Essential University Physics: Volume 1 (4th Edition)

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

Chapter 15 - Exercises and Problems - Page 291: 57



Work Step by Step

According to Bernoulli's equation $P_{atm}+140KPa=P_{atm}+\rho gh$ This simplifies to: $140KPa=\rho gh$ This can be rearranged as: $h=\frac{140KPa}{\rho g}$ We plug in the known values to obtain: $h=\frac{140\times 10^3}{9.81\times 1000}$ $h=14.27m$
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