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Ira Lisetskai [31]
2 years 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]2 years 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

You might be interested in
Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
Pachacha [2.7K]
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
5 0
2 years ago
Read 2 more answers
Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler
Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

          p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

         pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

        n m v₀ = (n m + M) v

        n m (v₀ - v) = M v

        n = M/m    v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

          F = m a

          a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

          v₁² = v₀² + 2 a x

          v₁² = 0+ 2 (100/1) 1

          v₁ = 14.14 m / s

This is the initial speed for the crash

         v₀ = v = 14.14 m / s

  Let's calculate

           n = M/m   v/ (v₀-v)

           n = 10/1   3 / (14.14 -3)

          n = 2.7 balls

you must throw 3 snowballs

7 0
2 years ago
An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac
Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

       = (50 - 0)/4

       = 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0
2 years ago
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt
Volgvan

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

<em>Lets explain how to solve the problem</em>

- The speed ramp has a length of 121 m and is moving at a speed of

 2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

  the ground

<em>Lets find your speed on the ground</em>

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

8 0
2 years ago
You are provided with three polarizers with filters making angles of (A) 90 ​∘ ​​ , (B) 180 ​∘ ​​ and (C) −45 ​∘ ​​ with respect
irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) 90^{\circ}:

The light intensity after passing through the first polarizer is I_{o} and this intensity does not depend on the angle of the polarizer.

Consider 90^{\circ} with the vertical, the intensity is given by:

I = I_{o}cos^{2}90^{\circ}

I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}

(B) 180^{\circ}:

Suppose the second polarizer is  45^{\circ} with the vertical.

Now, intensity through the second polarizer:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)

I' =  \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}

Now, if we consider the second polarizer to be 180^{\circ},

I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0

(C) - 45^{\circ}:

Now,

Intensity through the third polarizer, if it is 180^{\circ} with the vertical:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))

I' = \frac{I_{o}}{8}

5 0
2 years ago
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