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across
Group Title
Let's do some cool math:
A tank in the form of a rightcircular cylinder standing on end is leaking water through a circular hole in its bottom. Suppose the tank is \(10\) feet in height and has radius \(2\) feet and the circular hole has radius \(1/2\) inch. If the tank is initially full, how long will it take to empty?
Whoever gets this right first gets a cookie.
 one year ago
 one year ago
across Group Title
Let's do some cool math: A tank in the form of a rightcircular cylinder standing on end is leaking water through a circular hole in its bottom. Suppose the tank is \(10\) feet in height and has radius \(2\) feet and the circular hole has radius \(1/2\) inch. If the tank is initially full, how long will it take to empty? Whoever gets this right first gets a cookie.
 one year ago
 one year ago

This Question is Closed

across Group TitleBest ResponseYou've already chosen the best response.0
Hint 1: Torricelli's law states that\[v=\sqrt{2gh},\]where \(v\) is the speed of the water leaving at the bottom of a tank of height \(h\). \(g\) stands for the acceleration due to gravity, as usual.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
I really want to work this out... but, unless I'm missing something, using Bernoulli's, it'd take too long, and I don't think I have the attention span...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
~30 min
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
@Algebraic!, how did you get that? ;P
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
math&stuff
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
That's not a valid explanation! ;O
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
where's my cookie?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Roughly 30.275 mins.
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.0
idk differential equations, but i can try setting up one :p
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yeah, I forgot about Torricelli's law... I had started trying to derive it from Bernoulli's and was like... nope.
 one year ago

sara12345 Group TitleBest ResponseYou've already chosen the best response.0
flowrate outside = \(\pi (1/2)^2\sqrt{2gh} \) cubic inch per sec
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
using this differential equation : dh/dt = Ah/Aw sqrt (2gh) take g= 32 ft/s^2 Ah=Area of hole , Aw is area of water ,so Aw/Ah = 576 so (h^(1/2))dh = (8/576)dt so 2 (sqrt h) = (8/576) t put h= 10 to get t = 30.36 minutes.
 one year ago

across Group TitleBest ResponseYou've already chosen the best response.0
You guys are right. The answer is roughly \(30\) minutes. I guess I'll have to start posting Millennium Prize problems. That way, you guys will take at least five minutes longer to answer correctly. ;P
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
I mean, it's not hard to compute, just a bunch of numbers and converting feet to metres. So, converting everything, we get the very 'elegant' expressions of: \[ f_\text{rate}=(0.0127)^2\pi\sqrt{2\cdot3.048g}\approx0.003918\text{ m}^3/\text{s}\\ vf_\text{rate}\approx30.275\text{ mins.} \]
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
I meant: \[ \frac{v}{f_\text{rate}} \]... oops.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
still waiting...
 one year ago
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