The horizontal door of a submarine at a depth of 500m has an area of 0.16. calculate the force exerted by the sea water on the door at this depth. [Relative density of sea water=1.03] [Atmospheric pressure=1.0*10^5Nm^-2] [Density of pure water=1000kgm^-3] [g=10m/s^2].?
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Davian640
2 years ago
Pressure=force/area
cross multiply
Force=pressure * area
note: pressure=hpg
where;
h=500m
p=relative density of water*density=1.03*1000
g=10m/s^2
A=0.4m^2
Force on the door=Ahpg
=0.4*500*1.03*1000*10
=2.06*10^6N
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