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murzikaleks [220]
3 years ago
12

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

letely fills the pipe.
(a) At one point in the pipe the radius is 0.260 m. What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.20 m^3/s?
(b) At a second point in the pipe the water speed is 3.60 m/s. What is the radius of the pipe at this point?
Physics
1 answer:
slamgirl [31]3 years ago
6 0

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

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Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
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From the free body diagram, the tension in the chain is 450 N.
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The components of the tension are
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Vertical: 351.0 N, acting upward.

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