Answer:
a) θ = 65º , b) the light is refracted
Explanation:
When a ray of light passes from a material with a higher index to one with a lower index, the ray separates from the normal one, so there is an angle for which the ray is refracted at 90º, the refractive equation is
n₁ sin θ₁ = n₂
where θ₁ is the incident angle, n₂ and n₁ are the indexes of incident and refracted parts
let's calculate
sin θ = n₂ / n₁
sint θ = 1.3333 / 1.470
sin θ = 0.907
θ = sin⁻¹ 0.907
θ = 65º
For the incident angle of 50.2º it is less than the critical angle, so the light is refracted according to the refraction equation
Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).
Korolev's name became known in the West. I hope this helps.
Answer:
1) an observer in B 'sees the two simultaneous events
2)observer B sees that the events are not simultaneous
3) Δt = Δt₀ /√ (1 + v²/c²)
Explanation:
This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems
1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events
Consequently an observer in B 'sees the two simultaneous events
2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.
Consequently observer B sees that the events are not simultaneous
3) let's calculate the times for each event
Δt = Δt₀ /√ (1 + v²/c²)
where t₀ is the time in the system S' which is at rest for the events
