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anonymous
 5 years ago
a rectangular water tank has a length 20 ft, width 10 ft , and depth 15 ft . if the tank is full , how much work does it take to pump all the water out (using calculus)
anonymous
 5 years ago
a rectangular water tank has a length 20 ft, width 10 ft , and depth 15 ft . if the tank is full , how much work does it take to pump all the water out (using calculus)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this is another combination of mathematical calculus and the physical energy problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this one is much easier than the previous one, don't you think so? because there's no dripping water here~! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but im still confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is what i got 62.4(density of water)(200,the volume) integral 15h limit 15 to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the measurement is ft , not meter.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know that why u use 62.4 as the density

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youd better change to ISU, more convenient to calculate, as 1 g/cm^3 the density of the water/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry i had to go to bed, its midnight here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay thank you for ur help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anwar di u think u can help me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the water leaving the tank from the top or the bottom? I'll assume it's leaving from the bottom. W = F * d; draw the tank with the dimensions. Find the density of an infinitely thin "slab" of water. Its volume is going to be 20*10*dy = 200dy. The force is going to be the volume times the density, so 200dy * 62.4 = 12480dy. The distance that this slab is going to have to travel to leave the tank is y, so the work done in moving that slab is 12480ydy. Sum all the work done by all the slabs with... \[\int\limits_{0}^{15}12480ydy\] I got 1404000 footpounds of work.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank u for coming to the rescue , but the answer is 220,500 ft lb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrate 12480ydy from 0 to 15 is what I meant to say...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The distance would be 15  y if the water was leaving from the top... I don't think that wouldn't get the amount of work to be 220500, though...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it just says how much work would it take to pump the water, it doesn"t say top and bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, for water to be pumped out of a tank it needs to be pumped out from somewhere. The water needs to move to a part of the tank to be pumped out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here what I have, I am not sure about it though.. haven't done physics in a while: we have the following formula: \[W=\int\limits_{v_i}^{v_f}pdv\] where p is pressure of water (can be found known depth and temperature of water), vi and vf are the initial and final volume respectively (both known)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it doesn't matter if we use y or 15y to represent the distance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well we have the initial volume=20*10*15, final volume=0.. we will get a negative result for the work.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey im a little confused. can someone please explain to me how the distance become 15y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have no idea why my answer is incorrect. If you draw the height of the tank and pick an arbitary point along the height, the distance between the top of the tank and that point is 15  y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and when is it just y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Distance betwen the arbitary point and the bottom of the tank.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but in the triangle is the top is just y, and the bottom is 15y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is another question , when u slice a triangle and want to use similar triangles the top portion is y and the bottom half is 15y ,( this just an example)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I suppose you could think of it that way if you positioned the tank such that the xaxis was the top of the tank and 15 feet below that is the bottom of the tank.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey i have a question can i ask
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