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Zanzabum
3 years ago
7

50POINTS PLEASE HELP SPACE QUESTION

Physics
2 answers:
Vilka [71]3 years ago
7 0

The seasons in the Northern Hemisphere are the opposite of those in the Southern Hemisphere. Seasons occur because Earth is tilted on its axis relative to the orbital plane, the invisible, flat disc where most objects in the solar system orbit the sun.

leonid [27]3 years ago
4 0
Regardless of the time of year, the northern and southern hemispheres always experience opposite seasons. This is because during summer or winter, one part of the planet is more directly exposed to the rays of the Sun than the other, and this exposure alternates as the Earth revolves in its orbit.
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Tap on the photo. For each diagram, explain why the light behaves in the way that it does.
dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

n_1 sin \theta_1 = n_2 sin \theta_2

where n_1,n_2 are the refractive index of the 1st and 2nd medium

\theta_1, \theta_2 are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (n_1), this means that \theta_2, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (n_1>n_2), so it bends away from the normal (\theta_2>\theta_1).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0
4 years ago
A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?
diamong [38]
C theres a light house near by

8 0
3 years ago
Read 2 more answers
Bob and Lily are riding on a merry-go-round. Bob rides on a horse near the outer edge of the circular platform, and Lily rides o
dimaraw [331]

Answer: the same as Lily's

Explanation:

Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.

Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.

Based on the above analysis, Bob's angular speed will be thesame as that of Lily.

5 0
3 years ago
Which statement or question is a good hypothesis
Kisachek [45]
I’m pretty sure it’s A)
4 0
3 years ago
Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas
lorasvet [3.4K]

Answer:

Let’s assume that the projectile has the same initial velocity in both cases. It is necessary to find that velocity.

u = u[x]i +u[y]j = [u cos 15]i + [u sin 15]j

The p.v. ; s = s[x]i +s[y]j , get the time of flight by letting s[y]=0

s[y] = [u(y)].t -[1/2]gt^2 =0 , so t = 2u[y]/g

t = [ 2 u sin 15]/g eq 1

s[x] = [ u cos 15] .t eq 2 , substitute in eq q into eq2

range =r = [ u cos 15 ][ 2u sin 15 ]/g =[u^2/g][ 2 sin 15 . cos 15],

r =[ u^2 sin 30]/g =1.5 km

( from this we see that ,in general, r = [u^2 .sin (2a)]/g

u^2 = [ 3g] , eq 1

Now do the 2 nd part

r[2] = [u^2 sin (2a)]/g , eq 2 Let a =45 and substitute eq 1 into eq 2

r[2] = [3g sin 90]/g

r[2] =3 km.

Explanation:

7 0
3 years ago
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