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uranmaximum [27]
3 years ago
11

If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is

Physics
1 answer:
klio [65]3 years ago
6 0
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
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Answer:

d. equal to one-fourth the acceleration at the surface of the asteroid.

Explanation:

The explanation is attached as a picture with this answer

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as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

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4 0
4 years ago
Read 2 more answers
10points asap <br><br> A force of 30 N acts upon a 7 kg block. Calculate its acceleration.
nekit [7.7K]
Hello! Assuming that the only force acting on the mass is 30N...

Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
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3 years ago
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
mel-nik [20]

Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

Explanation:

Given

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and solving problem in Vertical and horizontal direction

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Conserving Momentum in Vertical direction

m(2v)+m(-vsin\phi )=2m(ucos\theta )

2ucos\theta =v(2-sin\theta )------1

Conserving momentum in x direction

mvcos\phi =2musin\theta-----2

squaring and adding 1 &2

(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2

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3 years ago
A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4
timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0
3 years ago
What formula is used to find an objects acceleration
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Acceleration = velocity / time.


7 0
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