1

PROBLEM:Two forces of 700 N each applied 500 mm from the axis of the shaft, is transmitted to a 100 mm diameter shaft by means of a shear key, 30 mm long and 10 mm wide and 15 mm deep in cross section. What is the shearing stress of the key?Solution:

shearing stress= PA

since, Torque=Pk (radius of shaft )

700 (500+500 )=Pk (50 )

Pk=14000 N

and, A=(10 ) (30 )=300mm2

therefore, shearing stress=14000 N

300mm2

shearing stress=46.67 MPa

PROBLEM:A cylindrical pressurized air freshener container of 100 mm height and 50 mm diameter has flat ends and carries an allowable stress of 150 MPa. How thick is the container if it is subjected to a maximum internal pressure of 200 MPa?Solution:

σ= pDL2t (D+L )

t= pDL2σ (D+L )

t=0.200 (50 )(100)2 (150 )(50+100)

t=0.02mm

PROBLEM 3:A 25 mm diameter bolt is subjected to a tensile load and is allowed to a maximum tensile stress and maximum shear stress of 160 MPa and 120 MPa, respectively. How much is the safe tensile load if its nuts is 10 mm thick and diameter and pitch at the root of the thread is 20 mm and 2 mm, respectively?Solution:P shearing:

P= (shearing stress ) A

P= (120 ) (10 )(20)π

P=75398.22 MPa

Note: 1 MPa=1N

mm2

P bearing:P= (shearing stress ) A

P= (160 ) (10 )(25)π

P=125663.71 MPa

Therefore, Safetensile load=Pbearing−Pshearing

Safetensile load=125663.71−75398.22

Safetensile load=50265.49MPa

QUESTION:Refers to the measurement based on qualitative law of chemical combination

2

ANSWER: Stoichometry

PROBLEM:A “point-souce” unshaded electric lamp has a luminous intensity of 100 candle 4 ft on top of the table. What is the illumination of the table?Solution:

E= I

s2

E=100candle

(4 ft )2

E=6.25candle / ft2

PROBLEM:A steel wire of 100 mm diameter is suspended at one end and has no load except its own weight. What is the maximum shearing stress if its length is 2m? Assume specific gravity of steel to be 7.85.Solution:

maximum shearingstress=γLwhere,

γ=density of wire

γ=density of water x specific gravity

L=length

therefore,

maximum shearingstress=(9.81kN

m3 ) (7.85 )(2m)

maximum shearingstress=154.017kPa

QUESTION:What is the mole fraction of the solute in a solution of 40% ethanol in water if one mol of ethanol is 46 gram/mol while water is 18 gram/mol?ANSWER: 0.21 mol

PROBLEM:A steel wire of 100 mm diameter is suspended at one end and has no load except its own weight. What is the maximum length if its maximum shearing stress is 100 kPa. Assume specific gravity of steel to be 7.85.Solution:

maximum shearingstress=γL

100kN

m2 =(9.81kN

m3 ) (7.85 )L

L=1.30m

PROBLEM:A 22.5 mL sample of 5.20 M solution of nitric acid is diluted to 40.5 mL. What is the molarity of the diluted solution?Solution:

M 1V 1=M2V 2

M 2=M 1V 1

V 2

M 2=(5.20M )(22.5mL)

40.5mL

M 2=2.89 M

PROBLEM:A container contains three gases: oxygen, carbon dioxide and helium. The partial pressures of the gases are 2.0 atm, 3.0 atm and

3

4.0 atm, respectively. Using Dalton’s law of partial pressure, what is the total pressure in the container?Solution:

PT=P1+P2+P3

PT=2.0+3.0+4.0

PT=9.0atm

PROBLEM:

A ray of light in water (nw=43 ) is incident upon the plate of a crown

glass (ng=1.517 ) at the angle of 45°. What is the angle of refraction for the ray of the glass? What is the index of the glass in reference to water?Solution:

n1 sin i=n2 sin r

r=sin−1( n1 sin i

n2)

r=sin−1( 43

sin 45 °

1.517 )r=38.43°

nr=n2

n1

nr=1.517

43

nr=1.14

PROBLEM:How much heat must be supplied to a pan of 500 g to increase the temperature from 20°C to 100°C if the pan is made up of aluminum? Specific heat capacity of aluminum is 0.22 cal/g-°C.Solution:

Q=mc∆T=500 (0.22 ) (100−20 )

Q=8800cal

PROBLEM:Mr. Reyes would like to earn 7% of his income after payment of his taxes due. If the income of his investment will be taxed 42%, what is the minimum rate of return, before payment, must his investment has to attain his goal?Solution:

Let: P = profitC = investment capital

rate of return= PC

0.07=P−0.42 PC

0.07=P (1−0.42)

C

0.1207= PC

therefore,

4

rate of return= PC

=12.07 %

PROBLEM:A 1.0 kg ball falls from a height of 1.0 m. What is the velocity as the ball hits the floor?Solution:

v=√2gh=√2 (9.81 ) (1 )

v=4.43m / s

PROBLEM:A flywheel rotating at 200 rev/min slows down at a constant rate of 2 rad/s2. What is the time when the flywheel stops and how many revolutions were made during the process?Solution:

ω−ωo=αt

but, ωo=200revmin ( 2π rad

1 rev )( 1min60 sec )=20.94

therefore, 0−20.94=(−2 ) t

t=10.47 sec

ω2−ωo2=2∝θ

02−20.942=2 (−2 )θ

θ=110rad ( 1 rev2π rad )

θ=17.51rev

PROBLEM:A flywheel in a form of uniform disc with a diameter of 4 ft, weighs 600 lb. What is the angular acceleration when the disc is acted upon a net torque of 225 lb-ft?Solution:

m=Wg

=60032

=18.75 slug

For disc: l=12m R2=1

2(18.75 ) (2 )2=37.50

Since, L=lα

α= Ll= 225

37.50

α=6 rad /s2

PROBLEM:A solid cylinder of 30 cm diameter at the top of incline 2 m high is released and rolls down without loss of energy due to friction. What is the linear and angular speed at the bottom?Solution:

v=√ 43gh=√ 4

3( 9.81 ) (2 )

v=5.11m /s

ω= vR

=5.110.15

ω=34.07 rad / s

PROBLEM:A heat engine operates with a supplied heat of 65 kcal and exhausts 40 kcal. How much work is done by the engine?Solution:

5

W=J (QH−QL)=4184J

kcal(65−40 )

W=104600J

PROBLEM:The human ear can distinguish reflected sound pulse from the original sound pulse if 0.10 sec or more elapses between the two sounds. What is the minimum distance to a reflecting surface from which we can hear an echo if the speed of sound is 343 m/s?Solution:

d=vt=343 (0.10 )=34.3m

dmin=34.3

2

dmin=17.15m

PROBLEM:Electron is held in the orbit by the force of electrical attraction at a typical distance of 1x10-10 m. What is the force of electrical attraction between the proton and electron?Solution:

F=kQ1Q2

d2 =9×109 N m2

c2 [ (1.6×10−19 C ) (1.6×10−19C )

(1×10−10m )2 ]F=2.304×10−8 N

( pakicheck kasiung sagot dun×10−11)

PROBLEM:What is speed of light in diamond? Index of refraction of diamond is 2.42.Solution:

v= cn=3×108

2.42

v=1.24×108m / s

PROBLEM:How much rate an elevator accelerates upward to cause a 80 kg man to exert a pressure of 1200 N at the floor?Solution:

P=W (1+ ag )

but, W=mg=80 kg (9.81 )=784.8 N

therefore, 1200 N=784.8 N (1+ a9.81 )

a=5.19m /s

PROBLEM:A car with a mass of 1500 kg moves along a uniform curve at a constant speed of 72 kph. The coefficient of friction between the tire and road is 0.65. What is the minimum radius for the car not to skid?Solution:

∑ F x=0 :man=0.65 N

∑ F y=0:W=N=mg

man=0.65mg

an=0.65 g= v2

r

6

but, v=72kmhr ( 1000

3600 )=20

therefore, r= v2

0.65 g=

(20)2

0.65(9.81)

r=62.73m

PROBLEM:Henry’s constant of carbon dioxide in water at 25°C is 0.031. What is the solubility of carbon dioxide in a bottle of cola drink if the pressure of CO2 is 4 atm over the liquid at the same temperature?Solution:

S=kP=0.031(4 )

S=0.124

PROBLEM:A Van de Graaff generator accelerates electron such that it has a potential difference of 150 megavolts. How much work is done on the electron by the machine?Solution:

W=qV= (1.6×10−19 ) (150×108 )=2.4×10−9 J

PROBLEM:A charge A of +250 statcoulomb is placed in between 2 charges B of +50 statcoulomb and C of -300 statcoulomb. A is 5 cm from B and 10 cm from C. What is the force on A?Solution:

F1=kQ 1Q2

d2 =1dyne−cm2

statcoulomb2 [ (250 ) (50 )(5 )2 ]=500 N

F2=kQ 1Q2

d2 =1dyne−cm2

statcoulomb2 [ (250 ) (300 )(10 )2 ]=750 N

F A=F1+F2=500+750

F A=1250 N

PROBLEM:A two charge of 50 µC and 10µC is 10 cm apart. What is the force between the two charges in a vacuum?Solution:

F=kQ1Q2

d2 =9×109[ (50 μ ) (10μ )(0.1 )2 ]

F=450 N

PROBLEM:Two balls are projected simultaneously at the same velocity at the top of the building, one vertically upward and the other vertically downward. If they reach the ground 15 sec. and 20 sec. respectively, find the time for the balls to reach the ground if they simply dropped at the top of the building.Solution:

t=√ t1 t 2=√ (15 ) (20 )

t=17.32 sec .

PROBLEM:An object is projected such that its horizontal distance is thrice the maximum height. What is the angle of projection?Solution:

R=vo

2

gsin 2θ

h=v o

2

2 gsin2θ

but, R=3h

7

therefore, vo

2

gsin2θ=

3 vo2

2gsin2θ

sin 2θ=32

sin2θ

2 sinθ cosθ=32

sin2θ

21.5

= sinθcosθ

=tanθ

θ=tan−1( 21.5 )

θ=53.13°

PROBLEM:What is the molality of the solution containing 75.5 g sucrose (molar mass = 342 g/mol) in 400.0 g of water?Solution:

Molality=N solute

Mass solvent=

75.5g342g /mol

0.4 kg

Molality=0.5519M

PROBLEM:Two apples were thrown by 2 boys at the same point, one with an angle of 60° and the other of 30°. What is the ratio of their velocities if they have the same horizontal distance?Solution:

R=vo

2

gsin 2θ

but, R1=R2

therefore, v1

2

gsin 2(30)=

v22

gsin 2(60)

( v1

v2)

2

= sin 120sin 60

=1

v1

v2

=√1=1

PROBLEM:The man on a ledge of the 5th floor throws a ball vertically upward on the top of the building. On its way up, the time the ball to reach the 7th floor is 1 sec. and on its way down, the time of the ball to reach back the 7th floor is 1.5 sec. What is the distance of the 5th and 7th floor of the building?Solution:

h=12g t 1t 2=

12

(9.81 ) (1 ) (1.5 )

h=7.35m

PROBLEM:A lift engine raises a 500 mass at a distance of 10 m at 5 sec. how much power is needed?Solution:

P=mght

=500 x 9.81x 105

=9810 watts

PROBLEM:What is the volume of 1.27 mole of helium gas at STP? Gas constant = 0.0821 L-atm/mol-K.Solution:

PV=nRT

8

V=1.27mol (0.0821

L−atmmol−K )(273 )

1atm

V=28.46LPROBLEM:A hotel invested an additional of Php 1,800,000 for fitness gym for an additional attraction to client. They expect an annual gross revenue of Php 500,000 and will incur annual expense of Php 50,000 for the operation and maintenance of the gym. What is the payback period?Solution:

Payback Period=Total InvestmentNet Annual Cost

= 1,800,000500,000−50,000

Payback Period=4 years

PROBLEM:During a race, a sports car has 250 Hp at 4000 rpm. What is the torque exerted?Solution:

torque=Pω

=(250 Hp )(550

lb− fts

Hp )(4000 rpm )( 2πrad

rev )( min60 sec )

torque= 328.26 lb-ft

PROBLEM:

40 g of methanol is mixed with 60 g of ethanol. One mole methanol and ethanol has a mass of 32 g and 46 g respectively. Find the mole-fraction of the solvent. (Solvent: Ethanol)Solution:

MoleEthanol=60g

46 g/mol=1.3mol

MoleMethanol=40g

32g /mol=1.25mol

MoleFraction=M Ethanol

M Methanol+MEthanol

= 1.31.25+1.3

MoleFraction=0.51M

PROBLEM:The machine cost Php 45,000 with a book value of Php 4,350, if it retires at the end of 6 years, what is the annual depreciation?Solution:

k=1− n√ Cn

Co

=1−6√ 4,35045,000

k=0.3225

PROBLEM:What is the molarity of a solution with 400.0 mL solution containing 8.0 g NaOH (Molar mass = 40 g/mol)?Solution:

Molarity=

8.0 g40g /mol

0.4 L

Molarity=0.5 M

PROBLEM:

9

What is the molarity of a solution with 3.0 mole of solute dissolved in 2000 mL of solution?Solution:

Molarity=3.0mol2 L

=1.5 M

PROBLEM:Determine the exact simple interest of Php 5,000 invested from period of January 16, 1900 to October 12, 1900, if the annual rate is 18%.Solution:

To determine if leap year, 1900400

=4.75, not exact so not leap year

I=Pin=Pi ( d365 )

January 16 = 15 May = 31 September = 30

February = 28 June = 30 October 12 = 12

March = 31 July = 31 d = 270

April = 30 August = 31

I=Pin=5000 (0.18 )( 270365 )

I=665.75

PROBLEM:A 20,000 railroad car coast 3 m/s collides and hit an identical car at rest. What will be the combined velocity?Solution:

m1 v1+m2 v2=(m1+m2 ) v(20 000 ) (3 )+(20 000 )(0)=(20 000+20 000 ) v

V = 1.5 m/s

PROBLEM:If a man invested 10,000 at 15% for 5 years, how much the money in peso will be account at maturity with annual inflation rate of 6%?Solution:

F=P(1+i )n

(1+r )n=10,000

(1+0.15 )5

(1+0.06 )5

F=15,030

PROBLEM:A solid cylindrical shaft has a torque of 20 kN-m which is 5 m long. What is the maximum torsional stress so that it will not twist more than 2 degrees? G = 80 GPa.Solution:

θ=2( π180 )= π

90

J= TLθG

=(20kN−m ) (5m )

(80×109 kNm )( π

90 )=27.93= π

32d4

therefore, d=0.246m;r=0.123m

δmax=TrJ

=(20kN−m ) (0.123 )

27.93

δmax=88.07

PROBLEM:A helical spring with mean coil diameter of 150 mm is consists of 14 turns of 20 mm diameter. What is the maximum shear stress with a force equal to 5 kN? Use G = 84 GPa.Solution:

τ=16 PR

π d3=

16 (5000 N ) (75mm )π (20mm )3

10

τ=238.73

PROBLEM:Two moles of ideal gas is to be compressed slowly from 4 to 1 ft3. How much work is done?Solution:

W=2.303nRT logV 1

V 2

W=2.303 (2mol )(8.317J

mol−K ) (300 K ) log41

W=6900J

PROBLEM:A helical spring has a mean coil diameter of 150mm and consists of 14 turns of 20mm diameter rod. What is the deflection stress with a force in it equal to 5kN. Use G = 84GPa.Solution:

δ = (64PR3n)/(Gd4)δ = [(64)(5000N)(75mm)3(14)]/ [(84,000MPa)(20mm)4]δ = 141mm

PROBLEM:How much heat is needed to melt 5 kg of ice at 0 0C?Solution:

Q=mLf=(5kg )(80calkg )=400cal

Q= (400cal )(4.184Jcal )

Q ¿1673.6J

PROBLEM:What must be the length of a pendulum in feet with a period equal to 2s.Solution:

L = (gT2)/(4π2)L= [(9.81)(22)] / [(4π2)] = 0.99m (just convert to ft.)

PROBLEM:What is the interest arte compound daily of 10% effective?Solution:

Effective Rate=(1+i )m−1

0.10=(1+i )365−1

1.10=(1+ i)365

1.101

365=(1+i )

i=0.000261 %

PROBLEM:A plano-convex lens of focal length of 12 cm is to be made a glass of refractive index of 1.50. What is the radius of curvature of the surface?Solution:

112

=( 1.501.00

−1)( 1R

− 1∞ )

6 cm

PROBLEM:What is the power of diverging lens whose focal length is 20cm?Solution:

p=1f= 1

0.20m=5.0 diopters

PROBLEM:What is the accumulated amount P10,000 after 10 years if invested 3% compounded continuously?

