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natulia [17]
3 years ago
9

How can we use variations in luminosity to set limits on the size of an active galactic nucleus?

Physics
1 answer:
morpeh [17]3 years ago
4 0

Answer:

If we see an object brighten in 1 hour, we know that the object cannot be longer than 1 light-hour in size. If it were, light from the far side of the object would start arriving at our telescopes more than an hour after the light from the front side and we would see the brightening take longer time than an hour to occur.

Explanation:

To understand the question above, let us discuss what variations in luminosity is:

variations in luminosity are two distinct word coming together to form a phenomenon. They are concepts that help us to bring to our understanding how brightness of the Sun significantly affects or impacts on global warming.

Variations as a concept can be describes as the total energy the sun generates in term of output or emission which also can be refereed to as the Luminosity ratio. Luminosity can also be refereed to as a concept caused by changing dark (sunspot) that bright structures on the solar disk during the 11-year sunspot cycle.

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What describes why ocean currents are considered convection currents
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The smallness of the critical angle θc for diamond means that light is easily "trapped" within a diamond and eventually emerges
hoa [83]

Answer:

26.6°

Explanation:

refractive index of diamond, n = 2.23

When a ray of light passes from denser medium to the rarer medium and refracts at an angle of 90 degree from the normal of the surface, such angle of incidence in the denser medium is called the critical angle.

By the Snell's law

\frac{Sini}{Sin r }= n

For critical angle, angle of incidence is critical angle, i = θc and angle of refraction, r = 90

So,

Sin θc / Sin 90 = 1 / 2.23

Sin θc = 0.448

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3 years ago
Karen travels 10 mile in 1 hour. How many
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5 0
3 years ago
A large cube has a mass of 25kg. It is being acclerated
soldier1979 [14.2K]

Answer:

p= 400.29N ........the horizontal force

Explanation:

Given data

mass=25 kg

small cube mass mass=4 kg

Us (The Coefficient of static b/w two cubes) = 0.71

to find

The horizontal force to keep the small cube  from sliding downward

Solution

F=ma.........................from Newton Second law

Where F=force

a=acceleration

m=mass

we can write equation in form of acceleration  

a=F/m

The acceleration on small box is same as that on the large box.

Let P be force to find.

then:

a=p/(25kg+4kg)

a=p/(29kg)m/s²

The force acting on small box:

F=ma

f=4*(p/29)N........................normal force

friction force= Us*(normal force).........where Us is coefficient of static friction.

friction force= 0.71*(4*p/29)

Now to find weight

weight= mg

weight= 4*9.8

for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.

0.71*(4*p/29)=4*9.8

solving for p(force)

p= 400.29N ........the horizontal force

3 0
3 years ago
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