The speed v of an object falling with a constant acceleration g can be expressed in terms of g and the distance traveled from the point of r

Question

The speed v of an object falling with a constant acceleration g can be expressed in terms of g and the distance traveled from the point of release, h, as v = kgp hq, where k, p, and q, are dimensionless constants. What must be the values of p and q?

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Xavia 6 months 2021-08-23T19:57:14+00:00 1 Answers 5 views 0

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    2021-08-23T19:59:10+00:00

    Answer:

    So the values are  p= \frac{1}{2} , q= \frac{1}{2}

    Explanation:

    From the question we are told that

             The equation is  v =  k [g^p][h^q]

    Now dimension of  v (speed ) is

              v  = m/s =  LT^{-1}

    Now dimension of  g (acceleration  ) is

            g=  m/s^2 =  LT^{-2}

    Now dimension of  h  (vertical distance  ) is        

            h= m  = L

    So  

             LT^{-1} =   [ [LT^{-2}]^p][[ L]^q]

           LT^{-1} =   [ [T^{-2p}][[ L]^{p +q}]

    Equating powers

           1 =p+q

           -1 = -2p

    =>      p= \frac{1}{2}

    and

            q= 1 -\frac{1}{2} = \frac{1}{2}

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