To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
Answer:
700 mL or 0.0007 m³
Explanation:
P₁ = Initial pressure = 2 atm
V₁ = Initial volume = 350 mL
P₂ = Final pressure = 1 atm
V₂ = Final volume
Here the temperature remains constant. So, Boyle's law can be applied here.
P₁V₁ = P₂V₂

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³
The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
Answer:
I think it is D but don't count on it
Charles' Law: The Temperature-Volume Law. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. As the volume goes up, the temperature also goes up, and vice-versa.