Answer
$p = 2.4\times 10^9~N/m^2$
Work Step by Step
We can find the required stress:
$p = B~\frac{\Delta V}{V}$
$p = (1.4\times 10^{11}~N/m^2)~\frac{(0.855~m)^3-(0.850~m)^3}{(0.855~m)^3}$
$p = 2.4\times 10^9~N/m^2$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 12 - Equilibrium and Elasticity - Problems - Page 352: 67
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