A16.0 m long, thin, uniform metal rod slides north at a speed of 21.0 m/s. the length of the rod maintains an east-west orientation while sliding. the vertical component of the earth's magnetic field at this location has a magnitude of 42.0 µt. what is the magnitude of the induced emf between the ends of the rod (in mv)?

14.112 mV

Explanation:

L = 16 m, v = 21 m/s, B = 42 μ T = 42 x 10^-6 T

The formula for the induced emf is given by

e = B x v x L

e = 42 x 10^-6 x 21 x 16 = 14.112 x 10^-3 V = 14.112 mV

Thus, the induce emf is 14.112 mV.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.