## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 20 - Traveling Waves - Exercises and Problems - Page 586: 3

#### Answer

The length of the string is 2.0 meters.

#### Work Step by Step

We can find the length of the string as: $v = \sqrt{\frac{F_T}{\mu}}$ $\frac{L}{t} = \sqrt{\frac{F_T}{(m/L)}}$ $\frac{L}{t} = \sqrt{\frac{F_T~L}{m}}$ $\frac{L^2}{t^2} = \frac{F_T~L}{m}$ $L = \frac{F_T~t^2}{m}$ $L = \frac{(20~N)(50\times 10^{-3}~s)^2}{0.025~kg}$ $L = 2.0~m$ The length of the string is thus 2.0 meters.

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