Answer to Question #176122 in Mechanics | Relativity for Akinola

Question

The force of a hatchdoor of a submarine of area 400cm² is 45kcak. Calculate the depth of the door below the sea level of the atmospheric pressure is 4atm

Answer 1

$Area=400 cm^2=0.04m^2$

$F=45 kCal=45 \times4186.80=188406 N$

Pressure at the door, $P_1=\frac{F}{A}=\frac{188406}{0.04}=4710150 \frac{N}{m^2}$

Atmospheric pressure, $P_2=4 atm=405300 \frac{N}{m^2}$

The depth of of the door can be found as below

$P_1+P_2=pgh$

$h=\frac{P_1+P_2}{pg}$

$h=\frac{4710150+405300}{997 \times9.81}=523.022 m$