A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20e5 N. What is the maximum depth in a lake to which the submarine can go without damaging the window?
As the submarine moves downward, the water pressure increases. Let’s determine the maximum pressure the window can stand.
Pressure = Force ÷ Area
Area = π * 0.15^2 = π * 0.0225
Pressure = 5.2 * 10^5 ÷ (π * 0.0225) = 7.356495147 * 10^6 N/m^2
Water pressure = Density * g * depth
Let’s assume the density is 1000 kg/m^3
Water pressure = 9800 * depth
9800 * depth = 5.2 * 10^5 ÷ (π * 0.0225)
depth = [5.2 * 10^5 ÷ (π * 0.0225)] ÷ 9800 = 750.6627701 meters
Comments
As the submarine moves downward, the water pressure increases. Let’s determine the maximum pressure the window can stand.
Pressure = Force ÷ Area
Area = π * 0.15^2 = π * 0.0225
Pressure = 5.2 * 10^5 ÷ (π * 0.0225) = 7.356495147 * 10^6 N/m^2
Water pressure = Density * g * depth
Let’s assume the density is 1000 kg/m^3
Water pressure = 9800 * depth
9800 * depth = 5.2 * 10^5 ÷ (π * 0.0225)
depth = [5.2 * 10^5 ÷ (π * 0.0225)] ÷ 9800 = 750.6627701 meters