Answer
$h=0.581~m$
Work Step by Step
$P_G=5980~Pa$
$\rho=1050~kg/m^3$
$g=9.8~m/s^2$
$P_G=\rho g h$
Solve for h
$h=\frac{P_G}{\rho g}=\frac{5980Pa}{(1050~kg/m^3)~\times~(9.8~m/s^2)}=0.581~m$
The reservoir will have to be 0.581 m above the arm.
University Physics with Modern Physics (14th Edition)
by Young, Hugh D.; Freedman, Roger A.
Published by
Pearson
ISBN 10:
0321973615
ISBN 13:
978-0-32197-361-0
Chapter 12 - Fluid Mechanics - Problems - Exercises - Page 391: 12.11
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