Question Number 209846 by peter frank last updated on 23/Jul/24
Answered by mr W last updated on 23/Jul/24
$${the}\:{result}\:{is}\:{too}\:{large}. \\ $$$$\mathrm{20}×\left(\mathrm{15}−\mathrm{10}\right)×\mathrm{12}×\mathrm{10}^{−\mathrm{6}} =\mathrm{0}.\mathrm{0012}{m} \\ $$$$=\mathrm{1}.\mathrm{2}\:{mm} \\ $$$$\Rightarrow{the}\:{error}\:{is}\:\mathrm{1}.\mathrm{2}{mm}. \\ $$
Commented by peter frank last updated on 23/Jul/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Spillover last updated on 23/Jul/24
$$\alpha=\frac{\bigtriangleup{L}}{{L}\bigtriangleup\theta} \\ $$$${Error}=\bigtriangleup{L} \\ $$$$\alpha=\frac{\bigtriangleup{L}}{{L}\bigtriangleup\theta} \\ $$$$\bigtriangleup{L}=\alpha{L}\bigtriangleup\theta=\alpha{L}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{0}} \right) \\ $$$$\theta_{\mathrm{1}} =\mathrm{15}^{°} {C}=\mathrm{288}{K} \\ $$$$\theta_{\mathrm{0}} =\mathrm{10}^{°} {C}=\mathrm{283}{K} \\ $$$$\bigtriangleup{L}=\alpha{L}\bigtriangleup\theta=\alpha{L}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{0}} \right) \\ $$$$\bigtriangleup{L}=\alpha{L}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{0}} \right)=\mathrm{12}×\mathrm{10}^{−\mathrm{6}} ×\mathrm{20}×\mathrm{5}=\mathrm{0}.\mathrm{12}{cm} \\ $$$$\bigtriangleup{L}=\mathrm{0}.\mathrm{12}{cm}=\mathrm{1}.\mathrm{2}{mm}\:\:\:\left({Too}\:{large}\right) \\ $$$$ \\ $$