Answer:
[tex]E = 71,22 MPa[/tex]
Explanation:
According to hookes law:
elongation = [tex]e = \frac{dL}{L} = \frac{F}{A*E}[/tex]
Where F is the for aplied, A the cross section area, and E de modulus of elasticity. Solving for E:
[tex]E = \frac{F+L}{A*dL}[/tex]
here [tex]A= (0,0165m)^{2} = 2,7225*10^{-4} m2[/tex]
[tex]L = 0,125m\\dL = 0,00043 m\\F = 66700 N[/tex]
[tex]E = \frac{66700N+0,125m}{2,7225*10^{-4} m2 * 0,00043 m}[/tex]
[tex]E = 71,22 MPa[/tex]
