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Alchen [17]
3 years ago
12

For the radial speed of an astronomical object to be determined, what must the object’s spectrum contain?

Physics
2 answers:
Ainat [17]3 years ago
5 0

Answer:

Either absorption or emission lines

Explanation:

For the radial speed of an astronomical object to be determined with a Doppler shift, what must we be able to see in the object's spectrum, Either absorption or Emission lines.

Asorption and Emission line explained below;

  • Absorption lines are usually seen as dark lines, or lines of reduced intensity, on a continuous spectrum.
  • An emission line will appear in a spectrum if the source emits specific wavelengths of radiation.

Alona [7]3 years ago
4 0

Answer:

The object's spectrum must contain a Doppler shift

Explanation:

This is because, the radial speed of the object is determined from the absorption and emission spectral lines.

The object can either be "blue shifted" or "red shifted". This blue or red shift helps us determine the radial speed of the object.

If the astronomical object moves towards us, it is blue shifted and if it moves away from us, it is red shifted.

This extra shift in its wavelength can thus be used to determine the radial speed of the astronomical object.

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3 years ago
Why is Jupiter so much larger than Earth? Check all that apply
Liono4ka [1.6K]

Answer:

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Explanation:

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3 0
3 years ago
Read 2 more answers
You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r
sesenic [268]

Answer:

0.139 rad

Explanation:

We use Snell's law n_1sin\theta_1=n_2sin\theta_2, where if n_1 is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, \theta_1 would be the <em>incident angle</em>, and if n_2 is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, \theta_2 would be the <em>refraction angle</em>, which we want, so we do:

sin\theta_2=\frac{n_1}{n_2}sin\theta_1

And finally:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)

We then insert our values:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686&#10;)=0.139 rad

6 0
3 years ago
A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
8090 [49]

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0
3 years ago
The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim
Vsevolod [243]

Answer:

The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

Explanation:

According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

σx . σpx ≥ h/4π

where,

h is the Planck´s constant

If σx = 5 × 10⁻¹²m,

5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π

σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹

4 0
3 years ago
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