Question

The power P of a jet of water is jointly proportional to the cross sectional area A and to the cube of the velocity v.

(a) the equation is P=kA(V)^3

(b) If the velocity is doubled and the cross sectional area is tripled, by what factor will the power increase?

Answer 1

Hello,

A: P=k*A*v^3

B:

v'=2v
A'=3A

P'=k*A'*V'^3
=k*3A*(2v)^3=24*k*A*v^3

Factor=24