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Hitman42 [59]
3 years ago
9

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

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Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
2 years ago
single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

4 0
3 years ago
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