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labwork [276]
2 years ago
5

You obtain a spectrum of an object in space. The spectrum consists of a number of sharp, bright emission lines. Is this object a

cloud of hot gas, a cloud of cool gas, or a star?
Physics
2 answers:
Andreyy892 years ago
6 0

Answer:

In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon

weqwewe [10]2 years ago
4 0

Answer:

The object is a cloud of hot gas.

Explanation:

Kirchhoff’s laws establish that:

   • A solid, liquid or dense incandescent gas emits a continuous spectrum.  

     

   • A hot and diffuse gas produces bright spectral lines (emission lines).

     

   • A gas of lower temperature against a source of continuum spectrum, produces dark spectral lines (absorption lines) superposed in the continuum spectrum.

     

Stars are perfect examples for Kirchhoff’s laws. Since in the case of the stars, the photons that are received are not directly from the nucleus, but those that have traveled hundreds of thousands of years to reach the stellar atmosphere. Due to the stars are not at homogeneous temperature, density and pressure, but have gradients in different layers because of the nuclear reactions, superficial gravity or to their constant exchange of heat with their surroundings in an attempt to reach the thermodynamic equilibrium, the continuum observed in the stellar spectra comes from the inner layer of the photosphere, while absorption lines are formed in the outer layer of the photosphere and the stellar atmosphere. More accurately, a photon of the inner layer of the photosphere will be absorbed by an electron of an atom or ion that is in the outer layer, generating an electronic transition¹. The electron, upon returning to its base state will emit a photon or a series of photons that will not necessarily go in the same direction of the incident photon, creating an absorption line in the stellar spectrum.

On the other hand, in the case where the stars have surrounding material (diffuse gas), the atoms, molecules or ions in the medium are excited by the radiation that comes from the stellar atmosphere, thus producing an emission spectrum.

For this particular case, the object will be a cloud of hot gas, in which the source of continuum spectra is not in the same field of view that the cloud.

Key terms:

¹Electronic transition: When an electron passes from one energy level to another, either for the emission or absorption of a photon.

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The Answer is 3.0uc. I took the quiz.
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A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me
never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

I=\frac{Q}{t}

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

I=\frac{420 C}{4800 s}=0.0875 A

2) Drift velocity of the electrons: 1.78\cdot 10^{-6} m/s

The drift velocity of the electrons in the wire is given by:

u = \frac{I}{nAq}

where

I = 0.0875 A is the current

n=5.8\cdot 10^{28} is the number of free electrons per cubic meter

A is the cross-sectional area

q=1.6\cdot 10^{-19} C is the charge of one electron

The radius of the wire is

r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m

So the cross-sectional area is

A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2

So, the drift velocity is

u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s

4 0
1 year ago
Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc
lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0
2 years ago
5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o
grigory [225]

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

7 0
1 year ago
A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta
yuradex [85]

Answer:

1.25 kgm²/sec

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Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

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Jc = 3.7 * 0.40² = 0.592 kgm²

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Ekd = 1/2 * Jd * ωd²

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Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

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ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

5 0
1 year ago
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