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nika2105 [10]
3 years ago
14

What evidence supports the ages of our planets, moon, sun, and asteroids

Physics
1 answer:
Pie3 years ago
6 0

It is indeed true that scientists have known about the background radiation (commonly known as the Cosmic Microwave Background) since the early 60s. It was first discovered quite by accident by Penzias and Wilson working at Bell Labs, who detected it as an unexplainable interference in their precision radio equipment. When people finally figured out exactly what it was they were seeing, they won the Nobel Prize for their discovery. Only a few years before, George Gamow had predicted that if the Big Bang theory were correct, we should observe just such a background radiation. The CMB is not the only evidence in favor of the Big Bang, but it is one of the most important. It is a natural consequence of the theory, and is pretty unexplainable in steady-state cosmology.

The 15-20 billion year number comes not from the CMB, but rather predominantly from measurements of nearby and distant galaxies, particularly their rates of expansion away from us. We find that the distance to a galaxy is proportional to its recessional velocity. The constant of proportionality is the Hubble Constant, H, which turns out to be (approximately) the reciprocal of the age of the universe. So we measure the age by measuring recessional velocities. T = 1/H is only true, however, if the universe is not significantly accelerating or decelerating its expansion rate. If the rate of expansion is rapidly accelerating, the universe may be older than 1/H = 15 billion years, give or take. Such an acceleration would be caused by a large value of the Cosmological Constant, a sort of anti-gravity force predicted by General Relativity. There is some evidence that this might be the case.

So finally, yes, the age of the universe, being based on the empirical determination of H, is based on the observed evidence.

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stealth61 [152]

Answer:

D) diffraction

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5 0
2 years ago
The volume of an object as a function of time is calculated by v = At3+B÷t where t is time measured in seconds and v is in cubic
madam [21]

Answer:

A = m³/s³ = [L]³/[T]³ = [L³T⁻³]

B = m³s = [L³T]

Explanation:

We have the equation:

V = At³ + B/t

where, the dimensions of each variable are as follows:

V = m³ = [L]³

t = s = [T]

substituting these in equation, we get:

m³ = A(s)³ + B/s

for the homogeneity of the equation:

A(s)³ = m³

<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>

Also,

B/s = m³

<u>B = m³s = [L³T]</u>

3 0
3 years ago
A charge of 8.4*10^-4 moves at an angle of 35 degrees to a magnetic field that has a field strentgh of 6.7
12345 [234]

The force on a charged particle in a magnetic field is given by

the speed of the charged particle = 10842 m/s.

Explanation:

F= q V B sinθ

F=force=3.5 x 10⁻²N

q= charge= 8.4 x 10⁻⁴ C

B= magnetic field= 6.7 x 10⁻³ T

θ=35⁰

Thus the velocity is given by V=\frac{F}{q B sin35}

V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]

V=10842 m/s

3 0
3 years ago
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