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!
Answer:
Each piece will have a north pole and a south pole
Explanation:
Answer:
Explanation:
I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.
The relation between the linear acceleration and the angular acceleration is
a = R α .... (1)
According to the diagram,
T x R = I x α
T x R = I x a / R from equation (1)
T = I x a / R² .... (2)
mg - T = ma .... (3)
Substitute the value of T from equation (2) in equation (3)


T is the acceleration in the system
Substitute the value of a in equation (2)


This is the tension in the string.
The answer is b
300,000 km