Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
Answer:
12387N
Explanation:
1 atm = 101325 Pa
Let g = 9.8 m/s2. The pressure of the sea water at 25 m below the surface is:
[tex]P = P_0 + \rhogh = 101325 + 1030*9.8*25 = 353675 Pa[/tex]
So the pressure difference between inside the submarine and outside is
[tex]\Delta P = P - P_0 = 353675 - 101325 = 252350 Pa[/tex]
The area of the hatch which is subjected to this pressure is:
[tex]A = \pi (d/2)^2 = \pi (0.25/2)^2 = 0.05 m^2[/tex]
So the force required to win over the pressure difference is:
[tex]F = \Delta P A = 252350 * 0.05 = 12387 N[/tex]
