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Strike441 [17]
1 year ago
11

The light from the sun has higher frequencies from one side of the sun than from the other side. What does that tell you about t

he sun?
Physics
1 answer:
morpeh [17]1 year ago
8 0

If the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

Doppler effect states that, if a person is standing still and a source ( sound / light ) is moving towards him, the frequency of the wave emitted from the object will increase and if the source ( sound / light ) is away from him, the frequency of the wave emitted from the object will decrease.

So, if the light from the sun has higher frequencies from one side of the sun than from the other side, it means that the Sun is rotating. The higher frequencies points are the points that rotating towards Earth and lower frequencies points are the points that rotating away from Earth.

Therefore, if the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

To know more about Doppler Effect

brainly.com/question/15318474

#SPJ1

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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
Why do aeroplanes take longer to fly west than east?
PolarNik [594]

Every planet/moon has global wind that are mostly determined by the way the planet/moon rotates and how evenly the Sun illuminates it. On the Earth the equator gets much more Sun than the poles. resulting in warmer air at the equator than the poles and creating circulation cells (or "Hadley Cells") which consist of warm air rising over the equator and then moving North and South from it and back round.

The Earth is also rotating. When any solid body rotates, bits of it that are nearer its axis move slower than those which are further away. As you move north (or south) from the equator, you are moving closer to the axis of the Earth and so the air which started at the equator and moved north (or south) will be moving faster than the ground it is over (it has the rotation speed of the ground at the equator, not the ground which is is now over). This results in winds which always move from the west to the east in the mid latitudes.

8 0
3 years ago
What is a planet’s period of rotation?
Annette [7]


The answer is the letter "C" ( I have honors science I am good at this type of stuff )

Hope I helped :) ( ask me for help when u need it :)

5 0
3 years ago
Read 2 more answers
PLEASE HELP ITS DUE TODAY ILL POST THE OTHER HALF
IgorLugansk [536]
2.B
4.C
3.D
1.C
5.C
Theses are the right answer
4 0
2 years ago
Read 2 more answers
A soccer ball is kicked 8 m north. Then a teammate kicks it 6 m east into the goal. What is the soccer ball’s displacement from
dimulka [17.4K]

Answer:

the displacement is zero and the distance is 100 meters

Explanation:

7 0
2 years ago
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