across
  • across
Let's do some cool math: A tank in the form of a right-circular 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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across
  • across
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.
anonymous
  • anonymous
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...
anonymous
  • anonymous
~30 min

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across
  • across
@Algebraic!, how did you get that? ;P
anonymous
  • anonymous
math&stuff
across
  • across
That's not a valid explanation! ;O
anonymous
  • anonymous
where's my cookie?
anonymous
  • anonymous
Roughly 30.275 mins.
anonymous
  • anonymous
idk differential equations, but i can try setting up one :p
anonymous
  • anonymous
Yeah, I forgot about Torricelli's law... I had started trying to derive it from Bernoulli's and was like... nope.
anonymous
  • anonymous
flowrate outside = \(\pi (1/2)^2\sqrt{2gh} \) cubic inch per sec
hartnn
  • hartnn
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.
across
  • across
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
anonymous
  • anonymous
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: \[ f_\text{rate}=(0.0127)^2\pi\sqrt{2\cdot3.048g}\approx0.003918\text{ m}^3/\text{s}\\ vf_\text{rate}\approx30.275\text{ mins.} \]
anonymous
  • anonymous
I meant: \[ \frac{v}{f_\text{rate}} \]... oops.
anonymous
  • anonymous
still waiting...

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