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elena-14-01-66 [18.8K]
2 years ago
12

What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem

ent is negligible
Physics
1 answer:
Step2247 [10]2 years ago
3 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

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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
RUDIKE [14]

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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