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# Traditional Old Midterm 2 - Version 080 Midterm 02...

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Version 080 – Midterm 02 – Yao – (58180) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In another solar system is planet Driff, which has 5 times the mass of the earth and also 5 times the radius. How does the gravitational acceleration on the surface of Driff compare to the gravita- tional acceleration on the surface of the earth? 1. It’s 25 times as great. 2. There is no gravity on Driff because 5 times 4000 miles (the radius of the earth), is 20000 miles, far beyond the pull of gravity. 3. It’s 5 times as much. 4. It’s 1 5 th as much. correct 5. It’s 1 25 th as much. 6. It’s the same, 10 m / s 2 . Explanation: Let : M D = 5 M e and R D = 5 R e . Gravitational force is F = m g = G M m r 2 g = G M r 2 M r 2 , so g D g e = m D r 2 D m e r 2 e = M D r 2 e m e r 2 D = (5 m e ) r 2 e m e (5 r e ) 2 = 1 5 g D = 1 5 g e . 002 10.0 points Two satellites A and B, where B has twice the mass of A, orbit the earth in circular orbits. The distance of satellite B from the earth’s center is twice the distance of satellite A from the earth’s center. What is the ratio of the orbital period of satellite B to that of satellite A? 1. T B T A = 1 / 8 2. T B T A = 2 3. T B T A = radicalbig 1 / 2 4. T B T A = 2 5. T B T A = 8 correct 6. T B T A = radicalbig 1 / 8 7. T B T A = 1 8. T B T A = 1 / 4 9. T B T A = 1 / 2 10. T B T A = 8 Explanation: According to Kepler’s third law, the square of the orbital period is proportional to the cube of the orbital radius. Therefore, T 2 A R 3 A = T 2 B R 3 B . Therefore, T B T A = parenleftbigg R B R A parenrightbigg 3 / 2 = 2 3 / 2 = 8 . 003 10.0 points The force F exerted on the block pushes the block against the ceiling and at the same time accelerates the block to the right.

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Version 080 – Midterm 02 – Yao – (58180) 2 m F θ μ a Find the kinetic friction between the block and the ceiling. 1. f k = μ ( F sin θ + m g ) 2. f k = μ m g 3. f k = μ F sin θ 4. f k = μ ( F sin θ m g ) correct Explanation: The net normal force which pushes against the ceiling is F sin θ m g . So the kinetic friction is f k = μ ( F sin θ m g ). 004 10.0 points When an automobile moves with constant ve- locity, the power developed is used to over- come the frictional forces exerted by the air and the road. If the engine develops 50 hp, what total frictional force acts on the car at 160 mph? One horsepower equals 746 W, and one mile is 1609 m. 1. 521.597 2. 205.781 3. 222.548 4. 238.444 5. 606.95 6. 1854.57 7. 1804.44 8. 513.573 9. 455.212 10. 596.111 Correct answer: 521 . 597 N. Explanation: Apply P = E t = Fd t = Fv F = P v = 50 hp · 746 W hp 160 mph · 1609 m mi · h 3600 s = 521 . 597 N 005 10.0 points Pretend you are on a planet similar to Earth where the acceleration of gravity is approxi- mately 10 m / s 2 . The pulley is massless and frictionless. A massless inextensible string is attached to the masses. The objects are initially held at rest.
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