Online Homework Answers 11 (Nov. 26, 2006)

Online Homework Answers 11 (Nov. 26, 2006) - Hood Charles...

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Unformatted text preview: Hood, Charles – Homework 11 – Due: Nov 26 2006, 11:00 pm – Inst: Jan Opyrchal 1 This print-out should have 5 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 Water flows over a section of Niagara Falls at a rate of 1 . 31 × 10 6 kg / s and falls 40 m. The acceleration of gravity is 9 . 8 m / s 2 . How many 80 W bulbs can be lit with this power? Correct answer: 6 . 419 × 10 6 . Explanation: Let : r = 1 . 31 × 10 6 kg / s , and h = 40 m . The power of the water is equal to the change in potential energy per unit time, so P = U t = mg h t = m t g h = (1 . 31 × 10 6 kg / s)(9 . 8 m / s 2 )(40 m) = 5 . 1352 × 10 8 W . The power required to light n = 80 W light bulbs is P = n (80 W) . Solving for n , n = P 80 W = 5 . 1352 × 10 8 W 80 W = 6 . 419 × 10 6 . keywords: 002 (part 1 of 1) 10 points A bead slides without friction around a loop- the-loop. The bead is released from a height of 8 . 5 m from the bottom of the loop-the-loop which has a radius 2 m. The acceleration of gravity is 9 . 8 m / s 2 . 8 . 5 m 2 m A What is its speed at point A ? Correct answer: 9 . 39149 m / s. Explanation: Let : R = 2 m and h = 8 . 5 m . From conservation of energy, we have K i + U i = K f + U f 0 + mg h = mv 2 2 + mg (2 R ) v 2 = 2 g ( h- 2 R ) ....
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Online Homework Answers 11 (Nov. 26, 2006) - Hood Charles...

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