contestada

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.

Respuesta :

Answer:

    [tex]F_{int}[/tex]= 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

        [tex]F_{int}[/tex] - [tex]F_{ext}[/tex] = 0

       [tex]F_{int}[/tex] = [tex]F_{ext}[/tex]

       [tex]F_{ext}[/tex] = P A

Water pressure pressure plus atmospheric pressure

       P = [tex]P_{atm}[/tex] + ρ g h

The hatch is a circle of radius = 0.18 m

       A = π r²

     [tex]F_{ext}[/tex] = ([tex]P_{atm}[/tex] + ρ g h) π r²

Let's calculate

      [tex]F_{int}[/tex] = (1,013 105 + 1025 9.8 80) π 0.18²

      [tex]F_{int}[/tex] = 9.049 105    0.1018

     [tex]F_{int}[/tex] = 0.921 105 N

     [tex]F_{int}[/tex]= 9.21 104 N

Otras preguntas