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Anettt [7]
3 years ago
14

If you were to divide the presentday universe into cubes whose sides are 10 million lightyears long, each cube would contain, on

average, about one galaxy similar in size to the Milky Way. Now suppose you travel back in time, to an era when the average distance between galaxies is 0.26 of its current value, corresponding to a cosmological redshift z = 2.7. How many galaxies similar in size to the Milky Way would you expect to find, on average, in cubes of that same size?
Physics
1 answer:
vovikov84 [41]3 years ago
8 0

Answer:

Explanation:

density of galaxies would be \frac{1}{0.27^3} times higher which is equal to 50.81.

It means  in a cube that today contains one galaxy the size of the Milky Way, we would instead find 50.81 galaxies this size.

You can round this off to 52

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a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
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Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

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Therefore, the centripetal acceleration of the stone found as follows;

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Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

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L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

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K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

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