College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 12 - Problems - Page 465: 53


The depth of the lake is $403~m$

Work Step by Step

We can use $v = 1493~m/s$ as the speed of sound in freshwater at a temperature of $25^{\circ}C$. We can find the total distance the sound travels in the water: $d = v~t = (1493~m/s)(0.540~s) = 806~m$ Since the sound reached the bottom of the lake and returned to the boat, the depth of the water is $\frac{806~m}{2} = 403~m$
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