A submarine window has 1730 N of force on it, due to water pressing it with 2.45 atm of pressure. What is the area of the wndow? (Unit \( =m^{\wedge} 2 \) ) As always, keep 3 sig figs.

To find the area of the submarine window, we'll use the relationship between force, pressure, and area: \[ \text{Pressure} (P) = \frac{\text{Force} (F)}{\text{Area} (A)} \] Rearranging for area: \[ A = \frac{F}{P} \] **Given:** - Force, \( F = 1730 \, \text{N} \) - Pressure, \( P = 2.45 \, \text{atm} \) **Step 1: Convert Pressure to Pascals (Pa)** \[ 1 \, \text{atm} = 101325 \, \text{Pa} \] \[ P = 2.45 \times 101325 \, \text{Pa} = 248,246.25 \, \text{Pa} \] **Step 2: Calculate the Area** \[ A = \frac{1730 \, \text{N}}{248,246.25 \, \text{Pa}} \approx 0.00697 \, \text{m}^2 \] **Rounded to three significant figures:** \[ A \approx 6.97 \times 10^{-3} \, \text{m}^2 \] **Final Answer:** The area of the window is approximately 6.97 × 10⁻³ m².