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Sauron [17]
1 year ago
5

Some scientists speculate that our universe is not the only universe that exists, but the only one we can gather evidence about.

This idea is called
Physics
1 answer:
o-na [289]1 year ago
6 0

This idea is called the multiverse theory

<h3>What is the Multiverse theory?</h3>

According to multiverse hypothesis, our universe, which spans tens of billions of light-years and contains hundreds of billions of galaxies and almost innumerable stars, may not be the only one. Instead, there may be a whole distinct world, far away from ours, and then another, and another. In fact, there may be an infinite number of universes, each with its own set of physical laws, stars and galaxies (if they can exist in those worlds), and perhaps even sentient civilizations.

The idea of the multiverse appears in a few branches of physics (and philosophy), but inflation theory provides the best-known example. According to inflation theory, a hypothetical occurrence took place when our universe was really young—less than a second old. According to NASA, the cosmos had a brief period of fast expansion before "inflating" to become several orders of magnitude larger than it had been.

According to Heling Deng, a cosmologist at Arizona State University and an authority on the multiverse hypothesis, the expansion of our universe is assumed to have stopped some 14 billion years ago. However, Deng noted in an email to Live Science, "inflation does not disappear everywhere at the same time." It's possible that if inflation slows down in one area, it picks up in another.

to learn more about the multiverse theory go to - brainly.com/question/12210965

#SPJ4

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If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is
Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

I'm taking the K12 Unit Test: Energy - Part 1 right now

7 0
2 years ago
You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,
telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0
3 years ago
The speed of light in a material is found to be 2.07 x 108 m/s. What is the most likely material from the options below if the s
Paraphin [41]

Answer:

1.3636

Explanation:

Write the expression for the refractive index.

n=c/v

c= speed of light in air

v= speed of light in material

=(3×10^8 m/s)/(2.2×10^8 m/s)

 =1.3636

4 0
3 years ago
Read 2 more answers
What is the calculation for elastic potential?
grin007 [14]
Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
3 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
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