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inysia [295]
3 years ago
7

A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v

1 = 0.84 m/s in the larger section of pipe. 1) What is the volume flow rate of the water?
Physics
1 answer:
Aleks [24]3 years ago
8 0

Answer:

0.116 m^3/s

Explanation:

The volume flow rate of a fluid in a pipe is given by:

Q=Av

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2

So, the volume flow rate is

Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s

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<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

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