It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel . The potential energy is stored in the capacitor will be 2.7×10⁻⁵.
<h3>What is parallel plate capacitor ?</h3>
It is an type capacitor is an in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates. helps to stop the flow of electric current through it due to its non-conductive nature .
The given data in the problem is;
Q is the charge= 1.5 µC
V is the change in voltage across the plates is = 36 V.
U is the potential energy=?
The formula for the potential energy is given by;
Hence the potential energy is stored in the capacitor will be 2.7×10⁻⁵.
To learn more about the parallel plate capacitor refer to the link;
brainly.com/question/12883102
Answer:
f = 1 / [ 2*pi*R*C*Sqrt(2*N)
Explanation:
f = 1 / [ 2*pi*R*C*Sqrt(2*N) is the formula. Sorry I answered in English.
<h3><u>Answer;</u></h3>
A. wind, tidal, geothermal, and hydroelectric
<h3><u>Explanation</u>;</h3>
- Renewable energy sources are the energy sources that are constantly being replenished, such as sunlight, wind, and water. This means that we can use them as much as we want, and we do not have to worry about them running out.
- Additionally, renewable energy sources are usually much more environmentally friendly than fossil fuels. Examples of renewable sources of energy includes; solar energy, wind energy, hydroelectric power, tidal energy, geothermal energy, etc.
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
