[Something descriptive here will yield more responses]

  1. 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.
     
  2. jcsd
  3. 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:

    https://www.physicsforums.com/forumdisplay.php?f=152
     
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