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.
Answer: ME= E total - E thermal
Answer:
I think the answer is b am sorry if it is wrong
Explanation:
To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.
The electric field in terms of the Force can be expressed as
Where,
F = Force
E= Electric Field
q = Charge
Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que
KE = W
KE = F*d
In the First Case,
In Second Case,
The total energy change would be subject to,
Therefore the Kinetic Energy change of the charged object is 27.976J
Answer:
if this is on odyssey ware then i can edit my answer to help you
Explanation:
