The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
![a \ student \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\ \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!](https://tex.z-dn.net/?f=%20a%20%5C%20student%20%20%5C%20is%20%5C%20able%20%5C%20to%20%5C%20lift%20%5C%20a%20%5C%20500%28n%29%20%5C%5C%20weight%20%5C%20by%20%5C%20applying%20%5C%20only%20%20%5C%20100%28n%29%20%5C%5C%20%5C%5C%20so%2C%20%5C%20this%20%5C%20info%20%5C%20here%2C%20%5C%20we%20%5C%20simply%20%5C%20divide%20%5C%20by%20%5C%5C%20how%20%5C%20much%20%5C%20this%20%5C%20kid%20%5C%20lifted%2C%20%5C%20by%20%5C%20the%20%5C%20weight%20%5C%20he%2Fshe%20%5C%5C%20%5C%20is%20%5C%20applying.%20%5C%5C%20%5C%5C%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B500%28n%29%2F100%28n%29%7D%7D%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%5C%5C%20and%20%5C%20from%20%5C%20this%2C%5C%20your%20%5C%20answer%20%5C%20would%20%5C%20%5C%5C%20conclude%20%5C%20to%20%5C%20be%20%5C%20%285%29%20%5C%5C%20%5C%5C%20%5Cboxed%7B5%7D%20%5C%20would%20%5C%20be%20%5C%20your%20%5C%20answer%21)