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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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If a force is 100 N and is pointing 37 degrees north of east. (a) Draw a diagram of this force. (b) Draw the force's x and y com

sasho [114]

Answer

given,

force = 100 N

Point 37 degrees north of east

a) and b) part is shown in the diagram attached below.

c) to find the x and y component of the force

x- component of the force

$F_x = F cos \theta$

$F_x = 100\times cos 37^0$

$F_x = 79.86 N$

y- component of the force

$F_y = F sin \theta$

$F_y = 100\times sin 37^0$

$F_y = 60.18 N$

3 0

3 years ago

Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

These three members of the Noble gas family all have one property in common because they are gases at room temperature. That is

AlekseyPX

B) are compressible.

gasses are compressible

3 0

3 years ago

Read 2 more answers

Please help me to do this problem

Novosadov [1.4K]

Answer:

we got time and velocity over time.

so the distance is again the area underneath the graph

for a triangle with known base and height it's

4*10 / 2

distance traveled is 20

deceleration occurs when velocity decreases. that happens from t=2 till t=4

in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time

a (from t=2 to t=4) = -5v/t

7 0

3 years ago

Determine the minimum size of glass tubing that can be used to measure water level. If the capillary rise in the tube does not e

Marina CMI [18]

Answer:

Explanation:

The formula to determine the size of a capillary tube is

h = 2•T•Cos θ / r•ρ•g

Where

h = height of liquid level

T = surface tension

r = radius of capillary tube

ρ = density of liquid

θ = angle of contact = 0°

g =acceleration due to gravity=9.81m/s²

The liquid is water then,

ρ = 1000 kg / m³

Given that,

T = 0.0735 N/m

h = 0.25mm = 0.25 × 10^-3m

Then,

r = 2•T•Cos θ / h•ρ•g

r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81

r = 5.99 × 10^-3m

Then, r ≈ 6mm

The radius of the capillary tube is 6mm

So, the minimum size is

Volume = πr²h

Volume = π × 6² × 0.25

V = 2.83 mm³

The minimum size of the capillary tube is 2.83mm³

8 0

3 years ago