SOL20 - Thevalingam Donald Homework 20 Due Mar 4 2007...

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Thevalingam, Donald – Homework 20 – Due: Mar 4 2007, midnight – Inst: Eslami 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Hey guys, instead oF giving me ten defni- tions For a little grade improvement in the last minute, you need to solve fve problems, but not For me, For yourselves. I do not need your solving problems; you need it! I have solved mine years ago when you were still crawling :)) 001 (part 1 oF 1) 10 points A bobsled slides down an ice track starting (at zero initial speed) From the top oF a(n) 194 m high hill. The acceleration oF gravity is 9 . 8 m / s 2 . Neglect Friction and air resistance and de- termine the bobsled’s speed at the bottom oF the hill. Correct answer: 61 . 6636 m / s. Explanation: By conservation oF energy: mgh = mv 2 2 hence the velocity at the bottom oF the hill is: v = p 2 gh = q 2(9 . 8 m / s 2 )(194 m) = 61 . 6636 m / s keywords: 002 (part 1 oF 1) 10 points A(n) 662 kg elevator starts From rest. It moves upward For 4 . 3 s with a constant acceleration until it reaches its cruising speed oF 1 . 91 m / s. The acceleration oF gravity is 9 . 8 m / s 2 . ±ind the average power delivered by the elevator motor during the period oF this accel- eration. Correct answer: 6

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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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SOL20 - Thevalingam Donald Homework 20 Due Mar 4 2007...

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