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Math Help - I need help solving an equation

  1. #1
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    I need help solving an equation

    I'm trying to do an air resistance lab problem for physics class.
    Basically, I had two sets of data (mass and velocity) and plugged them in to this equation: m(9.8) = kv^n

    I solved up to a certain point (here):
    31.16 = k(2.70)^((20.78 ln(k))/(.419))


    In the end, I need to find n, but I figure that if I can solve for k in that equation then I can easily find n.

    My teacher said we can just guess and check for this problem, but I'd rather have a solution for the equation.
    (And of course I will credit whoever helps me out)

    This is due on Tuesday, so I hope someone can help me out sometime tonight or tomorrow.

    Thanks
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  2. #2
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    Quote Originally Posted by Musturd View Post
    I'm trying to do an air resistance lab problem for physics class.
    Basically, I had two sets of data (mass and velocity) and plugged them in to this equation: m(9.8) = kv^n

    I solved up to a certain point (here):
    31.16 = k(2.70)^((20.78 ln(k))/(.419))


    In the end, I need to find n, but I figure that if I can solve for k in that equation then I can easily find n.

    My teacher said we can just guess and check for this problem, but I'd rather have a solution for the equation.
    (And of course I will credit whoever helps me out)

    This is due on Tuesday, so I hope someone can help me out sometime tonight or tomorrow.

    Thanks
    31.16 = k(2.70)^b , where b = \frac{20.78-\ln{k}}{.419}<br />

    you're not going to be able to solve the given equation using elementary algebraic methods.


    solving your equation using a calculator ...

    k \approx 3.308 \times 10^{14}

    which leads me to believe something may be in error.

    Correct me if I'm mistaken, but isn't the equation mg = kv^n valid only when the falling mass is at terminal velocity?

    do you have some representative data that you can share?
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  3. #3
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    Wow.
    Here is my data:
    (2.12g, 1.52 m/s)
    (3.18g, 2.7 m/s)

    I forgot to convert the grams to kilograms -- maybe I didn't screw up my math.

    So with the conversion the original equation would be:
    .3112 = k(2.70)^b , where b = \frac{.2078-\ln{k}}{.419}<br />

    Anyway, since this can't be solved with algebra (like you said), I plugged it into wolfram|alpha: http://www.wolframalpha.com/input/?i=(.3312)+%3D+k*(2.70)^((.2078+%E2%80%93+ln(k))/(.419)))

    k \approx 3.208
    looks promising
    ...
    so n \approx -2.28
    Which is not good. n should be around +2.28 not -2.28
    Maybe I did do my beginning steps wrong.

    Here are my beginning steps:
    (.0212)(9.8) = k*(1.52)n
    .2078 = ln(k*1.52n)
    .2078 = ln(k) + n*ln(1.52)
    (.2078 ln(k))/(ln(1.52)) = n
    Then
    (.0318)(9.8) = k(2.70)^n , where n = \frac{.2078-\ln{k}}{ln{1.52}}<br />
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  4. #4
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    Wow.
    Here is my data:
    (2.12g, 1.52 m/s)
    (3.18g, 2.7 m/s)

    I forgot to convert the grams to kilograms -- maybe I didn't screw up my math.

    So with the conversion the original equation would be:
    .3112 = k(2.70)^b , where b = \frac{.2078-\ln{k}}{.419}<br />

    Anyway, since this can't be solved with algebra (like you said), I plugged it into wolfram|alpha: http://www.wolframalpha.com/input/?i=(.3312)+%3D+k*(2.70)^((.2078+%E2%80%93+ln(k))/(.419)))

    k \approx 3.208
    looks promising
    ...
    so n \approx -2.28
    Which is not good. n should be around +2.28 not -2.28
    Maybe I did do my beginning steps wrong.

    Here are my beginning steps:
    (.0212)(9.8) = k*(1.52)^n
    .2078 = ln(k*1.52^n)
    .2078 = ln(k) + n*ln(1.52)
    (.2078 ln(k))/(ln(1.52)) = n
    Then
    (.0318)(9.8) = k(2.70)^n , where n = \frac{.2078-\ln{k}}{ln{1.52}}<br />
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  5. #5
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    mg = kv^n

    setting up a ratio ...

    \frac{.00212g = k(1.52)^n}{.00318g = k(2.7)^n}<br />

    \frac{.00212}{.00318} = \left(\frac{1.52}{2.7}\right)^n

    now solve for n
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  6. #6
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    Thanks for an easy way to calculate it.
    I can't believe I still was off by a zero on the kilogram conversion.
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