Answer:
The electric field always decreases.
Explanation:
The electric field due to a point charge is given by :
Where
k = electric constant
q = charge
r = distance from the charge
It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".
Answer:
Electric potential energy at the negative terminal: 
Explanation:
When a particle with charge 
travels across a potential difference 
, then its change in electric potential energy is
In this problem, we know that:
The particle is an electron, so its charge is
We also know that the positive terminal is at potential
While the negative terminal is at potential
Therefore, the potential difference (final minus initial) is
So, the change in potential energy of the electron is
This means that the electron when it is at the negative terminal has of energy more than when it is at the positive terminal.
Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is
Convex.
Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
