All categories

3 years ago

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.

You might be interested in

A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas

Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the block

F₁ = 65 N, horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight of the block

W= m*g = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force (FN)

We apply the formula (1)

∑Fy = m*ay ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N= 0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0

3 years ago

Can someone please help

coldgirl [10]

Answer:

Making a quick cut left to intercept a pass

Explanation:

It takes more energe to do than running

7 0

3 years ago

Why do some things float but others sink if they have the same mass?

Vsevolod [243]

It's called buoyancy. It is the tendency of an object to float

7 0

3 years ago

Read 2 more answers

!???!?!?!?!?????????????

snow_lady [41]

Answer:

can you type the question I can't click the

Explanation:

6 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

4 years ago