Elfi1. A football player kick the ball who has the inertial

moment 4,0 kgm2 and kinetic energy 1800 J. The angular velocity is . . . rad/sa.30b.60c.40d.70e.90

Answer

Known : I = 4,0 Kgm2

Ek = 1800 J

Ask : = ?

Answer: Ek =

1800=

= 900

= 30

2. A hollow cylinder have the mass about 125 g. If R1 10 cm and R2 30 cm, calculate the inertial moment if axis passing through the center.a.6,25 x 10-1 b. 6,25 x 10-2 c. 6,25 x 10-3

x 10-4 e.6,25 x 10-5

Answer

Known : m = 125 x 10-3 kgR1 = 10-1

R2 = 3 x 10-1

Ask : I = ?

Answer: I = m(R12+R2

2)

I= .125.10-3((10-1)2 + (3 x 10-1)2)

=6,25 x 10-5

3. T1 T2 T1 + T2 = ?

300 600

300 N

a. 300 Nc. 500 N d. 150

b.150(1 + ) N e. 300

If the weight the object are 200 N, and the angular is 300,calculate the force wich work at the hinge !

a.4.

5. The kinetic energy of 2 moles of monoatomic gas in 8 liter tube is 2,4 x 10-22 J. What is the pressure of the gas in the tube?a.2,408 x 104

b.8,56 x 104

c. 5,069 x 104

d.4,23 x 104

e.2,225 x 104

6. A container with temperature of 47 0 C and pressure of 1,2 x 105 Pa contain 2 gr of helium gas with molecule mass 4 g/mole. Calculate the inertial energy of the gas.

a. 1995b. 1998c. 1997

d. 1994e. 1912

7. If a body with 6000 cm2 in volume and 1200 kg/m3in density completely immersed in water. The magnitude of body in water is (W) ?a. 6b. 96 c. 48d. 24e. 12

8. If the velocity of air flow at the down side of the plane’s wings is 50 m/s. What if the velocity of at the upside of the plane’s wings if the upward pressure obtain is 10 N/m2.

a.

b.

c. 12d. 35

e.

9. Two mole of gas in compressed at a constant temperature of 270 C so that its volume increase 2 time from the initial volume .Calculate the work down by the gas. (ln 2 =0,69)

a. 3933b. 3586c. 2317d. 5342e. 5324

10. A gas is compressed so that the volume decrease from 5 liter to 2 liter. At a constant pressure of 1 x 105 Pa, calculate the external work applied to the gas !

a. - 300b. 300c. 900d. -900e. 450

Akbar1. A pulley with radius of 20 cm can rotate with the axis passing

the center of mass O. around the pulley is wrapped by a string. The end of the string is pulley with a constant force of 12 N. what is the magnitude of moment of force experienced by the pulley ?a. 2.6 N c. 2.0 N e. 3.0 Nb. 2.4 N d. 2.5 N

Solution Known : r = 20cm = 0.2 m

F = 12N Moment arm = radius of pulley= 2m

Ask : τ = ?Answer : τ = f . d

= 12 N . 0.2 m = 2.4 N ( b )

2. A wheel with 6 kg in mass and radius of 30 cm rotates at velocity of 300 rotation/ minute. Determine its inertial moment and kinetic energy !a. 0.45 kg m2 , 2.7 π2 joule c. 0.54 kg m2, 2.7 π2 jouleb. 0.1 kg m2 , 2.0 π2 joule d. 0.54 kg m2 , 7.2 π2 e. 2.0 kg m2 , 0.1 π2 joule

solutionm = 6 kgr = 30 cm = 0.3 mw = 300 rotation / minute

=

Its inertial momentI = mr2 = (6kg)(0.3m)2 = 054 kg m2

Its rotary kinetic energyEk rotation = ½ I W2

Ek rotation = ½ . (0.54 kg m2 ) . ( 10π2rad/s)= 2.7 π2joule ( c )

1

3. A homogeneous bar with 9 m in length and 8 N in weight is held by support 3 m from the left end. If a load weighing 10 N is hung on the left end. Determine the magnitude of fulling force downward on the right end.

3 m 6m

w 45m

10 N R

Solution Because the bar is homogeneous, then weight point is in the

middle of the bar.

Equilibrium condition :∑ = 0 ( x – axis direction )

∑fy = 0R-10N-w-f = 0R-10N-8N-F = 0

R-18N-F=0 …………. (1)

∑τh=0

(-10N)(3M)+(R)(0)+(8N)(1.5m)+(F)(6m)=0-30Nm+12Nm+F6m = 0

-18Nm+F6m = 0F6m = 18Nm F = 3N

…………………. (2)

Insert the value F = 3 N in equation (I)

R-18N-3N = 0R = 21N

4. A homogeneous bar is 6m in length and 8N in weight. At the right end of the bar is inserted a small load weighing 4N, as seen in the figure. Determine weight point of system of bar and load!A. 4 C. 4.1 E. 7B. 3 D. 6

6m

A

W1 = 8N w2 = 4 N

SolutionTake point A as reference pointW1 = 8N, x1 = 3m, w2=4N, x2 = 6m

Weigh point of system to y-axis

(A)

5. A swimming pool is 2 meters in depth. If the presser of external air (atmosphere) above water level is 72 CmHg. Water density 1000 kg/m3 and gravitational acceleration 10 m/s2. calculate the total pressure at the pool baseA. c.

e. 1.16 x 10-4

b. d.

Solution

Known : h = 2m p air = 1000 kg/m3

Answer: Po = 72 CmHg

=

= 9,6 X 104 N/m2

g = 10 m/s2

p = po + p air gh= 906x104 N/m2 + 2 x 104 N/m2

= 11,6 X 104 N/m2

= 1,16 X 105 N/m2 (a)

6. A hydraulic car jack as shown in the figure below, has two piston with each section area is A1 = 100 cm2 and A2 = 2000 cm2 what is the minimum magnitude of force F1 that must be given on section A1. so that the car weighing 1500 N can lifted?a. 350 N c. 550N e. 750Nb. 450N d. 570N

Solution.A1 = 100 Cm2

A2 = 2000 Cm2

W = F2 = 1500 N

According to Pascal’s lawP1 = P2

=

= 750 N ( e )

7. A closed vessel contains 20 l oxygen gas if the gas is at a temperature 0f 27oC and atmosphere pressure of 1 atm (105 Pa). determine the number of moles of the oxygen in the vessel

a. 0.802 c. 0.208 e. 0.902b. 0.812 d. 0.128

SolutionV = 20l = 20x10-3m3

T = 27 + 273 = 300 kP = 105 Pa

PV= n RT

n = .

…. ( a )

8. Determine the effective velocity of a gas particles at normal state, if the gas density is 10 kg/m3 and it’s pressure is 3X105N/m2

a. 30 m/s c. 3000m/s e. 3m/sb. 218 m/s d. 300 m/s

Solutionρ = 10kg/m3

P = 3X10N/m2

= =

300m/s . .. .. ..( d )

2

9. A gas is compressed so that the volume decrease from 5.0L to 3.5L at a constant pressure of 1.0X105Pa. calculate the external work applied to the gas.

a. 150J c. 15 J e. -1.5 Jb. -150J d. -1500 J

Solution

V1 = 5.0LV2 = 3.5LP = 1.0X105Pa

AnswerW = PXΔV = P (V2-V1) = 1.0X105(3.5L-5.0L)

= 1.X105PA(-1.5L) = -150J . . .. . .. ( b )

10. A monoatomic ideal gas is compressed adiabatically

and the volume decrease to its half. Determine the ratio of the

final pressure to the initial pressure.

a. 2 c. e.

b. d.

Solution

=

=

=

=

= . . .. . .. . ( a )

Anis

1. Known torsion magnitude of thing is 20 Nm and force that use is 15 N. If force’s working line which torsion’s arm, then determine torsion’s arm magnitude the thing.

A. 3 mB. 4 mC. 5 mD. 6 mE. 7 m

Solution

= 4 m

2. A thing in mass 2 kg rotate surrounding an axis have distance from the thing is 4 m. determine moment of inertial that thing.

A. 16 kgm

B. 24 kgm

C. 32 kgm

D. 64 kgm

E. 72 kgm

Solution

= 2(4)

= 32 kgm

3. Determine magnitude of Ek rotation’s plate in mass 4 kg and have radius 6 m rotate at 300 rad/s.

A. 3,24 . 10

B. 3,24 . 10

C. 3,24 . 10

D. 3,24 . 10

E. 3,24 . 10

Solution

kgm

x 10 J

4. A rod lean on at glossy wall and rest on rough floor, so frictional coefficient in A before that rod eventually falls is …

A. 1/8B. 3/8C. 6/8D. 4/8E. 7/8

Solution

= = =

x = 25 – 16 = = 3m

5. in mass 5,6 kg have 4 legs. Area each leg 2 . 10 m . Determine pressure of chair…

A. 7 x 10 Pa

B. 7 x 10 Pa

C. 7 x 10 Pa

D. 7 x 10 Pa

E. 7 x 10 Pa

SolutionF = m g = 5,6 x 10 = 56 NA(4) = 4 x 2. 10 = 8. 10 m

P = Pa

6. A vessel have the form of U contents of fluids like a picture. Weight A = 200 N and weight B = 500 N. If Area in a = 5 cm , so area in B is…

A. 2,0 x 10 m

B. 2,5x 10 m

C. 5,0 x 10 m

D. 1,25 x 10 m

E. 2,5 x 10 m

SolutionAA = 5 cm = 5 x10 m

3

200 = 2500 x 10

=

= 1,25 x 10 m

7. At normal situation, 4 gr O gasses have volume? (T = 0

, P = 1 atm, M = 32, R = 8341 J/ kmol K)

A. 1,4 x 10 m

B. 2,8 x 10 m

C. 22,4 x 102,0 m

D. 2,8 m

E. 22,4 m

Solution

n = = 0,125 mol

= = 2,8 x 10 m

8. If K = 1,38 . 10 J/K and temperature is 27 C so, Ek atom is…

A. 4,14 x 10 J

B. 2,07 x 10 J

C. 12,42 x 10 J

D. 5,59 x 10 J

E. 6,21 x 10 J

Solution

= 3/2 x 1,38. 10 .300

= 6,21 x 10 J

9. 2m helium gases with temperature 27 C heating with

isobaric until temperature 77 C. If the pressure 3 . 10

N/m so, determine work that do gases is…A. 100 kJB. 140 kJC. 200 kJD. 260 kJE. 320 kJ

Solution

= = 100 kJ

10. Determine efficiency maximum termal from heat engine between temperature O C and 100 C

A. 13 %B. 25 %C. 27 %D. 33 %E. 42 %

Solution

= 1- %

Anne

1. A solid sylender with mass of 3 kg,and radius 20 cm move at velocity of 50 m/s while rotating.Determine the kinetic energy of cylinder?a. 6525 J d. 1875 Jb. 5625 J e. 250 Jc. 3750 J

2. In the figure bellow.a pulley is the form of solid cylinder with mass 8 kg and radius 6 m,while the mass of A=4 kg and mass B=8 kg (g=10 m/s2).If the system moves and pulley follows to rotate then the acceleration of the system is….

a. 0,4 m/s2

b. 4 m/s2

c. 0.2 m/s2

d. 2 m/s2

e. 0,5 m/s2

3.A balok with mass in m kg,The magnitude of string tension P1 is….

a. √3 mg 37b. 1/3√3 mgc. ½ mgd. 1/3 mge. ¼ mg

4. A tank contains water of 1,25 m m in high.There a leak hole in tank 2 m from the bottom of tank.How far the place of water falls measured from the tank….

a. 10 m d. 4 mb. 5 m e. 1 mc. 50 m

5. A hydrolic car jack has two pistons with each section area is A1=200 cm2 and A2=1500 cm2.What is the minimum magnitude of force F1 that must be given on section A1,so that the car weighting 30000 N,can be lifted….

a. 1500 N d. 400 Nb. 30000 N e. 4500 Nc. 4000 N

6. The kinetic energy of 2 moles of monoatomic gas in a 5 L tube is 2,3 x 10-22 J.What is the pressure of the gas in tube……

a. 11,07 x 103 Pa d. 35 Pab. 4,6 x 10-22 Pa e. 12.04 Pac. 10 x103 Pa

7. Two moles ideal gas initially has temperature of 127oC.Volume of V1 and preassure of P1 = 5 atm.Gas expands isothermicality to volume V2 and pressure P2 = 3 atm.Calculate Eksternal work gas has done……(R= 8,31 J/mol K)

a. 6648 ln (5/4) d. 6638 (ln 3/5)b. 6648 (ln 3/5) e. 6628 (ln 3/5)c. 6648 (ln 4/5)

8. Carnot egine work at high temperature of 400 K for the mechanic work result.If the carnot egine absorbs heat of 500 J with minimum temperature is 200 K.So,magnitude of work done by carnot egine …….

a. 500 J d. 250 Jb. 200 J e. 1000 Jc. 50 J

9. A child weighting a length 5 m in high.The weight of length and child are 200 N and 400 N.What is the magnitude of coefficient static fiction between staircase and floor…….

a. 134/240 d. 134/250

4

m

T1T2

b. 134/230 e. 134/260c. 134/220

10.

The total angular momentum at O is…..

a. 5 kg m2/sb. 7 kg m2/sc. 8 kg m2/sd. 9 kg m2/se. 6 kg m2/s

PEMBAHASAN

1. Diket : m = 3 kg r = 20 cm => 0,2 m v = 50 m/s

Dit : ek ?

Answer

I = ½ mr2 V = r w

= ½ 3 (0,2)2 W =

= 0,06 =

W = 250

Ek = ½ mv2 + ½ I W2

= ½ x 3 x (50)2 + ½ x 6,10-2 x (250)2

= 3750 + 1875 = 5625 J

2. Dket : M = 8 kg: m1 = 4 kg: m2 = 8 kg

Asked : a ???

Answer

a = (m2 – m1) m1 + m2 + ½ 8

a = 8 - 4 8+4+1/2 8a = 0,4 m/s2

3. Diket : W = m x g θ 1 W= sin 150

θ 1 = 120asked : T1answer :

=

T1 =

=

= mg

4. Diket : h1 = 10 mh2 = 5 m

Ditnya : X ????

Answer :V =

= = 10 m/s

Y = V0 sin t + ½ gt2

Y = 0 + ½ (10 m/s2) t2

= 5 t2

T = = 1

X = Vo ( cos ) t = 10 x 1 x 1 = 10 m

5. Diket : A1 = 200 cm 2

: A2 = 1500 cm2

: w = 30000 NDit : F1

Answer :

P1=P2

=

=

F1 = 4000 N

6. Diket : n = 2 mol: V = 5 L

: Ek = 2,3 x 10-22

Ditnya : P???

Anwern = N No

N = n x No

= 2 x 6,02 x 1023

= 12,04 x 1023

P = Ek( )

= x 2,3 x 10-22 x

= 11,07 x 103

7. Diket : n = 2 mol: T1 = 400 K: P1 = 5 atm: P2 = 3 atm: R = 8,31 J/kmol

Ask : W =

W = nRTln

=2mol . 8,31 J/mol K .400 ln

5

Bc

Ac

3 m

5 m

4m5/3 m

O 3 m

3 m/s2m

3 kg

5 kg 2 m/s

=6648 ln J

8. Diket : Q1= 500 J: T2= 200 K: T1= 400 K

Dit : w ???Answer

= = 1-

= =1 -

= =

w = 250 J

9. Diket : as a pictureAnwer

= 0

Nb – Fx = 0

=0

Na – Wo –Wt = 0 = Wo + Wt = 400 N + 200 N = 600 N

=0

Di poros A

Na L Cos -Wt ½ L cos - Na L sin = 0

600 x 5 - 200 x ½ 5 x 3/5 – 400 x 2/3 x 3/5- x 600 x 5 x 4/5 =

0 = 1340/240 = 134/240

10. Diket :m1 : 5 kg: r1 : 2 m: V1 : 2 m/s: m2 : 3 kg: r2 : 3 m: V2 : 3 m/s

Dit : L

Anwer :

L = m1r1v1 + m2r2v2 = -(5 kg x 2 m x 2 m/s) + (3 kg x 3 m x 3 m/s) = 7 kg m2/s

Aris1. Homoge bar of Length AB 80m weighing 18 N, heavy of

burden 30 N and BC is string . If distance of AC 60im, hence level of string tension….

36 N48 N50 N65 N80 N

2. homogeneous solid cylinder with radius 1,5 and mass 7 singk. menggelinding of inclined plane top with height 30 m, calculate speed of moment object arrive at floor base.

a. 20 m/sb. 16 m/sc. 19 m/s d.17 m/s

e.25 m/s

3. a doorstep with length 13m and heavy 50kg. its back part [in] leaning to diding and convergent other [at] harsh floor. Wall distance to doorstep foot/feet [is] 5m . determining precise friction koefesien [of] doorstep will sliped

a.

b.

c.

d.

e.

4. a pulley have mass 16 singk and, 7cm, put down with a string. at the end of - tip of string tied object which is its mass 40kg and 20kg. determining masin - masing acceleration of object if at smooth pulley. a. 2,9 mb. 3,3 m/s2c. 1 m/s2d. 1,25 m/s2e. 3,67 m/s2

5. pipe of U contain mercury ( 13,6 g / cm3). [at] one column, decanted by oil ( 0,8 g / cm3). if is high of oil 10 cm of boundary area.Calculate high difference between oil and mercury.

a. 0,59 cmb. 9,41 cm c. 2,5 cm d. 7,5cme. 9,5 cm

6. plane with certain speed so that air which passing tabletop and under wide [of] wing [is] its surface 80 m3. peripatetic fastly each 400m / s and 500m / s. determining the level of style lift [at] wing.

a. 846 × 104

b. 486 × 104

c. 468 × 104

d. 492 × 104

e. 429 × 104

7. Determining gas particle mean kinetic energy owning temperature 127oC.

a. 8,28 × 10-21

b. 82,8 × 10-21 c. 74,3 × 10-21

d. 743 × 10-21

e. 81,4 × 10-21 8. Determining effective fast comparison between gas

molecules of X ( M = 2 g / mole ) and gas of Y ( M = 128 g / mole) [at] certain temperature.

a. 8 : 4b. 4 : 8

c. 16 : 1d. 1 : 16e. 1: 4

9. N gas in place of closed to experience of process like at picture. Determining work is gas of situation of direct

6

40 kg

20 kg

BA

C

C

BA

115

6

8

a.b.c.d.e.

10. Determining energi change in if absorb gas of kalor 400 calorie and do work 200 a. 200b.-200c. 600d. -600e. 400

Solution

1. known : AB = 80 Cm W AB = 18 N WBeban = 30 N AC = 60 Cm

Answer: -2400-720+80T1.0,6=03120=48T1T1=65N

2. V =

=

==20

3. =

=

=

4. g

=

= 3,3 m/s2

5. =

=

8000 = 13600 x

h =

h = 0,56

x = 10 – 0,59

x = 9,41cm

6. f1 – f2 =

=

=4680000

7.

=

=

=

8. =

9.

10. = 400 – 200

= 200

Ayu1. A solid sphere with mass of 2 kg and diameter 20cm moves

at velocity 5m/s while rotating. Determine the kinetic energy of the solid sphere?a. 25 joule c. 75 joule e. 125 jouleb. 50 joule d. 100 joule

2. A ballerina with stretched arms rotated with angular velocity of won the smooth horizontal –Floor. If she folds her arms, the initial moment decrease 15% from initial. Determine the ratio of Ek before and after!a. 3:4 c. 3:2 e. 1:2b. 4:3 d. 2:3

3. A homogeneus bar with 10m in length and 6 N in weight is held by support 4m from the left end. If a load weighing 7N is hungon the left end, determine the magnitude of pulling down word on the right end!

4m 6m

5m 7N R W (6N) F

a. 1,66 N c. 16,6 N e. 11,66 Nb. 11,6 N d. 16,16 N

4. A Big vessel full with water 5m. If at the bottom of vessel, there is a hole with the diameter 2cm.Determine the volume of water comes out of the hole after 1 minute!a. 1,884 liter c. 188,84 liter e. 188,4 literb. 18,84 liter d. 18,44 liter

5. A body with 600cm3in volume and 1500kg/m3 in density is completely innersed in water. What is the magnitude of buoyancy force experience by and the weight of the body in water?a. 30 N, 60 N c. 60N, 30 N e. No answerb. 30 N, 30 N d. 60N, 60 N

6. 7 moles of monoatomic ideal has temperature of 227 C.

What is the magnitude of the average kinetic energy and the internal energy?

7

a.-20 and 43 J d. -22

and 34 J

b.-20 and 43 J e. -21 and

43 J

c.-20 and 43 J

7. Oxygen gas have volume 80 liters with temperature 300K

and atmospheric pressure at . Calculate the

number of moles of the !

a. 0,32 mol c. 32 mol e. 2,30 molb. 3,2 mol d. 23 mol

8.3 helium gas 27 C temperature is heated until 87 C

by isobaric. The pressure of helium gas is 5N/m2,

then calculateof the external work done by the gas!a. 60 kJ c. 6000 kJ e. 60000 kJb. 600 kJ d. 60000Kj

9. The coefficient of performance of a refrigerator is What is the electric energy used to transfer 2400J of heat from food in the refrigerator?a. 24J c. 800 J e. 2400 Jb. 720 J d. 7200 J

10. A children has weight 300 N gone up to staircase far as

from the end of staircase. What is magnitude of coefficient static friction between staircase and floor,if the weight of staircase is 250 N?

B

a.c. e.

4m

b.d.

3m A

THE KEY ANSWER

1.

w= = = 50 rad/s

.2.(5m/s)2 + . 2 w2

= 25 joule+25 joule= 50 joule

2. I2= 75% I1

L akhir= L awalI2w2= I1w1

¾ I1. w2= I1w1

W2= 4/3 w1

Ek rot= ½ Iw2

= =

3. ∑fx= 0∑fy= 0R-7N-W-F= 0R-7N-6N-F= 0R-13N-F=0……..(1)

∑

(-7N) (4m) +R (0)+ (6N) (1m)+F(6m)= 28Nm+6Nm+F6m= 0F6m= 28-6F= 22/6 = 3,6

R-13N-F= 0R-13N-22/6= 16,6 N

4. Hukum Bernauli

p1= p2h1= 5m V1= 0h2= 0

v2

= 10 m/s

Q= Av2Q= 3,14 (10-2)2.10= 3,14 liter/sAfter 1 minute= 3,14x60= 188,4 liter

5. Va=vb=6000cm3= 0,006m3

Fb= 1500 kg/m3 Fa= 1000kg/m3

FA=

= 1000.0,006.10= 60 N

W udara =

= 90 NW air= 90-60

= 30 N

6. Ek= 3/2 KT

=

=

U= 3/2 NkT= 3/2.7.6,02 x 1023. 1,38x10-23.500= 43.614j=43 Kj

7. n=

= 3,2 joule

8.

V2=

W= p(v2-v1)= 2.105 (1,8 – 1,5 )= 60000j= 60 kJ

9. Cp=

3 = 2400/wW= 800 J

10.

Nb4m 5m

wbt Na

3m w0

8

∑fx=0Nb-Fk= 0Nb= Fk

Nb=

∑fy-= 0Na-Wt-Wo= 0Na= Wt+Wo = 250+300 =550 N

Na.L.cos

550.5. – 250. .5. -300. .5. - .550.5. = 0

675-2200 = 0

=

Bayu

1. A body system consists of two balls of 6 kg which are connected by a light and stiff rod (of which the mass is neglected) of 1 m length. The moment of inertia of the balls against the perpendicular axis of the road, while it is rotated at the center of the road is….

a.2,5 kg.m2 d.3,0 kg.m2

b.2.0 kg.m2 e.3,5 kg.m2

c.1,8 kg.m2

2. Calculate the average kinetic energy of the earth if its moment of inertia is 9.83 × 1037 kg.m2 and it moves with angular speed of 7 × 10-5 rad/s is……

a.24 × 1027 d.24 × 1028

b.24 × 1026 e.24 × 1024

c.24 × 1025

3. In the system of rigid body equilibrium, AB is a homogenous rod of 80 cm length and 18 N weights. The weight of w is 30 N. Determine the rope BC stress, if AC is 60 cm…..

a.60 N d.65 N b.55 N e.70 N c.63 N

4. Same with no. 3. AB is 60 cm of length and 20 N weights. The weight of w is 40 N. Determine The rope of BC stress, if AC is 50 cm……

a.105 N d.189 N b.120 N e.100 N c.125 N

5. A drum of 1 meter high is filled with kerosene (p = 0.8 g/cm3) to ¾ of the drum’s height. If g = 10 m/s2 , the pressure at the bottom of the drum is……

a.6.101 Nm-2 d.6.104 Nm-2

b.6.102 Nm-2 e.6.105 Nm-2

c.6.103 Nm-2

6. Given that the density of ice is 920 kg/m3 and the density of sea water is 1.030 kg/m3 , what fraction of the ice’s volume will be dipped into the sea…….

a.0,845 portion d.0,893 portion b.0,345 portion e.0,500 portion c.0,450 portion

7. A rubber balloon with volume of 20 L is filled with oxygen at pressure of 135 atm and temperature of 27o C. Determine the oxygen mass given that R = 8.314 J/(mol.K)…..

a.70 mol d.66 mol b.90 mol e.150 mol c.110 mol

8. Five moles of monatomic ideal gas has temperature of 900 K. Determine the internal energy (k = 1,38 × 10-23)……

a.60 kj d.44 kj b.56 kj e.50 kj c.43 kj

9. Certain gas undergoes expansion at a constant pressure of 5 atm (1 atm = 105 N/m2). Its initial volume was 300 L and the volume gets doubled soon after the expansion. Determine the work done by the gas……

a.1,5 × 106 J d.1,5 × 109 J b.1,5 × 105 J e.1,5 × 102 J c.1,5 × 104 J

10. Two moles of hydrogen gas is heated from 270 K to 310 K. If the heating is made at constant pressure, the heat needed is 2,33 kj. But, if the heating is made at constant volume, the heat needed is only 1,66 kj. Determine the molar capacity of the hydrogen gas at constant pressure……..

a.20,00 J/mol.K d.21,67 J/mol.K b.26,50 J/mol.K e.24,00 J/mol.K c.20,75 J/mol.K

Pembahasan

1. I = m1r12 + m2r2

2

= (6)(0,5)2 + (6)(0,5)2

= 3 kg.m2

2. Ek = ½ Iw2 = ½ (9,83 × 1037)(7 × 10-5) = 24 × 1028 joule3. AC = 60 cm, sin α = 0,6 AB = 80 cm, cos α = 0,8 AD = ½ AB = 40 = 0,4

w × AB + wAB × ½ AB – T sin α × AB = 0 (30 N)(0,8) + (18 N)(0,4) = T (0,6)° (0,8) 24 + 7,2 = 0,48 T 31,2 = 0,48 T T = 65 N

4. Cara untuk soal no 4 sama dengan soal no 3

5. Pbottom = Po + pgh = 101 N/m2 + (800 kg/m3)(10 m/s2)(0,75 m) = 6.101 N/m2

6. pfluida gVt = pbVbg Vt/Vb = pb/pfluida

= (920)(1.030) = 0,893 portion

7. n = PV/RT = (135 atm)(20 L)/ (0,0821 L.atm/mol.K)(300 K) = 109,6 = 110 mol

8. U = 3/2 NkT = 3/2 (5)(6,02 × 1023)(1,38 × 10-23)(900) = 56,076,3 = 56 kj

9. W = P(V2 – V1) = (5 × 105 N/m2)(0,6 – 0,3) = 1,5 × 105 J

10. Cp,m = Qp/n(T2 – T1) = 1,66 × 103/(2 mol)(40 K) = 20,75 J/mol.K

Bunga1.Determine the torque of each total force portion of P, on the image side? a.15Nm c.18Nm e.12Nm b.17Nm d.14Nm

20N 6N

9

Known : F1 = 10N ℓ 1 = 2,5m F2 = 8N ℓ2 = 0 F3 = 5N ℓ3 = -10m F4 = 4N ℓ4x = -10m ℓ4y = 2,5m

Ask : τt ?

Answer : τ1 = 10N . 2,5m = 25Nm τ2 = 8N . 0 = 0 τ3 = 5N . -10N = -50Nm τ4 τ4x = 6N . -10m . cos0 37 = 6N . -10m . 0,8 = -48 Nm τ4y = 6N . 2,5m . sin0 53 = 6N . 2,5m . 0,6 = 90Nm τt = τ1 + τ2+ τ3+τ4x + τ4y

τt = 25Nm + 0 + (-50Nm) + (-48Nm) + 90Nm = 17Nm 2. A solid cylinder has a mass of 8 kg with a radius of the rolling 40cm.Silinder solid with linear speed 5m/s.Determine momen inertialmomentum sudutl? a.0,34 kg m2 and 12 kg m2/s c.0,77 kg m2 and 15 kg m2/s e.0,64kg m2 and 18 kg m2/s b.0,65 kg m2 and 16 kg m2/s d.0,64 kg m2 and16 kg m2/s

Known : m = 8kg v = 5m/s R = 40cm = 0,4m Ask : I and L ?

Answer : I = 1/2mR2 L = m . v . R I = 1/2 . 8kg . (0,4m)2 =8kg . 5m/s .0,4m I = 1/2 . 8 . 0,16 = 16 kg m2/s I = 0,64 kg m2

3. Determine the stresses on the retaining strap to the two objects, with a heavy object on the image side of 800N? a.480N c.258N e.580N b.240N d.478N

Known : T1x = T1 . cos 53° = 0,8T1 T2x = T2 . cos 37° =0,8T2

T1y = T1 . sin 53° = 0,6T1 T2y = T2 . sin 37° = 0,6T2

W =800N

Ask : T2 ?

Answer : ∑ Fx = 0 ∑ Fy = 0 T2x – T1x = 0 T1y + T2y – W = 0 0,8T2 – 0,6T1 = 0 0,8T1 + 0,6T2 – W = 0 0,8T2 = 0,6T1 0,8T1 + 0,6 T2 = 800N T2 = 0,6T1 0,8 T2 = 4T2 3 0,8(4T2/3) + 0,6T2 = 800N 3,2T2/3 = 0,6T2 = 800N 3,2T2 +1,8T1 = 800N 3 3,2T2 + 1,8T2 = 2400N 5T2 = 2400N T2 = 2400N = 480N 4. Coordinates specify the composition of heavy point up as shown below: a.(22 , 37) c.(20 , 65) e.( 20,3 , 63,5 ) b.(20 , 63,5) d.(23 , 65) 7 2

5.

2 In a system like the picture below, the vacuum left 0,3dm2.dan B has a weight of cross-sectional area is negligible. And right cylinder R memilki 4dm2 sectional area and in the given load in the contents of X kg.Sistem liquid with a density of 900 kg/m3.bila balanced system to F for 20N.Berapa mass on X kg (g = 10 m/s2) a.242,88kg c.222,35kg fe.273kg b.242,67kg d.354,23kg.

Know : AB = 0,3 dm2 = 0,3 * 10-2 m2 ρ = 900 kg/m3 h = 6m AR = 4 dm2 = 4 * 10-2 m2 F = 20N Ask : Mass(nilai X kg) ?

Answer : ρ1 = ρ2 X . g = F + ρ . g . h AR AB X . 10 m/s2 = 20N + 900 kg/m3 . 10 m/s2 . 6m 4 * 10 -2 m2 0,3 * 1o-2 m2

X = 4 * 10-2 . 182

3 * 10-2

X = 728 = 242,67kg 3

6. An aircraft moving with the plane lift 280.000N.bergerak the rate of air flow at the top of the wing 310m / s and the bottom wing 280m/s.apabila Pud = 1.3 kg/m3.berapakah wing area? a.21 m2 c.22 m2 e.24,3 m2 b.24 m2 d.25,3 m2

Known : F1 – F2 = 280000N ρud = 1,3 kg/m3 V1 = 280 m/s V2 = 310N

Ask : A ? Answer : F1 – F2 = 1/2 . ρud(V2

2 – V12)A

280000 = 1/2 . 1,3 kg/m3{(310)2 – (280)2}A 560000 = 1,3(96100 – 78400)A A = 560000 = 24,3 m2 23010

7. Volume of a bubble of air at the bottom of a pond is 2cm3.berapakah volume of air bubbles when it was directly under the surface of the pool water? if the depth of the pool 100m.(= 800cmHg outside air pressures, the density of mercury = 13.6 g/cm3, the density of water = 1g/cm3) a.29,098 cm3 c.87,879 cm3 e.65,765 cm3

b.21,608 cm3 d.21 cm3

Known : V2 = 2 cm3 P0 = 75 cmHg h = 100m dianggap T1 = T2

Ask : V1 ?

Answer : Ph = 100 m air =10000 cm air * 1 cmHg = 735,3 cmHg 13,6 cm air

P2= P0 + Ph

= 75 cmHg + 735,3 cmHg = 810,3 cmHg P1V1 = P2V2 ( T1 = T2 ) (75 cmHg)V1 = (810,3 cmHg) 2 cm3

V1 = 1620,6 = 21,608 cm3 75

8. 4mol example methane gas which has a volume of 22.4 L at a pressure of 98 kPa.monatomic gas methane is on the state normal.maka, calculate the kinetic energies of molecules such as methane gas?

10

a.136,75 * 10-23 J c.143,25 *10-23 J e.136,75 * 10-24 J b.136,75 * 10-22 J d.136,75 * 10-21 J

Known : n = 4mol NA = 6,02 * 10-23 molekul/mol V = 22,4L P = 98 kPa = 98 * 10 3 Pa Ask : Ēk ?

Answer : N = n . NA

= 4mol . 6,02 * 10-23 molekul/mol = 24,08 molekul Ēk = 3 . PV = 3(98 * 103 Pa)22,4 * 10-3 m3

2 N 2 . 24,08 * 1023 molekul = 6585,6 = 136,75 * 10-23 J 48,16 * 1023

9. 8* 103 Ja business is given to adiabatically compress the 1 mole of an ideal absolute monoatomik temperature.tentukan 2 times the initial temperature of gas ( R = 8,31 J) a.642K c.641K e.564K b.546k d.654K

Known : R = 8,31 J T2 = 2T1

W = -8 * 103 J n =1mol

Ask : T1 ?

Answer : U = 3/2nRT U = Q – W (Q = 0 karena prorses adiabatik) U = -W 3/2nRT = -W T = -2W 3nR T2-T1 = -2W 3nR 2T1 – T1 = -2(-8 * 103 J) 3(1mol)(8,31 J) T1 = 16 * 103 = 641K 24,93

10. 3 mol of oxygen gas is heated at constant pressure and begins at a temperature known 127C(T2 = 2T1).Jika R = 3 cal/mol . K.Maka,the amount of heat that volume gas required for the traffic system is 3 time the initial volume? a.19kkal c.10kkal e.8,5kkal b.9kkal d.14kkal

Known : n = 3mol R = 3 cal/mol . K T1 = 127C = 400K V2 = 3V1

Ask : Q ? Answer : Cp = 5/2 nR Q = Cp . T = 5/2 nRT V1 = V2 T1 T2 V1 = 3V1 (T = T2 – T1=2T1 – T1 = T1) T1 T2

Q = 5/2 nRT1 = 5/2 . 3mol . 3 cal/mol . K . 400K = 9000kal = 9kkal

Erick1.A solid cylinder with 10 kg in mass and radius of 10 cm can rotate on a horizontal axis. A string that has a neglectable mass is wrapped on the cylinder.Then the end of string is pulled with static force of 20 N.If the mass of axis and friction between axis and its rotary place are neglectable,determine: A.Angular acceleration of solid cylinder B.Angular velocity of solid cylinder after rotating 1

Second C.The angle travelled by solid cylinder in 1 second A.40 rad/s,40 rad/s,20 rad B.30 rad/s,30 rad/s,10 rad C.40 rad/s,30 rad/s,20 rad D.30 rad/s,40 rad/s,30 rad E.40 rad/s,20 rad/s,30 rad

Solution: m= 10 kg I = ½ m r2

r = 10 cm= 0,1 m f = 20 N a= f . r = 20 N . 0,1 m =40 rad/s ½ mr2 ½ .10 kg.(o,1)2m

w= W0 + a . t = 0 + 40 rad/s2. 1,0 s = 40 rad/s

Q =w0 t + ½ a . t2

= 0 + ½ . 40 rad/s2.(1,o)2

= 20 rad

2. A solid cylinder with mass of 15 kg and radius 20 cm moves at velocity of 40 m/s while rotating.Determine the kinetic energy of the cylinder. A.16000 J B.17000 J C.18000 J D.19000 J E.20000 J Solution

m= 15 kg r = 20 cm = 0,2 m V = 40 m/s Inertial moment of solid cylinder

I = ½ mr2

= ½ 15 kg . (0,2 m)2

= 0,3 kg m2

Angular Velocity of Solid Cylinder

V = r . w w= V / r = 40 m/s = 200 rad/s 0,2 m

Ek tot = ½ mv2 + ½ I W2 = ½ (15 kg)(20 M/s)2+ ½ (0,3kg m2) (200 rad/s)2

= 18000 J

3.If the system below is in eQuilibrium ,determine the String stress TI,T2,and T3

30

0 60

0

A.40 ,50 ,60 N T1 T2 B.40, 40,8 , 80 N C.30 , 40 , 50 N T3 D.20 , 30 , 40 N E.10 , 20 , 30 N m=8 kg

Fx = 0 T2 cos 600-Ti cos 300=0 Fy = 0 TI sin 300 + T2 sin 600-T3=0 1/2TI + 3 TI / 2 = 80 N TI=40 N T2=40,8N

11

T3=w=80N

4.In the system of equilibrium of rigid body equilibrium as depicted in the picture,AB is a homogenous rod of 60 cm length and 16 N the weight of w is 30 N. Determine the rope BC stress,if AC is 40 cm

A.50 N D.70 N B.60 N E.80 N C.68 N Solution AC=40cm,sin A = 0,4 AB=60cm.cos A = 0,6 AD=1/2 AB =30

Using the second of condition for equilibrium at point A,

W . AB + W ab . ½ AB – T sin A . AB = 020 . o,6 + 16. 0,3 = T 0,4 . 0,612 + 4,8 = 0,24 T16,8 = 0,24 TT= 70N

5.The average velocity of water flow in a pipe with diameter 6 cm is 6 m/s.calculate the amount of fluida (water) Following per second ( Q ) A.16,95.10-3 D.18,95.10-3

B.15,95.10-3 E.19,95.10-3

C.14,92.10-3

Solution: d = 6 cm , r = 3 cm = 3 .10-2 m v = 4 m/s

Q = A v = 3,14.(3 . 10-2).6 = 3,14 . 9.10-4 .6 = 169,56.10-4

= 16,95.10-3 m3/s

6.A drum of 1 meter high is filled with kerosene (density=0,8 g/cm3 )to ½ drum’s height.if g=10 m/s-2 find the pressure at the bottom of the drum.

A.3150 N/m2

B.3182 N/m2

C.3386 N/m2

D.4204 N/m2

E.4101 N/m2

Density=0,8 g/cm3=800 kg/m3,h=0,5 m

P =101 N/m2+800kg/m3.10m/s2.0,5m =4101 N/m2

7.Oxygen gas at temperature of 270C and pressure of 105 Pa has volume 20 Liter.Determine the volume Of the oxygen given that the pressure becomes 2, 5.105 Pa and the temperature become 1270C A.2L B.3L C.4L D.5L E.6L Solution T1 = 270C = 300K T2 =1270c=400K P1=105Pa P2=2, 5.105 Pa V1=20 L

P1V1= P2V2

T1 T2 V2 = P1 V1 T2 P2T1

= 10 5 .20.300 2, 5.105.400

= 6 . 10 7 107

= 6 L

8.Four moles of monoatomic ideal gas has temperature Of 800 K.Determine: A.The Average Kinetic Energy B.The Internal Energy(k=1,38.10-23J/K) A.1,65.10-20 & 39 kJ B.1,65.10-20 & 40 kJ C.1,65.10-20 & 41 kJ D.1,65.10-20 & 42 kJ E.1,65.10-20 & 43 kJ Solution

A.Ek,av=3/2 k T =3/2 (1,38.10-23 J/K)(800K) =1,65.10-20

B.V =3/2 NkT =3/2(4 mol)(6,02.1023 Partikel/mol) (1,38.10-23)(800K) =39876,4 J =39 kJ

9.A certain gas at constant pressure of 7,2 .104 Pa is Compressed in such a way that volume changes from 6L to 1L.In this process,the gas released heat of 180 joule.Determine: A.The work performed by the gas B.The change of its internal energy. A.-386 & 182 J B.-360 & 180 J C.-386 & 204 J D.-360 & 182 J E.-360 & 183 k Solution: P = 7,2 . 104 Pa V1= 6L V2=1L Q = -180 J

The isochoric compression occurred: A. W = P( V2-V1 ) = (7,2 . 104 Pa)(1.10-3-6.10-3m3) = -360 J

U = Q – W = -180-(-360J) = 180 J

10.Certain gas undergoes expansion at a constant Pressure of 5 atm(1 atm=105 N/m2).Its initial Volume was 200L and the volume gets doubled Soon after the expansion.Determine work done By the gas. A.1105 J B.1001 J C.1100 J D.1900 J E.1182 J Solution

P=5 atm = 5.105N/m2

V1=200L=0,2 m3

12

V2=2V1=2(0,2)=0,4 m3

W =P(V2-V1) =(5.105 N/m2)(0,4m3-0,2m3) =5.105N/m2.0,2m3

=1.105J

Erin1. A wheel with radius in R1 , and radius in R2 give 2

force like this picture . if R1 =1m , and F1 = 4N , and R2 = 60cm and F2 = 6N , calculate moment and total force and the direction of turning wheel.

AnswerJari-jari dalam

= - (4N * 1m) = - 4 Nm (keluar)

Jari-jari , luar , = 6N * 0,6= 3,60 N (kedalam)

= - 4Nm + 3,60 Nm= 0,40 Nm (keluar)= -0,4 Nm

2. A homogen ball mass = 20 Kg , radius 0,4 turning to Z axis pass center. If the angular velocity of ball 4r/s , determine angular of momentum.

Answer

I= 2/5 mr² = 2/5 . 20 . 0,4 = 1,28L= I . ω =1,28 . 4 =5,12kg m²/s

3. A lamp hang at string . if m = 200 g and mass of rope is neglected. calculate tension of string.

Answer Σfy = 0T-W = 0T= mgT=(0,2kg . 9,8 m/s) = 19,6 w

4. Calculate the lie center of gravity from this body

Answer Luas bujur sangkar = 4m . 4m= 16m² Luas persegi panjang = (8m-0m) . 1m =8m² Dengan demikian hitungan absis 8 ordinat titik berat X0 = A1X1 + A2X2 A1 + A2

X0 = 16m² . 2m + 8m² . 6m 16m + 8m = 3,33m

Y0 = A1Y1 + A2Y2 Y1 + Y2 = 16m . 2m + 8m² . 1m =1,67m Koordinat = ( 3,33m,1,67)

5. A farmer make a pool with (4m*5m) , and hight 3m . if pool we fill with density 100kg/m³ (1 atm , 20˚C) and full . calculate mass and weight of water (g = 9,8 m/s²

Answer V= p . l = 4m . 5m . 3m = 60m³

M air = P air V = (1000kg/m³) (60m³) = 60000kg Berat air W air = M air . g =60000kg . 9,8m/s² = 580000 N

6. A thing input to the water with 40% . and it floads at water. Calculate density .

Answer Vs= 40% Vb ℓb < ℓf Vb= 100% Fs = w Adapun 40% benda mengapung berarti Vs Vs = 40% Vs = 0,4 Vb

V yang tercelup berarti dikurangi 100%

HK 1 Newton ∑fy = 0 Fs- w = 0 Fs = w ℓfVg = ℓbvbg ℓ = Vt . ℓf Vb = 0,6vb . 1,9 = 0.69/cm³ Vb

7. Graph p-v it's calor engine siclus calculate work .

Answer Berdasarkan grafik A-B = (4-1) m³ = 3m³ B-C = (400000- 200000) Nm‾² = 200000 Nm‾²

W = luas ABCD = AB . BC = 3m³ . 200000 Nm‾² = 600000 J

8. A carnot engine use high reservoir 800K, have efficiency 40%, calculate the temperature high reservoir if efficiency 60%.

AnswerT2 = n 40 % T1 = 800KŊ = (1- T2) . 100% T1(1- Tl ) = ŋ

T1 100% = 40% = 2/5 100%

13

2F

1F 1R

2R

1W

g200

0

4

4 8

)(my

)(mx

40%

AB

C D

5104x

5102x

T2 = 1 – 2/5 = 3/5 T1T2 = 3/5 . 800 = 480 K

T1 efficiency 60%Ŋ = ( 1 - T2 ) . 100% T160% = 1 – T2 = 1 – 0,6 = 0,4 T1 = T2 = 1.2 K100% T1 0,4

9. A nitrogen gas at standard condition have mass l0 g . Calculate the volume if weight of molecule nitrogen 28g/mol . And R = 8,314j/mol k

AnswerPv = nrt V = nrt P = mRt mRp V = 0,01 . 8,314 . 273 0.028 . 100000 = 8,11 . 10‾³

10. A gas at 27˚c mr = 32kg/k mol k = 1,38x10ˉ²³ J/k calculate ejective velocity.

Answerm= mr = 32 Na 602,2 . 10²³ mol /kmolm= 0,00531 10‾²³ Vrms = √3kt = √3 . 1,38 . 10‾²³ . 300 M 0,00531 . 10‾²³= 483,63 m/s²

Ezy1. Three particles which mass each other 0,6 kg roped on a wood with mass is neglected, like showed by the picture. Determine the moment inertia of the system against edge as the axis!

20 cm

15 cm a. 1,65 x 10-1 kg m2

1,65 x 10-4 kg m2

10 cm b. 1,65 x 10-2 kg m2

1,65 x 10-5 kg m2

axis c. 1,65 x 10-3 kg m2

Solution:

I = ∑miri2 = m1r1

2 + m2r22 + m3r3

2

= m(r12+r2

2+r32)

= 0,6[(10 x 10-2)2 + (25 x 10-2)2 + (45 x 10-3)2] = 1,65 x 10-1 kg m2

2. A grinding stole does translation and rotation moving with angular velocity and angular acceleration w and v. Determine total kinetic energy solid cylinder in m and v!a. 2/4 mv2

b. 4/3 mv2

c. 3/4 mv2

d. 4/2mv2

e. 1/3 mv2

Solution:I = ½ mR2

Total Ek = ½ mv2 + ½ Iw2

= ½ mv2 + ½(1/2mR2)v2/R2

= ½ mv2 + ¼ mv2

= ¾ mv2

3. Look at the following picture. Determine the resultant of forces from the four forces who has showed. And calculate the point of the resultant against A as the axis!

F1 F2 F3 F4

-2 A = 0 +3 +5a. 2 kN and 11mb. 3 kN and 12mc. 2 kN and 12md. 3 kN and 11me. 3 kN and 12m

Solution:Determine the up or down forces which positive mark or negative so we get:R = -4 – 2 + 12 – 4 = 2 kN

For A as the axis of zero, so absis the four forces showed on the picture:x1 = -2mx2 = 0x3 = +3mx4 = +5m

x = ∑Fxxi

R

= (-4)(-2) + 0 + 12(3) + (-4)(+5) 2

= (8 + 36 – 20) : 2 = 12 m

4. A homogen ladder AB has weight 100N on sleepy wall like showed on the picture. Determine the us!

a. 2/3 b. 3/2 c. 4/3 d. 1/3e. 3/5

Solution:

BC2 = 52 - 32

Tanα = AC/BC = ¾

So us = 1 2(3/4)

= 2/3

5. A place on the ass of the lake has 8 cm high.

Q

water h = 8 m

A

Calculate the total pressure in that point. (water density 1000 kg/m3, and g = 10 m/s2)

a. 1,81 x 104 Pab. 1,81 x 103 Pac. 1,81 x 105 Pad. 1,81 x 106 Pae. 1,81 x 102 Pa

Solution:

Patm = 1.01 x 105 Pa

Ph = ρgh

14

= (1000)(10)(8) = 8,0 x 104 PaPA = Ph + Patm

= 8,0 x 104 + 1,01 x 105

= 1,81 x 105 Pa

6. A piece of glass in the air has 25N weight and density 2,5 x 10 3

kg/m3. If water density = 1 x 103 kg/m3 and g = 10 m/s2, so the weight of glass in the water is…..

a. 30 Nb. 15 Nc. 32 Nd. 40 Ne. 35 N

Solution:wa = 25 Nρg = 2,5 x 103 kg/m3

ρw = 103 kg/m3

g = 10 m/s2

vw = vg = mg

ρw

= wa/g ρw

= 2,5 : 2,5 x 103

= 10-3 m3

Fw = 103 x 10 x 10-3 = 10 NFw = wa – ww

ww = wa – Fw

= (25-10) N = 15 N

7. How much are the molecules Oxygen per m3 in the air at temperature and pressure 273 K and 105 Pa? ( Na = 6,02 x 1026

molecule/kmol)a. 265 x 1026 molecule/m3 b. 2,65 x 1025 molecule/m3 c. 265 x 1024 molecule/m3 d. 265 x 1023 molecule/m3 e. 26,5 x 1026 molecule/m3

Solution:

PV = nRTN/V = PNa/RT = (105)(6,02 x 1026)/(8314)(273) = 26,5 x 1026 molecule/m3

8. Mr Nitrogen is 28, Oxigen 32. How are the temperatures when the effective value(vrms) the molecules of Nitrogen same with Oxigen at 300K?

a. 262,5b. 263,5c. 562.3d. 652,3e. 456,5

Solution:

VRMS N2 = TN2 x MO2

VRMS O2 MN2 TO2

1 = TN2 x 32 300 x 28

TN2 = 300 x 7 8

= 262,5 K

9. p-V graphic above is heat machine cycles. Determine work perform!

P(N/m2)4 x 105 A B

2 x 105 D C

1 2 3 4 v(m3) a. 6 x 103 J d. 6 x 106 J b. 6 x 104 J e. 6 x 107 J c. 6 x 105 J

Solution:From the graph we get:AB = (4-1) m3 = 3m3

BC = (4x 105) – (2 x 105 N m-2) = 2 x 105 N m-2

W = ABCDA’s area = 3 x 2 x 105

= 6 x 105 J

10. On this Graph of Carnor machine, known W = 10.000 J. Determine heat produce by the machine per cyclist?

a. 10.000/3 Jb. 10.000 Jc. 30.000 Jd. 20.000 Je. 10.000/7 J

Solution:

η = 1 – T2/T1

= ¾η = W/Q1

= 10.000: ¾ = 40.000/4

η = 1 – Q2/Q1Q2 = (1- η)Q1

= (1-3/4) 40.000/3 = ¼(40.000/3) = 10.000/3 J

Fhe_a4

1. A U-shaped pipe one of its foot filled with mercury, while the other one filled with alcohol. If the alcohol column in 20 cm in height and height difference between mercury and alcohol surface is 1 5 cm. What is the alcohol destiny if mercury destiny is 13,6 g/cm .

A. 54,4 g/cm

B. 3,4 g/cm

C. 40,8 g/cm

D. 45,34 g/cm

E. 44,4 g/cm

Solution= 20 cm

15 cm

13,6 g/cm

= 20 cm -15 cm = 5 cm

54,4 g/cm

15

2. A pipe has two difference cross sectional areas with diameter of 20 m and 10 cm. If the speed of small area is 12 m/s. Determine the speed of large area.

A. 48 m/sB. 3 m/sC. 24 m/sD. 36 m/sE. 44 m/s

Solution

m/s

= 4. 12 m/s = 48 m/s

3. A body system consist of 3 balls of 0,6 kg which connected by a light and stiff rod of 45 cm length. Determine the moment inertial of the ball at the poros of the rod.

A. 1,65 x 10 kg m

B. 1,65 x 10 kg m

C. 1,65 x 10 kg m

D. 1,65 x 10 kg m

E. 1,65 x 10 kg m

Solution

= 0,6 (10.10 ) + 0,6 (25.10 ) + 0,6 (4,5.10 )

= 0,6 .10

= 1,65 .10 kgm

4. 3 N 5 N 10 N 2m 3 m 2 m

4 NDetermine of torsion if poros in A.

A. 60 NmB. 30 NmC. 67 NmD. 70 NmE. 44 Nm

5. Which one the p-h graph for isochoric process.A.

B.

C.

D.

E.

6. 10L ideal gas with temperature 127 C and pressure 1665,6

N/m . Determine amount of gas particle.

A. 2,1 . 10

B. 4,0 . 10

C. 4,3 . 10

D. 3,0 . 10

E. 3,8 . 10

Solution

7. Determine kinetic energy helium with temperature 277 C.

A. 12,35 . 10 J

B. 10,00 . 10 J

C. 3,44 . 10 J

D. 10,21 . 10 J

E. 11,25 . 10 J

J

8. A gas ideal its heated in constant pressure 2.10 N/m . So the volume decrease from 20 L until 30 L. Determine the work done by the gas.

A. 2,0 . 10 J

B. 4,0 . 10 J

C. 6,0 . 10 J

D. 7,0 . 10 J

E. 10,0 . 10 J

Solution

J

9. An object weighing 20 kg is hung at a rest as illustrated in the figure below. Determine the tensions at both strong that retard it.

A. 100 N and 100 NB. 100 N 1and 100 NC. 100 N and 100 ND. 90 N and 100 NE. 100 N and 90 N

SolutionW = 20 x 10 = 200 N

16

060 030

W

1T 2T

N

N

10. Determine the coordinate in Y-axis and X-axis.

A.

B.

C.

D.

E.

SolutionI = 3m (1,3)

II = 2m (2,1)

x =

y =

Hendry1.) givmen that a pulley with mass M and radius R and two objects with mass

each

and ( > ) are set as in the figure.

If the coefficient of friction of object 1 to able is and

the pulley is assumed as solid cylinder, determine the acceleration of each object .

A.

B.

C.

D.

E.

Solution

2.) A solid object with mass M and radius R has inertial moment = k M is rolling on an inclined plan with inclination angle

illustrated in the figure

Determine the acceleration experienced by the solid object.

A. C.

17

M

1m

k

2m

R

1T

2T

gm2

2T

a

B.

D.

E.

Solution

…(1)

3.)And object weighing 80 N is hung at a rest state as illustrated in the figure bellow. Determine the tensions at both strong that retard

A. dan

B. dan =

C. dan

D. dan

E. dan

Solution

= =

=

18

80 N

6030

w

T

TT

Y

T 60 T30

T

= 0

1

4.)

.

The length of a stair case is 6 m and its weight 270 N. It leans on a smooth wall and form and angle of as seen in the figure.A moment before its slips, the person weighing 300 N has gone up as far as 4 m from the end of stairscase. What is the magnitude of coefficient static friction between staircase and floor?

A. 0,46B. 0,47C. 0,48D. 0,44

.E 0,43

Solution

w = 270 N

w = 300 NAB = 6 m

= AC = 4 m

F = 0

N - f = 0

N = f …..(1)

F = 0

N - w - w = 0

N - 270 N - 300 N = 0

N = 570 N…..(2)

= 0

N AB sin - w AB cos - - w AC cos 0 N = 0

N (6) (0,8) – 270 (3) (0,6)- 300 (4) (0,6) = 0

48N - 486 – 720 = 0

48N = 1206

N = 251,25 N…..(3)

f = 251,25 N

= = = 0,44 N

5.) A U-shape pipe, one of its foot filled with mercury, while the other one filled with alcohol. If the alcohol column is 30 cm in height and height difference between mercury and alcohol surface is 15 cm, what is the alcohol density,if mercury density = 13,6g/ !

A. 6,7g/cm

B. 6,8g/ cm

C. 6,9g /cm

D. 6,6g/ cm

E. 6,5g/ cm

Solution h = Alcohol height = 30 cm

h = mercury height

h = h - h =15 cm

h = h - h = 30 cm- 15 cm = 15 cm

p = p

h = h

=

= 13,6 g/cm

19

N

53

Af A

N

ww

B

C

+

B

C

53 A

= 6,8 g/cm

So, the alcohol density 6,8 g/cm

6.) A tank contains water of 1,25 m high. There is a leak hole in the tank 45 cm fromthe bottom of tank. How far the place of water falls measured from the tank (g = 10m/s)?

h = 1,25 m

h = 45 cm = 0,45 m

v =

v =

v =

v = 4 m/s

y = 0,45 mv = 4 m/s

g = 10 m/s

y = v sin t + gt

0,45 m = 0 + (10m/s ) t

t =

t = 0,3 sx = v ( cos ) t = (4 m/s)(1)(0,3s) = 1,2

7.) The kinetic energy of 2 moles of monoatomic gas in a 10 liter tube is 2.3 x 10 J. What is the pressure of the gas in the tube?

A .18461.3 Pa

B. 18461.2 Pa

C. 18471.2 Pa

D 19471.2 Pa .

E 19571.2 Pa .

Solution

n = 2 molesv = 10 L =

8.). A container with temperature of 60 and pressure of

1,8x10 contains 4g helium gas with molecule mass being of 4g/mol. Calculate the intenal energy of the gas.

A. 1766,4308 J

B. 1765,4308 J

C. 1756,4308 J

D. 1866,4308 J

E. 1766,5308 J

Solution

9. one mole of gas is compressed at a contans temperature of -23C so so that its volume decreases to half of its initial volume.

Calculate the work done by the gas.

A. 1,4 JB. 1,3 JC. 1,6 JD. 1,5 JE. -1,4 J

Solution

10.) 0,1 mole of air with 2 L in volume at 27 C and pressure

1x10 Pa is heated at constant pressure to 87 C. calculate the final volume.

A.2,5L

B. 2,4 L

C. 2,3 L

D. 2,8 L

20

E. 2,9 L

Solution

T =87

Isil1. A key weared to open mor

found on motorbike. The key have length 20 cm given force equal to 60 N which angular shape 450 to vertical direction. Determining the magnitude of moment of force.

Known: r=20 cm=0,2 mF=60 Nα=450

Ask: Answer: =r. F sin α

=0,2 . 60. Sin 450

= 12 . ½ √2

= 6 √2 mN

2. Bar of AB have a mass 2 Kg is turned around by A,in the

reality moment of inertia is 8 kg m2.If turned around by O

(AO=OB),the moment inertia be.....

