Answer:
The speed of proton when it emerges through the hole in the positive plate is
.
Explanation:
Given that,
A parallel-plate capacitor is held at a potential difference of 250 V.
A A proton is fired toward a small hole in the negative plate with a speed of, 
We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

So, the speed of proton when it emerges through the hole in the positive plate is
.
<span>braking and returning suddenly to the roadway</span>
Answer:
Energy is stored in the bonds between atoms
The answer is distressing
With constant angular acceleration
, the disk achieves an angular velocity
at time
according to

and angular displacement
according to

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

b. Under constant acceleration, the average angular velocity is equivalent to

where
and
are the final and initial angular velocities, respectively. Then

c. After 1.00 s, the disk has instantaneous angular velocity

d. During the next 1.00 s, the disk will start moving with the angular velocity
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to

which would be equal to
