Posted by **ToT_Alex** on Wednesday, June 6, 2012 at 8:41am.

A research submarine is operating at a depth of 115m in sea water of density 1025kg/m^3.

The absolute atmospheric pressure at sea level is 100kPa. The absolute air pressure inside the submarine is twice that at sea level. Find the net force exerted by the water outside and the air inside on a circular window of diameter 300m.

- Physics -
**MathMate**, Wednesday, June 6, 2012 at 1:22pm
The idea is to calculate the pressure difference ΔP between the outside and the inside. Multiply ΔP by the area to get the force.

One pascal = 1 N-m^{-2}.

Atmospheric pressure

Pa= 100 kPa

= 100*10^{3} Pa

Density of sea water

= 1025 kg-m^{-3}

= 1025*9.81 N-m^{-3}

= 10.055*10^{3} N-m^{-3}

At 115m deep, hydrostatic pressure

Ph= 115*10.055*10^{3} Pa

= 1156.3*10^{3} Pa

Net pressure difference

=Atm.Press.+hydrost.pres.-internal press.

ΔP

= Pa+Ph-2Pa

=Ph-2Pa

=(1156.3-100)*10^{3} Pa

=1056.3*10^{3} Pa

Area of window (diameter 300 cm ?)

Aw= π*0.3²/4

=0.0707 m²

Force on window

= Aw*ΔP

= 0.0707 m² *1056.3*10^{3} Pa

= 74669 N

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