Answer:
x_total = 600 m
Explanation:
This is an exercise and kinematics, let's find the time it takes to reach the velocity 20 m / s
v = v₀ + a t
as part of rest v₀ = 0
t = v / a
t = 20/2
t = 10 s
let's find the distance traveled in this time
x₁ = vo t + ½ a t2
x₁ = 0 + ½ 2 10²
x₁ = 100 m
The remaining time is
t₂ = 35 - t
t₂ = 35 - 10
t₂ = 25 s
as in this range it has a constant speed
v = x₂ / t₂
x₂ = v t₂
x₂ = 20 25
x₂ = 500 m
the total distance traveled is
x_total = x₁ + x₂
x_total = 100 + 500
x_total = 600 m
terminal velocity ... greater speed ... acc is 10m/s/s
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
