When a submarine dives to a depth of 5.0 * 102 m, how much pressure, in pa, must its hull be able to withstand? how many times larger is this pressure than the pressure at the surface?

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;

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.

So:

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.

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.

The ratio pt/p0 = 1 + ρ*g*h/p0

p0=101325 Pa

h = 500 m

g = 9.8 m/s^2

ρ = 1.025*10^3 kg/m^3.

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.

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.