A collision from above such as a meteor and an increase in volcano activity.
D.all of the above is the answer for this question
Answer:
The angle between the blue beam and the red beam in the acrylic block is

Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is 
The wavelength of the blue light is 
The refractive index of the transparent acrylic plastic for red light is 
The wavelength of the red light is 
The incidence angle is 
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
![r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_F%7D%20%5D)
Where
is the refractive index of air which have a value of
So
![r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.497%7D%20%5D)

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
![r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_C%7D%20%5D)
Where
is the refractive index of air which have a value of
So
![r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.488%7D%20%5D)

The angle between the blue beam and the red beam in the acrylic block

substituting values


Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m