Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 17 - Exercises and Problems - Page 315: 73


a) $y^2 = .25(L_0^2-d^2)+\frac{1}{2}L_0^2 \alpha \Delta T$ b) $\alpha = 2.35 \times 10^{-6} ; d=.8 \ meters $ c) Aluminum

Work Step by Step

a) Looking for a linear relationship between some function of y and T, we find: $y^2 = .25(L_0^2-d^2)+\frac{1}{2}L_0^2 \alpha \Delta T$ Note, this is linear because the function of y, in this case $y^2$, is equal to a constant, $.25(L_0^2-d^2)$, plus T raised to the first power multiplied by a constant, $\frac{1}{2}L_0^2 \alpha \Delta T$. b) We plot the data in Microsoft Excel, and we plot the best fit line. Doing this, we find that: $\alpha = 2.35 \times 10^{-6}$, and $ d=.8 \ meters $ c) Using the tables for the values of $\alpha$, we find that the substance is Aluminum.
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