Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 15 - Exercises and Problems - Page 279: 23


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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