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3 years ago

Which of these galaxies would you most likely find at the center of a large cluster of galaxies?

Physics

1 answer:

Arada [10]3 years ago

4 0

Answer:

b. a large elliptical galaxy

Explanation:

In elliptical galaxies the stars are grouped in an elliptical shape, it has a low quantity of gas and dust in comparison to spiral galaxies, and its stars belong to an old population, there is not new stellar formation in it.

The stars orbit in a messy way which made to believe that they form from the merger of galaxies.

They are also really massive (around $10^{4}$ solar masses).

The most massive and luminous can be found in the center of cluster of galaxies.

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Explain what is happening to an object that has an acceleration of 2ms-2

joja [24]

It is gaining speed at the rate of 2 m/s per second.

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3 years ago

Show which of the following functional forms work or do not work as solutions to this differential equation (known as 'the wave

Akimi4 [234]

Answer:

E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Explanation:

The wave equation is

d²y / dx² = 1 /v² d²y / dt²

Where v is the speed of the wave

Let's review the solutions

In this case y = E

a) E = A sin Bt

This cannot be a solution because the part in x is missing

dE / dx = 0

b) E = A cos (Bt + c)

There is no solution missing the Part in x

dE / dx = 0

c) E = A x² t²

The parts are fine, but this solution is not an oscillating wave, so it is not an acceptable solution of the wave equation

d) E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Both are very similar

Let's make the derivatives

dE / dx = -A B sin (Bx + Ct + D)

d²E / dx² = -A B² cos (Bx + Ct + D)

dE / dt = - A C sin (Bx + Ct + D)

d²E / dt² = -A C² cos (Bx + Ct + D)

We substitute in the wave equation

-A B² cos (Bx + Ct + D) = 1 / v² (-A C² cos (Bx + Ct + D))

B² = 1 / v² (C²)

v² = (C / B)²

v = C / B

We see that the two equations can be a solution to the wave equation, the last one is the slightly more general solution

8 0

3 years ago

A 0.5 kg ball moving at a speed of 3 m/s rolls up a hill. How does the ball roll before it stops

sergij07 [2.7K]

It will move 1.5 minutes before it comes to a stopping point.

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3 years ago

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krok68 [10]

Answer:

i & 2

Explanation:

U alr finished it so there's no point

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2 years ago

A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then st

sukhopar [10]

Answer:

Explanation:

angular frequency of resulting SHM ω = √(k/m

ω =√(100x32/1

ω = 56.57

Period T

ω = 2π /T = 56.57

T = 2π / 56.57

= .11 s .

spring is stretched by 1 inch so it will become the amplitude of frequency

A = 1 inch

when t = 0 , x ( displacement ) = A = 1 inch

equation of motion

x ( t ) in inch = A cosω t

= 1 cos 56.57t

x ( t ) in inch = 1 cos 56.57 t .

6 0

3 years ago