You design a vessel to withstand the pressure of seawater at a depth of 350 m.
1. At this depth, what is the net force due to the water on the outside and the air inside the bell on a circular glass window 30.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? Ignore the small variation of pressure over the surface of the window.

F_net = 249.726 KN

Explanation:

We are given;

Depth; h = 350m

Diameter; d = 30cm = 0.3m

So,Radius; r = 0.3/2 = 0.15

To solve this problem, we'll say that the total pressure on the glass of the bell under this depth is equal to the pressure under this depth minus the pressure of air inside the glass window.

So,

P_portal on glass window = P - P_surface

Now, P is expressed as P = P_surface + ρgh

Thus,

P_portal on glass window = P_surface + ρgh - P_surface

So, P_portal on glass window = ρgh

Now, we want to calculate the net force. We know that, the formula for pressure is P = F/A.

Thus,

P_portal on glass window = F_net/A

Thus,

F_net/A = ρgh

F_net = ρghA

Where A is area = πr²

Density of sea water has a constant value of 1030 kg/m³

Thus, plugging in the relevant values,

F_net = 1030 x 9.8 x 350 x π x 0.15²

F_net = 249.726 KN

no. dot product   =0 when vecs at right angles