<span>Depth = 5.0 Ă— 10^2 m
Density of sea water = 1.025 x 10^3
Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa Since the normal pressure is retained in the hull, no need to bother about Po
Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
So it is 49.56 times larger.</span>
Pressure is force over surface. p=F/S. In this case the force is given by the weight of the water G, so p = G/S; <span> G=m*g; m=ρ*V, V=S*h since we are interested only in the volume of water which presses on the surface of the submarine S, at depth h. </span><span> So: </span><span> p= ρ*S*h*g/S = ρ*g*h. This is the pressure at a certain depth h, exerted only by the water so we must add the atmospheric pressure p0. </span> <span>Therefore the total pressure is pt = p0 + ρ*g*h. We can see that when h=0, pt=p0 which is correct since at the surface we have only atmospheric pressure. </span><span> The ratio pt/p0 = 1+ ρ*g*h/p0 </span><span> p0=101325 Pa </span><span> h = 500 m </span><span> g= 9.8 m/s^2 </span><span> ρ = 1.025*10^3 kg/m^3. </span><span> We substitute in the equation and get pt/p0 = 50.56 . So the pressure at 500m depth is 50.56 times greater than at the surface. </span>
