Answer:
Waves; wavelength; electromagnetic energy; ultraviolet light.
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases.
Light wave can be defined as an electromagnetic wave that do not require a medium of propagation for it to travel through a vacuum of space where no particles exist.
Hence, sound and light are both found as waves, with a variety of wavelength. The sun, a source of light waves specifically, releases a type of electromagnetic energy. It can be found as UVA or UVB types. These lights give off different levels of ultraviolet light, some of wich can be harmful.
Additionally, the ultraviolet spectrum is divided into three categories and these are; UVA, UVB and UVC.
Answer:
1.195 m
2.8375 s
2.21433 rad/s
Explanation:
d = Distance = 2.39 m
N = Number of cycles = 8
t = Time to complete 8 cycles = 22.7 s
Radius would be equal to the distance divided by 2
![r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%5C%5C%5CRightarrow%20r%3D%5Cfrac%7B2.39%7D%7B2%7D%5C%5C%5CRightarrow%20r%3D1.195%5C%20m)
The radius is 1.195 m
Time period would be given by
![T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7Bt%7D%7BN%7D%5C%5C%5CRightarrow%20T%3D%5Cfrac%7B22.7%7D%7B8%7D%5C%5C%5CRightarrow%20T%3D2.8375%5C%20s)
Time period of the motion is 2.8375 s
Angular speed is given by
![\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%5C%5C%5CRightarrow%20%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7B2.8375%7D%5C%5C%5CRightarrow%20%5Comega%3D2.21433%5C%20rad%2Fs)
The angular speed of the motion is 2.21433 rad/s
Um student a because they were there a few seconds ahead
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.
![I = MR^2](https://tex.z-dn.net/?f=I%20%3D%20MR%5E2)
Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as
![\Omega = \frac{Mgd}{I\omega}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7BMgd%7D%7BI%5Comega%7D)
Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia
![\Omega= \frac{MgR}{MR^2 \omega}](https://tex.z-dn.net/?f=%5COmega%3D%20%5Cfrac%7BMgR%7D%7BMR%5E2%20%5Comega%7D)
![\Omega = \frac{g}{R\omega}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7Bg%7D%7BR%5Comega%7D)
The value for our angular velocity is not in SI, then
![\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})](https://tex.z-dn.net/?f=%5Comega%20%3D%201000rpm%20%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%29%28%5Cfrac%7B1min%7D%7B60s%7D%29)
![\omega = 104.7rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20104.7rad%2Fs)
Replacing our values we have that
![\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7B9.8m%2Fs%5E2%7D%7B%288%2A10%5E%7B-2%7Dm%29%28104.7rad%29%7D)
![\Omega = 1.17rad/s](https://tex.z-dn.net/?f=%5COmega%20%3D%201.17rad%2Fs)
The precession frequency is
![\Omega = \frac{2\pi rad}{T}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7B2%5Cpi%20rad%7D%7BT%7D)
![T = \frac{2\pi rad}{\Omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20rad%7D%7B%5COmega%7D)
![T = \frac{2\pi}{1.17}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B1.17%7D)
![T = 5.4 s](https://tex.z-dn.net/?f=T%20%3D%205.4%20s)
Therefore the precession period is 5.4s
Dispersion angle = 0.3875 degrees.
Width at bottom of block = 0.09297 cm
Thickness of rainbow = 0.07038 cm
Snell's law provides the formula that describes the refraction of light. It is:
n1*sin(θ1) = n2*sin(θ2)
where
n1, n2 = indexes of refraction for the different mediums
θ1, θ2 = angle of incident rays as measured from the normal to the surface.
Solving for θ2, we get
n1*sin(θ1) = n2*sin(θ2)
n1*sin(θ1)/n2 = sin(θ2)
asin(n1*sin(θ1)/n2) = θ2
The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
Red:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.641) = θ2
asin(1.00029*0.653420604/1.641) = θ2
asin(0.398299876) = θ2
23.47193844 = θ2
Violet:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.667) = θ2
asin(1.00029*0.653420604/1.667) = θ2
asin(0.39208764) = θ2
23.08446098 = θ2
So the dispersion angle is:
23.47193844 - 23.08446098 = 0.38747746 degrees.
Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
tan(θ) = X/11.6
11.6*tan(θ) = X
So for Red:
11.6*tan(θ) = X
11.6*tan(23.47193844) = X
11.6*0.434230136 = X
5.037069579 = X
And violet:
11.6*tan(θ) = X
11.6*tan(23.08446098) = X
11.6*0.426215635 = X
4.944101361 = X
So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm
The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
cos(θ) = X/0.092968218
0.092968218*cos(θ) = X
0.092968218*cos(40.80) = X
0.092968218*0.756995056 = X
0.070376481 = X
So the distance between the red and violet rays is 0.07038 cm.