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The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i =
where R is the resistance
R =
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i =
q =
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
Answer:
- E = 11.55J
- Q = 0.17C
- E' = (1/4)E
Explanation:
- To calculate the amount of energy stored in the capacitor, you use the following formula:
C: capacitance = 3800.0*10^-6F
V: potential difference = 78.0V
The energy stored in the capacitor is 11.55J
- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:
The charge on the capacitor is 0.17C
- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:
A: area of the plates
d: distance between plates
If the distance between plates is increased by a factor of 4, you have:
Then, the stored energy in the capacitor is decreased by a a factor of (1/4)