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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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You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment
Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0
3 years ago
A metal spoon can be classified as which of the following
LenaWriter [7]
What are the options
4 0
3 years ago
How is the acceleration of falling objects affected by gravity
Oliga [24]

Gravity is the CAUSE of 'falling', and the cause of acceleration
while falling. 

Objects falling near the Earth's surface all have the same acceleration ...
about 9.8 m/s².

4 0
3 years ago
10 PTS.
gayaneshka [121]

1.2 x (2.2 x 10⁵) = 264,000 Ω

0.8 x (2.2 x 10⁵) = 176,000 Ω

With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .

8 0
3 years ago
At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.9 m/s toward the h
forsale [732]

Answer:

17.10 m

Explanation:

When the toboggan stops, we have:

N-w_y=0\\F-w_x=0

The x-component of weight is the product of the weight and the sine of the angle above the horizontal, so the y-component of W is the product of the weight and the cosine of the angle above the horizontal.

N-mgcos(25^\circ)=0\\F-mgsin(25^\circ)=0\\F=mgsin(25^\circ)(1)

The work-energy principle states that the change in the kinetic energy of an object is equal to the net work done on the object.

\Delta K=W\\K_f-K_i=F\cdot h\\\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}=Fhcos(180^\circ)

Recall that v_f=0, replacing (1):

-\frac{m(v_i)^2}{2}=mgsin(25^\circ)h(-1)\\h=\frac{(v_i)^2}{2gsin(25^\circ)}\\h=\frac{(11.9\frac{m}{s})^2}{2(9,8\frac{m}{s^2})sin(25^\circ)}\\h=17.10 m

7 0
3 years ago
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