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# 09 - fryer(vdf96 Assignment 6 Universal Gravitation...

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fryer (vdf96) – Assignment 6 Universal Gravitation / Circular Motion – graves – (6) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two spheres have equal densities and are sub- ject only to their mutual gravitational attrac- tion. Which quantity must have the same mag- nitude for both spheres? 1. kinetic energy 2. acceleration 3. displacement from the center of mass 4. gravitational force correct 5. velocity Explanation: Two spheres with the same density have different masses due to their relative sizes. Using Newton’s third law, vector F 1 = - vector F 2 . All of the other quantities (acceleration, ve- locity, kinetic energy, and displacement from the center of mass) have different magnitudes because the two spheres have different masses. 002(part1of2)10.0points 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 decreasing gravitational force that acts downward correct 2. Both a constant gravitational force that acts downward and a decreasing force that acts upward 3. Only an increasing gravitational force that acts downward 4. No forces act on the ball. 5. Only a constant gravitational force that acts downward Explanation: There is no friction in the system, and the ball doesn’t have any contact with other ob- jects, so the only force acting on the ball is the attractive gravitational force, which acts downward. From vector F = - G M m r 2 ˆ r , the force will de- crease as the ball rises. 003(part2of2)10.0points The acceleration of the ball at the top of its path is 1. zero. 2. equal to the acceleration at the surface of the asteroid. 3. at its maximum value for the ball’s flight. 4. equal to one-fourth the acceleration at the surface of the asteroid. correct 5. equal to one-half the acceleration at the surface of the asteroid. Explanation: F = m a 1 r 2 , so a 1 r 2 and a 1 (2 r ) 2 = 1 4 1 r 2 1 4 a . 004 10.0points In introductory physics laboratories, a typical Cavendish balance for measuring the gravita- tional constant G uses lead spheres of masses 0 . 85 kg and 17 . 2 g whose centers are separated by 3 . 51 cm. Calculate the gravitational force between

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fryer (vdf96) – Assignment 6 Universal Gravitation / Circular Motion – graves – (6) 2 these spheres, treating each as a point mass located at the center of the sphere. The value of the universal gravitational constant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 7 . 91822 × 10 10 N. Explanation: Let : m 1 = 0 . 85 kg , m 2 = 17 . 2 g = 0 . 0172 kg , r = 3 . 51 cm = 0 . 0351 m , and G = 6 . 67259 × 10 11 N · m 2 / kg 2 .
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