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Lana71 [14]
8 months ago
6

Calculate two stars appear to you to have the same brightness. if star a is 7.00 light-years away from you, while star b is 15.0

0 light-years away, how many times more luminous is star b than star a?
Physics
1 answer:
vagabundo [1.1K]8 months ago
7 0

It  is eight times more than the star A.

<h3>What is luminosity  and on which it depends?</h3>

The luminosity of an object is a measure of its intrinsic brightness and is defined as the amount of energy the object emits in a fixed time.

luminousity depends upon the two factors are:

1) The star's actual brightness

Some stars are naturally more luminous than others ,so the brightness level from one star to next star is significantly different.

2) The star distance from us

The more distance of an object the dimmer it appears.

Energy emitted = sAT⁴

where s is stefan constant

A is surface area and T is temperature  .  

to learn more about Luminosity click here brainly.com/question/14140223

#SPJ4

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Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h
Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

1 ly = 63000 AU

also we know that

1AU = 1.5 \times 10^8 km

1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km

So the distance of Betelgeuse = 640 ly

d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m

distance to the star VY Canis Majoris: 3.09 × 10^8 AU

d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km

d_2 = 4.64 \times 10^{16} km

distance to the galaxy Large Magellanic Cloud: 49976 pc

1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km

1pc = 3.08 \times 10^{13} km

now we have

d_3 = 49976 \times 3.08 \times 10^{13}

d_3 = 1.54 \times 10^{18} km

distance to Neptune at the farthest: 4.7 billion km

d_4 = 4.7 \times 10^9 km

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0
3 years ago
What is a small portable device that counts every step taken throughout the day?
Valentin [98]

Answer: B

Explanation: i learned it last year

7 0
1 year ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

7 0
3 years ago
If C is 1kg and D is 100kg, and the initial velocities of both balls are 5m/s, how would the magnitude of the forces exerted by
kupik [55]

Answer:

Explanation:

The forces exerted by each mass is best understood in terms of their momentum.

Momentum is a sort of compelling force or impulse. It is given as:

                  Momentum  =  mass x velocity

Let us consider the momentum of the balls;

Substance C;

    Mass  = 1kg

     Velocity  = 5m/s

    Momentum of C = 1 x 5  = 5kgm/s

Substance D:

      Mass  = 100kg

      Velocity  = 5m/s

     Momentum of D  = 100kg x 5m/s  = 500kgm/s

Body D has a higher momentum compared to Body C. This suggests that body D will exert a higher force than C when they collide.

The higher the momentum, the more the force of impact it has.

3 0
3 years ago
Astronaut X of mass 50kg floats next to Astronaut Y of mass 100kg while in space, as shown in the figure. The positive direction
jonny [76]

Answer:

C

Explanation:

The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.

5 0
2 years ago
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