The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
<h3>What is electric potential energy?</h3>
Electric potential energy can be defined as the energy needed to move a charge against an electric field.
It is calculated using the formula;
U = Kq1 q2 ÷ r
Where Q = electric potential energy
k = Coulombs constant
q1 and q2 = charges
r = distance of separation
Electric potential energy is inversely proportional to the distance of separation of the charges.
If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.
Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
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Answer:
Utilization, effects
Explanation:
The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.
Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
