Answer:
40 s
Explanation:
After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start
Let t be the number of seconds after the second skater starts will the second skater overtake the first skater
The distance traveled by the first skater after t seconds is

Similarly the distance traveled by the 2nd skater after t seconds is

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

We can substitute 



Answer:

Explanation:
As we know that the charge per unit length of the long cylinder is given as

here we know that the electric field between two cylinders is given by

now we know that electric potential and electric field is related to each other as





The amplitude of the wave is the 'full height of the wave.' Amplitude is measured in m (meters) and is measured over the change of a single period.
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