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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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1 point
s2008m [1.1K]

Answer:

The person has no displacement

Explanation:

The given parameters are

The location of the person = The equator

The distance covered in one revolution = Total distance around the body

The total distance around the Earth = The circumference of the Earth = 40.075 kilometres

The total distance moved by the person standing at the equator during the Earths complete revolution = 40,075 kilometres

The initial location of the person in relation to a fixed point in space outside Earth at the start of the revolution = x km

The final location of the person in relation to the fixed point in space outside Earth at the completion of the revolution = x km

The displacement = Change in position = Final location - Initial location  

∴ The displacement = x km - x km = 0 km.

5 0
4 years ago
A simple pendulum with a length of 2.23 m and a mass of 6.74 kg is given an initial speed of 2.06 m/s at its equi- librium posit
solniwko [45]

Answer:

a)   T = 2.997 s

b)   K = 14.3 J

c)   φ = 0.444 rad

Explanation:

a) Determine its period

  The pendulum simple’s period is:

 

  T = 2π\sqrt{\frac{l}{g} }

        Where l: Pendulum’s length

                    g = 9.8 m/s2

  T = 2π\sqrt{\frac{2.23}{9.8} }

  T = 2.997 s

b) Total energy

  Initially his total energy is kinetic

  K = \frac{mv^{2} }{2}

  K = \frac{(6.74)(2.06)^{2} }{2}

  K = 14.3 J

c) Maximum angular displacement

  φ = cos^{-1}(1-\frac{E}{mgl} )

  φ = cos^{-1}(1-\frac{14.3}{(6.74)(9.8)(2.23)} )

  φ = 0.444 rad

4 0
3 years ago
PLEASE ANSWERRRRR ASAPPPPPP
Bad White [126]
D Valence
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^^^answer
5 0
3 years ago
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Which two conditions explain why special technology is required to study the
enot [183]

Answer:

A. Extremely high pressure

5 0
2 years ago
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Please help. Having a hard time figuring out
Goryan [66]
Yeah that’s is correct
5 0
3 years ago
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