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m_a_m_a [10]
3 years ago
13

The rotating loop in an AC generator is a square l on each side. It is rotated at frequency f in a uniform field of B. The gener

ator is loaded with a resistor of resistance R. Give formulas in terms of l, B, f , and R for the following quantities:
a. the flux Φ (t) through the loop as a function of time
b. the emf Σ(t) induced in the lop
c. the current I(t) induced in the resistor
d. the power R(t) delivered to the resistor
e. the torque τ(t) that must be exerted to rotate the loop.
Physics
1 answer:
spin [16.1K]3 years ago
8 0

Answer:

Explanation:

Area of square loop = L²

Flux Φ = area x magnetic field

= L²B

Frequency = f

angular velocity ω = 2πf

a )

Let at time t = 0 , the magnetic field is making 90 degree with the face of the loop

flux through loop = L²B

After time t , coil will turn by angle ω t = 2πft

Flux through the loop = L²B cosω t

Φ (t) = L²B cosω t

= L²B cos2πft

b )

emf induced e

= - d/dt [Φ (t)]

= - d/dt [ L²B cosω t]

= L²B ω sinω t

= L²B 2πf sin2πft

c )

current = e / R

(L²B ω/ R ) sinω t

Power delivered

P(t) = VI ,

VOLT X CURRENT

= AB ω sinω t X ( AB ω/ R ) sinω t

= L⁴B² 4π²f²/R sin²2πft

e )

torque = MB sinω t

τ(t) = i(L²B ) sinω t

= (L²B ω/ R ) sinω t x (L²B ) sinω t

= (L²B  )²ω/ R sin²ω t

= (L²B  )² 2πf/ R sin²2πft

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AURORKA [14]

Answer:

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Explanation:

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So work is one of the forms of energy transmission between bodies. To perform a job, you must exert a force on a body and it moves.

In the International System of Units, work is measured in Joule. A Joule is the work that a constant force of 1 Newton does on a body that moves 1 meter in the same direction and direction as the force. Then, Joule is equivalent to Newton per meter.

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves:

Work= Force*Distante* cos (θ)

In this case:

  • Force= 1,400 N= 1,400 kg*\frac{m}{s^{2} }
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Replacing:

Work= 1,400 N* 1,000 m* cos (30°)

Work= 1,212,435. 565 Joule≅ 1,212,436 J

<u><em> The work is done by the truck pulling the car 1 km is 1,212,436 J</em></u>

7 0
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Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant:  k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

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3 0
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MatroZZZ [7]
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vfiekz [6]

Answer:

a=(v-u)/t

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a= 8/3

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