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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 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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Physics
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2021-08-23T19:57:14+00:00
2021-08-23T19:57:14+00:00 1 Answers
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Answers ( )
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]