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#49 Part 4-Fundanental Problems INSTRUCTIONS: Circle the correct answer given fo

ID: 1580811 • Letter: #

Question

#49

Part 4-Fundanental Problems INSTRUCTIONS: Circle the correct answer given for each problem ( 5 points) PHYSICAL DATA: g 9.8 /n 32 ft/st sin 373/5: co 374/5; 50 mi/hr - 88 ft/s; 45, **Consider an Atwood nachine having a mass 3M on the left and a nass 6H on the right. When the system is released, what is ita acceleration in the absunce of friction? B. 2g C. g/2 D. 9/3 E. g/4 P. 9/5 46. The engines of a rocket of mass 3x10 sl produce a thrust of 7.56x1os ibe What is the vertical acceleration of the rocket in the FPS system? Note that the "net" vertical force is a ccebination of the rocket's thrust and weight A 7.84 B. 220 c. 243 D. 252 B. 291 2. 5760 47.**A 6 g bullet inwing with an initial speed of 4000 /s strikes, and passes copatay through, an 800 g wooden duck decoy initially at rest on smooth ice. What is the speed in/s of the decoy just after to collision, if the bullet has a pod of 2000 cn/s imediately after passing through the decoy? A. 15 B 30 c. 45 D. 150 300 E. 450 48. What is the radius of a planet in terns of the Earth'sTadius (R.?M fora planet whoeo nass and minimum escape velocity are six tires that of the Barth's? A. 1/9 B. 1/6 c, 1/3 D. 1 F. 6 49.A It an object weighs 100 1bs on the Earth and 50 lbs on another planet, how far in teet will this object fall on the other planet during the first 3 s of free fall? A. 18 B. 36 C. 72 D. 144 E. 288 F. 576 50. A skier has a speed of 40 ft/s at the botton of a nlopa. In the absence of friction, what in his speed at the top of the slope, if he ascends 3/8 ft? D. 8 E. 16 F. 64 A. 4 B. 5 C. 7

Explanation / Answer

49) C) 72 ft

from the given data,

Weight of the object on another planet = Weight of the object on the earth/2

m*g' = m*g/2

g' = g/2 (here g is the acceleration due to gravity on the earth and g's the acceleration due to gravity on another planet)

g' = 9.8/2

= 4.9 m/s^2

distance travelled in free fall in t seconds,

y = (1/2)*g'*t^2

= (1/2)*4.9*3^2

= 22.05 m

= 22.05/0.3048 ft

= 72.3 ft

so, Option C is the correct answer.