 1 at the top of the tank and point 2 at the level of the hole. With P1=

P2= Patmoand, 1 0. This gives

21 2 2

1

2gy gy = +v

or

( )

( )

222 2

1 22

1.33 m s9.00 10 m 9.00 cm

2 2 9.80 m s

h y y

g

= = = = =v

9.54 (a) Apply Bernoullis equation with point 1 at the open top of the tank and point 2 at the opening of the hole.

Then,P1= P2= Patmoand we assume 1 0. This gives1 2

2 2 12 gy gy + =v , or

( ) ( )( )22 1 22 2 9.80 m s 16.0 m 17.7 m sg y y= = =v

(b) The area of the hole is found from

3 36 2

22

2 .50 10 m min 1 min2 .35 10 m

17.7 m s 60 s

flow rateA

= = = v

The diameter is then

Page 9.37


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Chapter 9

( )6 22 32

4 2.35 10 m41.73 10 m 1.73 mm

Ad

= = = =

9.55 First, determine the flow speed inside the larger portions from

( )

4 3

1 221

1.80 10 m s0.367 m s

2.50 10 m 4

flow rate

A

= = =

v

The absolute pressure inside the large section on the left is P1= P0+ gh1, where h1 is the height of the water in the

leftmost standpipe. The absolute pressure in the constriction is P2= P1+ gh2, so

( ) ( )1 2 1 2 5.00 cmP P g h h g = =

The flow speed inside the constriction is found from Bernoullis equation withy1 =y2. This gives

( ) ( )2 2 22 1 1 2 1 1 22

2P P g h h

= + = + v v v

( ) ( )( )2 2

2 0.367 m s 2 9.80 m s 5.00 10 m 1.06 m s= + =v

The cross-sectional area of the constriction is then

4 34 2

22

1.80 10 m s1.71 10 m

1.06 m s

flow rateA

= = =

v,

and the diameter is

( )4 22 22

4 1.71 10 m41.47 10 m 1.47 cm

Ad

= = = =

9.56 (a) For minimum pressure, we assume the flow is very slow. Then, Bernoullis equation gives

2 2

river rim

1 1=

2 2P gy P gy

+ + + + v v

Page 9.38


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Chapter 9

( ) ( )river rim rivermin 0 1 atm 0P g y y+ = + +

or

( ) ( )5 3river 3 2min kg m1.013 10 Pa 10 9.80 2096 m 564 mm sP = +

( ) ( )5 7 7river min 1.013 10 1.50 10 Pa =1.51 10 Pa 15.1 MPaP = + =

(b) The volume flow rate is flow rate =A=(d2/4). Thus, the velocity in the pipe is

( ) ( )

( )

3

22

4 4500 m d4 1 d

2.95 m s86 400 s0.150 m

flow rate

d

= = = v

(c) We imagine the pressure being applied to stationary water at river level, so Bernoullis equation becomes

( ) 2river rim river rim1

0 1 atm2

P g y y + = + + v

or

( ) ( )

( )

2

2 3river river rim rivermin min

river min

1 1 kg m10 2.95

2 2 m s

4.35 kPa

P P P

P

3

= + = +

= +

v

The additional pressure required to achieve the desired flow rate is

P =4.35kPa

9.57 (a) For upward flight of a water-drop projectile from geyser vent to fountain-top, ( )2 20 2y y ya y= + v v , with y

= 0, when y = ymax, gives

( ) ( )( )20 max0 2 2 9.8 m s 40.0 m 28.0 m sy ya y= = =v

Page 9.39


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Chapter 9

(b) Because of the low density of air and the small change in altitude, atmospheric pressure at the fountain top

will be considered equal to that at the geyser vent. Bernoullis equation, with top= 0, then gives

( )2vent top vent1

02

g y y = + v

or

( ) ( )( )2vent top vent2 2 9.80 m s 40.0 m 28.0 m sg y y= = =v

(c) Between the chamber and the geyser vent, Bernoullis equation with chamber0 yields

( ) 2atm vent ventchamber1

02

P gy P gy + + = + +v

or

( )

( )( )

