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aksik [14]
2 years ago
10

You are an astronomer and are making observations about a visible but faraway galaxy. Describe what evidence you could gather to

gain more information about (1) the galaxy's elemental composition and (2) it's motion relative to the Milky Way galaxy.
Physics
1 answer:
Pavel [41]2 years ago
6 0

Answer:

Chemical Composition

The chemical composition of the Universe is dominated by the hydrogen and helium produced in the Big Bang. The remaining 90 or so chemical elements are produced in stars and constitute only a few percent of the overall mass. Astronomers refer to these elements (all except hydrogen and helium) as metals, even though this includes elements such as carbon and oxygen which are not considered metals in the normal sense.

The abundance of metals with respect to hydrogen is known as the metallicity. While hydrogen and helium are found in high abundance throughout the Universe, the metallicity varies depending on the history of star formation in the region. The chemical composition of the Sun gives us some idea of the chemical composition of the solar neighbourhood:

Chemical composition of the Sun

Hydrogen 73%

Helium 25%

Oxygen 0.80%

Carbon 0.36%

Iron 0.16%

Neon 0.12%

Nitrogen 0.09%

Silicon 0.07%

Magnesium 0.05%

Sulphur 0.04%

Others combined 0.04%

This indicates that metals constitute only about 2% of the Sun’s mass.

The highest metallicities are found in the centres of galaxies. For example, near the centre of the Milky Way, stars with metallicities of up to three times the solar value have been observed. However, there are also stars with only 1/10,000th of the solar value. These stars formed early in the history of the Galaxy, before the interstellar medium (and subsequent generations of stars) became enriched in metals through the actions of other stars.

Although never more than a few percent by mass, the metals content of stars has a significant effect on their stellar evolution, with metal-rich stars being cooler, larger and longer-lived than metal-poor stars of the same mass. Both the length of time spent on the main sequence and the detail of post-main sequence evolution are significantly affected by a star’s metallicity.

To fully describe the chemical composition of stars (or galaxies) it is also necessary to define abundance ratios. These relate the relative abundances of metals to each other (e.g. the abundance ratio of magnesium to iron or carbon to oxygen). Astronomers use these abundance ratios to measure how long the object in question has been forming stars.

Both metallicities and abundance ratios are usually expressed in terms of the values for the Sun, and normally on a logarithmic scale.

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If a seismic wave has a period of 0.0202s, find the frequency of the wave.
maw [93]

Answer:

49.5 Hz.

Explanation:

From the question given above, the following data were obtained:

Period (T) = 0.0202 s

Frequency (f) =?

The frequency and period of a wave are related according to the following equation:

Frequency (f) = 1 / period (T)

f = 1/T

With the above formula, we can obtain the frequency of the wave as follow:

Period (T) = 0.0202 s

Frequency (f) =?

f = 1/T

f = 1/0.0202

f = 49.5 Hz

Therefore the frequency of the wave is 49.5 Hz.

7 0
3 years ago
Define force ? ☹❣️️ Degine weight? ​
mylen [45]

Explanation:

force is anything that allows a body to be in motion..

6 0
3 years ago
Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?
Rzqust [24]
If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration  the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer  is the option D. only at constant accelerations.
3 0
3 years ago
A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
solniwko [45]

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

a_t = R\alpha

2) Centripetal acceleration

a_c = \omega^2 R

here we know that

\omega = \alpha t

a_c = (\alpha t)^2 R

now we know that net linear acceleration is given as

a = \sqrt{a_c^2 + a_t^2}

so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

a = R\alpha\sqrt{1 + \alpha^2t^4}

4 0
3 years ago
Which region of a longitudinal wave is this?
ipn [44]
I think the correct answer is is D.
8 0
3 years ago
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