Answer:
Young's modulus of this tendon is [tex]9.03\times 10^6\ N/m^2[/tex].
Explanation:
Given that,
Length of the tendon, l = 19 cm
It is stretched by 4.5 mm, [tex]\Delta l=4.5\ mm[/tex]
Force, F = 11.3 N
Average diameter, d = 8.2 mm
Radius, r = 4.1 mm
The formula of Young's modulus of this tendon is given by :
[tex]Y=\dfrac{Fl}{\Delta l A}\\\\Y=\dfrac{11.3\times 0.19}{4.5\times 10^{-3}\times \pi (4.1\times 10^{-3})^2}\\\\Y=9.03\times 10^6\ N/m^2[/tex]
So, the Young's modulus of this tendon is [tex]9.03\times 10^6\ N/m^2[/tex]. Hence, this is the required solution.
