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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.

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One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic fiel

bagirrra123 [75]

Answer:

The voltage induced in the coil is 1.25 V.

Explanation:

Given;

number of turns, N = 100 turns

cross sectional area of the copper coil, A = 0.1 m²

initial magnetic field, B₁ = 0.5 T

final magnetic field, B₂ = 1.00 T

duration of change in magnetic field, dt = 4 s

The induced emf in the coil is calculated as;

$emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts$

Therefore, the voltage induced in the coil is 1.25 V.

6 0

3 years ago

The planet Gallifrey has 2 times the gravitational field strength and 2 times the radius of the Earth.

Alexeev081 [22]

The mass of the planet Gallifrey is 8 times the mass of the Earth.

let the gravitational field of Earth = g
let the radius of the Earth = R
gravitational field of Gallifrey = 2g
radius of Gallifrey = 2R

<h3>What is gravitational potential energy?</h3>

This is the work done in moving an object to a certain distance against gravitational field.

The gravitational field strength of the Earth is given as follows;

$g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}$

The gravitational field strength of the Planet Gallifrey is calculated as follows;

$g_2 = \frac{GM_2}{r_2^2}$

$G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M$

Thus, the mass of the planet Gallifrey is 8 times the mass of the Earth.

Learn more about gravitational field strength here: brainly.com/question/14080810

4 0

2 years ago

EZ QUESTION!!!<br> Why are bananas curved?

Ksenya-84 [330]

Answer:

Bananas go through a process called ‘negative geotropism’ so they can reach the sun

5 0

3 years ago

Read 2 more answers

A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea

tangare [24]

Given:
u = 6.5 m/s, initial velocity
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v = the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s

Answer: 18.5 m/s

5 0

3 years ago

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?

kirill115 [55]

Using the "v = f. λ" <span>equation...

Your "v" or </span>velocity = 156.25 meters/second

5 0

3 years ago