<h2>MARK BRAINLIEST</h2>
For this assignment, you will develop several models that show how light waves and mechanical waves are reflected, absorbed, or transmitted through various materials. For each model, you will write a brief description of the interaction between the wave and the material. You will also compose two <u><em>typewritten</em></u> paragraphs. The first will compare and contrast light waves interacting with different materials. The second will explain why materials with certain properties are well suited for particular functions.
<h2><u>Background Information</u></h2>
A wave is any disturbance that carries energy from one place to another. There are two different types of waves: mechanical and electromagnetic. A mechanical wave carries energy through matter. Energy is transferred through vibrating particles of matter. Examples of mechanical waves include ocean waves, sound waves, and seismic waves. Like a mechanical wave, an electromagnetic wave can also carry energy through matter. However, unlike a mechanical wave, an electromagnetic wave does not need particles of matter to carry energy. Examples of electromagnetic waves include microwaves, visible light, X-rays, and radiation from the Sun.
Um what are the ten objects..?
Answer: The box was moving with a velocity of 0.256m/s when it hit the spring
Explanation: Please see the attachments below
First figure out how many atoms you have with Avogadro's number. Since there are 63.5 grams/mol and you have 50.6 grams, you have (50.6/63.5)6.022E23=4.7986E23 atoms. Since there are 29 protons per atom, there are also 29 electrons per atom, so you should have a total of
29*4.7986E23=1.3916E25 electrons.
Since there is a positive charge you know some of these electrons are missing. How many are missing can be found by dividing the charge you have by the charge on the electron: 1.6E-6/1.6022E-19 = 9.98627E12 electrons are missing.
Now take the ratio of what is missing to what there should be:
9.98627E12/1.3916E25 = 7.1760873E-13
To solve this problem it is necessary to use the concepts related to Snell's law.
Snell's law establishes that reflection is subject to
Where,
Angle between the normal surface at the point of contact
n = Indices of refraction for corresponding media
The total internal reflection would then be given by
Therefore the 
would be equal to
Therefore the largest value of the angle α is 30.27°
