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Homework Help: Air sacs fish problem

  1. Nov 2, 2007 #1
    I don't know why I can't set up this proportion correctly.

    A fish maintains its depth in fresh water by adjusting air contents in air sacs in its body. With its air sacs fully collapsed the fish has a density of 1.08 g/cm^3.

    To what fraction of its expanded body volume does it need to inflate its air sacs to reduce its density to that of the water.

    I can't seem to set this up. It is asking for [tex]\frac{V_{empty}}{V_{max}} [/tex] I know that [tex]\rho =m/V[/tex] [tex]\rho_w=\frac{1g}{cm^3}[/tex] and [tex]\rho_{fish}=1.08\frac{g}{cm^3}[/tex]

    I am not sure what my problem is right now. Anyine have a hint?


    Am I an idiot for not getting this?
    Last edited: Nov 2, 2007
  2. jcsd
  3. Nov 2, 2007 #2
    Work with two variables. Yo should get it.
  4. Nov 2, 2007 #3
    I'm sorry? I don't follow.

  5. Nov 2, 2007 #4
    Okay one way could be:
    Assume final volume (after expansion) be y, and the volume of air inhaled be x. You need to find x/y.
    Set up relations from given data.
    Last edited: Nov 2, 2007
  6. Nov 2, 2007 #5
    The volume of air inhaled IS the volume of the sacs after expansion isn't it?

  7. Nov 2, 2007 #6
    Assuming density of air to be negligible, mbefore expansion = mafter expansion.
    Relate x and y using densities and above assumption.
  8. Nov 2, 2007 #7
  9. Nov 2, 2007 #8
    Okay I know that the answer is .074 from the back of the book.

    I used [tex]\frac{m/V}{m/V'}=\frac{1}{1.08}[/tex] which gets me[tex]\frac{V'}{V}= .9259[/tex] but that is not quite it..I have to subtract that number from 1 to get .074. So I am clearly misunderstanding the question.

    Why do I need to do 1-V'/V ? Is it because V'/V is the part of the sacs IN USE and the entire air sac is 1?

    Last edited: Nov 2, 2007
  10. Nov 2, 2007 #9
    Think and realize, what is your V and V' in the formula?
    It is mis-typed I guess.. not 11.08.. it is 1.08
    Last edited: Nov 2, 2007
  11. Nov 2, 2007 #10
    V' is the volume after inhalation and V is volume before inhalation.

    oops I forgot \frac. I edited it. I used 1/1.08 since density of h20= 1g/cm^3
  12. Nov 2, 2007 #11
    Really?? If V' is total volume after inhalation, and V before it.. how come V'/V is less than 1?? After inhalation, volume of the fish should increase isnt it??
  13. Nov 2, 2007 #12
    Pkay...let me start over.
    [tex]\rho[/tex] before inhale=1.08 [tex]\rho'[/tex] after inhale =1
    [tex]\Rightarrow \frac{m/V}{m/V'}=\frac{1.08}{1}[/tex]
    [tex]\Rightarrow \frac{V}{V'}=\frac{1}{1.08}=.9259[/tex]

    This is the fraction of V'/V
  14. Nov 2, 2007 #13
    now, what you are asked?
  15. Nov 2, 2007 #14
    I am asked "To what fraction of its expanded body volume does it need to inflate its air sacs to reduce its density to that of the water.

    I am still trying to make the conection between V'/V-->1-(V'/V)

    Can you put 1-(V'/V) into words.....I would like to understand this better for future problems.

    Thanks for your help thus far!

  16. Nov 2, 2007 #15
    It is not 1 - (V'/V) .. rather it is 1 - (V/V'). (Note that V/V' = 0.9259, see your second last post.)
    Now 1 - (V/V') = (V' - V)/V'.
    V' = Final Volume of the fish after inhaling (i.e., expanded body volume)
    V' - V = Final Volume of the fish (after inhaling) - Initial Volume of the fish (before inhaling) = Volume of air inhaled.

    So, it is 1 - (V/V'), that you were asked!
  17. Nov 2, 2007 #16

    Ahhh. I think I see it now. I need to let this soak in a little. Thanks saket!

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