Question

Crew members attempt to escape from a damaged submarine 80 m below the surface. What force must they apply to a pop-out hatch of radius of 18 cm to push it out? Assume the density of ocean water 1025 kg/m3.

Answer 1

`F_(int)`= 9.21 104 N

Step-by-step explanation:

The definition of pressure is

P = F / A

In addition, the pressure varies with the depth by the equation

P = ρ g h

Let's write the second law and Newton for the hatch

`F_(int)` -
`F_(ext)` = 0

`F_(int)` =
`F_(ext)`

`F_(ext)` = P A

Water pressure pressure plus atmospheric pressure

P =
`P_(atm)` + ρ g h

The hatch is a circle of radius = 0.18 m

A = π r²

`F_(ext)` = (
`P_(atm)` + ρ g h) π r²

Let's calculate

`F_(int)` = (1,013 105 + 1025 9.8 80) π 0.18²

`F_(int)` = 9.049 105 0.1018

`F_(int)` = 0.921 105 N

`F_(int)`= 9.21 104 N