## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 15 - Exercises and Problems - Page 280: 49

27 meters

#### Work Step by Step

We have to find the ratios between the two heights, which we will call $h_1$ and $h_2$. Thus, we find: $\frac{h_2A\rho_wg}{h_1A\rho_wg}=\frac{3mg}{mg}$ Thus, we see that the ratio is 3 to 1. This means that the minimum depth is: $h= 9(3)=27m$

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