What is the electric potential energy of a charge that experiences a force of 3.6 x 10^-4N when it is 9.8 x 10^-5 from the sourc
1 answer:
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
(1)
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
The potential is given by the magnitude of the electric field, E, times the distance, d:
So we have
However, the electric field is equal to the electrical force F divided by the charge q:
Therefore (1) becomes
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
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Answer:
Kinetic energy of diver at 90% of the distance to the water is 9000 J
Explanation:
Let d is the distance between the position of the diver and surface of the pool.
Initially, the diver is at rest and only have potential energy which is equal to 10000 J.
As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.
Energy before diving = Energy during diving
(Potential Energy + Kinetic Energy) = (Kinetic Energy + Potential Energy)
When the diver reaches 90% of the distance to the water, its kinetic energy
is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,
K.E. =
K.E. = 9000 J
a) 22.5 m/s
The initial vertical velocity is given by:
where
u = 35 m/s is the initial speed
is the angle of projection of the ball
Substituting into the equation, we find
b) 26.8 m/s
The initial horizontal velocity is given by:
where
u = 35 m/s is the initial speed
is the angle of projection of the ball
Substituting into the equation, we find
c) 2.30 s
The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation
where
is the vertical velocity at time t
is the acceleration of gravity (negative because it is downward)
At the maximum height, the vertical velocity becomes zero, ; substituting, we find the time t at which this happens:
d) 25.8 m
The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:
where
s is the vertical displacement at time t
Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:
e) 123.3 m
In order to find how far does the ball lands, we have to consider the horizontal motion.
First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:
Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:
Therefore, the horizontal distance travelled during the whole motion is
So, the ball lands 123.3 m far from the initial point.