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Unformatted text preview: homework 04 – FONTENOT, BRIAN – Due: Feb 11 2008, 4:00 am 1 Question 1, chap 4, sect 5. part 1 of 3 10 points An athlete swings a 6 . 72 kg ball horizon tally on the end of a rope. The ball moves in a circle of radius 0 . 995 m at an angular speed of 0 . 787 rev / s. What is the tangential speed of the ball? Correct answer: 4 . 92015 m / s (tolerance ± 1 %). Explanation: ω = 0 . 787 rev / s = 4 . 94487 rad / s . v t = r ω = (0 . 995 m) (4 . 94487 rad / s) = 4 . 92015 m / s . Question 2, chap 4, sect 5. part 2 of 3 10 points What is its centripetal acceleration? Correct answer: 24 . 3295 m / s 2 (tolerance ± 1 %). Explanation: a r = r ω 2 = (0 . 995 m)(4 . 94487 rad / s) 2 = 24 . 3295 m / s 2 . Question 3, chap 4, sect 5. part 3 of 3 10 points If the maximum tension the rope can with stand before breaking is 54 . 6 N, what is the maximum tangential speed the ball can have? Correct answer: 2 . 8433 m / s (tolerance ± 1 %). Explanation: F = mv 2 r . If the maximum value of T is 54 . 6 N, then we have 54 . 6 N = (6 . 72 kg) v 2 (0 . 995 m) . From which, v max = 2 . 8433 m / s . Question 4, chap 4, sect 5. part 1 of 2 10 points A moon of a planet like the Earth is 486000 km distant from the planet’s center, and it completes an orbit in 32 . 5 days. Determine the moon’s orbital speed. Correct answer: 1087 . 47 m / s (tolerance ± 1 %). 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 (486000 km) 32 . 5 days = 1087 . 47 m / s . Question 5, chap 4, sect 5. part 2 of 2 10 points How far does the moon ”fall” toward the planet in 0 . 942 s? Correct answer: 1 . 07963 mm (tolerance ± 1 %). Explanation: The centripetal acceleration a of the motion of the moon is a = v 2 R = (1087 . 47 m / s) 2 4 . 86 × 10 8 m = 0 . 00243333 m / s 2 . Hence, in time t the moon falls a distance d = 1 2 a t 2 = 1 2 (0 . 00243333 m / s 2 ) (0 . 942 s) 2 = 0 . 00107963 m = 1 . 07963 mm . homework 04 – FONTENOT, BRIAN – Due: Feb 11 2008, 4:00 am 2 Question 6, chap 4, sect 6. part 1 of 2 10 points Consider a river with a uniform current velocity vectorv c . A boat crosses the river. When viewed from the shore, the boat’s motion is perpendicular to the river’s banks. However, the heading of the boat — the direction of its motion relative to the water — is at angle θ = 38 ◦ upstream from the perpendicular. Use these data to determine the boat’s speed v b relative to the water, or rather its ratio v b /v c to the current speed: 1. v b v c = 1 sin 2 θ 2. v b v c = sin θ 3. v b v c = tan θ 4. v b v c = 1 cos θ 5. v b v c = 1 tan θ 6. v b v c = cos θ 7. v b v c = 1 cos 2 θ 8. v b v c = cos 2 θ 9....
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 Spring '08
 Turner
 Physics, Work, General Relativity, Velocity, Correct Answer, Fontenot, BRIAN

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