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 l

Question

Mila

1 month ago

12

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 l

arger is this pressure than the pressure at the surface?

Physics

2 answers:

kicyunya [3.2K] · Answer 1 · 2026-03-01T20:17:21+03:00

Pressure is calculated as force per unit area.
p=F/S.
In this scenario, the force is due to the weight of the water G, thus
p = G/S;
G=m*g;
m=ρ*V;
V=S*h
Here, we focus on the water volume applying pressure at surface S, at depth h.
Therefore:
p= ρ*S*h*g/S = ρ*g*h.
This is the water pressure at depth h, and we must also account for atmospheric pressure p0.
Consequently, the total pressure would be pt = p0 + ρ*g*h.
At h=0, pt=p0 holds true, affirming only atmospheric pressure at the surface.
Now, the ratio pt/p0 = 1+ ρ*g*h/p0
where p0=101325 Pa,
h = 500 m,
g= 9.8 m/s^2,
and ρ = 1.025*10^3 kg/m^3.
Plugging into the equation yields pt/p0 = 50.56.
Thus, the pressure at a depth of 500m is 50.56 times higher than at the surface.

ValentinkaMS [3.4K] · Answer 2 · 2026-03-01T20:19:16+03:00

Depth = 5.0 × 10^2 m
Density of seawater = 1.025 x 10^3
Pd = Po + pgh
Standard atmospheric pressure is Patm = 1.01325 x 10^5 Pa
Since the pressure inside the hull is normal, we can disregard Po.
Thus, Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
Now calculating Pd / Patm gives us 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
This indicates the pressure is 49.56 times greater.