College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 360: 1


The average pressure on that area is $49~atm$

Work Step by Step

We can find the pressure: $P = \frac{F}{A} = \frac{500~N}{1.0\times 10^{-4}~m^2} = 5.0\times 10^6~N/m^2$ We can convert the pressure to units of atm: $(5.0 \times 10^6~N/m^2) \times \frac{1~atm}{1.013\times 10^5~Pa} = 49~atm$ The average pressure on that area is $49~atm$
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