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: 20

Answer

$892.7 \ Pa$

Work Step by Step

The gauge pressure at the bottom is equal to the sum of the pressures. Thus, it follows: $ P = (820)(9.81)(5\times10^{-2}) + (1000)(9.81)(5\times10^{-2})=892.7 \ Pa$
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