Step up transformer is a device which is used to step up the voltage which is input with some value.
This is based upon the principle of mutual inductance and in this the voltage input and voltage output is different because of number of turns.
Here if ideal transformer is given then power input and power output of the transformer must be same as there is no power loss in ideal transformer.
So we can write
here
= 73.37 A
= 4623 V
= 105033 A
now using above equation we will have
solving above we will have
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
The speed is 10.5 miles per minute, or 630 miles per hour.
We don't know the velocity, because you didn't tell us anything about the direction I'm going.
Answer:
áp dụng công thức í, mình thấy câu này có rắc rối gì đâu
