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nataly862011 [7]
3 years ago
7

If dark energy were to significantly increase or decrease, what outcome could this have on the universe?

Physics
2 answers:
Soloha48 [4]3 years ago
6 0

Answer: A) In case of dark energy increasing significantly,  the universe would accelerate its expansion at an ever-increasing rate which later on may result in "Big Rip".

B) In case of dark energy decreasing significantly, the universe would decelerate its expansion which later on may result in "Big Crunch".

Explanation:

In case of significant increase in dark energy, the universe would accelerate its expansion at more increasing rate. Ultimately the expansion would result in scale factor  becoming infinite within a finite period of time.  As a result due to the repulsive dominance of dark energy the whole universe  would come apart resulting in a situation called "Big Rip"

If dark energy were to decrease significantly enough with time, the accelerating expansion of the universe that is observed now would  start to decelerate as the mass once again dominates over dark energy. It will result in universe reversing its expansion resulting in its collapse, a situation called "Big Crunch".

Hope it helps!!!

Debora [2.8K]3 years ago
4 0

Answer:

Dark energy cannot be detected directly because it does not interact with electromagnetic radiation. It is known to exist because the observable matter in the universe is too small to cause its accelerated expansion.

If the dark energy were to increase or decrease significantly, it will lead to forever expansion that too more rapidly or the big crunch respectively.

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What happens when infrared hits a shiny surface
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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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