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Harlamova29_29 [7]
2 years ago
5

In our Solar System, the inner planets are rocky because Choose one: A. warm temperatures in the inner disk caused the inner pla

netesimals to be formed of mostly rocky material. B. the original cloud had more rocky material near the center. C. the hydrogen and helium atoms were too low mass to remain in the inner disk. D. the inner disk filled a smaller volume, making it denser.
Physics
1 answer:
xeze [42]2 years ago
3 0

The inner planets are rocky because The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

What are rocky planets?

  • Rocky planets are the planet in which constituents are mostly silicate rocks or metal. They are also regarded as a planet with a solid surface.
  • The formation of rocky planets is said to have occurred billions of years ago and its process of formation is termed accretion. Through accretion are its constituents formed as the more it goes bigger, the higher the rising temperature and pressure in its core and the elements which have to undergo accreted heat up, melt, and spread. Through this process, heavier elements go deeper into the core of the planet and lighter elements float toward the surface.
  • In the formation of rocky planets, the inner portions of the disk are said to be warm from the protostar thereby resulting in the production of the heavy elements that stay there.
  • Examples of rocky planets are Earth or Mars

Hence, from the above, we can say that,

The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

Here,

Option A is correct.

Learn more about rocky planets here:

<u>brainly.com/question/22392798</u>

#SPJ4

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Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol
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Answer:

R \approx 2.418\times 10^{9}\,km

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

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T = \frac{2\pi}{\omega}

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

\omega = \sqrt{\frac{a_{r}}{R} }

Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}

Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

R \approx 2.418\times 10^{9}\,km

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Answer:

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