Answer:
θ_p = 53.0º
Explanation:
For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º
Let's write the transmission equation
n1 sin θ₁ = ne sin θ₂
The angle to normal (vertcal) is
180 = θ2 + 90 + θ_p
θ₂ = 90 - θ_p
Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray
We replace
n1 sin θ_p = n2 sin (90 - θ_p)
Let's use the trigonometry relationship
Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p
In the law of reflection incident angle equals reflected angle,
ni sin θ_p = ns cos θ_p
n₂ / n₁ = sin θ_p / cos θ_p
n₂ / n₁ = tan θ_p
θ_p = tan⁻¹ (n₂ / n₁)
Now we can calculate it
The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water
n₂ = 1.33
θ_p = tan⁻¹ (1.33 / 1)
θ_p = 53.0º
n₂ = 1.40
θ_p = tan⁻¹ (1.40 / 1)
Tep = 54.5º
Answer:
the correct answer is C
Explanation:
When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.
Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.
Applying this to our case if there is
10 boulder on the chart
in reality there are #_boulder = 10 30
#_boulder = 300 boulder
so the correct answer is C
Answer:
Due to application of heat on a saturated solution, the interparticle space increases due to increase in the kinetic energy of the particles which allows more solutes to dissolve in the solution thereby making it unsaturated.
I believe that the answer is A but correct me if i’m wrong
Answer:
α = 395 rad/s²
Explanation:
Main features of uniformly accelerated circular motion
A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.
There is tangential acceleration (at) and is constant.
at = α*R Formula (1)
where
α is the angular acceleration
R is the radius of the circular path
There is normal or centripetal acceleration that determines the change in direction of the velocity vector.
Data
R = 0.0600 m :blade radius
at = 23.7 m/s² : tangential acceleration of the blades
Angular acceleration of the blades (α)
We replace data in the formula (1)
at = α*R
23.7 = α*(0.06)
α = (23.7) / (0.06)
α = 395 rad/s²
