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Reika [66]
2 years ago
13

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

ute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2. Part A Find the mass flow rate. M
Physics
1 answer:
Harman [31]2 years ago
4 0

Answer:

1.302\ \text{kg/s}

Explanation:

\rho = Density of water = 1\ \text{kg/L}

dV = Change in volume = 220\times 0.355\ \text{L}

dt = Time elapsed = 1 minute = 60 seconds

Mass flow rate is given by

\dot{m}=\rho\dfrac{dV}{dt}\\ =1\times \dfrac{220\times 0.355}{60}\\ =1.302\ \text{kg/s}

The mass flow rate is 1.302\ \text{kg/s}.

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A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
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Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

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let's calculate

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3 years ago
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