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Ira Lisetskai [31]
1 year ago
8

Suppose you see two main-sequence stars of the exact same spectral type. Star 1 is dimmer in apparent brightness than Star 2 by

a factor of 100. What can you conclude? (Neglect any effects that might be caused by interstellar dust and gas.)
Physics
1 answer:
katrin2010 [14]1 year ago
3 0

Options:

A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.

B. Star 1 is 100 times more distant than Star 2.

C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.

D. Star 1 is 10 times more distant than Star 2.

E. Star 1 is 100 times nearer than Star 2.

Answer:

D. Star 1 is 10 times more distant than star 2

Explanation:

For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.

Luminosity, L = 1/r²

Where r is the distance of the star to the earth

Since star 1 is dimmer in brightness than star 2 by a factor of 100,

L₁/L₂ = 1/100

i.e. L₁ = 1, L₂=100

L₁ = 1/r₁² ............(1)

1 =  1/r₁²

L₂ = 1/r₂²

100 =  1/r₂² .........(2)

divide equation (2) by equation (1)

100/1 = ( 1/r₂² )/ (1/r₁²)

100 = (r₁/r₂)²

r₁/r₂ = √100

r₁/r₂ = 10

r₁ = 10r₂

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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

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so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

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Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

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v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

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3 0
1 year ago
A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier
kolbaska11 [484]

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

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1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

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alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

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