2atm vent vent chamber

2

33 2

1

2

28.0 m skg m10 9.80 175 m 2.11 MPa

m 2 s

P P g y y = +

= + =

v

or

( ) ( )6 5gauge atmo 2.11 10 Pa 1 atm 1.013 10 Pa 20.8 atmospheresP P P= = =

9.58 (a) Since the tube is horizontal,y1 =y2and the gravity

terms in Bernoullis equation cancel, leaving

2 2

1 1 2 2

1 1

2 2P P + = +v v

or Figure 9.30(a)

( ) ( )31 22 22 1 2 3

2 1.20 10 Pa2

7.00 10 kg m

P P

= =

v v

Page 9.40


41/55

Chapter 9

and

2 2 2 22 1 3.43 m s =v v [1]

From the continuity equation,A11=A22, we find

2 2

1 12 1 1 1

2 2

2.40 cm

1.20 cm

A r

A r

= = = v v v v

or

2 14=v v [2]

Substituting Equation [2] into [1] yields

2 2 2

1

15 3.43 m s=vand 1= 0.478 m/s.

Then, Equation [2] gives 2= 4(0.478 m/s) = 1.91 m/s.

(b) The volume flow rate is

( ) ( ) ( )2

2 2 4 31 1 2 2 2 2 1.20 10 m 1.91 m s 8.64 10 m sA A r

= = = = v v v

9.59 From Fy= TmgFy = 0, the balance reading is found to be T= mg+ Fy where Fyis the vertical component of

the surface tension force. Since this is a two-sided surface, the surface tension force is F= (2L) and its verticalcomponent is F= (2L)coswhere is the contact angle. Thus, T= mg+ 2Lcos.

0.40 NT = when 0 = 2 0.40 Nmg L+ = [1]

0.39 NT = when 180 = 2 0.39 Nmg L = [2]

Subtracting Equation [2] from [1] gives

( )2

2

0.40 N 0.39 N 0.40 N 0.39 N8.3 10 N m

4 4 3.0 10 mL

= = =

9.60 Because there are two edges (the inside and outside of the ring), we have

Page 9.41


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Chapter 9

( )

total

22

2

= =2 ( )

1.61 10 N= = 7.32 10 N m

4 4 1.75 10 m

F F

L circumference

F

r

=

9.61 From h= 2cos/gr, the surface tension is

( ) ( ) ( ) ( )2 3 2 42

2cos

2.1 10 m 1 080 kg m 9.80 m s 5.0 10 m5.6 10 N m

2 cos0

h gr

=

= =

9.62 The height the blood can rise is given by

( )

( ) ( ) ( )2 2 62 0.058 N m cos 02 cos

5.6 m1 050 kg m 9.80 m s 2.0 10 m

hgr

= = =

9.63 From the definition of the coefficient of viscosity, = FL/A, the required force is

( ) ( ) ( ) ( )3 23

1.79 10 N s m 0.800 m 1.20 m 0.50 m s8.6 N

0.10 10 m

AF

L

= = =

v

9.64 From the definition of the coefficient of viscosity, = FL/A, the required force is

( ) ( ) ( ) ( )3 23

1 500 10 N s m 0.010 m 0.040 m 0.30 m s0.12 N

1.5 10 m

AF

L

= = =

v

9.65 Poiseuilles law gives( ) 41 2

8

P P Rflow rate

L

=

and P2= Patmin this case. Thus, the desired gauge pressure is

Page 9.42


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Chapter 9

( ) ( )( )( )

( )

2 5 3

1 atm 442

8 0.12 N s m 50 m 8.6 10 m s8

0.50 10 m

L flow rateP P

R

= =

or

61 atm 2.1 10 Pa 2.1 MPaP P = =

9.66 From Poiseuilles law, the flow rate in the artery is

( ) ( ) ( )( ) ( )

434

5 3

3 2 2

400 Pa 2 .6 10 m 3.2 10 m s

8 8 2 .7 10 N s m 8.4 10 m

P Rflow rate

L

= = =

Thus, the flow speed is

( )

5 3

23

3.2 10 m s1.5 m s

2.6 10 m

flow rate

A

= = =

v

9.67 If a particle is still in suspension after 1 hour, its terminal velocity must be less than

( ) 5max

cm 1 h 1 m5.0 1.4 10 m s

h 3 600 s 100 cmt

= = v .

Thus, from 1= 2r2g(f)/9, we find the maximum radius of the particle:

( )

( )

( ) ( )( ) ( )

maxmax

3 2 5

62 3

9

2

9 1.00 10 N s m 1.4 10 m s 2.8 10 m 2.8 m2 9.80 m s 1 800 1 000 kg m

t

f

rg

=

= = =

v

9.68 From Poiseuilles law, the excess pressure required to produce a given volume flow rate of fluid with viscosity

through a tube of radiusRand lengthLis

Page 9.43


44/55

Chapter 9

( )4

8 L V tP

R

=

If the mass flow rate is (m/t) = 1.0 103kg/s, the volume flow rate of the water is

36 3

3 3

1.0 10 kg s1.0 10 m s

1.0 10 kg m

V m t

t

= = =

and the required excess pressure is

( ) ( ) ( )

( )

3 2 6 3

54

3

8 1.0 10 Pa s 3.0 10 m 1.0 10 m s1.5 10 Pa

0.15 10 mP

= =

9.69 With the IV bag elevated 1.0 m above the needle, the pressure difference across the needle is

( ) ( )( )3 3 2 31.0 10 kg m 9.8 m s 1.0 m 9.8 10 PaP gh = = =

and the desired flow rate is

( )( )

3 3 6 3

7 3500 cm 1 m 10 cm

2.8 10 m s30 min 60 s 1 min

V

t = =

Poiseuilles law then gives the required diameter of the needle as

( )

( )

( ) ( ) ( )( )

1 41 4 3 2 7 3

3

8 1.0 10 Pa s 2.5 10 m 2.8 10 m s82 2 2

9.8 10 Pa

L V tD R

P

= = =

or

44.1 10 m 0.41 mmD = =

9.70 We write Bernoullis equation as

22out out out in in in

1 1

2 2P gy P gy + + = + +v v

Page 9.44


45/55

Chapter 9

or

( ) ( )22gauge in out out in out in1

2P P P g y y

= = + v v

Approximating the speed of the fluid inside the tank as in0, we find

( ) ( ) ( )( )23 3 2

gauge

11.00 10 kg m 30.0 m s 9.80 m s 0.500 m

2P

= +

or

5gauge 4.55 10 Pa 455 kPaP = =

9.71 The Reynolds number is

( )( )( )3 23 2

1 050 kg m 0.55 m s 2 .0 10 m

2 .7 10 N s m

dRN

= = =

v

4.3 103

In this region (RN > 3 000), the flow is turbulent.

9.72 From the definition of the Reynolds number, the maximum flow speed for streamlined (or laminar) flow in this

pipe is

( ) ( )( )

( ) ( )

3 2max

max 3 2

1.0 10 N s m 2 0000.080 m s 8.0 cm s

1 000 kg m 2.5 10 m

RN

d

= = = =

v

9.73 The observed diffusion rate is 8.0 1014kg/15 s = 5.3 1015kg/s. Then, from Ficks law, the difference in

concentration levels is found to be

( )

( )( )

( ) ( )

2 1

15

3 3

10 2 4 2

5.3 10 kg s 0.10 m1.8 10 kg m

5.0 10 m s 6.0 10 m

diffusion rate LC C

D A

=

= =

9.74 Ficks law gives the diffusion coefficient asD = diffusion rate/A.(C/L), where C/Lis the concentration gradient.

Page 9.45


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Chapter 9

Thus,

( ) ( )

1510 2

4 2 2 4

5.7 10 kg s9.5 10 m s

2.0 10 m 3.0 10 kg mD

= =

9.75 Stokess law gives the viscosity of the air as

( ) ( )

135 2

6 4

3.0 10 N1.4 10 N s m

6 6 2 .5 10 m 4.5 10 m s

F

r

= = =

v

9.76 Using 1=2r2 g(f)/9, the density of the droplet is found to be = f+ 9t/2r

2g. Thus, if r= d/2 = 0.500

103m t= 1.10 102m/s and when falling through 20 C water (= 1.00 103N.s/m2) the density of the oil is

( ) ( )( ) ( )

3 2 2

3 323

4 2

9 1.00 10 N s m 1.10 10 m skg1 000 1.02 10 kg mm 2 5.00 10 m 9.80 m s

= + =

9.77 (a) Both iron and aluminum are denser than water, so both blocks will be fully submerged. Since the two blocks

have the same volume, they displace equal amounts of water and the buoyant forces acting on the two blocks

are equal.

(b) Since the block is held in equilibrium, the force diagram at the right shows that

0yF T mg B = =

The buoyant force B is the same for the two blocks, so the spring scale

reading T is largest for the iron block, which has a higher density, and

hence weight, than the aluminum block.

(c) The buoyant force in each case is

( ) ( ) ( ) ( )3 3 3 2 3water 1.0 10 kg m 0.20 m 9.8 m s 2.0 10 NB V g= = =

For the iron block:

( ) ( ) ( ) ( )3 3 3 2iron iron 7.86 10 kg m 0.20 m 9.8 m sT V g B B= =

or

Page 9.46


47/55

Chapter 9

4 3 3iron 1.5 10 N 2.0 10 N 13 10 NT = =

For the aluminum block:

( ) ( ) ( ) ( )3 3 3 2

aluminum aluminum 2.70 10 kg m 0.20 m 9.8 m sT V g B B= =

or

3 3 3aluminum 5.2 10 N 2.0 10 N 3.3 10 NT = =

9.78 In going from the ocean surface to a depth of 2.40 km, the increase in pressure is

( ) ( ) ( )3 2 2 3 70 1.025 10 kg m 9.80 m s 2.40 10 m 2.41 10 PaP P P gh = = = =

The fractional change in volume of the steel ball is given by the defining equation for bulk modulus,

P=B(V/V), as

74

10steel

2.41 10 Pa1.51 10

16.0 10 Pa

V P

V B = = =

9.79 (a) From Archimedess principle, the granite continent will

sink down into the peridotite layer until the weight

f the displaced peridotite equals the weight of the continent.

Thus, at equilibrium,

( ) ( )g pAt g Ad g =

or g pt d = .

(b) If the continent sinks 5.0 km below the surface of the peridotite, then d= 5.0, and the result of part (a) gives

the first approximation of the thickness of the continent as

( )3 3

p

3 3g

3.3 10 kg m5.0 km 5.9 km

2.8 10 kg mt d

= = =

9.80 (a) Starting with P= P0+ gh, we choose the reference level at the level of the heart, so P0= PH. The pressure at

Page 9.47


48/55

Chapter 9

the feet, a depth hHbelow the reference level in the pool of blood in the body is PF= PH+ ghH. The pressure

difference between feet and heart is then

PF= PH+ ghH .

(b) Using the result of part (a),

( ) ( )( )3 3 2 41.06 10 kg m 9.80 m s 1.20 m 1.25 10 PaF HP P = =

9.81 The cross-sectional area of the aorta is2

1 1 4A d= and that of a single capillary is 22 4cA d= If the

circulatory system hasNsuch capillaries, the total cross-sectional area carrying blood from the aorta is

22

24

c

N dA NA

= =

From the equation of continuity,

12 1

2

A A

= v

v, or

2 22 1 1

24 4

N d d =

v

v,

which gives

2 22

1 1 72 6

2 2

1.0 m s 0.50 10 m2.5 10

1.0 10 m s 10 10 m

dN

d

= = =

v

v

9.82 (a) We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then

P= P0+ gh, with the reference level at the water surface inside the straw and Pbeing atmospheric pressure

on the water in the cup outside the straw, gives the maximum height of the water in the straw as

( ) ( )

5 2

atm atmmax 3 3 2

water water

0 1.013 10 N m 10.3 m1.00 10 kg m 9.80 m s

P Phg g = = = =

(b) The moon has no atmosphere so Patm= 0, which yields .hmax= 0.

