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Dennis_Churaev [7]
3 years ago
6

What is the apparent magnitude of the brightest star in the Big Dipper?

Physics
2 answers:
alex41 [277]3 years ago
4 0
The magnitude of Alioth ( the brightest star in the big dipper ) is 1.76 and it is about 81 light years distant from Earth. 
ollegr [7]3 years ago
3 0

Answer: +1.77

Explanation:

Big dipper is the  asterism made of seven bright stars in Ursa Major constellation. An asterism a pattern formed by joining the bright stars in the night sky. On the other hand, whole night sky is divided into 88 areas called constellations. These asterisms are used to identify the constellations.

The brightest star of the big dipper is Alioth which is the 32nd brightest star in the night sky. Its apparent magnitude is +1.77.

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This really calls for a blackboard and a hunk of chalk, but
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The energy dissipated by the circuit is the energy delivered by
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-- R₃ and R₄ in series make 6Ω.
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-- R1:  1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts. 
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-- R₂:  3 volts across it = V₂. 
           Current through it is  I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4:  6Ω in the series combination
                     3 volts across it
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--  Remember that the current is the same at every point in
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-- R₃:  1/2 Ampere through it = I₃ .
           1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄:  1/2 Ampere through it = I₄
           1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that  I₂  is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
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Concerning energy, we could go through and calculate the energy
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The energy dissipated by the resistors has to come from the battery,
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The power supplied by the battery  = (voltage) · (current)

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"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
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             (10 joules/second) · (1,800 seconds)  =  18,000 joules  in 30 min

The power (joules per second) dissipated by each individual resistor is

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They're not asking for that.  But if you did it and you actually got the same
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