91影视

Two students each build a piece of scientific equipment that

uses a 655-mm-long metal rod.

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through$鈭�T$, as well as length changes through $鈭�L$and the length's fractional changes $鈭�L/L$are effective to the temperature's change$鈭�T$.

The equation (18.8) in the form is

$\frac{鈭�L}{L}=伪鈭�T\left(1\right)$

In this case, $伪$is the coefficient. Table 18.4 effectively gives the coefficient expression of linear for common materials, for instance

$伪=1.2脳{10}^{-5鈭�}{C}^{-1}$

The equation for $鈭�L$is

$鈭�L=伪L鈭�T\left(2\right)$

In this case, L is the initial length. The temperature effectively changes through $鈭�T=5掳C$for brass $鈭�{伪}_{brass}=1.9脳{10}^{-5掳}{C}^{-1}$

For Brass, plug the values of ${伪}_{brass,}鈭�L,T$into equation (2) in order to get $鈭�L_{brass}$

$\begin{array}{rcl}鈭�L_{brass}& =& {伪}_{brass}L鈭�T\\ & =& (1.9脳{10}^{-5掳}{C}^{-1})(655mm)\left(5^{掳}C\right)\\ & =& 0.06222mm\end{array}$

plug the values of $鈭�L,T,{伪}_{invar}$ into equation (2) in order to get $鈭�L_{invar}$

role="math" localid="1649070713758" $\begin{array}{rcl}鈭�L_{invar}& =& {伪}_{invar}L鈭�T\\ & =& (0.09脳{10}^{-5掳}{C}^{-1})(655mm)\left(5^{掳}C\right)\\ & =& 0.00295mm\end{array}$

The calculation depicted the brass changes through a larger value of invar and the difference is

role="math" localid="1649070956077" $\begin{array}{rcl}鈭�L& =& 鈭�L_{brass}-鈭�L_{invar}\\ & =& 0.06222mm-0.00295mm\\ & =& 0.05927mm\\ & =& 5.927脳{10}^{-2}mm\end{array}$

Therefore, the answer is$\begin{array}{rcl}& & 5.927脳{10}^{-2}mm\end{array}$