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Mrrafil [7]
3 years ago
10

Suppose that you wish to construct a simple ac generator having 64 turns and an angular velocity of 377 radians/second (this is

the frequency point of 60 Hz). A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 0.01 m 2, what is the maximum output voltage?
Physics
1 answer:
Trava [24]3 years ago
3 0

Answer:

\epsilon_{max} =12.064\ V

Explanation:

Given,

Number of turns, N = 64

angular velocity, ω = 377 rad/s

Magnetic field, B = 0.050 T

Area, a = 0.01 m²

maximum output voltage = ?

We know,

\epsilon_{max} = NBA\omega

\epsilon_{max} = 64\times 0.05\times 0.01\times 377

\epsilon_{max} =12.064\ V

The maximum output voltage is equal to 12.064 V.

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Answer:

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The large block on the X axis has an applied P force and as it moves feels a force from the small block.  In the Y axis has the weight (W1) and the reaction to normal (N1)

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Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

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       fr = W2

       

The definition of friction force is

       fr = μ N

       

Let's replace and calculate

       μ (m2 a) = m2 g

       μ (P / (m1 + m2)) = g

       P = g /μ  (m1 + m2)

Let's calculate the value of this force

       P = 9.8 / 0.710 (28.9 +4.4)

       P = 13.80 (33.3)

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Answer:

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