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serg [7]
3 years ago
5

How are galaxies organized and distributed within the universe

Physics
2 answers:
den301095 [7]3 years ago
3 0
The galaxies aren't distributed randomly throughout the universe but they are grouped in gravitationally bound clusters.
Mars2501 [29]3 years ago
3 0

Answer:

<u><em>The answer is</em></u>:<em><u> </u></em><u>Many are associated in pairs, trios, groups or clusters, and their distribution in filamentary structures.</u>

<u></u>

Explanation:

<em>Galaxies rarely appear isolated and their distribution in the Universe is not uniform</em>. Many are associated in pairs, trios, groups of some tens or clusters of up to a few thousand. <em>These groups are held together by gravitation. </em>

The large-scale distribution of galaxies seems to indicate that they are concentrated along large filamentary structures.

<u><em>The answer is</em></u>:<em><u> </u></em><u>Many are associated in pairs, trios, groups or clusters, and their distribution in filamentary structures.</u>

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Answer:

B

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because atoms make up an element.

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What is the most widely accepted model used to predict the future of the universe?
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Explain why a parachute would be useless if you went to the Moon?
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If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p
Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0
2 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
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