11

Solution:F=Pe(NR)N=10,000e0.03 (10)=P13,498.59

PROBLEM:At what height above the surface of the earth will your weight be half of your weight in the surface of the earth?Solution:

h=3960√ WW2

h=5600 mi

PROBLEM:A gas sample has a volume of 50 ml at 25 0C at 1 atm pressure. If the pressure remains constant, what will be the volume at 0 0C?Solution:

V 1

T 1

=V 2

T2

50ml25+273

=V 2

0+273V2 = 45.8 ml

PROBLEM:A steam engine receives steam from a boiler at 180 0C and exhausts in air at 100 0C. What is the efficiency?Solution:

efficiency=(180+273 )−(100+273)

(180+273)x100 %

efficiency=17.6 %

PROBLEM:What is the energy of a photon of red light with a freq. 4.00x1014Hz? Planck’s constant h = 6.63x10-34Js.Solution:

E = hf = (6.63x10-34Js)( 4.00x1014/s)E = 2.62 x10-19J

PROBLEM:The steel rail …, cannot expand. What is the stress in the rail with a change of temperature equal to 20°C.? Coefficient of thermal expansion is equal to (nklimutan ko yung value). Modulus of rigidity of steel = 200GPa.Solution:

E = 200GPa = 200,000MPa

δ = ά L ∆Tδ = ( coef of thermal expansion)(L)( 20°C)δ = value in terms of L

ε = δ / L where: δ is from the 1st soln. (value in terms of L) therefore: L will be cancelled out.

σ = εEσ = _____ (200, 000 MPa)σ = _____ MPa

PROBLEM:The F. pendulum swings to a height of 3 in above the ground which is its max. pt. and it touches the ground at its min. What is its max. velocity?Solution:

KEgain =PElost

12

½ mv2 = mghv = √2ghv = √2(32ft/s)[(3in)(1ft/12in)]v = 2m/s

PROBLEM:How much is supplied to 100 g of boiling water to change it to water vapor at 100 0C?Solution:

Q=mLV= (0.1kg )(540calkg )=54cal

Q= (54 cal )(4.184Jcal )

Q ¿226 J

PROBLEM:A solid cylindrical shaft has a torque of 20kN-m which is 5m long. What is the max. diameter required so that it will not twist more than 2 degrees?Solution:

θ = 2 (π /180) = π / 90

θ = (TL) / (GJ)π / 90 = (20kN-m)(5m) / (80x106kN/m) JJ=27.93x103

J = (π/32) (d4)27.93x103 = (π/32) (d4)d = 23.1m (pero pagka2tanda kong answer is 0.246m)

PROBLEM:Human eye is most sensitive to light with wavelength 5.60x10-7m, which is the yellow green color. What is its frequency?Solution:

λ = c/ff = c/ λf = (3x108m/s) / (5.60x10-7m)f = 5.36 x1014Hz

PROBLEM:What is the wavelength of violet line (n=6) in the hydrogen line spectra accdg. to Balmer’s eqn.? Constant R = 1.097x107

Solution:1/ λ = R [(1/22)-(1/n2)]1/ λ =(1.097x107 )[(1/22)-(1/62)]λ= 410.21 x10-9m

PROBLEM:What is the Broglier wavelength of the baseball, 0.150kg and with velocity equal to 50m/s? Planck’s constant h = 6.63x10-34Js.Solution:

λ= h /mvλ= 6.63x10-34Js / (0.150kg)(50m/s)λ= 8.84 x10-35m

PROBLEM:A 60lb. box is being pulled through an incline plane 5oft long and 30ft high by a man exerting a force of 45lb along the plane. Solution:a. What is the efficiency?

EFF=(60lb)(30ft)/(45lb)(50ft) x 100%EFF= 0.80 x 100% = 80%

b. What must be the force of the man if plane is frictionless?

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F = (45lb)(0.8)F = 36lb

PROBLEM:Mr. Dela Cruz borrow P2,000 from Mr. San Juan and promise to pay in one year. Mr Dela Cruz received P1920 after Mr. San Juan collects advance interest P80. What is the rate of discount?Solution:

rate ofdiscount= total interestamount borrowed

= 802000

=0.04=4 %

PROBLEM:A car and a truck travelling at the same direction at a constant velocity of 25m/s and 40m/s, respectively. Determine the distance between the car and truck after 8sec at the instant that they are 400m apart.Solution:Since they were travelling at constant velocity, acceleration=0.

Distance at 400m apart:SCAR=0STRUCK=400m

After 8 seconds:SCAR=0+25 (8 )=200

STRUCK=400+40 (8 )=720

Distance between the truck and car after 8 seconds:D=720−200D=520m

PROBLEM:

A gas sample has a volume 10mL, 60°C and 567 mmHg. What is the volume of this gas at STP?Solution:

V 1P1

T 1

=V 2P2

T 2

V 2=10mL (567mmHg)(273K )760mmHg(273 K+60 K)

V 2=6.21mL

PROBLEM:The bases of trapezoid are 3m and 5m respectively. It has an altitude of 4m. Where is the center of gravity measured from the 5m base?Solution:

y=(2a+b )a+b (h3 )

y=(2 (3 )+5 )3+5 (4

3 )y=1 . 83m

PROBLEM:

14

A man has a concave shaving mirror whose focal length is 20in. how far is his face from the mirror in order to have a two-fold magnification.Solution:

M = q/p = 2

q = -2p

(1/p) + (1/q) = 1/f

(1/p) + (1/(-2p)) = 1/20

p = 10inch

PROBLEM:A light passes through a 60°00’ prism has an angle of deviation 51°20’. What is the angle of refraction of the glass?Solution:

n=1.00( sin12(60 ° 00 ’+51 °20 ’ )

sin12(60° 00 ’) )

n=1.65

PROBLEM:A basketball is drop at a height of 10m on a horizontal ground. The ball rebounds 7m high. Find the coefficient of restitution.Solution:

PROBLEM:A lathe machine was purchased P100,000. Annual cost of P18,000. Using interest of 8%, what is the depreciation rate?Solution:

d=C0+Ai=100,000+ 18,000

0.08=325,000

PROBLEM:A steel tape 100m long is 2mm too short. The effective dimension of the steel tape is 40mm, 20mm thickness, and 20mm width. How much tensile pull should be applied at the edge of the steel tape to have 100m distance? The modulus of elasticity for steel is 200000MPa. Solution:

e2=h2

h1

e=√710

e=0 . 837

15

δ=PLAE

P=δ AEL

P=(2 ) ( 40x 20 ) (200000 )(100 x 1000 )

P=3200 N=3 .2kN

PROBLEM:What is the effective force constant of a spring of a 1500 lb car with a period of vertical oscillation of 2sec?Solution:

k=4 π 2WgT 2

k=4 π 2 (1500 )(32.2 ) (2 )2

k=459 . 76

PROBLEM:A 40m hemispherical cylinder is created to flange the tanks. How many 25mm diameter bolt should be used to withstand a tensile strength of 85MPa with an internal pressure of 200KPa?Solution:

p=200KPa=0 . 2 MPa

F=pπD2

4

F=0 .2π (4 . 0 x1000 )2

4F=2513274 NF=2 .51 N

σ=FnA

n=F

σ (πD2

4 )n=2513274

85π4

(25 )2

n=60 .2353n=61bolts

PROBLEM:One of the steel supports of the San Francisco Bridge has a height of 150m. The upper 50m has a cross sectional area of 1 square meter, the remaining has a cross sectional area of 1.25 square meter. It

16

supports an axial compressive load of 200MN. What is the unloaded height of the steel support? Modulus of Elasticity of steel is 20GPa.Solution:

Upper 50m:

δ=PLAE

δ=(200 x 106 ) (50 )(1 ) (200 x109)

δ=0 .05m

Lower 100m:

δ=PLAE

δ=(200 x 106 ) (100 )(1 .25 ) ( 200x 109 )

δ=0 .08m

Unstrained height H:H=150+0 .05+0 .08H=150 . 13m

PROBLEM:A drum diameter of 300mm has a sheared key of with a diameter of 50mm. The force is 20kN. Determine the shearing stress of the key.

Solution:

T 2

T 1

=eμπ

20kNT 1

=e0 .3 π

T 1=7 .79kN

∑MCENTER=0

P (25 )+T 1 (150 )=T 2 (150 )P (25 )+7 .79 (150 )=20 (150 )P=73. 26kN

Shearing Stress=PA

Shearing Stress=73 . 26kN

(π (25 )2

4 ) Shearing Stress=149 .24 MPa

PROBLEM:What is the critical angle between carbon disulfide and air?

