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In Fig , water flows through a pipe that...

In Fig , water flows through a pipe that tapers from radius `r_(1) = 0.020` m where the water has a uniform speed `v_1 = 1.5` m/s , to a radius `r_(2) = 0.010` m . What is the speed in the smaller pipe ?

Text Solution

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Pick a cross-sectional area through point 1 in the wider pipe and another one through point 2 in the narrower pipe . Then we can write
`v_(2) A_(2) = v_(1) A_(1)`
Solving for `v_(2)` and substituting for the area of a circle lead to
`v_(2) = v_(1) ""(pi r_(1)^(2))/(pi r_(2)^(2))`
`= (1.5 m//s) ((0.020 m)^(2))/((0.010 m)^(2)) = 6.0` m/s
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