Question

A research submarine has a 20-cm-diameter window that is 8.5 cm thick. The manufacturer says the window can withstand forces up to 1.2×106 N .

Part A What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm.

Answer 1

Given is :-

Diameter of the window d = 20 cm

thus the radius of the window is r = 20/2 = 10cm

The capacity of the window to withstand force is

F = 1.2 x 106N

The pressure inside the submarine is to be maintained at  

Pe=latm

Now,

we know that   $P = \frac{F}{A}$

by plugging the values we get

1.2 x 106 (10 x 10-2m)

which gives us

P = 3.81972 x 10' N/m

Using

P - P.= pgh

by plugging all the values we get the safe depth for submarine, which is

3.81972 x 10' N/m² – 1.013 10 N/m2 = (10kg/m)(9.8m/s)

which gives us

h = 3887.34m