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vodomira [7]
4 years ago
14

If a 12-V battery is connected to a circuit that has a current of 3.0 A,?what is the total resistance in the circuit

Physics
2 answers:
Simora [160]4 years ago
6 0
Hi, hope this helps :)

Montano1993 [528]4 years ago
4 0

Answer:

4 ohms

Explanation:

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Soils with a pH of 6.5 are considered slightly acidic true or false
Novay_Z [31]

Answer:

that is true

Explanation:

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4 years ago
A wave has 25 wavelengths in 5 secs. The frequency of the wave is _.
kati45 [8]
Heyyyy I wish I could help but I don’t know
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3 years ago
Please help meh its due
Fed [463]
-0.000393025 light years
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3 years ago
Read 2 more answers
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
Does a comets tail always trail along behind it in its orbit?
Marat540 [252]
No, it only does when entering an atmosphere
4 0
4 years ago
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