17

Solution:

sin ic=n2

n1

ic=sin−1( 1.001.643 )

ic=37.49

PROBLEM:A water tank has a small window which is a part of spherical tank 6 inches radius. The convex side of the window is on the outside of the tank. A small light is place 30 inches from the window. What is the image (a) outside (b) inside the tank?Solution:a)

n1

p+n2

q=

n2−n1

R

1.00030

+ 1.333q

=1.333−1.0006

q=60∈¿

Image is real and inside the water

b)n1

p+n2

q=

n2−n1

−R

1.33330

+ 1.000q

=1.000−1.333−6

q=90∈¿

Image is real and outside the water

PROBLEM:A car enters a curve of radius 300m at a velocity of 60kph. If the car accelerate at 3m/s2, what is the total acceleration of the car (in meter per square second) when it travels 300m?Solution:

a=√(a¿¿n)2+(a¿¿ t )2 ¿¿

V 2=V i2+2aS

V i=60 kph( 1000m1km

x1hr

3600 s )V i=16.667 m/ s

V=√16.6672+2 (3)(300)=45.58m /s

an=V 2

r=45.582

300=6.925m /s2

a=√6.9252+32=7.547m /s2

PROBLEM:An equipment has a salvage value of P10,000.00 for 10yrs. . What is the first cost of the equipment if it is depreciated 0.8%.Solution:

Let d – depreciation annually

(Depreciation rate) = d / Co

d = (depreciation rate)Co equation 1

d = (Co – Cn) / n equation 2

solving for Co:

(0.08)Co = (Co – 10,000) / 10

18

Co = P5,000

PROBLEM:A sticky object has a mass of 5kg, moving to the left at a speed of 12m/s. it strikes another object with a mass of 4kg moving to the right at a speed of 10m/s. if the two object lump together after collision, how much is the kinetic energy dissipated.Solution:

KE1 = (1/2)m1v12 + (1/2)m2v2

2

KE1 = (1/2)(5)(-12)2 + (1/2)(4)(10)2

KE1 = 560

KE2 = (1/2)MV2

(1/2)MV2 = (1/2)m1v12 + (1/2)m2v2

2

MV = m1v1+m2v2

(5+4)V = (5)(-12) + (4)(10)

V = -2.22

KE2 = (1/2)MV2

KE2 = (1/2)(5+9)(-2.22)2 = 22.1778

KEDISSIPATED = KE1 – KE2

= 560 – 22.1778

= 537.822

QUESTION:Efren bata Reyes cue a ball and rolls to the billiard table and touch the object ball no.9. if the cues ball stops rolling after collision, what will be the velocity of the object ball after collision?

ANSWER: the same initial velocity of the cue ball before collision

PROBLEM:An angle of reflection is 30 degrees, from glass to air. What is the angle of incident if the index of refraction of glass is 1.50?Solution:

n1sin1θ = n2sin2θ

sin1θ = (n2sin2θ) / n1

sin1θ = (1.5)sin30 / 1.00

θ = 48.59 degrees

PROBLEM:Tiger Woods hit a golf ball at a velocity of 80m/s. the head of the club was in contact to a 50g at 0.04ms. What is the average force when the ball was hit?Solution:

F = (∆mv)/∆t

F= (80m/s × 0.05kg)/(0.0004sec)

F= 10,000N = 10kN

PROBLEM:A spherical shell of a small electrostatic generator has a charge of 2.0microColoumb. What is the potential difference at the point 10cm from the center of the sphere?Solution:

V = k(Q/S2)

V = 9×109 x 2×10−6

(0.10)2

19

V = 200microVolt

PROBLEM:An object is placed 20in of a certain lens. A virtual image was formed at 10in. determine the focal length of the lens.Solution:

(1/p) + (1/q) = 1/f

(1/20) + (1/(-10)) = 1/f

f= 20 inch

PROBLEM:A candle is held 3.0 in of a concave mirror whose radius is 24in. where is the image of the candle.Solution:

(1/p) + (1/q) = 1/f

(1/20) + (1/q) = 1/24

q= -4inch ; the (-)negative sign indicate that the image is virtual

PROBLEM:A glass is 1 inch thick. A paper with a dot is 0.640 inch thick. The glass is put on top of the paper and looks it into the microscope. What is the index of reflection?Solution:

nr = 1.00 / 0.640

nr = 1.5625

PROBLEM:

A 120-V potential difference is maintained between each end of a long high resistance wire. The center of the wire is grounded. What is the potential of the center and each end of the wire?Solution:

Potential at the center = 0. Potential at one end = 60V. Potential at other end = -60V.