# Step 1: Calculate the Cross-Sectional Area The tendon is approximately round, so its cross-sectional area \(A\) is: \[ A = \pi r^2 \] where \( r = \frac{7.91\, \text{mm}}{2} = 3.955\, \text{mm} = 3.955 \times 10^{-3}\, \text{m} \). \[ A = \pi (3.955 \times 10^{-3})^2 \approx \pi \times 1.564 \times 10^{-5} = 4.913 \times 10^{-5}\, \text{m}^2 \] **Explanation:** We first convert the diameter to radius (in meters), square it, and multiply by \(\pi\) to get the area. --- # Step 2: Calculate Young's Modulus Young's modulus \(E\) is given by: \[ E = \frac{F L}{A \Delta L} \] where: - \( F = 13.3\, \text{N} \) - \( L = 12.5\, \text{cm} = .125\, \text{m} \) - \( A = 4.913 \times 10^{-5}\, \text{m}^2 \) - \( \Delta L = 2.21\, \text{mm} = 2.21 \times 10^{-3}\, \text{m} \) \[ E = \frac{13.3 \times .125}{4.913 \times 10^{-5} \times 2.21 \times 10^{-3}} \] \[ = \frac{1.6625}{1.085 \times 10^{-7}} \] \[ \approx 1.53 \times 10^7\, \text{Pa} \] **Explanation:** We substitute all values into the Young's modulus formula, being careful with unit conversions. --- # Final Answer **Young's modulus for the tendon is approximately \(1.53 \times 10^7\, \text{Pa}\) (Pascals).**