# F=MA 2012 Exam #19: Find Pipe Radius

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In summary, a 1,500 Watt motor is used to pump water from a flooded basement at a height of 2.0 meters through a cylindrical pipe, ejecting the water at a speed of 2.5 m/s. Ignoring friction and assuming all the energy goes to the water, the closest radius of the pipe is 10 cm. This can be calculated using the power formula and considering the gravitational force and cross-sectional area of the pipe.
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## Homework Statement

19. A 1,500 Watt motor is used to pump water a vertical height of 2.0 meters out of a ﬂooded basement through a
cylindrical pipe. The water is ejected though the end of the pipe at a speed of 2.5 m/s. Ignoring friction and
assuming that all of the energy of the motor goes to the water, which of the following is the closest to the radius
of the pipe? The density of water is ρ = 1000 kg/m3
.
(A) 1/3 cm
(B) 1 cm
(C) 3 cm
(D) 10 cm ← CORRECT
(E) 30 cm

## Homework Equations

Pressure = ρgh
Bernoulli's Eq: p + ρgh + 1/2ρv^2 = Constant
Power = F dot v, Power = Work / t = dW/dT
Flow Rate Continuity:
A_0v_0 = Av

## The Attempt at a Solution

First, I said that:
A_0v_0 = pir^2*2.5
Thus, v_0 = pir^2*2.5 / A_0
Then, I used Bernoulli's Eq, but I'm confused as to how I make use of the power? I need some guidance on how to use the power in the problem or how to turn one of these equations into
one I can handle.

How much energy does it take to lift a mass through a given height?
What is the expression that tells you the mass of water that is ejected through the pipe each second?

It takes mgh to lift an object height h. Are you talking about the mass flow rate equation:
That's m/t which is equal to (rho)Av.
Here's an approach:
W = PE + KE
W = mgh + 1/2mv^2
W/t = Power =m/t(gh + 1/2v^2)
1500 = pAv(gh+ 1/2v^2)
A = pi*r^2
1500 = r^2(pi)(rho)v(gh), since the velocity of the basement is negligible ( I would think )
r = sqrt (1500 /(pi*rho*v*gh)
so r = .099 m = 9.9 cm so
10cm -- This is the correct answer.

Last edited:
assuming that the v is almost negligible
How good is that assumption?
1500 = r^2(pi)(rho)gh
[color-red]r = sqrt( 1500 / ((pi)(rho)gh))[/color]
r = .49m = 49 cm? - This is wrong
What happened to the KE term?

Yep, I saw that error in the radius and fixed it. How can we do the problem without assuming v is negligible?

what happens to the step you made the assumption if you don't?
i.e. what problem did the assumption solve exactly?

Try writing down the rate mass flows through the pipe as ##\frac{dm}{dt}## and then ##P=\frac{dE}{dt}## ... remembering that we are told that only mass changes with time.

so:
dE/dt = dm/dt(gh + 1/2v^2)
if power is constant, does that suggest that
gh + 1/2v^2 = 0,
or that v= root(2gh)?
No I don't think I see how to use that.

Here's a thought that disregards the velocity:
Power = Fv
F in this case is the gravitational force,mg because that is what must be overcome to pump the water at a height of 2m
v is given as 2.5 m/s - that is the speed of pumping.
Thus:
P = mgv
m = ρV
P = ρVgv
V = h*A, where A is cross-sectional area
P = ρhAgv
A =∏r^2, sorry I can't find the little pi
P = r^2ρ∏hgv
r = √(P / ρ∏hgv)
All these are given
r ≈ 10 cm

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Can you assume the speed of the water is constant?

## 1. What is the formula for F=MA?

The formula for F=MA is force (F) equals mass (M) multiplied by acceleration (A).

## 2. How do I find the pipe radius using F=MA?

To find the pipe radius using F=MA, you will need to rearrange the formula to solve for the radius (r). The formula will then be r=F/(MA). Plug in the values for force, mass, and acceleration to calculate the radius.

## 3. What units should be used for F, M, and A?

Force (F) should be measured in Newtons (N), mass (M) should be measured in kilograms (kg), and acceleration (A) should be measured in meters per second squared (m/s^2).

## 4. Can F=MA be used to calculate the radius of any type of pipe?

Yes, F=MA can be used to calculate the radius of any pipe as long as the values for force, mass, and acceleration are known.

## 5. Are there any limitations to using F=MA to find pipe radius?

One limitation to using F=MA to find pipe radius is that it assumes the pipe is a perfect cylinder with uniform density. It may not be accurate for pipes with irregular shapes or varying densities.

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