CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do
1 answer:
Answer:
The elastic potential energy of the spring change during this process is 21.6 J.
Explanation:
Given that,
Spring constant of the spring,
It extends 6 cm away from its equilibrium position.
We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :
So, the elastic potential energy of the spring change during this process is 21.6 J.
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The diameter of the disks is 1.32 cm.
If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,
Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.
The potential difference V between the disks separated by a distance d is given by,
here, E is the electric field.
Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.
The capacitance C of the capacitor is given by,
The capacitance of a parallel plate capacitor is given by,
Here, ε₀ is the permittivity of free space and A is the area of the disks.
Rewrite the expression for A.
the area A of the disks is given by,
Here, D is the diameter of the disk.
The diameter of each disc is found to be 1.32 cm.
Answer:
Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.
Explanation:
We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.
<u>According to the Ohm's law, we find its resistance:</u>
So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.
Here the value of frequency is the number of times the current changes its direction or the polarity in one second.