Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2 
tends to replaced by 0.4 then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.
Force = (mass) x (acceleration)
= (0.025 kg) x (5 m/s²)
= 0.125 Newton
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
A simple rule to bear in mind is that all objects (regardless of their mass) experience the same acceleration when in a state of free fall. When the only force is gravity, the acceleration is the same value for all objects. On Earth, this acceleration value is 9.8 m/s/s.
