Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 17 - Exercises and Problems - Page 326: 71

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.