A B

Known: m= 2 kg

IA = 8 kg m2

Ask: Io = ?

Answer: IO = 1/12 ML2 ; 1/ 3 ML2

So, IO/IA = 1/12 ML2/ 1/ 3 ML2 = ¼

IO = ¼IA = ¼(8 kg m2 ) = 2 kg m2

3. A advertisement board have mass 20 under arrest singk by log which is have mass 15.Determining string tension.

Answer: poros di tengah

-Wbt . ½ L – Wbn L + 2/3 L Ty = 0 -200 . ½ L – 150 L + 2/3 L ½ √2 T = 0

-100 L – 150 L + 2/3 L ½ √2 T = 0

2/3 L ½ √2 T= 250 L

√2/3 LT = 250 L

√2/3 T = 250

T = 250 . 3 / √2 from the floor,as

T = 375 √2

4. A wheel with mass of 24 kg and radius 1 m on the floor and

leans on a stair that is 0,4 m in hight from the floor,as seen

in the figure.What is the magnitude of minimum horizontal

force F that is enought to lift the whell from top of floor?

(g=10 m/s2)

0,4 m

W = m.g=24.10=240

BC = -

=

= 0,8 m

B = 0

F(0,6)- w(0,8) = 00.6F – 240 . 0,8 = 00,6F – 192 =0 F=320 N

5. Determine density of oil....

Answer: ho = 22 cm

ha = 22cm – 10 cm = 12 cm

P0 + ρa . g .ha = P0 + ρo . g . ho

ρa . ha = ρo . ho

ρo = ρa . ha / ho

= 1000 kg/m3 . 12 cm / 22cm

21

O

1m

= 545,45 kg/m3

6. A hydraulic car jack,has two pistons with each section

area is A1 = 100 cm2 and A2 = 2000 cm2.What is the

minimum magnitude of force F1 that must be given on

section A1 so that the car weighing 15000 N can be lifted?

Answer: A1 = 100 cm2

A2 = 2000 cm2

W= F2= 15000 N

P1 = P2

F1 = A1/ A2 . F2

= 100 cm2/2000cm2 . 15000 N

= 750 N

7. Counted 40 oxygen L have temperature 200C pressure

and 2 atm depressed till reach volume 36 L and at the

same of oxygen temperature boosted up till reach 500C.

How much is oxygen gas pressure now?

Known: P1 = 2 atmT1 = 20 +273 =293 KT2 = 50 + 273 = 323 KV1 = 40 LV2 = 36 L

Ask: P2 = ?Answer: P1 V1/ T1 = P2 V2/ T2

P2 = P1 V1 T2

V2 T1

= 2 atm. 40 L. 323K36L.293K

= 25840 10548

= 2,4 atm

8. A spherical air bubble originating from a scuba diver at a depth of 18.0 m has adiameter of 1.0 cm. Assuming constant temperature, what will be the diameter of the bubble when it reaches the surface?Known: h = 18 m

d1 = 1cm

T1 = T2

Ask: d2= ?

Answer:

Situation 1 = d1= 1cm h = 18 m P1 = P0 + Ph;

Ph = 1800 cm water . 1 cm Hg/13,6 cm water =132,4 cm Hg

P1 = 207,4 cm HgSituation 2 =P2= P0 = 75 cm Hg P1 . d1

3 = P2 . d23

d23 =P1 . d1

3

P2

= 207,4 cm Hg/75 cm Hg . 1= 2,8cm

d2 =

= 1,4 cm

9. An ideal gas, consisting of n moles, undergoes an irreversible process in which the temperaturehas the same value at the beginning and end. If the volume changes from Vi to Vf , the changein entropy of the gas is given by:A. nR(Vf-Vi)B. nR ln(Vf-Vi)C. nR ln(Vi/Vf )

D. nR ln(Vf/Vi)E. none of the above (entropy can’t be calculated for an

irreversible process)Answer: D Cause S= Q/T

T=CONSTANT;and at cycle carnot U = 0 U = Q – w ; Q=w ; =nRT In(Vf/Vi)S= nRT In(Vf/Vi)/T= nR ln(Vf/Vi)

10. One mole of an ideal gas expands reversibly and isothermally at temperature T until its volume is doubled. The change of entropy of this gas for this process is:A. R ln 2B. (ln 2)/TC. 0D. RT ln 2E. 2R

Answer= A, Cause S= Q/T ; T=constant and at cycle carnot U = 0 U = Q – w ; Q=w ; =nRT In(V2/V1)S = nRT In(V2/V1)/TS = 1.R. In(2V1/V1) S = R In 2

Kurnia

1. A homogeneous plate in the form of and size as the figure beside,determine the coordinate of weight point.

Y

6

4

2

X2 4 6

SOLUTION: Y

6

4

2

X2 4 6

Area of every plane Weight pointof every planeA1 = 2 X 6 = 12 X1 = 1 Y1 =3 A2 = 2 X 4 = 8 X2 = 3 Y2 =2A3 = 2 X 2 = 4 X3 = 5 Y3 =1

So,weight point of plane = (X,Y)

X= Y=

= =

= =

22

2

1

= =

=2,34 =2,34 Answer = (X,Y) =(2,34;2,34)

a.2,34 and 3 d.3 and 3b.3 and 2,34 e.2,5 and 2,5c.2,3 and 2,34

2. Look at the picture!!!!!!!!!!The magnitude of string T3 and W is……..

a.4800N and d.1600N and N

b.800N and 1600N e.1600N and 4800N

c.800N and N

Based of the picture

T =800N

So,T =800N and W=1600N

3. The xylem in a plant can transport water from land to top of plant. If contact angel ,water surface tension 0,072

N/m and gravitational acceleration g=10 m/ what is the radius of the xylem ifthe height of raised passing through the xylem = 0,6 meter!

a.2,4 x 10 d.2,4 x 10

b.2,5 x 10 e.2,5 x 10

c.2,6 x 10

h= r =

r =

r =

=2,4 x 10

4. A pipe with area 1200 ,flows an ideal fluid with speed 3m/s. Calculat volume of fluid flowing 10 minute!

a.23000L d.30000Lb.21500L e.21600Lc.24000LQ=A.v

Q= V=Q.t

=216m =21600L5. Look at the picture!!!!!!

Beside of the picture ,where is the point with largest the pressure at the vessel?

a.A d.Db.B e.all is samec.C

Answer:ABecaues the longer the vessel location from the source of

water, then the smaller The pressureat the vessel.

6. A tank that is 24 m high from land is filled up with water .A valve (tap) placed 4 m under water level in the tank. If the valve is opened,what is velocity of water spray?

a.25m/s d.35m/sb.22m/s e.20m/sc.30m/s

Answer:v

=

==20m/s

7. Determine the effective velocity of a gas particles at normal state,if the gas density is and its pressure is

!!!!

a.450m/s d.600m/sb.540m/s e.650m/sc,800m/s

V =

=

23

W400N

60° 30°

T1 T2

source

Water

A BDC

flow

h

1h

2h

= = =600

8. In adiabatic process if the volumeof an ideal gas is enlarged 4 times of its initial volume, and its internal energy turn out to be 3 times its initial energy,than the pressure of the gas becomes?

a.2P d.4P

b. e.

c.

9. A carnot engine uses a high temperature reservoir with temperature book and efficiency 50%. In order to increas the efficiency to 60%.The temperature of the high reservoir must be increased to…….

a.400K d.450Kb.350K e.500Kc.300K

10. If a carnot engine work at temperature of 200 K and 800 K. Then its efficiency is…………

a.50% d.70%b.75% e.65%c.60%

=75%

Ihsan

1.Six of particles such as picture shown below, connected by a light rigid rod of negligible mass (massa diabaikan). Determine the moment of inertia of the particles to the axis : a.axis AB b.axis CD

A m 2m m C 2m m m

B 2a a a a 2a

Answer:

a.axis at AB: I=mr2

=0+2m.4a2+m.9a2+2m.16a2+m.25a2+m.49a2

=8ma2+9ma2+32ma2+25ma2+49ma2

=123ma2

b.axis at CD I=mr2

=m.16a2+2m.4a2+m.a2+0+m.a2+m.9a2

=16ma2+8ma2+ma2+ma2+9ma2

So, moment of inertia about the axis of AB and CD are 123ma2

and 35ma2

2.A solid cylinder with radius of 15 cm and 7 kg mass moving with a velocity of 40 m / s while rotating. Determine the total kinetic energy experienced by solid cylinders!

Answer:

Ek = Ektrans + Ekrot

= ½ mv2 + ½ Iω2

= ½ mv2 + ½. ½. mr2 .V/r2

= 3/2 mv 2

= ¾ . 7 . 1600 = 8400 J

Thus, the total kinetic energy experienced by solid cylinders is 8400 J

3. An object weighing 200 N homogeneous, AB whose length L and weight 400 N is in a balanced state in the given hinge rod as shown. Determine the magnitudes of the tension!

200 N 53o

¼ L

400 N

Answer: Στ = 0Ty.L-W. ¾ L-W batang.1/2 L = 0T sin θ. L = W. ¾ L + W batang. ½ L T sin 53. L = 400. ¾ L + 200 ½L T sin 53. L = 100. 3 L + 100 LT sin 53. L = 300 L + 100 LT sin 53. L = 400 L

T = 400/0,8 = 500 N

Thus, the magnitude of the tension force is 500 N

4. A ladder homogen weighing 400 N rests on a smooth wall to be climbed by a person that weighs 800 N.. If the length of 10 m already climbed the ladder the person is as far as 6 m from the floor and stairs with a floor slope of 37o, determine the coefficient of friction with the floor!

Answer:

24

W

800 N

4 m

400 N

1 m

37o

So, the coefficient of friction with the floor is 2/35. Calculate the absolute pressure at depths of 1000 m above sea level. Assume water density 1.0 x 105 kg/m3, the acceleration of gravitaty is 9.80 m/s2, and P0 = 1.01 x 105!

Answer:P=P0+ℓgh

=1,01x105 + (1,0x103) (9,80) (1000)

=0,101x106 + 9,80 x 10

=9,9 x 106 Pa

Thus, the absolute pressure is 9.9 x 106 Pa.

6. A long pipe has three distinct sections. Sequentially 1,2,3-sectional area is 200 cm2, 100cm2, dan 400cm2. If velocity of water through the section 1 is 6m / s, calculate the velocity of water through parts 2 and 3!

Answer:

A2v2=A1v1 A3v3=A1v1

v2=A1/A2.v1

v3=A1/A3.v1

=200 cm2/100cm2 x 6m/s =200cm2/400cm2 x 6m/s

=12m/s=3m/s

Thus, the velocity of water through parts 2 and 3 is 12ms and 2m / s.

7. A tube pump its volume 200 L contain nitrogen gas at temperature 0o C and pressure 10 ATM. At certain time, the piston 15 pushed down in a way that its volume 50 L. as a result temperature of gas 123o C what is current pressure?

Answer:

So, the current pressure is 58,60 ATM…

8. In a enclosed container there is gas whose molecule has effective velocity of 200 m/s. if the gas temperature altered into 8 time of its. Initial temperature, determine the final effective velocity!

Answer:VRMS2=…?

VRMS1/VRMS2= VRMS2

So, the final effective velocity is 560m/s

9. 4 mol of air with 5 L in volume at 400 K and pressure 2 ,02 x 105Pa. is heated 523K. Calculate the final volume!

Answer:

R=8,314 J/mol

P∆V=nR∆T

2,02. 105(V2-5.10-3)=4.8,314.123

V2-5.10-3=4. 8,314.123 / 2,02 .105

V2=4. 8,314.123 / 2,02 . 105 +5.10-3

10. 4 grams oxygen as is heated from 260 K to 320 K. If it is at constant pressure. The heat needed is 4 kJ. if it is constant volume, the heat needed 1,5 kJ .if the relative mass of oxygen is 28 g/mol, determine the heat capacity of gas at constant pressure, constant volume, and Laplace constant…!

Answer:

Cp=Q p /∆t=4000J/60K =66,67 J/K

So, the heat capacity of gas at constant pressure is 66,67 J/K

Cv=QV/∆t=150J/60K

=2,5J/KSo, the heat capacity of gas at constant volume is 2,5 J/K

γ =Cp/Cv

=4000J/60K:150J/60K =26,6J/K

So, the Laplace constant is 26,6J/K

Agung1. A solid cylinder with mass of 20 kg and radius of 10 cm

moves at velocity 50 m/s while rotating. Determine the kinetic energy of the cylinder.

A. 15.000 NB. 14.000 NC. 13.000 ND. 9.000 NE. 7.000 N

Solution 20 kg

10 cm = 0,1 m 30 m/s

kgm

rad/s

N

2. A wood ball with mass 100 N and radius 100 cm. Determine its inertial moment.

A. 0,04 kgm

B. 0,2 kgm

C. 0,4 kgm

D. 2 kgm

E. 4 kgm

Solution100 cm = 1 m

25

kg

kgm

3. A homogenous bar with 9 m in length and 8 N in weight is force down ward o held by support 3 m from the left end. If a load weighing 10 N is hung on the left end. Determine the magnitude of pulling on the right end.

A. 44 NB. 21 NC. 22 ND. 42 NE. 41 N

Solution

R – 10 N – w – = 0R - 10 N – 8 N - = 0R – 18 N - = 0

(- 10 N) (3) + (R) (0) + (8 N) (1,5 m) + (F) (6 m) = 0 - 30 Nm + 12 Nm + F6Nm = 0

- 18 Nm + F6Nm = 0

F = 3 N

Now we seek the reaction force by the support (R) to the bar insert the value F = 3 N

R – 18 N – 3 N = 0 R = 21 N

4. A homogenous bar AB 4 m in length while its mass 8 kg. A is held by a string BD. If at C is given a load of 4 kg and the bar lies perpendicularly to the wall and the tension 62,5 N (g = 10 m/s )

A. 0,54B. 0,44C. 0,34D. 0,22E. 0,65

SolutionAB = 4 m

8 kg

4 kg

g = 10 m/sT = 62,5 N

m

N

N

N – (0,6) T = 0N – 0,6 (62,5) = 0N = 37,5 N

2

5. If the velocity of air flow at the down side of a plane’s wings up is 60 m/s. what is the velocity at the side of the plane’s wings if the upward pressure obtained is 10 N/m (

1,29 kg/m )A. 60,13 m/sB. 56,13 m/sC. 40,00 m/sD. 60,15 m/sE. 65,15 m/s

SolutionN/m

m/s

= 3615,5 m /s

6. A closed vessel contains 20 L oxygen gas. If the gas is at a temperature of 27C and atmospheric pressure 1 atm. Determine the number of moles of the oxygen gas in the vessel.