9.83 (a) ( )22 H O160 mm of H O 160 mmP g= =

Page 9.48


49/55

Chapter 9

( )33 2

kg m10 9.80 0.160 m

m s

= = 1.57 kPa

( )3

5

1 atm

1.57 10 Pa 1.013 10 PaP

= = 21.55 10 atm

The pressure is 2 2H O H O Hg HgP gh gh = =

( )22

3 3H O

Hg H O 3 3Hg

10 kg m160 mm 11.8 mm of Hg

13.6 10 kg mh h

= = =

(b) The fluid level in the tap should rise.

(c) Blockage of flow of the cerebrospinal fluid.

9.84 When the rod floats, the weight of the displaced fluid equals the weight of the rod, or displaced 0 rodfgV gV = .

But, assuming a cylindrical rod,2

rodV r L= . The volume of fluid displaced is the same as the volume of the rod

that is submerged, or ( )2displacedV r L h= . Thus, ( )2 20fg r L h g r L = , which reduces to

0f LL h

=

9.85 Consider the diagram and apply Bernoullis equation to

points A and B, takingy= 0 at the level of point B,

and recognizing that A 0. This gives

( )

2

0 sin

1 0

2

A w

B w B

P g h L

P

+ +

= + +v

Recognize that PA= PB=Patmsince both points

are open to the atmosphere. Thus, we obtain

Page 9.49


50/55

Chapter 9

( ) ( ) ( )22 sin 2 9.80 m s 10.0 m 2 .00 m sin 30.0 13.3 m sB g h L = = = v

Now the problem reduces to one of projectile motion with

0 sin 30.0 6.64 m sy B= =v v

At the top of the arc,y = 0, andy=ymax.

Then, ( )2 20 2y y ya y= + v v gives ( ) ( )( )2 2

max0 6.64 m s 2 9.80 m s 0y= + , or

max 2.25 m above the level of point By = .

9.86 When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the filled balloon

plus the weight of string that is above ground level. If msandLare the total mass and length of the string, the mass

of string that is above ground level is (h/L)ms. Thus,

air balloon balloon helium balloon s

hgV m g gV m g

L

= + +

which reduces to

( )air helium balloon balloons

V mh L

m

=

This yields

( ) ( )( )

33 31.29 kg m 0.179 kg m 4 0.40 m 3 0.25 kg2.0 m 1.9 m

0.050 kgh

= =

9.87 When the balloon floats, the weight of the displaced air equals the combined weights of the filled balloon and its

load. Thus,

air balloon balloon helium balloon loadgV m g gV m g = + + ,

or

Page 9.50


51/55

Chapter 9

( )balloon load 3 3

balloon 3air helium

600 kg 4 000 kg4.14 10 m

1.29 0.179 kg m

m mV

+ += = =

9.88

(a) Consider the pressure at points A and B in part (b) of the figure by applying P= P0+ fgh. Looking at the left

tube gives ( )atm waterAP P g L h= + , and looking at the tube on the right, atm oilBP P gL= + .

Pascals principle says that PB= PA. Therefore, ( )atm oil atm waterP gL P g L h + = + , giving

( )3

oil

3water

750 kg m1 1 5.00 cm 1.25 cm

1 000 kg mh L

= = =

(b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid

levels in the two tubes. First, apply Bernoullis equation to points A and B. This gives

2 2air air air air

1 1

2 2A A A B B BP gy P gy + + = + +v v

Since , , and 0A B A By y= = =v v v , this reduces to [1]

2air

1

2B AP P = v

Now use P= P0+ fghto find the pressure at points C and D, both at the level of the oilwater interface in the

Page 9.51


52/55

Chapter 9

right tube. From the left tube, waterC AP P gL= + , and from the right tube, oilD BP P gL= + .

