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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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When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc
gtnhenbr [62]

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07  = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

7 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe
Nata [24]

Answer:

Explanation:

Given

Each student exert a force of F=400 N

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students F=18\times 400=7200 N

therefore weight of car W=7200

mass of car m=\frac{W}{g}

m=\frac{7200}{9.8}=734.69 kg

(b)7200 N \approx 1618.624\ Pound-force

734.69 kg\approx 1619.71 Pounds                  

6 0
3 years ago
Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with
wlad13 [49]
It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.
4 0
3 years ago
How do electrons flow through a wire
aniked [119]

Every atom has electrons. When you add new electrons to the wire, they will be passed on to an atom. The electrons keep passing from atom to atom until it reaches the light source, basically. It's kinda like that one song "100 jugs of milk" or whatever it's called. Each atom passes the atom next to it an electron.

7 0
3 years ago
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