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Math Help - Basis Dimensional analysis Question

  1. #1
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    Basis Dimensional analysis Question

    Let v be a velocity, t a time, x and r lengths. Let [A]=L (length)

    Find the mks units of B and C for:

    r=\sqrt{Ax+Bt}C^2v

    What I found was that [A] could be length, B could be \frac{L^2}{T} and C could be the square root of a time over the square root of a lengh times the fourth root of 2. My question is whether or not the 1 over 4th root of 2 is acceptable.
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  2. #2
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    My question is whether or not the 1 over 4th root of 2 is acceptable.
    I agree with everything you've done, except the fourth root of two. In dimensional analysis, you typically don't include numerical constants anywhere except in exponents. Think of it this way: the fourth root of two, if necessary, will be included in C, but not the units of C.

    Make sense?
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  3. #3
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    Okay, so when I play the factoring game, I can also ignore the \sqrt{2} , that pops out of factoring the terms inside the radical out?
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  4. #4
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    If you're doing dimensional analysis, yes, you can ignore factors such as you mentioned. Otherwise, of course, you can't.
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  5. #5
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    Quote Originally Posted by HelloWorld2 View Post
    Let v be a velocity, t a time, x and r lengths. Let [A]=L (length)

    Find the mks units of B and C for:

    r=\sqrt{Ax+Bt}C^2v

    What I found was that [A] could be length, B could be \frac{L^2}{T} and C could be the square root of a time over the square root of a lengh times the fourth root of 2. My question is whether or not the 1 over 4th root of 2 is acceptable.
    I would disagree with your use of the words "could be"! You are told that A has units of length. So Ax has units of "length squared" and, to be able to add Bt to that, Bt must have units of "length squared" also. There is no "2" in the units. Since we are told that t has units of time, B must have units of "length squared over time". "Length squared" plus "length squared" has units of "length squared". The the entire square root has units of length.

    v is a velocity, so it has units of length over time and the square root times v has units of "length squared over time". In order that r have units of length only, C^2 must have units of "time over length" and so C itself must have units of square root of "time over length".
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