Pascals principle says that PD= Pc, and equating these two gives

oil waterB AP gL P gL + = + or ( )water oilB AP P gL

= [2]

Combining Equations [1] and [2] yields

( )

( )( ) ( )

water oil

air

2 2

2

2 1 000 750 9.80 m s 5.00 10 m13.8 m s

1.29

gL

=

= =

v

9.89 While the ball is submerged, the buoyant force acting on it isB= (wV)g. The upward acceleration of the ball while

under water is

( )( ) ( )

3

33 2 2

41

3

1 000 kg m 40.10 m 1 9.80 m s 31 m s

1.0 kg 3

y wy

F B mga r g

m m m

= = =

= =

Thus, when the ball reaches the surface, the square of its speed is

( ) ( )( )2 2 2 2 20 2 0 2 31 m s 2.0 m 125 m sy y ya y= + = + =v v

When the ball leaves the water, it becomes a projectile with initial upward speed of 0 125 m sy =v and

acceleration of29.80 m sya g= = . Then, ( )2 20 2y y ya y= + v v gives the maximum height above the

surface as

( )

2 2

max 2

0 125 m s6.4 m

2 9.80 m sy

= =

9.90 Since the block is floating, the total buoyant force

must equal the weight of the block. Thus,

Page 9.52


53/55

Chapter 9

( ) [ ]

( )

oil water

wood

4.00 cm

4.00 cm

A x g A x g

A g

+ =

whereAis the surface area of the top or bottom

of the rectangular block.

Solving for the distancexgives

( ) ( )wood oilwater oil

960 9304.00 cm 4.00 cm 1.71 cm

1 000 930x

= = =

9.91 A water droplet emerging from one of the holes

becomes a projectile with 0y= 0 and 0x= 0. Thetime for this droplet to fall distance hto the floor is

found from1 2

0 2y yy t a t = +v to be

2 ht

g=

The horizontal range is

2 hR t

g= =v v .

If the two streams hit the floor at the same spot,

it is necessary thatR1=R2, or

1 21 2

2 2h h

g g=v v

With h1= 5.00 cm and h2= 12.0 cm, this reduces to

21 2 2

1

12.0 cm

5.00 cm

h

h= =v v v or

1 2 2.40=v v[1]

Page 9.53


54/55

Chapter 9

Apply Bernoullis equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is

atmospheric pressure at both points and, if the tank is large in comparison to the size of the holes,30. Thus, we

obtain

2atm 1 1 atm 3

10

2P gh P gh + + = + +v or ( )21 3 12g h h= v . [2]

Similarly, applying Bernoullis equation to point 2 (the upper hole) and point 3 gives

2atm 2 2 atm 3

10

2P gh P gh + + = + +v or ( )22 3 22g h h= v . [3]

Square Equation [1] and substitute from Equations [2] and [3] to obtain

( ) ( )3 1 3 22 2 .40 2g h h g h h =

Solving for h3yields

( )2 13

2 .40 12 .0 cm 5.00 cm2.4017.0 cm

1.40 1.40

h hh

= = = ,

so the surface of the water in the tank is 17.0 cm above floor level.

9.92 When the section of walkway moves downward distance L, the cable is stretched distance Land the column is

compressed distance L. The tension force required to stretch the cable and the compression force required to

compress the column this distance is

st eel c abl ecable

cable

Y A LF

L

= and

Al columncolumn

column

Y A LF

L

=

Combined, these forces support the weight of the walkway section:

cable column 8 500 NgF F F+ = = orsteel cable Al column

cable column8 500 N

Y A L Y A L

L L

+ =

giving

Page 9.54


55/55

Chapter 9

steel cable Al column

cable column

8 500 NL

Y A Y A

L L

=+

The cross-sectional area of the cable is

( )2

22

cable

1.27 10 m

4 4

DA

= =

and the area of aluminum in the cross section of the column is

( ) ( ) ( )2 2

2222outer innerouter inner

cable

0.162 4 m 0.161 4 m

4 4 4 4

D DD DA

= = =

Thus, the downward displacement of the walkway will be

( ) ( )( )

( ) ( ) ( )

( )

2 22 1010 2

8 500 N

7.0 10 Pa 0.162 4 m 0.161 4 m20 10 Pa 1.27 10 m

4 5.75 m 4 3.25 m

L

=

+

or

48.6 10 m 0.86 mmL = =

chapter 9 solutions to serway's college physics

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