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 1 year 2021-08-23T19:57:14+00:00 1 Answers 23 views 0

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

    Answer:

    So the values are  [tex]p= \frac{1}{2}[/tex] , [tex]q= \frac{1}{2}[/tex]

    Explanation:

    From the question we are told that

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

    Now dimension of  v (speed ) is

              [tex]v = m/s = LT^{-1}[/tex]

    Now dimension of  g (acceleration  ) is

            [tex]g= m/s^2 = LT^{-2}[/tex]

    Now dimension of  h  (vertical distance  ) is        

            [tex]h= m = L[/tex]

    So  

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

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

    Equating powers

           [tex]1 =p+q[/tex]

           [tex]-1 = -2p[/tex]

    =>      [tex]p= \frac{1}{2}[/tex]

    and

            [tex]q= 1 -\frac{1}{2} = \frac{1}{2}[/tex]

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