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: 26


$892.7 \ Pa$

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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