The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
The average power output:
P = V * I * t
V = 220 V, I = 15 A;
t = 2 ms * 200 = 400 ms = 0.4 s
P = 220 V * 15 A * 0.4 s
P = 1320 W ≈ 1.3 kW
Answer:
b. 1.3 kW
Angel ! You have a formula, and you have an example that's
completely worked out. The ONLY POSSIBLE reason that you
could still need help is that you're letting math scare you.
I'll do 'A' for you, 'B' most of the way, and get 'C' set up.
If THAT's not enough for you to run with and finish them all,
then you and I should both be embarrassed.
Write the formula on the wall:
°F = (9/5) °C + 32°
A). Convert 35° C °F = (9/5)(35°) + 32°
(9/5)(35) = 63 °F = 63° + 32°
°F = 95°
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B). Convert 80°F to °C
The formula: °F = (9/5) °C + 32°
°F = 80 80 = (9/5)°C + 32
Subtract 32 from each side: 48 = (9/5)°C
Multiply each side by 5 : 240 = (9) C
Now you take over:
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C). Convert 15°C to °F.
The formula: °F = (9/5) °C + 32°
°C = 15 °F = (9/5) 15° + 32
(9/5) (15) = 27
Go ! °F =
