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3 years ago

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Star X has an apparent magnitude of 1. Star Y has an apparent magnitude of 4. Both stars are in the same star cluster. Which sta

r is more intrinsically luminous and by what factor

1 answer:

sashaice [31]3 years ago

7 0

Answer:

Explanation:

From the given information:

Since both stars are in the same cluster, the magnitude and luminosity relationship can be calculated as:

$m_1 - m_2 = -2.5 log _{10} (\dfrac{L_1}{L_2})$

Given that;

m_1 = 1 and

m_2 = 4

Therefore,

$1 - 4 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

$3 = -2.5 log _{10} ( \dfrac{L_1}{L_2})$

Making $\dfrac{L_1}{L_2}$ the subject of the formula:

$\implies \dfrac{L_1}{L_2}= 10^{(\dfrac{3}{2.5})}$

=15.84

≅ 16

Hence, we can conclude that star X is more luminous by a factor of 16

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Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

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<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

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$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

<em />

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Refraction brainly.com/question/12370040

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