hw29 - Muraj Hamza – Homework 29 – Due 4:00 am – Inst...

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Unformatted text preview: Muraj, Hamza – Homework 29 – Due: Apr 12 2006, 4:00 am – Inst: Florin 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: G = 6 . 67259 × 10- 11 Nm 2 / kg 2 . An apparatus like the one Cavendish used to find G has large lead balls that are 6 . 2 kg in mass and small ones that are 0 . 0345 kg. The center of a large ball is separated by 0 . 06 m from the center of a small ball. Mirror m Light source M r κ Schematic diagram of the Cavendish ap- paratus for measuring G . As the small spheres of mass m are attracted to the large spheres of mass M , the rod between the two small spheres rotates through a small angle. A light beam reflects from a mirror, fixed to the axis of the small masses. A torsion ribbon rotates (shown with torsion constant κ ) holding the mir- ror and the small masses. The skewed dashed line represents the position of the rod when the masses are at a large dis- tance. Find the magnitude of the gravitational force between the masses m and M . Correct answer: 3 . 96463 × 10- 9 N. Explanation: By Newton’s Universal Law of Gravitation, F = G m 1 m 2 r 2 = (6 . 67259 × 10- 11 Nm 2 / kg 2 ) × (6 . 2 kg)(0 . 0345 kg) (0 . 06 m) 2 = 3 . 96463 × 10- 9 N . 002 (part 1 of 2) 10 points A moon of a planet like the Earth is 299000 km distant from the planet’s center, and it com- pletes an orbit in 20 . 7 days. Determine the moon’s orbital speed. Correct answer: 1050 . 43 m / s. Explanation: Using the obvious formula v T = 2 π R , we obtain for the moon’s orbital speed v = 2 π R T = 2 π (299000 km) 20 . 7 days = 1050 . 43 m / s ....
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hw29 - Muraj Hamza – Homework 29 – Due 4:00 am – Inst...

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