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Reil [10]
3 years ago
15

What is proposed as evidence that supports the Big Bang Theory?

Physics
2 answers:
boyakko [2]3 years ago
7 0

Evidence that shows that the observed universe is expanding

The red move saw known to man was one of the primary types of proof for the Enormous detonation. The red move or doppler impact gives proof that the watched items known to mankind are moving far from one another. On the off chance that the items known to man are moving far from one another, the universe is extending.

The rationale is if the universe is growing now it probably been littler at one time. Extrapolating in reverse the universe is proposed to have been at one time a superdense wad of issue.

Such a superdense wad of issue would have been shaky. The Huge explosion recommends that this superdense ball detonated.

The possibility of the universe starting with a touchy extension predicts that there ought to be foundation radiation. This anticipated foundation radiation was found giving additional proof to the Theory of how things came to be.

The Theory of how things came to be was opposed in light of the fact that researchers used to trust that the universe was everlasting and existed in a relentless state. The predominant logical rationality of material authenticity ordered that issue and vitality are on the whole that there is and in this manner must be everlasting. Proof that the development of the universe is expanding in rate demonstrates that the present universe had a start as well as have a consummation.
Alexus [3.1K]3 years ago
7 0

That's not the way it works.

Real scientists don't gather evidence to support a theory that they made up.

Real scientists build a theory to explain the evidence that they observed.

The so-called "Big Bang" theory is the best explanation ... so far ... of:

-- the observed Cosmic Background radiation,

-- the observed probable expansion of the universe (not "proved"),

-- the observed probable existence of "dark matter" (not "proved"),

-- the observed success of predictions that come from standard theories of fundamental particles, and

-- the observed success of predictions (like GPS) that come from Einstein's general theory of relativity.  

BUT ! If you come up with a better explanation tomorrow ... one that explains the observed evidence more accurately, and makes more accurate predictions of things that we can observe or test ... then the "Big Bang" theory will go out the window, and the scientific world will adopt the "Sid4067" theory for the same evidence that we already have.

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An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
Hii I need help ASAP with this physics question
Readme [11.4K]

Between the top of the first and the top of the second loop, the coaster has lost potential energy = mgh, where h = 22.2 - 15 = 7.2m

This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.

4 0
3 years ago
After scientists analyze the results of their experiments they
Oksi-84 [34.3K]
They communicate their result to the scientific community- so to speak
8 0
3 years ago
The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object
Ann [662]

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object  is 16.4 N.

<h3>What is frictional force?</h3>

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal  force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

                         = 0.40 × 41.16 N

                         = 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

To find more on frictional force, refer here:

brainly.com/question/1714663

#SPJ1

Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

3 0
1 year ago
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i
spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

7 0
3 years ago
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