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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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What is 2m+n =k for M
kotykmax [81]
Answer : M= k-n/2
Subtract n from each side
2m=n-k
Divide out 2 from both sides to get M alone
m= k-n/2
5 0
3 years ago
Consider projectile thrown horizontally at 50 m/s from height of 19.6 meters. The projectile will take ______________ time to hi
aleksley [76]

Answer:

C)The Same

Explanation:

Kinematics equation:

y=v_{oy}*t+1/2*g*t^2

for both cases the initial velocity in the axis Y is the same, equal a zero.

So the relation between the height ant temps is the same for both cases (the horizontal velocity does not play a role)

C)The Same

3 0
3 years ago
Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i
leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are \dfrac{\lambda}{2}

Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)

A = 0.765a

7 0
3 years ago
Can any one answer these two
dalvyx [7]

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0
3 years ago
The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?​
Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}

Put all the values,

h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m

So, it will reach to a height of 2.5 m.

8 0
3 years ago
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