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alex41 [277]
3 years ago
15

How does the rotation of a galaxy result in spectral line broadening?

Physics
1 answer:
IRISSAK [1]3 years ago
4 0

Answer:

Explanation:

Normal galaxies are made up of stars and (in the case of spiral and irregular galaxies) gas and dust. Their spectra consist of the sum of the spectra of these components.

The optical spectra of normal stars are continuous spectra overlaid by absorption lines (Figure 1). There are two factors to consider when adding up the spectra of a number of stars to produce the spectrum of a galaxy:

Different types of star have different absorption lines in their spectra. When the spectra are added together, the absorption lines are 'diluted' because a line in the spectrum of one type of star may not appear in the spectra of other types.

Doppler shifts can affect all spectral lines. All lines from a galaxy share the red-shift of the galaxy, but Doppler shifts can also arise from motions of objects within the galaxy. As a result, the absorption lines become broader and shallower. We explain below how this Doppler broadening comes about.

HII regions in spiral and irregular galaxies (though not, of course, ellipticals) shine brightly and contribute significantly to the spectrum of the galaxy. The optical spectrum of an HII region consists mainly of emission lines, as in Figure 2. When the spectra of the HII regions and the stars of a galaxy are added together, the emission lines from the HII regions tend to remain as prominent features in the spectrum unless a line coincides with a stellar absorption line. There are Doppler shift effects, however, as described for stellar absorption lines, and hence emission lines too are broadened because of the motion of HII regions within a galaxy.

Box 1: Doppler Broadening

The Doppler effect causes wavelengths to be lengthened when the source is moving away from the observer (red-shifted) and shortened when the source is moving towards the observer (blue-shifted).

Light from an astrophysical source is the sum of many photons emitted by individual atoms. Each of these atoms is in motion and so their photons will be seen as blue- or red-shifted according to the relative speeds of the atom and the observer. For example, even though all hydrogen atoms emit H photons of precisely the same wavelength, an observer will see the photons arrive with a spread of wavelengths: the effect is to broaden the H spectral line - called Doppler broadening.

In general, if the emitting atoms are in motion with a range of speeds Δν along the line of sight to the observer (the velocity dispersion) then the Doppler broadening is given by

where c is the speed of light, and λ is the central wavelength of the spectral line.

Why would the atoms be in motion? An obvious reason is that they are 'hot'. Atoms in a hot gas, for example, will be moving randomly with a range of speeds related to the temperature of the gas. For a gas of atoms of mass m at a temperature T, the velocity dispersion is given by

where k is the Boltzmann constant (1.38 × 10−23 J K−1).

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

8 0
3 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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How much do a steel BB weight?
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6 0
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We know today that atoms cannot be divided into smaller parts true or false
djverab [1.8K]
Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.
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A girls throws a rock horizontaly with velocity=10m/s from a bridge .It falls 45 m​
kherson [118]

Answer:

the answer is 45 m plus 10

4 0
3 years ago
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