Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems - Page 289: 72

Answer

$1\times 10^4Pa$

Work Step by Step

We can find the required pressure as $\Delta P=\rho g\Delta h$ We plug in the known values to obtain: $\Delta P=(1.05\times 10^3)(9.81)(1)$ $\Delta P=1.029\times 10^4Pa=1\times 10^4Pa$
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