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Dimensional Analysis

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A=B^3C^1/2 where A has the dimensions L/M and C has dimensions L/T. What are the dimensions of B?


    2. Relevant equations


    3. The attempt at a solution
    When I worked the problem I got B=M/T but it is wrong. I'm not sure how to approach the question.
     
  2. jcsd
  3. Oct 7, 2013 #2

    Simon Bridge

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    You substitute the dimensions into the equation and work out what dimenstions B has to have to make the LHS match the RHS.

    Please show your working.
     
  4. Oct 7, 2013 #3
    So far I have B^3=TL/M^2

    L/M=B^3(L/T)^1/2
    with L/T then in the square root, square both sides and get T/L*L/M^2=B^3, cancel the L on bottom and one on top to get the above answer.
     
  5. Oct 7, 2013 #4
    This doesn't make a lot of sense to me since B is still B^3. Am I supposed to have it look similar to the other side with TLM?
     
  6. Oct 7, 2013 #5
    You have several errors in this.
    Squaring will produce B^6. And L^2 in L/M.
    Using parentheses will make the things clearer. For you as well as for the people reading your posts.
    You don't need to square. Just solve for B and put the dimensions.
     
  7. Oct 7, 2013 #6

    Simon Bridge

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    T/L*L/M^2=T/L^2/M^2=T/(L^2M^2) ...

    Yike... you need to use brackets more to group your terms.
    Use square brackets to represent when you mean "dimensions of"


    [B^3]=(T/L)(L/M^2)=T/(M^2)

    OK - but you need ... you've not finished.
    (And - check your arithmetic.)
     
    Last edited: Oct 7, 2013
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