## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 13 - Problems - Page 496: 12

#### Answer

The area of the hole increases by $~3.80~\times 10^{-10}~m^2$

#### Work Step by Step

We can find the increase in area of the hole when the brass plate is heated: $\Delta A = 2~\alpha~\Delta T~A$ $\Delta A = (2)(19\times 10^{-6}~K^{-1})(10.0~K)~(1.00\times 10^{-6}~m^2)$ $\Delta A = 3.80~\times 10^{-10}~m^2$ The area of the hole increases by $~3.80~\times 10^{-10}~m^2$

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