solution-hw5

solution-hw5 - lai (yl8859) HW05 Ross (15200) 1 This...

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Unformatted text preview: lai (yl8859) HW05 Ross (15200) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The orbit of a Moon about its planet is ap- proximately circular, with a mean radius of 6 . 08 10 8 m. It takes 18 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 2456 . 39 m / s. Explanation: Dividing the length C = 2 r of the trajectory of the Moon by the time T = 18 days = 1 . 5552 10 6 s of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 r T = 2 (6 . 08 10 8 m ) 1 . 5552 10 6 s = 2456 . 39 m / s . 002 (part 2 of 2) 10.0 points Find the Moons centripetal acceleration. Correct answer: 0 . 00992409 m / s 2 . Explanation: Since the magnitude of the velocity is con- stant, the tangential acceleration of the Moon is zero. The centripetal acceleration isis zero....
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solution-hw5 - lai (yl8859) HW05 Ross (15200) 1 This...

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