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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?

Respuesta :

okpalawalter8

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]