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erastovalidia [21]
3 years ago
8

Which best describes cosmic microwave background radiation? radiation thought to exist but not yet measured radiation predicted

to exist in millions of years radiation that existed before the big bang leftover radiation from the big bang
Physics
1 answer:
grigory [225]3 years ago
8 0

Answer:  leftover radiation from the big bang

Microwave background radiation is a form of electromagnetic radiation that is present in the whole universe and completely fills it.

Its frequency belongs to the microwaves range and is one of the main demonstrations of the of the Big Bang theory model.

It is important to note the Big Bang occurred 400,000 years before the events related to the microwave cosmic radiation, this means the Big Bang occurred first.

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In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
3 years ago
If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________
siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

To know more about escape velocity follow

brainly.com/question/14042253

#SPJ4

8 0
1 year ago
What is the amount of work done when a force of 10N moves a 20kg mass of 8 meters?
Nataly_w [17]

Yo sup??

we should know that

work done=force*displacement

W=F.s

F=10 N

s=8 m

therefore

W=10*8

=80 N

Hope this helps.

3 0
3 years ago
Give an example of a Claim.
Olin [163]

Answer:

Yeah

Explanation:

Look at the pic!!

5 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
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