Plastic is what they are made of
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is
constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?
4a. If the magnetic field lines are equally spaced apart, in other words share the same
density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density)
4b. The electric potential is measured in Volts(V) and is uniform along the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.
I really hope this makes sense to you and that my pictures help! :)
The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is

where c is the speed of light, f the frequency and

the wavelength of the wave.
Using

and

, we can find the value of the frequency:
The answer is C) an electromagnetic wave
An electromagnetic wave, which includes electromagnetic radiation such as visible light, moves the fastest of all of the options listed by a significant margin, especially through space. In fact, light travelling through space is technically the theoretical limit of how fast something can travel.
Answer:
80 amperes
Explanation:
Current in the circuit = ?
Voltage in the circuit = 160 Volts
Resistance = 2 Ω
Voltage = Current x Resistance
V = IR
160V = I x 2 Ω
I = 160V / 2 Ω
I = 80 Amperes
Therefore the current in the circuit is 80 amperes :)