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kirza4 [7]
2 years ago
13

According to the big bang theory, what did protons and neutrons form from?

Physics
1 answer:
Pani-rosa [81]2 years ago
3 0

Answer:

Explanation:

In the first moments after the Big Bang, the universe was extremely hot and dense. As the universe cooled, conditions became just right to give rise to the building blocks of matter – the quarks and electrons of which we are all made. A few millionths of a second later, quarks aggregated to produce protons and neutrons. Within minutes, these protons and neutrons combined into nuclei. As the universe continued to expand and cool, things began to happen more slowly. It took 380,000 years for electrons to be trapped in orbits around nuclei, forming the first atoms. These were mainly helium and hydrogen, which are still by far the most abundant elements in the universe. Present observations suggest that the first stars formed from clouds of gas around 150–200 million years after the Big Bang. Heavier atoms such as carbon, oxygen and iron, have since been continuously produced in the hearts of stars and catapulted throughout the universe in spectacular stellar explosions called supernovae.

But stars and galaxies do not tell the whole story. Astronomical and physical calculations suggest that the visible universe is only a tiny amount (4%) of what the universe is actually made of. A very large fraction of the universe, in fact 26%, is made of an unknown type of matter called "dark matter". Unlike stars and galaxies, dark matter does not emit any light or electromagnetic radiation of any kind, so that we can detect it only through its gravitational effects.  

An even more mysterious form of energy called “dark energy” accounts for about 70% of the mass-energy content of the universe. Even less is known about it than dark matter. This idea stems from the observation that all galaxies seems to be receding from each other at an accelerating pace, implying that some invisible extra energy is at work.

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a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun
zubka84 [21]

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before. 
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

               Distance = (1/2 acceleration) x (Time²).

                           D = 1/2 A T²

For your student who dropped an object out of the window,

     Distance = 19.6 m
     Acceleration = gravity = 9.8 m/s²

                                              D = 1/2 G T²

                                          19.6 =   4.9   T²

Divide each side by 4.9 :       4  =           T²

Square root each side:           2  =          T

When an object is dropped in Earth gravity,
it takes  2  seconds to fall the first 19.6 meters.

8 0
3 years ago
Read 2 more answers
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
State how you agree or disagree with the following statement. A good circuit cannot have internal resistance.​
Rashid [163]

Answer: I do

Explanation:

Resistance opposes current thereby reducing the amount of current that flows through a circuit. In other words, it leads to a loss of electrical energy.

Ideally speaking, a good circuit should have no internal resistance as this would lead to more energy having to be supplied to overcome that resistance. External resistance however, is not a bad thing. For instance, oxygen being removed from lightbulbs.

7 0
3 years ago
A racehorse is running with a uniform speed of 69 km/hr along a straightaway. what is the time it takes for the horse to cover 4
Alex
Hello there,
400 meters= 0.4 km
Time= Distance / speed
        = 0.4 / 69
        = 0.0057971014492754 hr
        = 0.35 min

Hope this helps :))

~Top
 
8 0
3 years ago
Read 2 more answers
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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