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Komok [63]
3 years ago
11

1. Define dark matter and dark energy, and clearly distinguish between them. What types of observations have led scientists to p

ropose the existence of each of these unseen influences?
Physics
1 answer:
Vlada [557]3 years ago
7 0

Explanation:

Dark Energy is a property of the entire universe. And the universe is roughly made up of 68% of this dark energy whereas dark matter makes up 27% of the universe matter.

   The influence of the dark matter shows in the individual galaxies whereas dark energy acts only on the entire universe. dark matter is responsible for the gravity effect on the plants and the heavenly bodies which produces gravity force. But the dark energy is known as the anti gravity force as it is responsible for the expansion of the universe. Dark matter attracts and produces an attractive force while dark energy repels and produces a repulsive force.

          Scientist believe that they exists to account for the very fact that gravity holds all the heavenly bodies and the galaxies together. They are even responsible for the fluctuation in Cosmic Microwave background. Researchers and scientist firmly believe that all the visible matter that exists may not have enough gravitational pull to hold everything together in the galaxies. And also the expanding universe strongly proves that dark energy exists in the universe.

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A neutral atom of an element has the same number of __________ and ________.
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The answer is protons and neutrons (C)
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Anyone know how to do this?
Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

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How to do this question?​
Zigmanuir [339]

Answer:

First.

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3 years ago
Constructive interference of two coherent waves will occur if the path difference is:___.
Olegator [25]

Constructive interference of two coherent waves will occur if the path difference is λ/2.

<h3>Constructive interference:</h3>

When two waves are in phase and their maxima add, a process known as constructive interference occurs where the combined amplitude of the two waves equals the sum of their individual amplitudes.

The resultant wave is created by adding the amplitudes of two waves that are in phase and traveling in the same direction. The waves in this instance are said to have experienced beneficial interference. The upward displacement of the medium is higher than the displacement of the two interfering pulses because upward displacement occurs when the waves experience constructive interference. When the phase difference between the waves is an even multiple of (180°), constructive interference happens.

Learn more about  constructive interference here:

brainly.com/question/17329186

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8 0
2 years ago
) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe
pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0
3 years ago
Read 2 more answers
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