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

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
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.