Why do astronomers use spectroscopes to analyze light from distant objects?

1 answer:

7 0

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. ... "It lets you see the chemicals being absorbed or emitted by the light source.

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Hydrogen has only one electron in its only (and outer) electron shell. If a hydrogen atom were to absorb a small amount of energ

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D. move up to another shell that would form

Explanation:

An atom has protons, neutrons and electrons. Protons and neutrons are present in the nucleus and electrons orbit the nucleus in fixed shells. An electron can jump to higher shell when it gains energy and lower one when it loses energy. Thus, when single electron in hydrogen atom is given a small amount of energy, it would jump to another higher shell.

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Which is true about surface waves?

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The disturbance does not have a specific motions

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What was the average velocity for the entire trip?

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Match the technology that uses radio waves to the field in which it is used

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Naturally occurring radio waves are made by lightning or by astronomical objects. Artificially generated radio waves are used for fixed and mobile radio communication, broadcasting, radar and other navigation systems, communications satellites, computer networks and innumerable other applications.

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

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3 years ago

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