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

@ksfishburne26 @CaseybethMtz @kisstherains please do help. i cannot figure out this problem.

Answer 2

Substutute these into p = kav^3:

p = k*a/2*(2v)^3

p = k*a/2*8v^3

p = k*4av^3

The power has increased by a factor of 4.

Answer 3

you had the equation right my friend c;

Answer 4

I don't know how to do this..Sorry ): I wish I could help.

Answer 5

i typed in 4 as the increase @plohrr and i was marked wrong for that answer

Answer 6

? thats impossible there is no rate here

Answer 7

It's okay thanks anyways @kisstherains

Answer 8

unless there is different values. is there any more info in the question

Answer 9

No problem. (:

Answer 10

no that is all the available info. so what i did was use random numbers as values for the variables k,a, and v but kept them constant when the area tripled and the velocity doubled @plohrr to figure out the factor it increased by.

Answer 11

i don't know how ...P=k*(A*v^3) maybe that is the correct equation?

Answer 12

maybe i've been stuck for quite some time now. and i have done a problem with the same set up as this one and did the same process. but the second answer was the only one incorrect. @plohrr

Answer 13

hm im afraid i won't be able to help then . this question has me stumped lol

Answer 14

hahah well thanks for your help anyway greatly appreciated @plohrr

Answer 15

:)