Pressure on a window of a submarine

[indietro]

Homework Statement

a research submarine has a 20 cm diameter window 8 cm think. the manufacturer says the window van withstand forces up to 1x10⁶N. What is the submarine's maximum safe depth?

Homework Equations

p = p_o + $$\rho$$gd

The Attempt at a Solution

p = F/A = 1X10⁶ / 0.03 = 3.2x10⁷ Pa

so now do i use p = p_o + $$\rho$$gd ??

but if i do what would be my $$\rho$$?
$$\rho$$ = m/v = m/$$\pi$$r²h
how would i find mass?

thanks in advance for help!

 

[indietro]

ok so i solved it with the help of a different post but i had a question:
she does p' = p₁ + $$\rho$$gd
p' - p₁ = $$\rho$$gd

and then she says that p₁ = 1x10⁵ Pa > atmospheric pressure. I was just wondering why?

can someone explain how
$$\Delta$$p = p₁ - p_o relates or is used in conjunction with p' = p₁ + $$\rho$$gd? is $$\Delta$$p = p'

 

[indietro]

anyone??

 

[indietro]

i really just can't seem to wrap my head around pressure concepts, if someone knows how to explain it clearly or knows a website where i can see examples and stuff that would be great!

 

[PhanthomJay]

ok so i solved it with the help of a different post but i had a question:
she does p' = p₁ + $$\rho$$gd
p' - p₁ = $$\rho$$gd

and then she says that p₁ = 1x10⁵ Pa > atmospheric pressure. I was just wondering why?

it gets confusing with so many different letter variables like $$p, p_1, p_o, p', \rho$$ (and Pa for Paschals to boot), so let's define the variables:

let

$$p_o$$ = atmospheric pressure = 1.01(10^5) Paschals
$$p$$ = total pressure at depth d
$$\rho$$ = density of water = 1000kg/m^3)

Now when the top of the water is not sealed off (that is, it is exposed to the atmosphere),
$$p = p_o + \rho gd$$, that is , the total pressure at depth d is the sum total of the atmospheric pressure and the water pressure at that depth.