College Physics (4th Edition)

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

Chapter 13 - Problems - Page 497: 27


$\frac{\Delta A}{A_0} = 2~\alpha~\Delta T$

Work Step by Step

Let the original area $A_0 = s_0^2$ We can find an expression for the new area after the material expands: $A = (s_0+\Delta s)^2$ $A = s_0^2+2s_0~\Delta s+(\Delta s)^2$ $A \approx s_0^2+2s_0~\Delta s$ We can find an expression for the ratio $\frac{\Delta A}{A_0}$: $\Delta A = A-A_0$ $\Delta A = (s_0^2+2s_0~\Delta s)-s_0^2$ $\Delta A = 2s_0~\Delta s$ $\Delta A = 2s_0~(\alpha~\Delta T~s_0)$ $\Delta A = 2s_0^2~\alpha~\Delta T$ $\Delta A = 2~A_0~\alpha~\Delta T$ $\frac{\Delta A}{A_0} = 2~\alpha~\Delta T$
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