Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 15 - Exercises and Problems - Page 289: 17

Answer

Please see the work below.

Work Step by Step

We know that $P=\rho gh$ This can be rearranged as: $\rho=\frac{P}{gh}$ We plug in the known values to obtain: $\rho=\frac{100000}{9.8(6.0)}$ $\rho=1700\frac{Kg}{m^3}$
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