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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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billiardballsnew2 A white billiard ball with mass mw = 1.49 kg is moving directly to the right with a speed of v = 3.09 m/s and
Kobotan [32]

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

7 0
3 years ago
What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?
arlik [135]

Answer: 0.25 m/s

Explanation:  Speed = wavelengt · frequency  

v = λf   and frequency is 1/period  f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0
3 years ago
A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
Trava [24]

Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
3 years ago
Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
Oksanka [162]

Answer:

1.58 Hz

Explanation:

The frequency of the simple pendulum is given by

f = 1/T

 = 1/2π√g/l  

In this problem, I = 10.0 cm = 0.1 m  

f = 1/2π√9.8/0.1

=  1.58 Hz

7 0
3 years ago
Which scientist began organizing elements into the periodic table?
Gre4nikov [31]

Answer:

B. Dmitri Mendeleev

Explanation:

he invented it

3 0
3 years ago
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