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A horizontal wire is stretched with a tension of 94.0 N,

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 23E Chapter 15

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 23E

A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 492 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be m order for the average power carried by the wave to be 0.365 W?

Step-by-Step Solution:

Solution 23E Step 1: Data given Tension T = 94.0 N Speed of waves v = 492 m/s Frequency f = 69.0 Hz Power P = 0.365 W We need to find the amplitude of traveling wave This is given by using 2 2 P = 1/2 (F A ) --------------(1) 2 We can obtain = T/v Substituting we get = 94.0 N/(492 m/s)2 = 3.89 × 104kg /m Step 2 : We can obtain angular velocity = 2 f Substituting we get = 2 × 69.0 Hz = 433.5 rad/s

Step 3 of 3

Chapter 15, Problem 23E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 492 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be m order for the average power carried by the wave to be 0.365 W?” is broken down into a number of easy to follow steps, and 50 words. The full step-by-step solution to problem: 23E from chapter: 15 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: Wire, wave, speed, frequency, horizontal. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 23E from 15 chapter was answered, more than 738 students have viewed the full step-by-step answer.

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