SEP Use Mathematics A wire with a length of 3.0 m and a cross- sectional area of \( 0.0020 \mathrm{~m}^{2} \) is stretched 0.10 cm by a \( 98-\mathrm{kg} \) hanging weight. Determine Young's modulus for the wire.

To determine Young's modulus (\( Y \)) for the wire, we'll use the following formula: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{F \cdot L}{A \cdot \Delta L} \] **Given:** - **Length of the wire (\( L \))**: 3.0 m - **Cross-sectional area (\( A \))**: \( 0.0020 \, \text{m}^2 \) - **Elongation (\( \Delta L \))**: 0.10 cm = 0.0010 m - **Mass hanging (\( m \))**: 98 kg - **Acceleration due to gravity (\( g \))**: \( 9.8 \, \text{m/s}^2 \) **Step 1: Calculate the Force (\( F \))** \[ F = m \cdot g = 98 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 960.4 \, \text{N} \] **Step 2: Plug the Values into Young's Modulus Formula** \[ Y = \frac{F \cdot L}{A \cdot \Delta L} = \frac{960.4 \, \text{N} \times 3.0 \, \text{m}}{0.0020 \, \text{m}^2 \times 0.0010 \, \text{m}} = \frac{2881.2 \, \text{N}\cdot\text{m}}{0.000002 \, \text{m}^3} = 1.4406 \times 10^9 \, \text{Pa} \] **Final Answer:** The Young’s modulus of the wire is approximately \( 1.4 \times 10^{9} \, \text{Pascal} \). **Expressed in Gigapascals (GPa):** \[ Y \approx 1.4 \, \text{GPa} \]