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lions [1.4K]
2 years ago
10

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

e building, 3.5 m higher, and the pressure remains unchanged. What is the speed of the water flow in the pipe on the second floor?
Physics
1 answer:
RUDIKE [14]2 years ago
4 0

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2} (1)

Where:

P_{1}, P_{2} - Pressures of the water on the first and second floors, measured in pascals.

\rho - Density of water, measured in kilograms per cubic meter.

v_{1}, v_{2} - Speed of the water on the first and second floors, measured in meters per second.

z_{1}, z_{2} - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]

\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]

v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) } (2)

If we know that P_{1}-P_{2} = 0\,Pa, \rho=1000\,\frac{kg}{m^{3}}, v_{1} = 15.5\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and z_{1}-z_{2} = -3.5\,m, then the speed of the water flow in the pipe on the second floor is:

v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}

v_{2} \approx 13.100\,\frac{m}{s}

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

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Answer:

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Replacing at the equation our values given we have that

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Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

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Answer:

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