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SIZIF [17.4K]
3 years ago
6

Water enters a shower head through a pipe of radius 0.0112 m at 3.25 m/s. What is it’s volume flow rate? (Unit= m^3/s)

Physics
2 answers:
kenny6666 [7]3 years ago
7 0

Answer:

1.28 \times 10 ^{-3} m^3/s

Explanation:

Given that the radius of the pipe, r =0.0112 m

So, the area of the cross-section, a= \pi (0.0112)^2 = 3.941\times 10^{-4} m^2

Speed of water in the pipe, v=3.25 m/s

Volume flow rate =av= 3.941\times 10^{-4} \times 3.25 = 1.28\times 10 ^{-3} m^3/s

Hence, the volume flow rate is 1.28 \times 10 ^{-3} m^3/s

tangare [24]3 years ago
4 0

Answer:

For Acellus the answer is 0.00128

Explanation:

Trust me

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The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

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-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

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2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

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<em>v</em>₀² = 9.73 m²/s²

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