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[Something descriptive here will yield more responses]I'm trying to solve the question b) in this linkhttp://i36.tinypic.com/2h3wmtx.jpg
My reasoning is [tex] dK = dm v^2 [/tex] [tex] dm = \rho dV = \rho A(x) dx [/tex] where A(x) is a function of how the cross sectional area varies along the tube from left to right. [tex] A(x) = \pi r(x)^2 [/tex] where r(x) is how the radius of the cross sectional area varies and is given by [tex] r(x) = r(1+x/l) [/tex] this gives [tex] dm = \rho \pi r^2 (1+x/l)^2 [/tex] the answer given gives the (1+x/l)^2 term in the denominator but the method i have used will give it in the numerator, i don't see the flaw in what i have done though unfortunately. |

Re: [Something descriptive here will yield more responses]Hi Narcol2000!
This is a kind of question that I think is better posted in the Homework area (see link below). And don't forget to give the post a title! :smile: http://www.physicsforums.com/forumdisplay.php?f=152 |

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