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.

    = 9.21 104 N

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

        - = 0

        =

        = P A

Water pressure pressure plus atmospheric pressure

       P = + ρ g h

The hatch is a circle of radius = 0.18 m

       A = π r²

      = ( + ρ g h) π r²

Let's calculate

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

      = 9.049 105    0.1018

      = 0.921 105 N

     = 9.21 104 N

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