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This is a three part question. I have answered two of them.

“A pump is spraying water into the air to form a fountain. The water emerges vertically from a jet of cross-sectional area 2.0*10^-4 m^2 at a rate of 1.0*10^-3 m^3 s^-1”

a) What is the velocity of the water as it emerges from the nozzle?

v=(1.0*10^-3m^3 s^-1)/(2.0*10^-4 m^2)

v=5m/s

b) What is the power of the pump?

W=E/t

W=1/2mv^2

W=1/2*1*5^2

W=12W

c)If the power is doubled, how much higher will the water rise (ignore resistance forces)

I am lost with this question. The answer is 0.75m.

Initially, I have worked out the height of the initial pump.

s=0.5(0+5)1

s=2.5m

But any calculation I try from here results in an incorrect answer

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$
    – user4552
    May 24 '18 at 1:10
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Before I given you the full solution I think it might be beneficial to try again replacing the mass term, m, with AVp where A is the area v is the velocity and p is the density.

Firstly, doubling the power. We have:

$P=\frac{1}{2}A\rho v^3$

Thus doubling the power will result in the velocity of the fluid being multiplied by $2^1/3$.

Now you can plug this into a SUVAT equation ($v^2=u^2-2as$) to get the height for the original case (which you calculated incorrectly. Try that formula) and the new velocity. Note v is not the same as the previous equation.

Now just find the difference!

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