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Harrizon [31]
3 years ago
10

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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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

p_f=p_i

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

p_{of}+p_{wf}=p_{oi}+p_{wi}

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so v_{oi}=v_{wi} = 0\frac{m}{s}:

m_o*v_{of}+m_w*v_{wf}=0

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for v_{wf}:

v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}

v_{wf}=-6.06\frac{m}{s}

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0
3 years ago
A gun is fired parallel to the ground. at the same instant a bullet of equal size and mass next to the muzzle is released and dr
jolli1 [7]

Both hits the ground <u>at the same time</u> because they have <u>same vertical acceleration</u>

<u></u>

<h3>What is vertical  acceleration?</h3>

A vertical acceleration is typically one for which the direction of the vector is vertically upward, usually aligned with and opposite to the gravity vector. But this is a descriptive term, not a rigorous or technical term. A car may accelerate along a road and that would generally be assumed to be a horizontal.

The vector perpendicular to this direction, as perhaps a suspension motion over a bump, would be described as vertical even if it is not strictly vertical.

Note that acceleration is defined as the rate of change of the velocity vector. But the gravitation vector, ‘g’, generally vertically downward, is often denoted by what acceleration a mass in free fall (absent air resistance) would experience, i.e. the relationship between mass and weight.

Learn more about vertical acceleration

brainly.com/question/19528199

#SPJ4

3 0
1 year ago
Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple
Finger [1]
Kepler derived his three laws of planetary motion entirely from
observations of the planets and their motions in the sky.

Newton published his law of universal gravitation almost a hundred
years later.  Using some calculus and some analytic geometry, which
any serious sophomore in an engineering college should be able to do,
it can be shown that IF Newton's law of gravitation is correct, then it MUST
lead to Kepler's laws.  Gravity, as Newton described it, must make the planets
in their orbits behave exactly as they do.

This demonstration is a tremendous boost for the work of both Kepler
and Newton.
5 0
3 years ago
Read 2 more answers
PLEASE HELP ME IM ON A TIMER
garri49 [273]

Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

Displacement = \sqrt{(3)^2+ (6)^2} \\Displacement = 6.70

And the angle will be defined by:

tan(beta)=3/6

beta = tan^-1(6/3)

beta = 63.43°

5 0
3 years ago
Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=
drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2

v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095

v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095

The final relative velocity of the satellite, v_f = v₁ + v₂

∴ v_f = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, v_f = 0.190 m/s

4 0
3 years ago
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