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# HW 8 - johnson(mjj622 – HW 8 – Coker –(56625 1 This...

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Unformatted text preview: johnson (mjj622) – HW 8 – Coker – (56625) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment covers the last material that will be included on Quiz 2, to be held the evening of October 13. Quiz 2 covers Chapters 5, 6, 7, 8 and 9 in the text. 001 (part 1 of 2) 10.0 points The gravitational acceleration on the surface of planet Zeb is the same as on the surface of the earth, but planet Zeb has twice the radius of the earth. What must be the ratio of the mass of planet Zeb to the mass of earth? 1. 1 (the same) since g is the same. 2. 4 correct 3. 1 4 4. 6 5. 10 6. 1 8 7. 2 8. 1 10 9. 8 Explanation: Let : R 2 = 2 R e . g = GM e R 2 e = GM z R 2 z , so M ∝ R 2 and M z M e = R 2 z R 2 e = (2 R e ) 2 R 2 e = 4 . 002 (part 2 of 2) 10.0 points What must be the ratio of the density of planet Zeb to the density of the earth? 1. 0.01 2. 0.1 3. 0.5 correct 4. 20 5. 4 6. 0.04 7. 1 (the same) since g is the same. 8. 100 Explanation: The volume of a sphere is 4 3 π R 3 , so the average density is ρ = M V = M 4 3 π R 3 ∝ M R 3 and ρ z ρ e = M z R 3 z M e R 3 e = M z M e parenleftbigg R e R z parenrightbigg 3 = 4 parenleftbigg 1 2 parenrightbigg 3 = 1 2 . 003 10.0 points How much less would you weigh on the top of Mount Everest (elevation 8850 m) than at sea level? The value of the universal gravita- tional constant is 6 . 67 × 10 − 11 N · m 2 / kg 2 . 1. 0.5% 2. 4% 3. 0.2% 4. 2% 5. 0.1% 6. 1% 7. 3% 8. 0.4% johnson (mjj622) – HW 8 – Coker – (56625) 2 9. 0.3% correct 10. 5% Explanation: Let : M E = 6 × 10 24 kg , R E = 6 . 4 × 10 6 m , h = 8850 m , and G = 6 . 67 × 10 − 11 N · m 2 / kg 2 . At sea level, g = GM E R 2 E = (6 . 67 × 10 − 11 N · m 2 / kg 2 ) (6 × 10 24 kg) (6 . 4 × 10 6 m) 2 = 9 . 77051 m / s 2 . At the top of Mount Everest, g ′ = GM E ( R E + h ) 2 = (6 . 67 × 10 − 11 N · m 2 / kg 2 ) (6 × 10 24 kg) (6 . 4 × 10 6 m + 8850 m) 2 = 9 . 74354 m / s 2 . Since W = mg , Δ W W = mg- mg ′ mg = 1- g ′ g = parenleftbigg 1- 9 . 74354 m / s 2 9 . 77051 m / s 2 parenrightbigg × 100% = . 27599% . 004 (part 1 of 2) 10.0 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. No forces act on the ball. 2. Both a constant gravitational force that acts downward and a decreasing force that acts upward 3. Only a constant gravitational force that acts downward 4. Only an increasing gravitational force that acts downward 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, so the only force acting on the ball is the attractive gravitational force, which acts...
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HW 8 - johnson(mjj622 – HW 8 – Coker –(56625 1 This...

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