Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
Answer:

Explanation:
We are given that







We have to find the exit temperature.
By steady energy flow equation



Substitute the values




The change in velocity is +4 m/s to the right (or -4 m/s to the left).
The object's mass is irrelevant.