An electron has a constant acceleration of +2.9 m/s2. At a certain instant its velocity is +8.6 m/s. What is its velocity (a) 2.1 s earlier and (b) 2.1 s later?

(a) = 2.51m/s

(b) = 14.69m/s

Explanation:

Given.

Velocity v = 8.6m/s

Acceleration a = 2.9m/s2

The final velocity of particle is given by

Vf = Vo+at

.(a) 8.6= Vo + (2.9)(2.1)

VP = 8.6 - (2.9)(2.1)= 2.51m/s

(b) Vf = 8.6+(2.9)(2.1) = 14.69m/s

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it