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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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Construction of a solar power plant is proposed for a desert area near a school. A student has hypothesized that the shade cast
hjlf

Answer:

4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period

Explanation:

The creosote bush depends on sunlight to produce the food they require through photosynthesis. The shade from the solar panels would reduce the amount of sunlight that the bush receives. This would increase the mortality of the bush.

In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.

Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.

4 0
2 years ago
The tire of your bicycle needs air so you attach a bicycle pump to it and begin to push down on the pump’s handle. If you exert
dusya [7]

Answer:

12.5 J

Explanation:

Force, F = 25 N

Distance, d = 0.5 m

The direction of force and the displacement is same.

Work is defined as the product of force in the direction of displacement and the displacement.

Work = Force x displacement x CosФ

Where, Ф be the angle between force and the displacement

Here, Ф = 0°

So, W = 25 x 0.5 x Cos0°

W = 12.5 J

6 0
2 years ago
An NFL place kicker is practicing his extra-point kicks. He kicks
Nezavi [6.7K]

Answer:

83 mph

Explanation:

We are told that the collision is perfect elastic, this means that no energy whatsoever was lost in the process.

We can apply the principle of conservation of momentum. It states that the total final momentum is equal to the total initial momentum in a system.

Hence, if the mass of the wall is assumed to be M and the mass of the ball is assumed to be m, we have that:

(M * V) + (m * v) = (M * U) + (m * u)

Where V = final velocity of Wall = 0 mph

v = final velocity of ball

U = initial velocity of wall = 0 mph

u = initial velocity of ball = 83 mph

Hence:

(M * 0) + (m * v) = (M * 0) + (m * 83)

=> mv = m * 83

=> v = 83 mph

The ball would have a final velocity of 83 mph.

5 0
2 years ago
Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr
Liula [17]

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}  

therefore : Ra = \frac{sin120^o * 750}{sin 35^o} = 1132 N

To determine the scalar component Rb

\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}

therefore : Rb = \frac{sin 25^o * 750}{sin 35^o}  = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos25^o = 679.7 N

6 0
2 years ago
If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e
ryzh [129]

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

K = \frac{A*E}{L}K=

L

A∗E

L = \frac{A*E}{K}L=

K

A∗E

A=\frac{\pi d^{2}}{4}A=

4

πd

2

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}A=

4

π(0.02)

2

A = 0.0003 m²

L = \frac{A*E}{K}L=

K

A∗E

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

5 0
2 years ago
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