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Nostrana [21]
1 year ago
6

We can learn a lot about the properties of a star by studying its spectrum. All of the followingstatements are true except one.

Which one?A. the peak of the star’s thermal emission tells us its temperature: Hotter stars peak at shorter (blue) wavelengths.B. The total amount of light in the spectrum tells us the star’s radius.C. We can look at Doppler shifts of spectral lines to determine the star’s speed toward or away from us.D. we can identify chemical elements present in the star by recognizing patterns of spectral lines that correspond to particular chemicals

Physics
1 answer:
Kisachek [45]1 year ago
3 0

Answer:

B. The total amount of light in the spectrum tells us the star’s radius.

Explanation:

A.

The effective temperature of a star can be determined by means of its spectrum¹ and Wien's displacement law.                    

Since stars behave in a local way as a blackbody, it will take the wavelength at which is the peak of emission greater in the continuum (see the image below).

Then, the maximum peak of emission (\lambda_{max}) will be replaced in the next equation of the Wien's displacement law:

T = \frac{2.898x10^{-3} m. K}{\lambda max}  (1)

Where T is the effective temperature of the star.

Bodies that are hot enough emits light as consequence of its temperature. For example, a iron bar in contact with fire will start to change colors as the temperature increase, until it gets to a blue color, which its know as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase.

The same scenario described above can be found in the stars, a star whit higher temperature will have a blue color and one with lower temperature, a red color.

B.

Since star does not have the same size, they have different brightness, That is because the photons have a free mean path greater in a bigger radius.

So a star brightness is a consequence of its radius.

               

C.  

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).        

By using that shift in the spectral lines, the Doppler velocity can be determined.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (2)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

   

D.

When a photon is absorbed by an electron in an atom of a particular element in the star photosphere, the electron will be pass to a higher state, when it comes back to the ground state, a photon will be emitted again. If the emitted photon does not go in the same direction of the incident photon an absorption line will be created in the spectrum of the star.          

This patterns of spectral lines in the spectrum of the star are compared with the patterns that are got by lamps of that element in a laboratory.

Key term:

¹Spectrum: decomposition of light in its characteristic colors (wavelengths).

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
2 years ago
1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and
saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

   Centripetal acceleration  = \frac{v^{2} }{r}

   v is the velocity of the body

    r is the radius of the circle

  putting in the parameters:

   Centripetal acceleration = \frac{4^{2} }{0.75}

    Centripetal acceleration = 10.67m/s²

Centripetal force = m  \frac{v^{2} }{r}

          m is the mass

 Centripetal force = mass x centripetal acceleration

                              = 3 x 10.67

                              = 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0
1 year ago
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
1 year ago
A metal disk weighing 1 N is resting on an index card that is balanced on top of a glass. When the index card is quickly pulled
Burka [1]
Answer: D



Step by step explanation:
8 0
2 years ago
Read 2 more answers
Consider the following:
pychu [463]

Answer:

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

Explanation:

The complete question is

Consider the following:

a) radio waves emitted by a weather radar system to detect raindrops and ice crystals in the atmosphere to study weather patterns;

b) microwaves used in communication satellite transmissions;

c) infrared waves that are perceived as heat when you turn on a burner on an electric stove;

d) the multicolor light in a rainbow;

e) the ultraviolet solar radiation that reaches the surface of the earth and causes unprotected skin to burn; and

f) X rays used in medicine for diagnostic imaging.

Which of the following statements correctly describe the various forms of EM radiation listed above?

check all that apply to the above

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through a vacuum depending on their frequency.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

They require different media to propagate.

All the above phenomena are due the electromagnetic wave spectrum. Electromagnetic waves travel at a constant speed of 3 x 10^8 m/s in a vacuum. Within the spectrum, the different types of electromagnetic waves exists in different band range of frequencies and wavelengths unique to each of the waves, and the energy they carry. When these waves enter a non-vacuum medium, their speed change, depending on the nature of the material of the medium, and the frequency or the wavelength of the incoming wave.

5 0
2 years ago
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