False. C + O --> CO not CO2. Carbonmonoxide
You<span> should </span>test<span> FC and PH as soon as </span>you<span> take the sample</span>
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
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
