Online Homework Answers 11 (Nov. 26, 2006)

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ) ....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online