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2 years ago

Why were quasars and active galaxies not initially recognized as being “special” in some way?

Physics

1 answer:

djyliett [7]2 years ago

8 0

Answer:

In the past, astronomers look into the sky and the universe as a whole with an idea that it is a place where stars are born, transition through their life stages and ultimately die and this is because they couldnt differentiate between stars, quasars and active galaxies because with advanced equipment, they all look similar but as they technological ages arrived, they were able see that they are not the same.

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The wet density of a gravel was found to be 2.32 Mg/m^3 and the field water content was 16%. In the laboratory, the density of t

Sholpan [36]

Answer:

the relative density of the gravel in the field is 0.7365 or 73.65%

Explanation:

Given that;

wet density of gravel $r_{b}$ = 2.3 Mg/m³

field water content w = 16%

density of solids in lab $r_{s}$ = 2.70 Mg/m³

Maximum void ratio $e_{max}$ = 0.59

Minimum void ratio $e_{mini}$ = 0.28

first we determine the dry density of gravel $r_{d}$ = $r_{b}$ / 1+ w

$r_{d}$ = 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³

we know that; $r_{w}$ = 1000 kg/m³ = 1 g/cm³

Specific Gravity of soil G = $r_{s}$ / $r_{w}$ = 2.70 Mg/m³ / 1 = 2.70 Mg/m³

eo = ((G× $r_{w}$ )/ $r_{d}$ ) - 1

eo = (( 2.70 × 1) / 1.9827 ) - 1

eo = 1.3617 - 1

Co = 0.3617

so Relative density $I_{D}$ will be;

$I_{D}$ = $e_{max}$ - eo / $e_{max}$ - $e_{mini}$

we substitute

$I_{D}$ = 0.59 - 0.3617 / 0.59 - 0.28

$I_{D}$ = 0.2283 / 0.31

$I_{D}$ = 0.7365 or 73.65%

Therefore; the relative density of the gravel in the field is 0.7365 or 73.65%

3 0

3 years ago

A disk ring of inner radius a and outer radius b lying on the y − z plane has a uniform surface electric charge density +σ(>

solmaris [256]

Answer:

a) V = k 2π σ (√(b² + x²) - √ (a² + x²)) ,

b) E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

Explanation:

a) The expression for the electric potential is

V = k ∫ dq / r

For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P

R = √(x² + r²)

The charge on a ring is

σ = dq / dA

The area of a ring is

A = π r

dA = 2π r dr

So the charge is

dq = σ 2π r dr

We substitute

V = k σ 2pi ∫ r dr / √(r² + x²)

We integrate

V = k 2π σ √(r² + x²)

We evaluate from the lower limit r = a to the upper limit r = b

V = k 2π σ (√(b² + x²) - √ (a² + x²))

b) the electric field and the potential are related

E = - dV / dx

E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))

E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))

7 0

3 years ago

Can someone help me answer this??

Bingel [31]

Answer:

7 0

3 years ago

If i have no swag and i dab what happens

vlabodo [156]

You earn swag points.

7 0

2 years ago

Read 2 more answers

"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th

lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.