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Unformatted text preview: homework 12 – RAHMAN, TARIQUE – Due: Mar 4 2008, 9:00 pm 1 Question 1, chap 9, sect 1. part 1 of 2 10 points A ball is tossed straight up from the surface of a small, spherical asteroid with no atmo sphere. The ball rises to a height equal to the asteroid’s radius and then falls straight down toward the surface of the asteroid. What forces, if any, act on the ball while it is on the way up? 1. Only a constant gravitational force that acts downward 2. Only an increasing gravitational force that acts downward 3. Both a constant gravitational force that acts downward and a decreasing force that acts upward 4. No forces act on the ball. 5. Only a decreasing gravitational force that acts downward correct Explanation: There is no friction in the system, and the ball doesn’t have any contact with other ob jects. Thus the only force acting on the ball is the attractive gravitational force, which means the force points downward. From the formula vector F = G M m r 2 ˆ r , we know the force will decrease as the ball rises. Question 2, chap 9, sect 1. part 2 of 2 10 points The acceleration of the ball at the top of its path is 1. equal to onehalf the acceleration at the surface of the asteroid. 2. zero. 3. equal to the acceleration at the surface of the asteroid. 4. at its maximum value for the ball’s flight. 5. equal to onefourth the acceleration at the surface of the asteroid. correct Explanation: Because the gravitational force is inversely proportional to the distance squared, the force at the top of its path is equal to onefourth the force at the surface of the asteroid. So the acceleration at the top is equal to onefourth the acceleration at the surface of the asteroid. Question 3, chap 9, sect 1. part 1 of 2 10 points Compare the gravitational force on a 30 kg mass at the surface of the Earth ( R E = 6 . 4 × 10 6 m, M E = 6 × 10 24 kg) with that on the surface of the Moon ( M M = 1 81 . 3 M E , R M = 0 . 27 R E ). What is it on the Earth? Correct answer: 293 . 115 N (tolerance ± 1 %). Explanation: Let : m = 30 kg , and R E = 6 . 4 × 10 6 m . F = GmM E R 2 E = (6 . 67 × 10 11 N · m 2 / kg 2 ) (30 kg) (6 . 4 × 10 6 m) 2 × (6 × 10 24 kg) = 293 . 115 N . Question 4, chap 9, sect 1....
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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