In an electromagnetic wave, the electric and magnetic portions are proprtional to each other. The ratio of electric field to magnetic field is , the speed of light.

Using this information, we can determine the strength of one part given the other.

We have  and , so now we multiply the two.

Therefore, the strength of the electric field is .

In an electromagnetic wave going through a vacuum, the ratio of  and  is the speed of light, ; however, this relation doesn't change when light goes through a medium. The ratio still is the speed of light, but instead of , it's the speed of light through the medium. Therefore, for velocity , 

In this problem, we're told that the speed of light through the medium is , so if we rearrange the equation to solve for  and use  in place of v, we'll get our answer.

Therefore, the strength of the magnetic field is .

Wavelength is indirectly related to the amount of energy in the wave (the frequency). The most energetic wves are gamma rays, while the least energetic are radio waves. This means that radio waves have the longest wavelength.

A helpful mnemonic for remembering the order of wavelengths in the electromagnetic spectrum from longest to shortest is "Raging Martians invaded Roy G Biv using x-ray guns."

In order, the letters stand for 

1. Radio
2. Micro
3. Infrared
4. Red
5. Orange
6. Yellow
7. Green
8. Blue
9. Indigo
10. Violet
11. Ultraviolet
12. X-ray
13. Gamma

When light is refracted, the average speed is lower if the index of refraction is higher (and vice versa). In order to refract, the medium light travels through must have a changing index of refraction. Because the amount of energy the light carries must be constant during refraction, the frequency (which is proportional to energy) cannot change. Therefore, since  and the frequency must stay the same when the speed changes, wavelength must also change.

When trying to find the minimum distance between two objects that makes them still distinguishable, we use the Rayleigh criterion equation.

 is the minimum angle at which two objects can be resolved,  is the wavelength of the wave being used, and  is the diameter of the circular aperture (in this case, the antenna would be the aperture). An important part of equations like these is the small-angle approximation, which states, as the angle approaches zero,

This means that, if the angle is small enough, . In our problem, we're trying to find , so we can substitute  for 

Next, since we're not given the wavelength, but we are given the frequency, we can find the wavelength using the following equation: 

Now, solve for .

Using

Where  is the speed of light, is frequency and is wavelength

Converting to and plugging in values

Solving for

For this question, we're given the distance between wave peaks (wavelength) as well as how long it takes for each wave to hit the shore (period). We're then asked to use this information to solve for the speed of the waves.

First, we'll need to use an equation for the speed of a wave.

We know from the question stem that we have the wavelength, which is the distance between the peaks of each wave. But we do not have the frequency. However, we can solve for frequency by using the period of the wave as follows:

Which we can alternatively write as:

The period of the wave, , represents the amount of time it takes for a single wave to pass a given point, and thus has units of time. The inverse of this quantity is the frequency, which describes how many waves pass by a given point in a certain amount of time. Usually, the unit of time in these cases is seconds.

Thus, for every one second that passes, one-fifth of a wave will pass by any given point.

Now, we can plug this value into the equation for the speed of the wave to solve for our answer:

To answer this question, let's first recall that on the visible spectrum, red light has the longest wavelength. Due to this, red light also has the lowest energy. We can show this with the equation

As the dimmer is gradually turned down, less and less energy is being provided to the light bulb. Consequently, since red light has the least amount of energy of all the colors on the visible spectrum, we briefly see red light the instant before the light turns off completely. Note that the speed of a wave is determined by the medium. In this case, regardless of the wavelength, the medium is air, which determines the speed of any electromagnetic wave. 

For this question, we need to consider wave characteristics. Specifically, we need to determine how a decrease in the frequency of a wave will alter its wavelength, period, and amplitude.

First, let's consider the period of the wave. The period is defined as the amount of time needed for one complete cycle of the wave to occur. Conversely, frequency is defined as the number of wave cycles completed within a given time frame. As such, period and frequency are inversely related to one another, as the following expression shows:

Therefore, the period of the wave will certainly increase as the frequency of the wave decreases.

Now, let's take a look at wavelength. We need to recall the speed of a wave is defined in terms of its wavelength and frequency according to the following equation:

As we can see from the above equation, frequency is inversely related to wavelength, just as it is with period. Therefore, as the frequency of a wave decreases, the wavelength will indeed rise.

Finally, let's look at amplitude. The amplitude of a wave is the magnitude of the difference between the extremes of the wave and its equilibrium position. In transverse waves, such as a rope, the amplitude is the maximum displacement of a particle of that rope from its equilibrium position in the direction perpendicular to the propagation of the wave. In longitudinal waves, such as sound, the amplitude is the maximum displacement of the medium from its equilibrium position in the direction parallel to the propagation of the wave.

Because there is no relationship between amplitude and frequency, a decrease in a wave's frequency will have no effect on that wave's amplitude. Thus, for this question, only wavelength and period are increased due to a decreased frequency.

Using the following equation:

Converting to and plugging in values: