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3 years ago

What are the characteristics of a nebulae? (Select all that apply.)

Physics

1 answer:

erica [24]3 years ago

7 0

Answer:

B. contain hydrogen

C. clouds of gas and dust

E. needed to create a star

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

Some of the examples of stars are; Vega, Sun (closest to planet Earth), Antares, Betelgeus, Canopus, etc.

Stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He). The chronological order in which the formation of a star occur are;

1. Gravity pulls gas and dust together to form dense cores.

2. A protostar forms as mass increases.

3. Nuclear fusion begins under high pressure.

Scientists have been able to understand and discover that, gravity pulled materials (low-density cloud of interstellar gas and dust known as a nebula) together forming the planetary bodies in our solar system.

A dark nebula can be defined as an interstellar cloud that is so dense as a result of high concentration of gas and dust and as such it obscures the visible wavelengths of light from stars behind it, thus appearing completely opaque (dark patch) in front of a bright emission nebula or in regions having plenty stars.

The characteristics of a nebulae are;

I. It contain hydrogen.

II. Clouds of gas and dust

III. It is needed to create a star.

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Explain how a star's absorption spectrum and the concept of red shift help support the Big Bang Theory.

marta [7]

Answer:

Red shift supports the big bang theory. ... The light from distant galaxies is red shifted (this tells us the galaxies are moving away from us) and the further away the galaxy the greater the red shift (this tells us that the more distant the galaxy the faster it is moving). Constellations look like they are moving because earth is rotating on it's axis.

May I have brainliest, please?

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3 years ago

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

Read 2 more answers

Which statement best explains why atoms form chemical bonds with other atoms?

Alex73 [517]

Because atoms is something that pops or has bubbles in it

7 0

3 years ago

A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bl

inna [77]

Answer:

a) T=1.35s

b) amplitude = 0.0923m

Explanation:

m=300 gr

k=6.5 N/m

first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m

6 0

3 years ago

In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3

Archy [21]

Answer:

Newton

Explanation:

The earth attracts every body towards its centre. The force with which the earth attracts any body towards its centre, is called its weight.

It is a vector quantity.

It always acts towards the centre of earth.

The SI unit of Newton.

4 0

3 years ago