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Allushta [10]
3 years ago
14

When a court has "___ ___" they can hear a case once a lower court has ruled on it.

Physics
1 answer:
notka56 [123]3 years ago
3 0
Your answer is going to be Appellate jurisdiction.
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A ringing bell sends sound waves in all<br> directions<br> places<br> sides
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Answer:

direction

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because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

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The centripetal force acting on a satellite in orbit
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3 years ago
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If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?
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The correct answer is B. <span>0.002010812m3. Good Luck! :)</span>
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3 years ago
Kenneth was preparing a meal for his family. As he was prepping the kitchen, his mother asked him, “Do you need help? I know coo
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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
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