Electromagnetic induction is the production of a voltage across a conductor when it is exposed to a varying magnetic field. This
2 answers:
Answer:
D) Electric power distribution.
Explanation:
As we know that the Faraday's law of electromagnetic induction says that rate of change in magnetic flux linked with a closed conducting loop will induce EMF in the loop.
This induced EMF is used for power production so here when we use it for the production of voltage across conductor.
it is used in AC generator when a coil is rotated at high speed between strong magnetic field of magnets.
This is induced EMF is given by
so here correct answer will be
D) Electric power distribution.
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Answer:
The answer to the question is;
The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m
Explanation:
To solve the question we first resolve the units of the given quantities as follows
The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute
1 revolution = 2π rad and
1 minute = 60 seconds
Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s
= 2010.619 rad/s
The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows
360 ° = 2π radians
1 ° = radians and
1.0×10⁻⁶ ° = radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad
To find the torque we note that the torque is given by
Precession angular speed × The moment of inertia × angular velocity
The precession angular speed is given by
The precession angle was determined in rad as 1.745×10⁻⁸ rad
The precession time is 5 hours which is equal to 5×60×60 = 18000 s
Therefore the precession velocity = = 9.696×10⁻¹³ rad/s
The moment of inertia is given by
Formula for the moment of inertia of a thin walled cylinder I = m·r²
Where:
r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = 0.025 m
m = Mass of each gyroscopes = 2.0 kg
Therefore I = m·r² = 2.0 kg × (0.025 m)² = 0.00125 kg·m²
Torque, τ = Ω·I·ω
Where:
Ω = Precession velocity = 9.696×10⁻¹³ rad/s
I = Moment of inertia = 0.00125 kg·m²
ω = Angular speed = 2010.619 rad/s
τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =
2.44 ×10⁻¹² kg·m²/ s² = 2.44 ×10⁻¹² N·m
The required torque is 2.44 ×10⁻¹² N·m.
Answer:
the tension of the rope is 34.95 N
Explanation:
Given;
length of the rope, L = 3 m
mass of the rope, m = 0.105 kg
frequency of the wave, f = 40 Hz
wavelength of the wave, λ = 0.79 m
Let the tension of the rope = T
The speed of the wave is given as;
Therefore, the tension of the rope is 34.95 N