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mojhsa [17]
3 years ago
11

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

e 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?
Physics
1 answer:
vlada-n [284]3 years ago
4 0

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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