When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling 
the capacitance of the capacitor in air, the charge Q, the capacitance and the voltage (
) are related by
(1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):
this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:
And since 
, substituting into the previous equation, we find:
Answer:
$ 1.1
Explanation:
From the question given above, the following data were obtained:
Cost per kWh = $ 0.1
Current (I) = 10 A
Voltage (V) = 220 V
Time (t) = 5 h
Cost of operation =?
Next, we shall determine the power the electric oven. This can be obtained as follow:
Current (I) = 10 A
Voltage = 220 V
Power (P) =?
P = IV
P = 10 × 220
P = 2200 W
Next, we shall convert 2200 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
2200 W = 2200 W × 1 KW / 1000 W
2200 W = 2.2 KW
Thus, 2200 W is equivalent to 2.2 KW.
Next, we shall determine the energy consumed by the electric oven. This can be obtained as follow:
Power (P) = 2.2 KW
Time (t) = 5 h
Energy (E) =?
E = Pt
E = 2.2 × 5
E = 11 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost $ 0.1
Therefore,
11 KWh will cost = 11 × 0.1
11 KWh will cost = $ 1.1
Therefore, the cost of operating the electric oven is $ 1.1
Answer: 90 km/hr
Explanation:
Speed= distance divided by time
540/6
= 90km/hr
Most of the problem depends on which object you observe. For the speed, take the absolute value of the derivative of the polynomial interpolation of position verses time.
