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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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A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

  • m is mass supported by the steel
  • g is acceleration due to gravity
  • A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

  • E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

  • ΔL is change in length
  • L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0
2 years ago
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