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?

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

Answers ( )

  1. Ladonna
    0
    2021-08-23T19:59:10+00:00
    Reply

    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}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )