Answer:
Velocity of the electron at the centre of the ring, 
Explanation:
<u>Given:</u>
- Linear charge density of the ring=

- Radius of the ring R=0.2 m
- Distance of point from the centre of the ring=x=0.2 m
Total charge of the ring

Potential due the ring at a distance x from the centre of the rings is given by

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

Let
be the change in potential Energy given by

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

So the electron will be moving with 
A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.
Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.
It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.
Diffraction patterns are due to interference<span>. Diffraction is a phenomena which occurs when a wave encounters an obstacle. It is the bending of light around the corners if the obstacle.</span>
Answer:
Speed= 6cm/s and velocity= 6cm/s in the negative direction
Explanation:
the change in position is from 45cm to 27 cm (moving towards the negative x direction)

And the change in time:

Now we must define the difference between speed and velocity:
Speed is a scalar quantity, which means that it is a number. Velocity is also a number but you must also indicate the direction of the movement.
Thus, the speed is:

An the velocity is:
6cm/s in the negative direction