We could take the easy way out and just say
(110 kW) x (3 hours) = 330 kilowatt hours .
But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
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" 110 kw " means 110,000 watts = 110,000 joules/second .
(3 hours) x (3600 sec/hour) = 10,800 seconds.
(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules
That's
==> 1,188,000,000 joules
==> 1,188,000 kilojoules
==> 1,188 megajoules
==> 1.188 gigajoules
Atsa nawfulotta energy !
It goes back to that "110 kw appliance" that we started with.
That's no common ordinary household appliance. 110 kw is something like
147 horsepower. In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole. Most houses
are limited to 100 or 200 Amperes, tops. And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.
Answer:
A) greater
Explanation:
acceleration is calculated by dividing velocity over time..so by calculating, you find acceleration of A is greater than that of B
Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).
M = V d
M = 1.4 * 2 = 2.8 kg
Answer:
1.843 x 10^-5 C
Explanation:
<u><em>Givens:
</em></u>
It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.
Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation
E = (Q/A)/∈o (1)
Where,
Q: total charge on the disk.
A: the area of the disk.
<u><em>Calculations: </em></u>
We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold
Q = EA∈o
= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)
= 1.843 x 10^-5 C
note:
calculations maybe wrong but method is correct
