Exam_3_practice - Version PREVIEW – Exam 3 Practice –...

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Unformatted text preview: Version PREVIEW – Exam 3 Practice – chelikowsky – (59005) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Conceptual 05 09 001 (part 1 of 2) 10.0 points Calculate the weight of a solid rocket booster of the space shuttle with mass 5 . 7 × 10 5 kg on Earth. The Earth has a mass of 6 × 10 24 kg and a radius of 6 . 4 × 10 6 m. 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 5 . 586 × 10 6 N. Explanation: Let : m = 6 × 10 24 kg and g = 6 . 67 × 10 − 11 N · m 2 / kg 2 . W = mg = (5 . 7 × 10 5 kg) (9 . 8 m / s 2 ) = 5 . 586 × 10 6 N . 002 (part 2 of 2) 10.0 points What would this weight be on Mars, with mass 0 . 11 m E and radius 0 . 53 R E ? The value of the universal gravitational constant is 6 . 67 × 10 − 11 N · m 2 / kg 2 . 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 2 . 18089 × 10 6 N. Explanation: Let : M E = 6 × 10 24 kg , R E = 6 . 4 × 10 6 m , M M = 0 . 11 m E , R M = 0 . 53 R E , and G = 6 . 67 × 10 − 11 N · m 2 / kg 2 . On Mars g ′ = GM M R 2 M = 6 . 67 × 10 − 11 N · m 2 / kg 2 [0 . 53 (6 . 4 × 10 6 m)] 2 × [0 . 11 (6 × 10 24 kg)] = 3 . 82612 m / s 2 . W ′ W = mg ′ mg W ′ = g ′ g W = 3 . 82612 m / s 2 9 . 8 m / s 2 ( 5 . 586 × 10 6 N ) = 2 . 18089 × 10 6 N . Finding G Like Cavendish 003 10.0 points An apparatus like the one Cavendish used to find G has a large lead ball that is 5 . 2 kg in mass and a small one that is 0 . 056 kg. Their centers are separated by 0 . 046 m. Find the force of attraction between them. The value of the universal gravitational con- stant is 6 . 67259 × 10 − 11 N · m 2 / kg 2 . 1. @@@ 2. @@@ 3. @@@ 4. @@@ Version PREVIEW – Exam 3 Practice – chelikowsky – (59005) 2 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 9 . 18269 × 10 − 9 N. Explanation: Let : d = 0 . 046 m , G = 6 . 67259 × 10 − 11 N · m 2 / kg 2 , m 1 = 5 . 2 kg , and m 2 = 0 . 056 kg . By Newton’s universal law of gravitation, F = G m 1 m 2 d 2 = (6 . 67259 × 10 − 11 N · m 2 / kg 2 ) (5 . 2 kg) (0 . 056 kg) . 046 m 2 = 9 . 18269 × 10 − 9 N . Suspended Over a Planet 004 10.0 points A “synchronous” satellite is put in orbit about a planet to always remain above the same point on the planet’s equator. The planet rotates once every 10 . 1 h, has a mass of 2 . 1 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given: G = 6 . 67 × 10 − 11 N m 2 / kg 2 . Calculate how far above the planet’s surface the satellite must be. 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 9 . 74957 × 10 7 m....
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This note was uploaded on 07/07/2010 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.

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Exam_3_practice - Version PREVIEW – Exam 3 Practice –...

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