A submarine is at a depth of 500 m under the water. The force on a circular hatch of diameter 1.00 m due to the seawater (density = 1,025 kg/m3) pressure from outside the submarine is

To solve this problem, we need to use the formula for pressure at a certain depth in a fluid, which is P = ρgh, where:

• P is the pressure,
• ρ is the density of the fluid,
• g is the acceleration due to gravity, and
• h is the height (or depth in this case).

Step 1: Substitute the given values into the formula

The density of seawater (ρ) is given as 1,025 kg/m³, the acceleration due to gravity (g) is approximately 9.81 m/s², and the depth (h) is 500 m. Substituting these values into the formula gives:

P = 1,025 kg/m³ * 9.81 m/s² * 500 m = 5,033,125 Pa (Pascal)

Step 2: Calculate the force on the hatch

The force exerted by a pressure on a surface is given by the formula F = PA, where:

• F is the force,
• P is the pressure, and
• A is the area of the surface.

The area of a circle (the hatch is circular) is given by the formula A = πr², where r is the radius of the circle. The diameter of the hatch is given as 1.00 m, so the radius is 0.5 m. Therefore, the area of the hatch is:

A = π * (0.5 m)² = 0.785 m²

Substituting the values for the pressure and the area into the formula for the force gives:

F = 5,033,125 Pa * 0.785 m² = 3,950,953.125 N

Therefore, the force on the hatch due to the seawater pressure from outside the submarine is approximately 3,951,000 N.