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 289: 15


Please see the work below.

Work Step by Step

We know that $F=mg$ $F=(4680)(9.8)=45864N$ $D=42.6cm=0.426m$ Now $A=\frac{\pi D^2}{4}=0.1425m^2$ We can find the pressure as $P=\frac{F}{A}$ We plug in the known values to obtain: $P=\frac{45864}{0.1425}$ $P=322KPa$
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