0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating
the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :
1 answer:
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Explanation :
It is given that,
Diameter of the coil, d = 20 cm = 0.2 m
Radius of the coil, r = 0.1 m
Number of turns, N = 3000
Induced EMF,
Magnitude of Earth's field,
We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :
So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.