The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is 
<h3>
How to calculate the speed of the electron?</h3>
We know, that the energy of the system is always conserved.
Using the Law of Conservation of energy,
U=0
Here, K is the kinetic energy and U is the potential energy.
Now, substituting the formula of U and K, we get:
=0------(1)
Here,
m is the mass of the electron
v is the speed of the electron
q is the charge on the electron
V is the potential difference
Let 
and 
represent the final and initial speed.
Here, =0
Solving for , we get:
=6.49
m/s
To learn more about the conservation of energy, refer to:
brainly.com/question/2137260
#SPJ4
Answer:
A- Greatest Kinetic Energy
B- Increasing Potential Energy
C- Increasing Kinetic Energy
D- Greatest Potential Energy
Explanation:
hope this is right.
Answer:
Kinda? Depends what the question is fully asking
Explanation:
Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.
So that -2 m/s thing after one second will be going -1 m/s.
After another second it'll be going 0 m/s.
After another itll be going +1 m/s and so on.
So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.
It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
hence
there is a force acting on the object, in the direction of the velocity
Answer:
Explanation:
Electric field between plates of a parallel plate capacitor is uniform .
In a uniform electric field , relation between electric field and potential gradient is as follows
electric field = potential gradient [ E = - dV / dl ]
in the given case ,
dV = 51 V ,
dl = 4 cm
= 4 x 10⁻² m
E = 51 / 4 x 10⁻²
= 12.75 x 10² V / m
= 1275 V / m