A. 0,80 moleB. 0,44 moleC. 0,15 moleD. 0,34 moleE. 1,12 mole

Solution20 L = 20 x 10 m = 2 x 10 m

K

Pa

Mole

7. A container with temperature of 67C and pressure of 1.2 x 10 Pa contains 2 g helium with molecule mass being of 4 g/mole. Calculate the internal energy of the gas.

A. 2118,4 JB. 2234,5 JC. 1561,5 JD. 2404,4 JE. 2640,1 J

Solution67 C K

a

2 g4 m/g

J/K

AN molecule/mol

molecule

26

m3 m6

RN10 w F

J

8. A monoatomic ideal gas is compressed

adiabatically and the volume decrease to its half. Determine the ratio the final pressure to the initial pressure.

A. 4

B. 2

C. 6

D. 4

E. 4

Solution

=

9. One mole of gas is compressed at a constant temperature of -23C so that its initial volume. Calculate the work done by the gas.

A. -1,4 ln ½ kJB. 4,40 ln 1 kJC. 2,6 ln ½ kJD. -2,0 ln 1 kJE. 2,08 ln ½ kJ

Solution1 mol

= 250 K

J/mol K

Putra1. Tom and Jerry launch wheel, bowling, cylinder barble,

basketball, and a wet soap at rest which the same height on a incline plane. The order of these matters which arrive on the floor are …

a. wet soap- wheel- barble- basketball- bowlingb. wet soap- bowling- barble- wheel- basketballc. bowling- wet soap- wheel- basketball- barbled. wet soap- wheel- basketball- bowling- barblee. basketball- wet soap- wheel- barble- bowling

SOLUTION

the more distributed the matter mass to its axis, k >>the greater value of k, 1+k >>

the greater value of 1+k, v <<so the velocity determined by the value of k matter,

Wheel , k = ½

Bowling , k = 2/5

Barble , k = ½

Basketball , k = 2/3

Wet soap so v is the greatestThus the order of these matters which arrive on the floor are: wet soap- bowling- barble- wheel- basketball

2. A hollow flimsy ball let slides from the top incline plane which length 24 m. To reach the bottom of incline plane it need 6 second, so the height of incline plane are …

a. 1 mb. 1,3 mc. 2,52 md. 3,24 me. 4 m

SOLUTION

3.

Weight of load B is …a .150 Nb. Nc. 600 Nd. 900 Ne.1200 N

SOLUTION

4. Choky has weight 600N and Yuanita has 500N. They sit on the wobble up-down (jungkat-jungkit) which lenghth 5 m and its weight 100N. If Yuanita sit on one edge, where is Choky must be sit so that their position are balance?

a. 0,42 mb. 1,08 mc. 2,08 md. 2,25 me. 2,5 m

SOLUTION

5. A spoon has weight 0,5N when it weighed by spring balance in air and it shows 0,25N when completely

27

A B

60°B

30°

T2T3

W = 300 N

T1

6 m

A

N N

immersed in oil (ρ = 800 kg/m3). Determine the density of spoon!

a. 6400Nb.1600Nc.1200Nd.800Ne.400N

SOLUTION

6. The xylem in a plant can transport water and mineral materials from land to top of plant. If the radius of the xylem is 10-5 m, contact angle 0°, water surface tension 0,072 N/m and gravitational acceleration, g = 10 m/s2, what is the height of water raised passing trough the xylem?

a. 0,144 mb. 144 mc.14,4 md. 0,0144 me.1,44 m

SOLUTION

Thus water in xylem will raise 1,44 meter high.

7. Six litre gas at 47˚C and pressure 1 atm compressed until the volume into 3 litre and boiled so the temperture into 87˚C. How is the final pressure of gas?

a.22,5 atmb.2,25 atmc.3,7 atmd.4,5 atme. 5,6 atm

SOLUTION

8. If in a closed room the gas boiled until point T° K, so …a. molecule potensial energy be smallerb. these molecules kinetic energy = 2/3 NkTc. these molecules kinetic energy = 3/2 NkTd. gas volume always increase because gas is expande. the pressure of gas is constantSOLUTION- Ep has no relation in it- these molecules kinetic energy = 3/2 NkT only if the gas

is monoatomic- P will increase because the volume is increase, Pv = cSo the true statement is gas volume always increase because gas is expand

9. An ideal gas trough cycle proccess like on the diagram P-V bellow. Work that produced are … kJ.

a. 200b. 400c. 600d. 800e. 1000

SOLUTIONWork that produced in the cycle is equal to wide area on graphyc P-v W = wide area a.b.c.d

10. A Carnot engine, if its high reservoir has temperature 400ºK it will has eficiency 40%. If its high reservoir has temperature 640ºK, its eficiency becomes … %.

a. 50,0b. 52,5c. 57,0d. 62,5e. 64,0

SOLUTIONCarnot engine with T1 = 400 ºK η1= 40% If T1 = 640ºK η2= ?

Reza1. A rotating wheel with the axis z axis and its angular

acceleration is given by α = (1.86 rad / s ²) t, with t in seconds. Determine the angular velocity at t = 2s, if the initial angular velocity of -2.9 rad / s.

A. 0,82 rad/s D. 1 rad/sB. 0,72 rad/s E. 1,2 rad/sC. 0,93 rad/s

Answer:For α = α (t) given the angular velocity ω is obtained by integration.

ω = ω 0 + ∫ α dt ; ω 0= -2,9 rad/s ω = -2,9 + ∫ 1,86t dt ω(t) = -2,9 + 0.93t² ω(t) = 2 = -2,9 + 0,93(2,0) ²

= 0,82 rad/s

2. Known torque value of an object is 20 Nm, and the force used 5N. If the force is perpendicular to the arm, then how large is the torque arm in metre?

A. 2 D. 5B. 3 E. 6C. 4

Answer:τ = l F then l = τ/Fτ = 20/5 = 4m

3. Which of the following items which are in stable equilibrium?

A. D.

E. B.

C.

Answer : Stable equilibrium is the balance of things, if given the force

on the object, then after a while the force was removed, then the object back to its original state. B

4. A homogeneous bar is 16 m in length and 8 in weight. At the right end of the bar is interested a small load weighting 4 N, as seen in the figure. Determine weight point of system of bar and load!

28

P (1

05 Pa)

3

13

013

2013

42013

a42013

ba42013

cba42013

dcba42013

V (m3)

A. 4m D. 7mB. 5m E. 8mC. 6m

Answer:

= 4m

5. A tube filled with water until full (ρ=1g/cm3) and oil (ρ=0,9g/cm3). What is the high ratio of water and oil to prevent hydrostatic pressure at the bottom of the tube = 9600 Pa? (G = 10 m / s²)

A. 2:3 D. 3:4B. 3;2 E. 4:3C. 2:1

Answer: Hydrostatic pressure at the bottom of the tube, ρ = 9600

Pa. Water height h1, oil height h2.Bottom of the tube gets heavy load of water and heavy oil so that :

ρ = ρ1 + ρ2

ρ = ρgh1 + ρgh2

9600 = (1000)(10) h1 + (900)(10) h2

9600 = (103)(10h1+9 h2

6. Heavy objects in the air 3000 N. If weighed in water weighs 200 N. If the type of water 1000 kg/m3, determine the volume of objects!

A. 2,9 x 10-2 D. 4,5 x 10-2

B. 3,5 x 10-2 E. 4,6 x 10-2

C. 2,8 x 10-1

Answer:wu = 3000Nwa = 200Nρ = 100kg

7. How many oxygen molecules per cubic meter of air at

normal temperature and pressure (OOC and 1 atm) (1atm = 105 Pa, R = 8314 J kmol-1 K-1, Na = 6.02 x 10 molecules per

A. 265 x 1025

B. 260 x 1025

C. 267 x 1026

D. 270 x 1026

E. . 263 x 1027

Answer:

8. A gas tank containing 8 kg of O2 (M = 32 kg / kmol). What is the number of molecules contained in the tank?

A.

9. A heat machine receives calor 2000 kalori from high temperature reservoir and remove the heat of 1600 calories in a low temperature reservoir. If the machine is ideal, then calculate the efficiency of these machines!

A. 0,1 D. 1,5B. 0,2 E. 1,2C. 0,3

Answer:

10. The coefficient of performance of a refrigerator is 4.5. How much energy is used to move the 3600 Joule heat from the food in refrigerator?

A. 7B. 8C. 9D. 10E. 11

Answer:

1. A wheel of mass 5 kg and 40 cm radius rotating with angular velocity 5 rad / s. Moment of inertia and kinetic energy wheel is ...

a. – 10 J d. 12 Jb. 10 J e. 0,1 Jc. – 12 J

Solution : m = 5 Kg r = 40 cm = 0,4 mw = 5 rad/s

29

600

T1 T2

300

4 m

h = 3 m

moment of inertiaI= mr

= (5 Kg)(0,4 m) = 0,8 Kgm2

Rotational kinetic energy Ek rotasi = ½ I

= ½ (0,8 Kg m2)(5 Rad/s) = 10 J

2. A solid wheel has a moment of inertia of 10 kgm at rest. A moment of a force of 50 Nm to work on these wheels. Large angular acceleration experienced by the wheel is ... rad/s

a. 5 d. 7b.6 e. 3c.8

Solution :I = 10 Kgm

= 50 Nm

= Iα 50 = 10α α = 5 rad/s

3. Which of the following items which are in labile equilibrium? A. D.

E. B.

C.

Solution : Labile, which, if given the force equilibrium, it can not return to its original state. B

4. A box was hung like the picture below. System in balance. Large rope tension T1 is ... N

a.500 d.800b.600 e.900c.700

Solution :W= 800 N

T1/sin(90+30)=W/sin(60+60)T1/ ½ = 800 N / ½T1 = 800 N

5. A length of pipe has three different cross-sectional area of 1.2, and 3 row is 200 cm , 100 cm , and 400

cm . If the velocity of water through a section is 6 m/s, the velocity of water through parts 2 and 3 is ...a. 3 m/s and 12 m/s d. 15 m/s and 10 m/sb. 10 m/s and 15 m/s e. 4 m/s and 8 m/sc. 12 m/s and 3 m/s

Solution :A1 = 200 cm A2=100cm A3=400 cm V1=6 m/s

V2= .V1

= 200/100 (6)

= 12 m/s

V3= . V1

= 200/400 (6) = 3 m/s

6. A water supply tank has a lid that given the hole so that outside air can enter the tank at the top. Bottom of the tank was 4 m below the surface of the water in the tank and a faucet located 3 m below the surface of the water in the tank. How much velocity the water when the tap in open?

a. 3 m/s d.

m/s

b. 4 m/s e. 2

m/s

c. 58 m/s

Solution :h = 3 m

V =

=

=

= 2 m/s

7. A volume of a cylinder which contains 5 mol 1 m3 of helium gas at a temperature of 77 C. If heliun considered an ideal gas. What is the gas pressure in the cylinder?a. 1,4x10 d. 1,4x10

b 1,4x10. e. 1,4x10

c. 1,4x10

Solution :T = 77 C = 273+77=350 K

Preasure of ideal air

Pa

8. Gas contained in a tube which was closed by the piston is heated up to 127

C temperature increases. If the

pressure tubes equipment 2.5 x 10 Pa

and the initial volume 3 m and initial

temperature 27 C. So the work done is…a. 1,5 Joule d. 4,5

Joule

b. 2,5 Joule e. 5,5

Joule

c. 3,5 Joule

Solution :T1 = 27 C = 273 +27 = 300 K

T2 = 17 C = 273+127= 400 K

V1= 3 m

W=

30

=

= =

= 4 m

= 2,5 Joule

9. A monatomic ideal gas (γ = 5 / 3) in adiabatic compressed and the volume is reduced by half. Final pressure of gas is ...

a. 2 P d.2 P

b. 2 P e. 2 P

c. 2 P

Solution :

V = ½ V

So :

= 2 P

10. In the P-V graph for the following carnot machine. Business note produced by the engine at each cycle is W = 5000 J.

When heat is absorbed at 15 000 J, so the quantity of heat released by the engine per cycle is ...

a. 10 J d. 10000 Jb. 100 J e. 100000 Jc. 1000 J

Solution : W = 5000 J Q = 15000 J

15000 - Q = 5000

Q = 10000 J

Riska1.A ballerina with stretched arms rotates with angular velocity of 1,2 rad/s on the smooth horizontal floor. The inertial moment conserved by the ballerina is 4,0kgm2 . If she folds her arms then the inertial moment decreases 10% from the initial. What is the angular velocity of the ballerina when both arms are folded.A. 1.33 rad/sB. 1.27 rad/sC. 3.6 rad/sD. 1.11 rad/sE. 2.22 rad/s

SolutionKnown:1=1,2RAD/S

I1=4kgm2

Ask :2=?Answer:

I2=I2-10%I1

=4-(10%*4)= 4-0,4=3,6kgm2

L1=L2

I11=I22

2=I1/I2*1

=4/3,6*12=1,33rad/s

2.Solid cylinder with 10kg in mass and radius of 10cm can rotate on a horizontal axis.A string that has a neglectable mass is wrapped on the cylinder.Then the end of the string is pulled with a static force of 20N.If the mass of axis and friction between axis and its rotary place are neglectable,determine:a.Angular acceleration of solid cylinderb.Angular velocity of solid cylinder after rotating in 1secondA. 40,20B. 40,30C. 40,40D. 40,25E. 40,35SolutionKnown:m=10kg

I=1/2mr2

r=10cm=0,1mF=20N

Ask :α,?Answer:

Α=τ/I=Fr/1/2mr2

=20*0,1/1/2*10*0,1=40rad/s

=0+α.t=0+40*1=40rad/s

3.An object wita mass 0,6kg is hung at a rest as illustrated in the figure below.Determine the tensions at both strongs that retard it.(g=10m/s)

300 a. 6√3, 12b. 12, 6√2c. 6√3, 11d. 6√3, 10e. 6√3,13

SolutionKnown: m=0,6kg G=10m/s2

Ask :T1 and T2Answer: W=m.g =0,6.10=6N α1=900+300=1200

α2= 900

α3= 360 – α1 – α2= 3600-1200=1500 sin α3= sin1500=1/2

T1 m.g T2 m.g= =

Sin α1 sin α3 sinα1 sinα3

T1 6 T2 = 12N=

1/2√3 1/2

T1= 6√3

4.A homogenous bar AB is 4m in length while is mass is 8kg.It is held by a string BD. If at C is given a load of 4kg and the bar lies perpendiculary to the wall(g=10m/s2).Calculate string tension!a. 52,5D b. 62,5 c. 42,5 d. 63,5

