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erastova [34]
3 years ago
5

What is the relevant similarity of the following argument?

Physics
1 answer:
meriva3 years ago
3 0
B. Tolerating a bully or dictator leads to further abuse
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Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newto
denis23 [38]

Answer:

Option (3)

Explanation:

Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.

He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.

This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.

Thus, the correct answer is option (3).

6 0
3 years ago
Arbeitsauftrag 2
kramer

Explanation:

<em>The height of the pendulum is measured from the lowest point it reaches (point 3). </em>

At 1, the kinetic energy of the pendulum is zero (because it is not moving), and it has maximum potential energy.

At 2, the pendulum has both kinetic and potential energy, and how much of each it has depends on its height—smaller the height greater the kinetic energy and lower the potential energy.

At 3, the height is zero; therefore, the pendulum has no potential energy, and has maximum kinetic energy.

At 4, the pendulum again gains potential energy as it climbs back up,  Again how much of each forms of energy it has depends on its height.

At 5, the maximum height is reached again; therefore, the pendulum has maximum potential energy and no kinetic energy.

Hope this helps :)

8 0
3 years ago
The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 x 10
antoniya [11.8K]

Answer:

4.3 * 10^28 kg

Explanation:

Given:

Period, T = 84s

Radius of satellite orbit, r = 8*10^6

Using the relation :

M = 4π²r³ / GT²

Where G = Gravitational constant, 6.67 * 10^-11

M = 4*π^2*(8*10^6)^3 / 6.67 * 10^-11 * 84^2

M = (20218.191872 * 10^18) / 47063.52 * 10^-11

M = 0.4295937 * 10^18 - (-11)

M = 0.4295937 * 10^29

M = 4.295937 * 10^28 kg

M = 4.3 * 10^28 kg

Mass of planet Nutron = 4.3 * 10^28 kg

8 0
3 years ago
when using parallax distance is calculated from the sun and not earth do you think this matters? explain your answer.
Elden [556K]

When using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

<h3>What is parallax distance?</h3>

Parallax enables astronomers to measure the distances of far away stars by using trigonometry.

<h3>Why does astronomers measure parallax distance from sun?</h3>

As the distance of star increases, the parallax angle decreases, and great degree of accuracy is required for its measurement.

So taking a refence from the earth instead of the sun will impact the accuracy of their measurement.

Thus, when using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

Learn more about parallax distance here: brainly.com/question/2128443

#SPJ1

7 0
1 year ago
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its
dolphi86 [110]

Answer: 4.487(10)^{11}m

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

<em />

This law states a relation between the orbital period T of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size a of its orbit:

T^{2}=\frac{4\pi^{2}}{GM}a^{3} (1)  

Where:

T=3.87Earth-years=122044320s is the period of the orbit of the exoplanet (considering 1Earth-year=365days)

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

M=3.59(10)^{30}kg is the mass of the star

a is orbital radius of the orbit the exoplanet describes around its star.

Now, if we want to find the radius, we have to rewrite (1) as:

a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}} (2)  

a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}} (3)  

Finally:

a=4.487(10)^{11}m This is the radius of the exoplanet's orbit

3 0
3 years ago
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