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solution 7 - reyes(mr29667 Homework 7 chelikowsky(59005...

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reyes (mr29667) – Homework 7 – chelikowsky – (59005) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If the earth were of uniform density, what would be the value of g inside the earth at half its radius? (The value of g at the surface of earth is 9.8 m/s 2 .) 1. 19.6 m/s 2 2. 9.8 m/s 2 3. 4.9 m/s 2 correct 4. 39.2 m/s 2 Explanation: Consider the spherical shell idea: only the mass of the earth in the inner shell con- tributes to g ; the mass of the sphere of ra- dius r 2 . To determine this smaller mass, con- sider the effect on the volume: V r 3 , so V parenleftbigg 1 2 r parenrightbigg 3 = 1 8 r 3 ; 1 8 of the original volume means 1 8 of the original mass. Now G m r 2 , so G 1 8 m parenleftbigg 1 2 r parenrightbigg 2 = 1 8 m 1 4 r 2 = 1 2 m r 2 , and the accleration due to gravity at r 2 is g 2 . 002 10.0 points The planet Mars has a mass of 6 . 1 × 10 23 kg and radius of 3 . 8 × 10 6 m. What is the acceleration of an object in free fall near the surface of Mars? The value of the gravitational constant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 2 . 81875 m / s 2 . Explanation: Let : M = 6 . 1 × 10 23 kg , R = 3 . 8 × 10 6 m , and G = 6 . 67259 × 10 11 N · m 2 / kg 2 . Near the surface of Mars, the gravitation force on an object of mass m is F = G M m R 2 , so the acceleration of an object in free fall is a = F m = G M R 2 = (6 . 67259 × 10 11 N · m 2 / kg 2 ) × 6 . 1 × 10 23 kg (3 . 8 × 10 6 m) 2 = 2 . 81875 m / s 2 . 003 10.0 points An apparatus like the one Cavendish used to find G has large lead balls that are 8 . 4 kg in mass and small ones that are 0 . 067 kg. The center of a large ball is separated by 0 . 058 m from the center of a small ball.
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