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Lesechka [4]
3 years ago
5

An astronomer examines an emission line spectrum for a star from data recorded by a spectroscope. There are four lines: two in t

he blue region, one in the green region, and one in the red region of the spectrum. How does the spectrum obtained from the spectroscope help the astronomer?
Physics
1 answer:
blondinia [14]3 years ago
3 0

The characteristics of the atomic spectrum allow to find the result so that the measured lines help the astronomer is:

  • The astronomer can identify the gases that are present in stars using emission lines.

The emission spectrum is the emission of light due to the atomic transition in atoms, consequently the spectrum of each chemical element is unique.

The astronomer was able to identify the chemical elements that are in stars by their emission spectrum.

In the case presented there are two lines in the blue region, a line in the green region and a line in the red region. We know that the most abundant element in stars is hydrogen.

Hydrogen has a series of well-defined spectral lines, the series that is in the visible part of the spectrum is called the Balmer series and is composed of the lines in the table.

Color     wavelength (nm)

red           656.3

blue          486.1

violet        434.1

The second element in quantity in stars is helium, which has the spectrum given in the table and in the attached.

color   wavelength (nm)

Red          667.8

orange     587.6

green       501.6

Blue         492.2

Blue         447.1

The astronomer compares his measured spectrum with the spectra of these two gases and finds that the element he is measuring is helium.

In conclusion, using the characteristics of the atomic spectra, we can find the result to help the astronomer with the measured lines is:

  • The astronomer can identify the gases that are present in stars using emission lines.

Learn more here:  brainly.com/question/14819167

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Answer:

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Radio waves have longer wavelength than visible waves.

Radio waves have lower frequency than visible waves.

Explanation:

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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
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The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
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<h2>Answer:</h2>

<h2>3m</h2>

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3 years ago
After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

5 0
3 years ago
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