Answer:
Power of the speakers = 1.57 
Explanation:
There are speakers setup in a park and we have to find the power it would require to have intensity of sound to be 
at a distance of 5 meter.
In one dimension the intensity is constant as the wave travels. In two or three dimensions, however, the intensity decreases as you get further from the source.
Intensity is inversely proportional to the square of the distance from the point of sound production.
Here area = 2
area = 157 
Power = ![10^{-8} \times 157Power = 1.57 [tex]\times 10^{-6} W](https://tex.z-dn.net/?f=10%5E%7B-8%7D%20%5Ctimes%20157%3C%2Fp%3E%3Cp%3E%3Cstrong%3EPower%20%3D%201.57%20%5Btex%5D%5Ctimes%2010%5E%7B-6%7D%20W)
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
Answer:
b
Explanation:
I'm not so sure but it makes sense
