Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
1 m = 1 000 000 ym
converted other way we can say that:
1 ym =

m
Now, since we have ym^2 which is ym*ym which means:
1 ym^2 =

m
we have 1,5 ym^2 which means that answer is:
Answer:
Angular acceleration, is 
Explanation:
Given that,
Initial speed of the drill, 
After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, 
We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

So, the drill's angular acceleration is
.
Answer:
8.3m/s
Explanation:
Given parameters:
mass of clay ball = 5kg
Speed of clay ball = 25m/s
mass of clay ball at rest = 10kg
speed of clay ball at rest = 0m/s
Unknown:
Velocity after collision = ?
Solution:
Since the balls stick together, this is an inelastic collision:
m1v1 + m2v2 = v(m1 + m2)
5(25) + 10(0) = v (5 + 10)
125 = 15v
v = 8.3m/s