Answer:
young's modulus[tex]=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2[/tex]
Explanation:
We have given length of tendon L= 13 cm =0.13 m
Change in length [tex]\Delta L=3.7mm=3.7\times 10^{-3}m[/tex]
Force = 12.1 N
Average diameter d =8.8 mm
So [tex]r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m[/tex]
Area [tex]A=\pi\times (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2[/tex]
Now stress [tex]=\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2[/tex]
Strain [tex]=\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}[/tex]
Now young's modulus[tex]=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2[/tex]
