Question

(11%) Problem 5: A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of sea water is 1.03 × 103 kg/m3. show answer No Attempt Calculate the force, in newtons, needed to open the hatch from the inside, given it is circular and 0.25 m in diameter. The air pressure inside the submarine is 1.00 atm.

Answer 1

F = 1.24*10^4 N

Step-by-step 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 2

12387N

Step-by-step explanation:

1 atm = 101325 Pa

Let g = 9.8 m/s2. The pressure of the sea water at 25 m below the surface is:

`P = P_0 + \rhogh = 101325 + 1030*9.8*25 = 353675 Pa`

So the pressure difference between inside the submarine and outside is

`\Delta P = P - P_0 = 353675 - 101325 = 252350 Pa`

The area of the hatch which is subjected to this pressure is:

`A = \pi (d/2)^2 = \pi (0.25/2)^2 = 0.05 m^2`

So the force required to win over the pressure difference is:

`F = \Delta P A = 252350 * 0.05 = 12387 N`