A research submarine has a 20 cm diameter window which is 8cm thick. The manufac

Question

A research submarine has a 20 cm diameter window which is 8cm thick. The manufacturer says the window can withstand forces up to 1.0 x 10^6N. What is the submarine's maximum safe depth? Express the answer in km. *Assume density sea water=1030kg/m^3

I just don't understand this concept, whatsoever! Please give me an explanation so I'm not so in the dark about this. My midterm is soon and I want to understand this concept.

Basically, what I thought I was supposed to do was the following...
Pressure = P atmosphere + Rho(g)(d)
I thought that by equating maximum force to the absolute pressure, I was going in the right direction.
So... Force = 1e6N = Pressure/Area => 1e6N/pi(.1)^2 = 31,830,988.62Pa
Then...I just solved it
31,830,988.62Pa = 101,300Pa + (1030)(9.81)(d)
d= 3140m or 31.40km
However...The answer I saw did it in a way different way and I don't get it!
P = P atmosphere + Rho(seawater)(g)(d)
at maximum safe depth, d, force on window is F
-> F/A = P - P atmosphere --> P-P atmospheric = Rho(seawater)gd = F/A
d = F/(A*Rho*g) = 1e6/[pi(0.1^2)(9.8)(1030)] = 3153m, 31,53km

My ONLY question is....why in the world is atmospheric pressure ignored...? I just want to know what I did incorrectly because I don't understand this logic.

Explanation / Answer

Pnet = Po + dgh Density of saltwater = 1030 kg/m^3. Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). d = 20 cm = 0.2 m r = d/2 = 0.1 m A = pi(r^2) A = 3.14159265(.1^2) A = 0.0314159265 m^2 p = F/A p = (1.1 x 10^6) / (0.0314159265) p = 35,014,087.5 Pa 1 atm = 101,325 Pa P = Po + dgh h = (P - Po) / dg h = (35,014,087.5 - 101,325) / (1030 x 9.81) h = 3 455.23812 m h = 3.5 km

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.