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Flauer [41]
3 years ago
12

How much voltage is in the primary coil if there are 3200 windings in the

Physics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

\frac{V_1}{V_2} = \frac{N_1}{N_2}

here we know that

V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V

N_1 = 500

N_2 = 3200

Now we have

\frac{V_1}{25} = \frac{500}{3200}

V_1 = 3.91 V

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Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

K_{1}  = K_{2} + W_{loss}

Where:

K_{1}, K_{2} - Initial and final kinetic energy, measured in joules.

W_{loss} - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

\frac{1}{2}\cdot m \cdot v_{1}^{2}  = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s

Where:

v_{1}, v_{2} - Initial and final speed of the skater, measured in meters per second.

m - Mass of the skater, measured in kilograms.

\Delta s - Length of the rough patch, measured in meters.

f - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

f = r \cdot m \cdot g

Where:

r - Ratio of friction force to weight, dimensionless.

g - Gravitational constant, measured in meters per square second.

Then,

\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s

The equation is simplified algebraically and patch length is cleared afterwards:

\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s

\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }

Given that v_{1} = 5\,\frac{m}{s}, v_{2} = 2.5\,\frac{m}{s}, r = 0.22 and g = 9.807 \,\frac{m}{s^{2}}, the length of the rough patch is:

\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta s = 4.345\,m

The length of the rough patch is 4.345 meters.

6 0
3 years ago
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