Answer:
The self-induced emf in this inductor is 4.68 mV.
Explanation:
The emf in the inductor is given by:
Where:
dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)
L: is the inductance = 0.260 H
So, the emf is:
Therefore, the self-induced emf in this inductor is 4.68 mV.
I hope it helps you!
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The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
It's a combination of all those things. probably because we are taught from an early age to write in an academic fashion, giving balanced arguments and a conclusion. When speaking from the heart, there is no opposing argument nor is there a conclusion, just emotion.
The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.
The time period T of the block with mass M attached to a spring of spring constant K is given by,
T = 2π(√M/K).
Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,
T' = 2π(√2M/K)
Putting T = 2π(√M/K) above,
T' =√2T
So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.
To know more about time period of mass, visit,
brainly.com/question/20629494
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Explanation:
According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely falling object.