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Nikitich [7]
3 years ago
7

How can astronomers infer wich elements are found in a star

Physics
2 answers:
alina1380 [7]3 years ago
8 0

Just recently, astronomers discovered a distant solar system, 127 light years away with up to seven planets orbiting a Sun-like star called HD 10180.

Like the very first exoplanet 51-Pegusus discovered in 1995, this new system was found using the science of spectroscopy.

In fact, most of the roughly 500 planets so far found orbiting other stars, were detected by the same method.

Spectroscopy — the use of light from a distant object to work out the object is made of — could be the single-most powerful tool astronomers use, says Professor Fred Watson from the Australian Astronomical Observatory.

"You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours.

"It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson.

When heated or when electrically charged, certain chemicals emit radiation at very specific colours or wavelengths called emission lines.

There are also absorption lines that appear as dark marks dividing the spectrum at specific wavelengths.

Absorption lines are created when light from something hot like a star passes through a cooler gas, cancelling out the emission lines the chemicals in the gas would normally create.

When you look at the spectrum of a star, for example, you can see absorption lines because the star's outer atmosphere is cooler than the central part, explains Watson.

Alchen [17]3 years ago
5 0
By previous elements found in other stars, accurate guesses and lots of studying
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A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the
babunello [35]

Answer:C

Explanation:

Given

mass m_1=400\ gm is at x=20\ cm mark

mass m_2=320\ gm is at x=75\ cm mark

Scale is Pivoted at x=50\ cm mark

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)

T_{net}=0.12g-0.08g=0.04g

Therefore a net torque of 0.04 g is required in CW sense which a mass 400\ gm can provide at a distance of x_o from pivot

0.04g=0.4\times g\times x_o

x_o=0.1\ m

therefore in meter stick it is at a distance of x=60\ cm

6 0
3 years ago
Which of these forms of radiation passes most easily through the disk of the Milky Way?
Rama09 [41]
<h2>Answer: 3 - infrared light</h2>

Explanation:

<u>There are certain areas of the Milky Way that cannot be observed using the visible range of the electromagnetic spectrum</u> (this includes blue light and red light). This is because these areas are covered or hidden behind columns of interstellar dust and dark matter.

However, using infrared light and sometimes radio waves, it is possible to observe the galaxy better, because this light manages to pass through all that interstellar dust.

4 0
3 years ago
What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field
ELEN [110]

Answer:

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m,  t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

3 0
3 years ago
A pendulum has 141 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Lubov Fominskaja [6]
Assume that energy losses (due to aerodynamic resistance or bearing friction are negligible).

At the highest point, the pendulum has its maximum potential energy (PE) and zero kinetic energy (KE). The total energy is equal to the PE = 141 J.

At the lowest point, the pendulum has its maximum KE, and zero PE. The total energy is equal to the KE.

Because energy is preserved, therefore
KE at the lowest point  = PE at the highest point  = 141 J.

Answer:  141 J
3 0
3 years ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
3 years ago
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