Online Homework Answers 2 (Sept. 21, 2006)

Online Homework Answers 2 (Sept. 21, 2006) - Hood Charles...

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Unformatted text preview: Hood, Charles – Homework 2 – Due: Sep 21 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 15 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 2) 10 points A particle moves according to the equation x = (10 m / s 2 ) t 2 where x is in meters and t is in seconds. Find the average velocity for the time in- terval from t 1 = 2 . 19 s to t 2 = 4 . 31 s. Correct answer: 65 m / s. Explanation: x = (10 m / s 2 ) t 2 v ave = Δ x Δ t = (10 m / s 2 ) ( t 2 2- t 2 1 ) t 2- t 1 = (10 m / s 2 ) [(4 . 31 s) 2- (2 . 19 s) 2 ] (4 . 31 s)- (2 . 19 s) = 65 m / s 002 (part 2 of 2) 10 points Find the average velocity for the time interval from t 1 = 2 . 19 s to t 3 = 2 . 29 s. Correct answer: 44 . 8 m / s. Explanation: v ave = Δ x Δ t = (10 m / s 2 ) ( t 2 3- t 2 1 ) t 3- t 1 = (10 m / s 2 ) [(2 . 29 s) 2- (2 . 19 s) 2 ] (2 . 29 s)- (2 . 19 s) = 44 . 8 m / s keywords: 003 (part 1 of 3) 10 points Consider the acceleration of gravity to be 10 m / s 2 . What is the magnitude of the instantaneous velocity (speed) of a freely falling object 26 s after it is released from a position of rest? Correct answer: 260 m / s. Explanation: k ~v k = g t = (10 m / s 2 )(26 s) = 260 m / s . 004 (part 2 of 3) 10 points What is its average speed during this 26 s interval?...
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Online Homework Answers 2 (Sept. 21, 2006) - Hood Charles...

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