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Nostrana [21]
2 years 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]2 years 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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Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

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2 years ago
Ram has power of 550 watt. What does it mean?
WARRIOR [948]
It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps
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2 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
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Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
2 years ago
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a
marysya [2.9K]

Answer:

The movable piston

Explanation:

Work is said to be done when a distance is been covered by a force . In this case kinetic energy will be change by an equal amount into work done.

Pushing the piston with a known mass of (m) and an accelarating rate from rest of ( a)   to cover a known distance of (d).The idea of work done is been achieved and can be mathematically represented by:

  • Work done = Force x distance (d)
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8 0
2 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
2 years ago
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