With the help of a transformer input voltage is transformed into an output voltage
<h3>What is induced voltage?</h3>
Electromagnetic induction is what causes the induced voltage. Electromagnetic induction is the process of generating emf (induced voltage) by subjecting a conductor to a magnetic field.
In this case, a magnet is pushed in and out of a wire coil attached to a high-resistance voltmeter.
Typically, a transformer's primary winding is attached to the input voltage source and changes electrical power into a magnetic field.
The secondary winding's role is to turn this alternating magnetic field into electricity, generating the necessary output voltage.
Hence with the help of a transformer input voltage is transformed into an output voltage.
To learn more about the induced voltage refer to the link;
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Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
Answer:
if he is chasing his tail faster with each circle, then that would be acceleration, If not then no