Question

The port hole in the side of a ship is in the

shape of a circle with a 2 foot diameter.
The top of the port hole is 6 feet below
the surface of the water and the density of
the water is 62.4 pounds per cubic foot.
Find the total force on this port hole due
to liquid pressure, accurate to the nearest
whole number.
[?] pounds

Answer 1

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.

Answer 2

Answer: 578490-=356478e

Explanation:

part of a circle is 75849=n so we use all the equations so 56748=4 we know the answer by the book page 356 then on chemestry rules.