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3 years ago

How much of the matter in the universe is comprised of atoms?

Physics

2 answers:

andreyandreev [35.5K]3 years ago

7 0

Answer:

4.6%.

Explanation:

Actually, the belief before few decades ago is that the whole of the universe comprises of atoms, you, me, stars and so on comprises of atoms that is protons, neutrons and electrons but now according to some researches made, the whole of the universe does not contain atoms naturally as their are some unseen things in the universe. According to the researches by astrophysicists the universe contains about 4.6 percent atoms and the others are dark energy which has the largest portion of the universe of about 71.4 percent and cold dark matter which contains about 24 Percent of the universe.

Deffense [45]3 years ago

4 0

Explanation:

Atoms are the components of ordinary matter, also called baryonic matter, which only represents 4% of the universe, while the remaining 96% would be formed by what is known as dark matter and dark energy which constitute two of the unsolved problems in physics.

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Is this answer correct?

alisha [4.7K]

Yes the answer is correct

4 0

3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes th

max2010maxim [7]

Answer:

its D

Explanation:

edg2020

7 0

3 years ago

Read 2 more answers

A spring has been compressed inside a toy what kind of energy is this

zlopas [31]

Answer:

Elastic Potential Energy

Explanation:

Elastic Potential Energy (“Spring Energy”) is the form of energy an object has when it is stretched, compressed, twisted, bent, or otherwise has its shape changed as long as the object resists and will try to return to its original state.

4 0

3 years ago

A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki

Bond [772]

A = .3*g = 2.94 m/s²

t = v/a = 9/2.94 = 3.061 sec 

W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts

4 0

3 years ago