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anonymous
 4 years ago
A cylindrical tank of radius 2 m and height 4 m is full of water. Find the
work required to pump all of the water to a point that is 2 m above the top of the tank.
The density of water is = 1000 kg/m3
and the acceleration of gravity is g = 9:8 m/s
Define clearly all the variables that enter into your solution and provide a drawing which
shows their meaning
anonymous
 4 years ago
A cylindrical tank of radius 2 m and height 4 m is full of water. Find the work required to pump all of the water to a point that is 2 m above the top of the tank. The density of water is = 1000 kg/m3 and the acceleration of gravity is g = 9:8 m/s Define clearly all the variables that enter into your solution and provide a drawing which shows their meaning

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saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0What part u dont understand?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok ... we have to make small cylinders with thickness dx. so the volume will be 4pi dx and to find the work you find the mass and the force and then to find the work you multiply it by the hieght and integrate it... now what are your limits of integration ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1W = Fd is something I remember about this

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1328655961951:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the distance changes as we integrate; the weight of the water is a factor of volume ....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1and i forgot the 2m above so that should be included in overall distance as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh.. ok , now I remembered , from the bottom to the top of the cylinder is 4m and we want to take it 2m above so we integrate from 0 to 6

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.youtube.com/watch?v=oWcN96kcbkM&feature=related yep, this might help, patrick is awesome

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1cable rope pull is same concept as water pumped

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://books.google.com/books?id=9ue4xAjkU2oC&pg=PA494&lpg=PA494&dq=calculus+work+pumping+water&source=bl&ots=XT_7RFQdg&sig=Vxg2w2jwUhhTeHdcCVYLkiZh5zk&hl=en&sa=X&ei=HbAxTzWIIvkggf70GbBQ&ved=0CF4Q6AEwBzgK#v=onepage&q&f=false im reviewing this to refresh my memory as well :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1volume of water = pi * (2m)^2 * (dy)m * (1000) = 4000pi(distance moved)dy distance moved is: 6y \[\int_{0}^{4}4000pi(6y)dy\]sounds right to me question is, is it right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1im not sure what the gravity is for; unless this is spose to be measured in newtons

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup, that's what I got , the gravity is for the force W=F.d F=mg m=(roh)V but the only thing I don't get is that when we bring the water 2m above the cylinder then why don't we integrate from 0 to 6 instead of 0 to 4 ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1because the weight of the water is not fluctuatung in that last little bit i assume. or rather, the amount of water that is above 4 is zero, so that extra 2 feet only affects the distance to move the water
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