Using the following equation:

Converting to and plugging in values:

We're given the wavelengths of two subsequent harmonics of a string fixed at both ends, and we're asked to solve for the length of the string.

First, we must used the equation for a standing wave that has a node at both ends.

Before we can solve for the length of the string, we need to first solve for the value of . We can do this by considering the wavelength of the two harmonics as follows.

Now, since the length of the string for either harmonic is the same length, we can set the two expressions above equal to each other.

Now, we can solve for the variable .

Now that we have the value for , we can plug this value into the original expression to calculate the length of the string.

We are given the distance the wave travels (twice the Earth's diameter), and the time it took for it to travel that distance. Using this we can determine the wave's speed.

Now we can use  to find the frequency.

-wavelength

-frequency

*Side note-Let's assume the index of refraction of the hole is constant. In order for the wave to travel at that speed, the index of refraction must be  .

The fundamental harmonic is when , and therefore the wavelength  can be given by:

, where  is the frequency of the fundamental harmonic. 

, where  is the velocity of the wave,  is the wavelength of the wave, and  is the frequency. When the speed is cut in half, the frequency does not change! However, that means the wavelength is directly proportional to the speed and will also be cut in half.

In a vacuum, waves that travel at the speed of light propagate. Therefore, we need to determine which of the three types of waves have that speed. All parts of the electromagnetic spectrum travel at the speed of light (so X-ray and radio are true). Sound waves travel at the speed of sound, which is less than the speed of light. Therefore, that one is incorrect. 

Wavelength and frequency are inversely related. We define the wavelength of a wave as the distance between the two subsequent waves measured at the same point on the wave. Therefore, we can measure from crest to crest or trough to trough.

In this question, we're told that a certain sound is increases by  decibals. We're asked to find how the intensity of the sound changes as a result.

To answer this, we need to understand that there is a logarithmic relationship between decibals and intensity. This can be shown by the following expression that relates the two.

We can rearrange this equation to isolate the intensity ratio of after and before.

This shows that the intensity increases by a factor of .

As a side note, for every  decibals added, the intensity of the sound increases by a factor of .

This question is asking us to identify a statement that is not consistent with the wave theory of light.

In its beginning, light was traditionally viewed as being only a wave. In time, however, certain observations came to light (no pun intended) that did not agree with the wave theory. As a result, scientists needed to turn to a new theory that could account for the new observations.

This led to what is called the particle nature of light. When shining a beam of light on a metal plate, it was found that only certain frequencies of light could eject electrons from the metal. However, the wave theory of light alone could not explain this phenomenon, which came to be called the photoelectric effect. Instead, if light were to be treated like a particle, in which there are many discrete photons that behave like particles, the photoelectric effect can be adequately explained.

It's important to note, however, that this did not usurp the wave model. In fact, the wave model is able to explain several things that the particle model cannot. For instance, the diffraction, refraction, and interference of light are all consistent with the wave model, but not the particle model. Consequently, both models are valid and are invoked depending on the circumstance. This is what is known as the wave-particle duality of light.