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2 years ago

Explain the origin of earth's motion based on the origin of the galaxy and its solar system?

1 answer:

5 0

There are numerous theories as to the origin of our galaxy and our solar system within our galaxy. I have attached two references that will detail several of them.

As far as motion of the earth, our planet has a prograde revolution around the sun. Prograde motion would be a counterclockwise motion. All the planets in our solar system have this prograde revolution. They all accomplish this revolution in the same plane. Most all the planets have a prograde roatation, like earth, as they perform this revolution.

One of the theories about how the universe and the galaxies and solar systems were formed after the Big Bang is a theory called the "nebular theory." A nebula is a gas and dust cloud. According to this theory, the gas and dust particles collided with each other, were gravitationally attracted to each other, and got progressively larger. As this formation happened, a centripetal force was created by the force of the collisions and the mass of planets began to spin and revolve. It should be noted there is room for the revolutions and rotations to go one way or another, depending on the angle of incidence in the collision of said gas and dust particles. This would be one accounting for the present motion of the planetary constituents in our solar system.

You might be interested in

Which sound waves in the electromagnetic spectrum are considered low energy waves

Mariulka [41]

Radio waves are the waves with the lowest energy in the electromagnetic spectrum. X-rays and gamma rays are the highest. Sound is not part of the electromagnetic spectrum.

7 0

2 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

I’m not sure how to solve this

spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0

2 years ago

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60

german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

$m_1$ = Mass of glider without person = 680-60 kg

$v_1$ = Gliders speed just after the skydiver lets go

$m_2$ = Mass of person = 60 kg

$v_2$ = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

$m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s$

The gliders speed just after the skydiver lets go is 34 m/s

3 0

3 years ago

The _____ of a mechanical wave is a direct measure of its energy.

KiRa [710]

I think it's amplitude

6 0

3 years ago