Answer:
Strain of the copper specimen will be equal to 0.0013
Explanation:
We have given tension in the wire [tex]F=44500N[/tex]
Area of rectangular copper specimen [tex]A=15.2\times 19.1\times 10^{-6}=290.32\times 10^{-6}m^2[/tex]
We have to find the strain of the copper specimen
Stress will be equal to [tex]stress=\frac{force}{area}=\frac{44500}{290.32\times 10^{-6}}=153.448\times 10^6N/m^2[/tex]
Elastic modulus of copper is given [tex]=11\times 10^10N/m^2[/tex]
Modulus of elasticity is equal to [tex]=\frac{stress}{strain}[/tex]
So [tex]11\times 10^{10}=\frac{153.448\times 10^6}{strain}[/tex]
Strain = 0.0013
So strain of the copper specimen will be equal to 0.0013
