Answer:
the total force on the port hole due to liquid pressure is approximately 7400 pounds.
Explanation:
The area of the circle is A = πr^2, where r = d/2 = 1 foot is the radius of the circle. So, the area is A = π(1 ft)^2 = π ft^2.
The port hole is submerged in water, with a height of 6 feet. The pressure of water at a depth of h feet is given by the formula P = ρgh, where ρ = 62.4 lb/ft^3 is the density of water, and g = 32.2 ft/s^2 is the acceleration due to gravity.
The total force on the port hole due to liquid pressure is the product of the pressure and the area of the circle, so we have:
F = P × A = ρgh × A = 62.4 lb/ft^3 × 32.2 ft/s^2 × 6 ft × π ft^2 ≈ 7400 lb
Therefore, the total force on the port hole due to liquid pressure is approximately 7400 pounds.