# College Physics Problem 2.15

#### SOLUTION:

A) The formula to be used is

$average\:speed=\frac{distance}{time}$

Rearraging the formula–solving for the distance

$distance=speed\times time$

Substituting the given values

$d=\left(2.20\times 10^6\:\frac{m}{s}\right)\left(1\:s\right)\:$

$d=2.20\times 10^6$

We know the total distance traveled in 1 sec. We know that in 1 revolution, the total distance is

$1\:revolution=2\pi r$

$1\:revolution=2\pi \left(\frac{d}{2}\right)=\pi d$

$1\:revolution=\pi \left(1.06\times 10^{-10}\:m\right)$

Therefore, the total number of revolutions traveled in 1 second is

$no.\:of\:revolutions=\frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)}$

$no.\:of\:revolutions=6.61\times 10^{15}$

B) In one complete revolution, the electrons will go back to its original position. Thus, there is no net displacement. Therefore,

$v=0\:m/s$

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