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3 years ago

What is the study of the cosmos? The study of the cosmos includes what topics?

Physics

2 answers:

densk [106]3 years ago

8 0

Answer:

Cosmos is a Greek word which is used for universe. Study of cosmos means study of universe which is known as cosmology. Cosmology is a branch of physics in which evolution of universe and composition of universe is studied.

Explanation:

Cosmos is a Greek word which is used for universe. Study of cosmos means study of universe which is known as cosmology. cosmology is a branch of physics in which evolution of universe and composition of universe is studied.

The topic covered in the study of the cosmos are as follows:

1. How universe is evolved

2. Composition of universe

3. Study of stars

4. Composition of stars

5. Geometry of universe

6. History and future of universe

7. Expansion of universe

Gennadij [26K]3 years ago

6 0

The study of cosmos is of the large scale structures & evolution of the universe, it encompasses the origin, evolution, composition, and the eventual face of the universe.

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The right eye and right lung are __________.

PilotLPTM [1.2K]

The correct answer is A) Ipsilateral

Explanation:

The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.

3 0

2 years ago

he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

3 years ago

Can someone please help me with this

SashulF [63]

There’s no picture or question

3 0

3 years ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration is given by;

α = (ωₓ - ωₐ) / T₁

α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

8 0

3 years ago

PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST!!!!!

umka2103 [35]

Answer:

i think its The movement of large pieces of ice from one place to another.

Explanation:

3 0

2 years ago