Answer
Given,
refractive index of film, n = 1.6
refractive index of air, n' = 1
angle of incidence, i = 35°
angle of refraction, r = ?
Using Snell's law
n' sin i = n sin r
1 x sin 35° = 1.6 x sin r
r = 21°
Angle of refraction is equal to 21°.
Now,
distance at which refractive angle comes out
d = 2.5 mm
α be the angle with horizontal surface and incident ray.
α = 90°-21° = 69°
t be the thickness of the film.
So,


t = 2.26 mm
Hence, the thickness of the film is equal to 2.26 mm.
The speed increases, because as the angle increases (the wing slants up more steeply), the air has to go farther to get over the wing.
Answer:
80.386 degrees
Explanation:
We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse
The adjacent side = 699ft
The hypotenuse = 1034ft
using cos∅ = Adjacent/hypotenuse
where ∅ is the unknown angle
cos ∅ = 699/1034 = 0.167
∅ = arccos 0.167 = 80.368°
As easy as one can imagine
According to KE = (3/2)kT
reducing temperature, in KELVIN, by half, average KE is reduced by half.
The answer is A for sure , hope it’s correct