31

a b

d c

m

e. 53,3A B

C

Known:AB=4kg mAB=8kg mC=4kg g=10m/s2 AC=1/4AB=1/4.4=1m WAB=80N WC=40NAsk :T=?Answer: ∑Fx = 0N-0,6T=0N= 0,6T

∑Fy= 0F+0,8T-40N-80N= 0F+0,8T= 120N

∑I= 040.1 + 80.2 – 0,8T.4 = 0200-3,2T = 0T = 62,5 N

5.A body is that floats on water surface with water density 900kgm-3. calculate water density of solution!a. 1200 kgm-3b. 1100 kgm-3c. 1120 kgm-3d. 1020 kgm-3e. 1002 kgm-3

Solution:Vbf-Vb-1/4Vb=3/4VbΡb/ρf = Vbf/Vb 3/4Vbf = ρb = 3/4

Vb ρf

Ρf = 4/3 ρb = 4/3 . 900 = 1200 kgm-3

6. A dorm that open is filled with water to height H. There is a hole in the thank which is located at a height h, with area 3,0 cm2 , so the water flowing from that with Q = 1,8 L/s. if distance water with land 1,5m, calculate H.a. 2,11 e. 12,5 b. 2,12c. 2, 13d. 11,2

Solution:V = √2ghV2 = 2ghh = v2/2gh= 62/2.10 = 1,8 m

R = 2 √h(H-h)R2 = 4h(H-h)R2/4h = H-h

H = h + R2/4h = 11,8 + 1,52/4 (1,8) = 1,8 + 1,5.1,5?4. 1,8 = 2,11m

7. A thank contains Argon gas with relative atomic mass of 40 kg/Kmol at temperature of 270C. Determine:a. The average translation at kinetic energy for moleculeb. affective velocity

a. 6,21.10-21(433,5)b. 6,21.10-21(342,5)c. 6,21.10-21(432,5)d. 6,21.10-21(243,5)e. 6,21.10-21(234,5)

Solution:Ek = 3/2 kT = 3/2 (1,38.10-23)(300= 6,21.10-21)

Veff = √3RTMr

= √3. 8,314.30040

= 432,5 m/s

8. The kinetic energy of 2 moles monoatomic gas in a container with volume of 2,5 L is 1,01 x 10-20J. Determine the gas pressure in the container!

a. 2,69.1021,5

b. 2,69.10-11

c. 2,69.10-19

d. 2,69.10-22

e. 2,69.10-21

Solution:

PV = 2 Ek?3V = 2. 1,01. 10-2 = 2,69.10-21 J/L3. 2,5

9. A carnot engine which uses 800K hot reservoir has efficiencies 90%. What will the temperature of hot reservoir if the efficiencies 50%?a. 462,4b. 465,4c. 426,4d. 466,4e. 467,4

Solution:

h = 1- T2/T1 x 100%2/5 = 1- T2/T1T2/T1 = 3/2T1 = 1600/3 = 533 K

50/100 = 1- T2/T10,2-1 = - T2/T10,8 = T2/533T2 = 426,4 K

10. Several foods in a refrigerator produce heat of 2.800 J. Given that the coefficient of the refrigerator is 2. Determine the electric energy!

a. 14.10-2

b. 14.10-3

c. 1,4.103

d. 1,4.104

e. 1,4.10-2

Solution:W = Qc/K = 2800/2 = 1400 J

Sahad1. A thin hollow cylinder weighs 60 N, the cylinders are

working moment of a force of 48 Nm, if the diameter of the cylinder is 180 cm, calculate the angular acceleration of the cylinder…

w = mg

m = = = 6 Kg

D = 180 cm = 1,8 mr = 0,9 m

= I

= = = = 9,87 rad/s

2. Mass of a solid ball 10 kg rolling on an inclined plane with a speed of 4 m/s, if the radius is 12 cm, calculate the kinetic energy of the solid ball…

32

=

3. Determine the mass of the box on the image if T =50 N…

4. A homogeneus bar 7 m in length. 9 N in weight. At right end bar is inserted a small load weighing 5 N. as seen in figure. Determine weigh of point of system of bar and load…

5. An object floating on the two types of fluid, namely alcohol and water, if the body is 20% volume in alcohol, 60% in water, and the rest on the surface of alcohol, calculate the density of the object… (Alcohol 800 Kg/m3 )

6. 56 grams diatomic gas at a temperature of 500 K, has a heat capacity of 3.5 nR at pressure constant, if the relative atomic mass of the gas is 28 kg/kmol, calculate the value of heat capacity at volume constant…

Diatomic gas (500K)

7.

If debit of fluid (water) in B is 250 m /s. Determine velocity of fluid in A

8. An ideal gas occupies space at a pressure of 8 atm at a temperature of 127 C, if the mass of the gas relatih 32 g/mol, calculate the mass of gas type...(R=0,082Latm/molK)

9. Nitrogen gas at a temperature of 27 C has a volume of 3.0 m3 and pressure of 6 x 104 N/m2. if the gas is compressed at a pressure of 8 x 104 N/m2 and its volume half of a first volume, calculate the gas temperature after the compression…

10. Carnot machine has efficiency 60% at a low temperature reservoir 300 K, determine the low temperature reservoir if the efficiency is 70%...

Tika1.A force of F=(6I+4J)N, woks on a point of a axis with positial vektor =(2i+5j).The torsio magnitude is….

a.6Nm d.9Nmb.7Nm e.10Nmc.8Nm

2.In a wheel of 15 kg.m2 momnt inertia , a constant torsi of 45 Nm is applied.If the radius of a wheel is 2m.Determine the acceleration…..

a.3 d.6b.4 e.7d.5

3.An objectof 300N wight is hung in equilibrium as seen in the picture below.The magnitude of rop stress is…..

a.150N d.400Nb.120N e.150Nc.240N

33

A B25m

090 060

2T

1T

300 450

450

1T 2T

4.Determine the coordinate of the center gravity from picture below…

a.(-6, 4,5) d.(2,3)b.(-4,5 , 6) e.(-2,3)c.(6, 4,5)

0 2 6 8

5. A pipe has two different cross sectional areas with diameter 20 cm and 40 cm. If the speed at the small area is 10 m/s, determine the speed at the large area….

a.2,5 m/s d.5 m/sb.3,5 m/s e.6 m/sd.4,5 m/s

6.A cylindral container having a lage cross sectional areas is filled with water heigh 100 cm..Ther is a narrow hole at a heigt 20 cm from th bottom.If G= 10 m/s2.Calculate the scape speed of water trough the hole…a.2m/s d.5 m/sb.3m/s e.6m/sc.4 m/s

7.Akind of ideal gas at pressure p and temperature 370C is compressed in a way that’s its volume becomes half of its initial value.If the tempratur 540C , the final pressure is…..a.0,5 p d.2,6pb.2,1 p e.4,5 pc.3,1 p

8. A rubber ballon with volume 40L is filled with oxygen at pressure 120 atm and temperature 270C. Determine the mass given that R=0.082 L.atm/mol.K…a.5 kg d.6,5 kgb.6 kg e.6,2 kgc.7kg

9.An amout of 2 m3 helium gas at temperature of 270C is hatd isobarically to 870C . If the pressure of the helium gas is 2.105 n/m2.Its external work will be….a.8kJ d.11kJ

b.9Kj e.12Kjc.10Kj

10.A coolant engine has performance coefficient of 5.If temperature In room is 270C the temperature in ngin is…a.-230C d.-500Cb.-250C E.-450cc.-300C

Pembahasan

1. torsi = F.r =(-6i+4j).(2i+5j)

=-12+20 = 8Nm

2.

= 45/15 =3

a= r.α = 2.3

=6m/s2

3. w/sin 90 = t1/sin 150 300/1 = t1/0,5 t1 = 150 N

4. A1=32 x1,y1 = 6,2 A2 = 24 x2,y2 = 6,8 X = A1.x1 +A2.x2

A1+A2

X = 32.6 +24.6 / 56 = 336 / 56 = 6

Y = A1.y1 + A2.y2A1 + A2

= 32.2 +24.8/ 56 = 256 / 56 = 4,5

5. V1 = D22

V2 D12

10/V2 = 0,42/0,22

V2 = 2,5 m/s

6. V =√2gh = √2.10.0,8 = 4 m/s

Uje1. A particle undergoes a rotational motion because the moment gaya30 obtain N m, if the angular acceleration happens adalh 3 rad / s, what is the moment of inertia of the particles.

a. 5d 0.5 kgm. 3 kgm b. 2.5 kgm e. 5 kgm c. 1 kgm Answer:

τ-I then I = α

Inertia moment of the particle:

I = 1 kgm

2. What is the kinetic energy of rotation of the disc with mass 2kg and has a radius of 2 m rotates at 300 rad / s?

a. 2.8 x 10 J d. 4 x 10 J b. 1.8 x 10 J e. 2 x 10 J c. 3.5 x 10 J

Answer:

Moment of inertia of these discs:

I = mr = (2 kg) (2 m)

= 4 kgm

Kinetic energy of rotation with = 300 rad / s is

EK = I

= (4 kg m )(300 rad/s)

= 1,8 x 10 J

3. Field of homogeneous objects in the picture next to have the following sizes. .

ac = ac = centimeters

34

ac = cd = de = eb = 4 cm What is the point where the weight towards the point of origin

4. When this system is known in equilibrium, how much rope tension.

a. 20 N b. 40 N c. 10 N d. 50 N e. 56 N Answer: According to the sine rule:

T =

= ( X 100) N

= 50 N

5. Basically a tube 20 cm wide, entered the water as deep as 5000 cm. a. What is the height of water in these tabungg b. How large hydrostatic pressure on the bottom of the tube?

a. 3.5 m & 1 x 10 N / m b. 1.5 m & 2 x 10 N / m c. 2.5 m & 2.5 x 10 N / m d. 3 m & 1.5 x 10 N / m e. 1 m & 2 x 10 N / m

Answer: A = 20 cm V = 5 x 10

a. High water V = A h

h = = cm = 2,5 m

b. Pressure based

h = = cm = 2,5 m

6.Sebuah tank containing water is placed on the ground, water surface height is 1.25 m from the ground, There is a high leakage hole 0.8 m from the ground, What is the velocity of water out of this hole?

a. 0.5 m / s b. 1.5 m / s c. 2m / s d. 3m / s

e. 5m / s

Answer: With the water level height H 1.25 m - 0.8 m = 0.45, g = 10 m / s. a. The speed of the water out. V =

V = = V = 3 m/s

7. 20,78 Lexample argon gas at temperatures 27 C and atmospheric pressure 1 atm (1 atm = 10, Pa). Determine the number of moles of argon gas in the sample. (R = 8.314 J mol).

a.0, 364 mol b.0, 608 mol c.0, mol 833 d.0, 196 mol e.0, 567 mol

Answer: Volume V = 20.7 L = 20.78 x 10 = 2.078 x 10; absolute temperature T = 27 + 273 = 300K; pressure p = 10 mol Pa.Banyak gas meter can be calculated using the ideal gas equation,  pV = nRT

n = ;

n =

=

= 0,833 mol

8.Sejumlah mass of gas is cooled so that volumnya reduced from 4 L to 2.5 L at a pressure of 1 x 10 Pa kanstan. what is out of business done by the gas.

a.1, 5 J b.15 J c.1500 A d. J-15 e. J-150

Answer: Isobaric process is calculated with the equation: W = p.ΔV So that W = 1 x 10 Pa(2,5 – 4)

= 1 x 10 Pa(-1,5 x 10

=-1,5 x 10 J = -150 J

Negative indicates the volume is reduced.

9.2,5 m neon gas temperature 52 ° C heated to 91 ° C. isobaric If the gas pressure is 4.0 x 10 N / m, Determine the work done neon gas.

a.1, 2 x 10 b.1, 3 x 10 c.1, 8 X 10 d.1, 5 x 10 e.1 4 x 10

Answer: Work done in the process of isobaric gas we can calculate with the equation:

W =

=

b

e d

a c

cm13cm13

4 cm4 cm 4 cm

35

=

10. A gas expands from 7.0 L to 8.2 L at constant pressure 2.5 bar (1 bar = 10 Pa) During the expansion of 500 J of heat added. Determine the energy change in the gas.

a.400 J b.200 J c.100 A d.500 J e.150 J

Answer:

We compute the gas business with the equation: W = p (V2-V1) = 2.5 x 10 (8.2 to 7.0) 10 = 300 J Q = +500 J, then the incoming heat is calculated with the equation Δu kesistem.Maka Q = W + Δu Δu = Q - W = 500-300 = 200 J

Yudi1.A solid ball rotates with angular velocity 6rad / s period of 2 kg and radius 50 cm. And the kinetic energy is ...

a.3,7 J c.3,2 J e.3,4 J

b.3,9 J d.3,6 J

solution:Ekrot = ½ x I x ω

=1/2 x 2/5 x 2 x (0,5)=1/5 x 2 x 0,25 x 36 = 3,6 J

2.

Determine the weight point of object

a.(1,2 , 5,6) c.(1,5 , 1,1) e.(1,5 , 2,8)

b.(1,5 , 2,4) d.(3,4 , 5,6)

3.A U pipe initially filled with water, then filled oil.tinggi other side of oil is measured from the boundary between water and oil is..(ρw=1000kg/m ,ρo=800kg/m ,hw=3cm)

a.2,4 cm c.2,0 cm e.3,75 cm

b.1,5 cm d.3,79 cm

solution:

pw x hw = ρo x ho

1000 x 3 = 800 x hh = 1000x3/800 =3,75 cm

4. A pipe has a cross-sectional area 500 cm . If the water velocity in the pipe 0.5 m / s. then discharge the water is ..

a.3,0 x 10 c.2,9 x 10 e.4,5 x 10

b.2,5 x 10 d.4,7 x 10

solution:

Q = A x V =5 x 10 x 0,5

=2,5 x 10

5. In the room-temperature 27 C and pressure 2 x 10 N/m , a gas mass 44 grams. If the number of moles of gas of 1 mole specify the type of gas ..a.7,6 kg/m c.5,5 kg/m e.1,8 kg/m

b.2,7 kg/m d.3,5 kg/m

solution:PV = nRT2.V = 1 . 0,082 . 3002V = 24,6V = 12,3p = m/V =44/12,3 = 3,5 kg/m3

6.A gas at a temperature of 400 K has a volume of 20 L and the pressure 5 x 10 N/m . If the gas is compressed so that the volume to 10 L at a temperature of 500 K. Determine the pressure ..

a.20 x 10 c.12,5 x 10 e.13,7 x 10

b.13,6 x 10 d.17,5 x 10

solution:

=

P2 = x P1 = x 5 x 10 = 12,5 x 10 N/m

Diatomic 7.Gas at 500 K has a degree of freedom 5/2. Calculate the average kinetic energy of each molecule ..a.1,25 x 10 c.2,5 x 10 e.3,5

x 10

b 1,75 x 10 d.1,5 x 10

solution:

Ek = x k x T

= x 1,38x10 x 500 = 1,72 x 10

8.Determine energy of a diatomic gas has a period of 1.4 grams, the temperature 4,81 x 10 and mr = 28..

a.4,5 x 10 c.6 x 10 e.6,7

x 10

b.7,5 x 10 d.8 x 10

solution:

36

U = x n x k x T

= x x NA x k x T

= x x 6,02x10 x 1,38x10 x 4,81x10 = 7 x 10

9.3 moles of monatomic ideal gas at constant pressure, the temperature increased by 300K.jika general gas constant R = 8.31 J / mol K, determine the effort that is required by the gas ..

a.6 x 10 J c.5,5 x 10 J e.8 x 10 J

b.7,5 x 10 J d.5 x 10 J1solution:

W = P . ΔV = n R ΔT =3 x 8,31 x 300 = 7,5 x 10

10.Efisiensy carnot engines that operate with low temperature T

and high temperature T is ...

a.10% c.50% e.100%

b40% d.90%

solution:

η = 1 - /1 x 100 % = 50 %

Iyan1. Calculate the value of torsion at point B as center that we

know L=12 m at system below.10N

1/6L 20N

1/3 L 1/3L 30N

a. 350 Nm40N

b. 340 Nmc. 368 Nmd. 358 Nme. 348 Nm

2. A solid ball has 10 kg in mass with 80 cm radius. The ball rotates by linear velocity 4 m/s.The value of inertial moment and angular momentum are…

a. 2,56 kg and 32 kg /s

b. 2,2 kg and 30 kg /s

c. 2,56 kg and 30 kg /s

d. 2,2 kg and 35 kg /s

e. 2,26 kg and 32 kg /s

3. In the system of rigid body equilibrium as depicted in the picture, AB is a homogenous rod of 40 cm length and 12 N weight. The weight of w = 20 N. the rope BC stress if AC = 30 cm is?

