Answer:

∆L = FL mg = AY AY =

= 0.0000262689 m

A) B) C) D) E)

2.6 x 10-5 m 1.3 x 10-5 m 4.8 x 10-5 m 7.2 x 10-5 m 3.5 x 10-5 m

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q2. In Fig. 2, PQ is a horizontal uniform beam weighing 155 N. It is supported by a string and a hinge at point P. A 245 N block is hanging from point Q at the end of the beam. Find the horizontal component of net force on the beam from the hinge.

Answer:

Toque about P implies:

0 (T sin 35 ) (1.35) − (155) 1.7 − (245)(1.7) = 2 o

= ⇒T

131.75 + 416.5 = 708 N (1.35)sin 35o

cos35o cos35o ) = T= 708 (= 579.98 N ≈ 580 N FH

The force on x-axis implies:

A) B) C) D) E)

580 N 310 N 491 N 164 N 200 N

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q3. A 20.0 m long uniform beam weighing 550 N rests on supports “A” and “B”, as shown in Figure 3. Find the magnitude of the force that the support “A” exerts on the beam when the block of weight 200 N is placed at D. Fig#

Answer:

The torque about point B implies:

− M D × 5 + M beam × 5 − FA × 12 = 0 = FA

A) B) C) D) E)

−200 × 5 + 550 × 5 = 145.8 N ≈ 146 N 12

146 N 241 N 501 N 315 N 185 N

Q4. At what height above earth’s surface would the gravitational acceleration be 0.980 m/s2?

Answer:

2 GM g GM / R E = = ⇒h = (R E + h ) 2 10 10

10R E − R E = 1.38 × 107

7 = 1.37737 10 m

A) B) C) D) E) 1.38 × 107 m 1.12 × 107 m 7.12 × 107 m 5.82 × 108 m 4.05 × 108 m

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q5. In Figure 4, what is the net gravitational force exerted on the 5.00 kg uniform sphere by the other two uniform spheres? Fig#

Answer:

The gravitational force is attractive: M M 0.1 10 F = F1 + F2 = GM 5 20.1 + 210 = 6.67 × 10−11 × 5 × 2 + 2 = +3.54 × 10-9 N r10 1 .4 r0.1 9 = 3.54344 10

A) + 3.54 × 10 −9 ˆ N i B) + 2.32 × 10 −11 ˆ N i C) − 2.32 × 10 −11 ˆ N i −13 ˆ D) − 1.45 × 10 i N E) +1.45 ×10−13 ˆ N i

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q6. A rocket is launched from the surface of a planet of mass M = 2.20 × 1028 kg and radius R = 5.35 × 106 m. What minimum initial speed is required if the rocket is to rise to a height of 6R above the surface of the planet? (Neglect the effects of the atmosphere).

Answer:

Conservation of total energy implies:

1 GmM GmM = 0− mv 2 − 2 7R R ⇒ v = 2GM 1 − ) (1= 7 R 2 × 6.67 × 10−11 × 2.2 × 1028 1 − ) (1= 6.86× 105 m/s 6 5.35 × 10 7

= 685708. A) B) C) D) E) 6.86 × 105 m/s 3.44 × 105 m/s 2.18 × 106 m/s 8.20 × 106 m/s 9.45 × 105 m/s

Q7. A satellite of mass 200 kg is placed in Earth orbit at height of 200 km above the earth surface. How long does the satellite take to complete one circular orbit?

Answer:

T = 4π 2 ( R E + h )3 GM E

= 1.47168 A) B) C) D) E) 1.47 hours 2.77 hours 8.14 hours 9.56 hours 7.38 hours

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q8. In a hydraulic press, shown in Figure 5, the large piston has a cross sectional area of A1 = 150 cm2 and mass m1 = 450 kg. The small piston has a cross sectional area of A2 = 10 cm2 and mass m2. If the height difference between the two pistons is 1.0 m, what is the mass m2? [Note: The fluid in the hydraulic press is water] Fig#

Answer:

The pressures at points A and B must be the same so that p (due to A1) = p (due to A2) + p (due to water height of 1.0 m)

450 × 9.8 m 2 × 9.8 = + (1.0 × 103 )(9.8)(1) −4 −4 150 × 10 10 × 10 ⇒ m 2 =kg 29

A) B) C) D) E)

29 kg 33 kg 15 kg 40 kg 11 kg

Phys101 Term: 111

Final Tuesday, January 10, 2012

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Q9. Figure 6 shows an open-tube manometer...