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svetoff [14.1K]
3 years ago
9

Which of the following pairs of terms directly relates to the actual brightness of a star?

Physics
2 answers:
Aloiza [94]3 years ago
7 0

Answer:

D. absolute magnitude and apparent magnitude

yan [13]3 years ago
5 0
D. Absolute magnitude and apparent magnitude
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A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter
lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

V_d=Ir

Therefore, the voltage across the terminals of the battery is given as:

V= E-V_d\\\\V=E-Ir

Now, rewriting in terms of 'r', we get:

Ir=E-V\\\\r=\frac{E-V}{I}

Plug in the given values and solve for 'r'. This gives,

r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms

Therefore, the internal resistance of the battery is 5 ohms.

5 0
3 years ago
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
3 years ago
when using parallax distance is calculated from the sun and not earth do you think this matters? explain your answer.
Elden [556K]

When using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

<h3>What is parallax distance?</h3>

Parallax enables astronomers to measure the distances of far away stars by using trigonometry.

<h3>Why does astronomers measure parallax distance from sun?</h3>

As the distance of star increases, the parallax angle decreases, and great degree of accuracy is required for its measurement.

So taking a refence from the earth instead of the sun will impact the accuracy of their measurement.

Thus, when using parallax, astronomers calculate distance from the sun and not earth to improve on the accuracy of their measurement, since parallax angle decreases as star distance increases.

Learn more about parallax distance here: brainly.com/question/2128443

#SPJ1

7 0
1 year ago
Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd
Inga [223]

Answer:Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd!

Explanation:

6 0
3 years ago
Read 2 more answers
Dr. Metallica wants to change the property of an alloy? How could she do this? SELECT ALL THAT APPLY. (Please help!)
Serhud [2]

Answer:

I think itd A

Explanation:

4 0
3 years ago
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