83,4 N80,4 N82,4 N84,4 N81,0 N

A D B

W AB 20 N

4. 6 the coordinate of center of

gravity from the picture is …

a. ( 4.6 , 2.3 )b. ( 2.3 , 4.6 )c. ( 5.6 , 2.3 ) d. ( 4.6 , 3.3 ) e. ( 2.3 , 5.6)

5. a drum of 1,5 m is filled with kerosene ( density = 0,8 g/cm3 ) to 3/5 of the drum’s height. If g= 10 m/s2 . find the pressure at the bottom of the drum…

a. 6200 N/m2

b. 7200 N/m2

c. 6400 N/m2

d. 7400 N/m2

e. 6300 N/m2

6. a cylindrical container having a large cross sectional area is filled with water to a height of 200 cm. there is a narrow hole at height of 150 cm from the bottom o the container. If g = 10 m/s2. the value of the escape speed of water through the hole is…

a. √10 m/sb. 10 m/sc. 1 m/sd. 5 m/se. 2 m/s

7. a rubber with volume of 30 L is filled with oxygen at pressure of 140 atm and temperature of 57 0C. the mass of oxygen that R = 8314 J/mol.K is….

a. 6,6 kgb. 6,0 kgc. 7,6 kgd. 6,3 kge. 5,5 kg

8. the kinetic energy of 4 moles monoatomic gas in a container with volume of 3 L is 2,02 x 10 -20 J. the gas pressure in the container is …

a. 4,4 x 10-21 atmb. 4,0 x 10-21 atmc. 5,4 x 10-21 atmd. 6,6 x 10-21 atme. 3,4 x 10-21 atm

9. a certain gas at constant pressure of 5,2 x 104 Pa is compressed in such a way that volume changes from 8 L to 4 L. in this process, the gas released heat of 250 J. the value of work that performed by the gas is …

a. 200,8 Jb. 196,8 Jc. 208,0 Jd. 198,8 Je. 189,8 J

10. an amount of 4 x 10 -3 mol gas expands isothermally from V1 = 30 cm3 to V2 = 60 cm3 at temperature of T = 300 K. if the universal gas constant is R = 8,3 J/mol.k. the work performed in the system is …

a. 6,9 Jb. 5,9 J

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c. 5,0 Jd. 6,4 Je. 5,4 J

Solution

1. ∑ =

= 30.L -10.sin . L + (40 sin 37 L -20 . L

= 30.12-10. . .12+40. . .12-20. .12

= 360-48-96-40= 368 Nm

2. known : m = 10 kg R = 80 cm = 0, 8 m V = 4 m/s

Question : I and L ??

I = 2/5 mR2 = 2/5 x 10 ( 0,8 )2

= 2, 56 Kg m2

L = mVR = 10 x 4 x 0,8 = 32 Kg m2/s C

3. AC = 30 cm T sin α = 0,6 AB = 40 cm T cos α = 0, 8 AD = ½ AB = 20 cm = 0,2 m

A D B

∑τ = 0 = w. AB + w.AB. ½ AB – T sin α. AB = 0 = 20 N. 0,8 + 12 N. 0,2 = T .0,6 .0,8

T = 40,o 0,48 = 83, 4 N

4. X1 = 5 X2 = 4 X3 = 5 Y1 = 1 Y2 = 3 Y3 = 5 A1 = 3 A2 = 2 A3 = 1

X0 = ∑An.Xn y0= ∑AnYn

∑An ∑An = A1Xi + A2X2 + A3X3

= A1Y1 + A2Y2 + A3Y3 A1 + A2 + A3

A1 + A2 + A3 = 3.5 + 2.4 + 1.5 = 3.1 + 2. 3 + 1.5

6 6

= 15 + 8 + 5 = 2,34 6 = 4,6( 4.6 , 2.3)

5. Known : ρ = 0,8 g/cm3 = 800 Kg/ m3 h = 3/5 x 1,5 = 0,9 m

question : P dasar ?

P = ρgh = (800)(10)(0,9) = 7200 N/m2

6. Known : h1 = 200 cm = 2 m h2= 150 cm = 1,5 m

P1=P2

V2 = √ 2g ( h1-h2) = √(2)(10)(2-1,5) = √10

7. known :

T = 570C = 330 KR = 8314 = 0,0821 L.atm/mol.KV= 40 LP= 140 atm

n = pV:RT = (140 atm)(40L) : (0,0821)(330) = 5600 : 27,093 = 206,6

m = n.Mr = 206,6 x 32 gr/mol = 6614 gr = 6,6 kg

8. known :

n=4 molv= 3 LEk= 2,02 x 10-25 J

PV= 2/3 Ek P = 2 Ek 3 V = 2 ( 2,02 x 10 -20 ) 3(3L)

= 4,04 x 10 -20 9 = 4,4 x 10 -21 atm

9. known : ISOKHORIKP = 5,02 x 10 4

V1= 8 L V2= 4LQ = - 250 J

W = P.∆V = P ( V2 – V3) = 5,02 x 104(4 x 10-3 - 8 x 10-3) = 5,02 x 104 x – 4 x 10-3

= 200,8 J

10. w = nRT ( ln 60)30

= 4 x 10-3 x 8,3 x 300 x 0,6 = 5,9 J

Rian1. 3 Balls are connected with 2m a bar as figure bellow with

mass 3,0kg/ balls. Determine the initial moment of system that rotates at point of O, if the distance O-B 0,4 m!

a O b c a. 5,48 kgm2 c. 6,48 kgm2

e. 8,48 kgm2

b. 4,48 kgm2 d. 7,48 kgm2

2. Two body system A and B at a pulley are connected by a rope, mass of A and B are 30 N and 50N. Pulley are smoothed. Determine the acceleration of system (g=10m/s2)!a. 1m/s2 c. 2,3m/s2

e. 3m/s2

b. 2m/s2 d. 2,5m/s2

3. Determine the coornidate of the center of gravity for an object as shown in the picture!

38

4

3

2

1 I ii iii

1 2 3 4 5

a.c.

e.

b.d.

4. A stair case gone up by someone in coordinate of the center of gravity of staircase. Determine the coeffient static friction, if the weight of human and staircase is 400 N and 100 N!

4m 5m

a. 0,253m c. 0,45e. 0,65

b. 0,55 d. 0,35

5. A vessel of 1,75m high is filled with mercury (

). Calculate the hydrostatics

pressure of the vessel! (g=10m/s2)!a. 238.000 Pa c. 240.000 Pa

e. 242.000 Pab. 239.000 Pa d. 241.000 Pa

6. An air plane has a total area of wings 50m2 move with velocity 450m/s and 300m/s. Calculate the lift force of Airplane’s wings!a. 3652000 N c. 3653000 N

e. 3656200 Nb. 3652500 N d. 3656250 N

7. Calculate the pressure of gas with volume 30 L from 3

mol monoatomic gas. Ek= J!

a. 1110 kPa c. 1310 kPae. 3210 kPa

b. 1210 kPa d. 2310 kPa

8. Calculate the internal energy of 2 mol gas of 1500K with has 4 translation, 2 rotation, and 2 vibrarion!a. 124.614 J c. 144.614 J

e. 421.130 Jb. 135.555 J d. 312.211 J

9. A system absorbs heat 2500 J from the environment and the system do work 4500 J. Determine the change of internal energy!a. -1500 J c. -2500 J

e. -3500 Jb. -2000 J d. -3000 J

10. Calculate the efficiency of engine that adsorb 600 J heat and released 1200 J !a. 75% c. 70,5%

e. 60,5%b. 70% d. 66,7%

THE ANSWER KEY

1. I=

= 3 (1)2 +3 (0,4)2 + 3 (1)2

=6,48kgm2

2.

=

= 2,5m/

3. I= 4 (1,1)ii= 8 (3,2)Iii= 3 (4,5 , 3,5)

= = =

= = =

NB

4. 4m 5m Na

Fk 3m wt wo

∑fy= 0

Nb-fk=0 -wt+wo+Na= 0Nb= 0 Na= wt+wo

= 500 NPoros di B

Na.L.cos

500.5. – 100. .5. -400. .5. - .500.5. = 0

750- .2000= 0

= 0,35

5. Ph = gh

= 13.600.10.1,75=238.000 Pa

6. .A

7.

2/3. 3,03x10-20( )

8. Ek=

= 10 (1/2. 1,38x10-23) 1500= 1,035x10-19 j U= N.Ek= n.Na.Ek= 2. (6,02x10-23) 1,035x10-19

= 124.614 J

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9. u= Q-W

= -2000J

10.= 1-1200/3600.100%

=

Wiwi1. A carrousel of 4 m diameter and 150 kg.m2 moment of

inertia rotates at angular velocity of 1,5 rotation per second. Seven kids of 30 kg mass each suddenly jumped in at sat at the hinges of the carousel.determine the angular velocity of the coursel!

a. 2,3 rpsb. 0,23 rpsc. 23 rpsd. 3 rps

Answer: r = 4 mI = 150kg/m2

W1 = 1,5 rotation/s = 1,5 rpsMass of the seven kids = 7(30 kg) = 210 kgIa = mr2

= 210 kg. 4 m2 = 840 kg.m2

L1 = L2

I1W1 = I1 + Ia W2

2. Seorang penari di lantai es memiliki momen inersia 4 kg.m2 4 kgm2, ketika kedua lengannya terentang dan 1,2 kgm2 ketika tangannya merapat ke tubuhnya. Penari mulai berputar pada kelajuan 1,8 putaran/s ketika kedua lengannya terentang. Berapa kelajuan sudut jika kedua lengannya merapat ke tubuhnya

a. 4 put/sb. 5 put/sc. 6 put/sd. 7 put/s

Answer:Tangan terentang

I1 = 40 kg m2

W1 = 1,8 put/sTangan merapat

I2 = 1,2 kg m2

L1 = L2

3. If the system below is in equilibrium, determine the string stresse of p!

45O

L P

300 N

a. 300 Nb. 350 Nc. 400 Nd. 450 N

Answer:

135o 45o

90o

P = W = 300 N

4. Determine the center of gravity from the picture! 1 m

2 m 2 m 1 m

a. (1,4 , 1,4)b. (1,3 , 1,3)c. (1,2 , 1,2)d. (1,1 , 1,1)

Answer:

3 x, y – (0,5 , 1,5) 2

1 0,5 m (2 , 0,5)

α benda I = 1x3 = 3 m2

α benda II = 2x1 = 2 m2

Coordinate the center of gravity = (1,1 , 1,1)

5. Sebuah dongkrak hidrolik memiliki pengisap kecil yang diameternya 5 cm dan pengisap besar yang diameternya 25 cm, jika pengisap kecil ditekan dengan gaya 400 n, berapa gaya yang dihasilkan pada pengisap besar?

a. 25.000 Nb. 20.000 Nc. 15.000 Nd. 10.000 N

Answer:Pengisap kecil D1 = 5 cmGaya F1 = 400 NPengisap besar D2 = 25 cm

6. Sebuah batu yang volumenya 2000 cm3 berada dalam air. Jika massa jenis air = 1 gr/cm3. Tentukan gaya apung benda!

a. 0 Nb. 10 Nc. 20 Nd. 30 N

Answer:ℓ air = 1 gr/cm3 = 1000 kg/cm3

Vb = 2000 cm3 = 2 x 10-3 m-3

g = 10 m/s2

F a = ℓ air. g. V batu = 1000 x 10 (2x10-3) = 20 N

7. Tentukan energi dalam (internal energy) 1 mol gas monoatomik pada suhu 27o C

a. 374 Jb. 375 Jc. 376 Jd. 373 J

Answer:N = 1 molR = 8,315 J/kmol KT = (27 + 273) K = 300 k

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8. Tentukan kecepatan efektif partikel partikel suatu gas dalam keadaan normal jika masa jenis gas tersebut 10 kg/m3 dan tekanannya 1,2 x 106 N/m2!

a. 242 m/sb. 424 m/sc. 324 m/sd. 323 m/s

Answer:ℓ = 10 kg/m3

P = 1,2 x 106 N/m2

9. Dua mol gas argon dengan volume awal 0,025 m3

memuai secara isoternal pada suhu 27o menjadi 0,05 m3

a. 2487 Jb. 2908 Jc. 3456 Jd. 3214 J

Answer:

W = nRT

= 2(8,31)(300)

= 4986 (Ln 2) = 3456 J10. Gas ideal yang menempati ruang 20 cm3 dipanaskan pada

tekanan tetap sehingga volumenya menjadi dua kali semula tekanan gasitu 2x105 N/m2. berapakah usaha yang dilakukan olehh gas?

a. 1 Jb. 2 Jc. 3 Jd. 4 J

Answer:P= 2 x 105 N/m2

V1 = 20 cm3 = 20 x 10-6 m3

V2 = 40 cm3 = 40 x 10-6 m3

W = P(∆V) = (2 x 105)(20 x 10-6) = 4 J

Ito

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