Answer:
The tension in the tendon is 2.9kN
Step-by-step explanation:
To solve the problem it is necessary to apply the concepts related to stress from the elongation of bodies by means of Young's Module.
The definition of Tension is given by
![F = AY(\Delta L)/(L_0)](https://img.qammunity.org/2020/formulas/physics/college/fv91ryzfn0vd47i7w21gtjgrtay8102mpa.png)
Where,
A =Area
Y = Young's modulus
Change in lenght
Initial Lenght
The change in Length for this case is given by
= 16.1-15 = 1.1cm
The area can be calculated as,
![A = \pi r^2](https://img.qammunity.org/2020/formulas/mathematics/college/cr8j38l09a4n3n5h6eay27m7d2yuucup99.png)
![A = \pi (3*10^(-3))^2](https://img.qammunity.org/2020/formulas/physics/college/tdvqnl13whvlm18nql6q3x31t57c5gb39w.png)
![A = 2.827*10^(-5)m^2](https://img.qammunity.org/2020/formulas/physics/college/u5ft5zmc4attmv8pmcoahim21pk5lycxf0.png)
Therefore replacing the tension would be,
![F = (2.827*10^(-5))(1.50×10^9)(1.1*10^(-2))/(16*10^(-2))](https://img.qammunity.org/2020/formulas/physics/college/v7ssm6t0j7j9t5rvp4xclzriqqhfgrdzup.png)
![F = 2915.34N](https://img.qammunity.org/2020/formulas/physics/college/3yx7gmcn6occfvdwk316r687x88n0101qn.png)
Therefore the tension in the tendon is 2.9kN