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Artemon [7]
3 years ago
13

How is it possible for dark matter to bend reality in space?

Physics
1 answer:
LuckyWell [14K]3 years ago
7 0

Answer:

A: Dark matter is called “dark” because it doesn’t give off or interact with light — including through scattering. It is simply the nature of dark matter and why it is so difficult to study. But some models of dark matter state that on rare occasions, dark matter particles could be capable of interacting with normal matter, including by scattering light.

Astronomers know that dark matter is largely situated in spherical halos that enclose galaxies (more on that in a moment). If the dark matter in that halo scatters the galaxy’s starlight, even rarely, it could create a dim glow, like the halo of a light seen in thick fog. Researchers have searched for that glow but so far have not seen it. One possibility is that the glow is difficult to see at optical wavelengths, which is where past studies have focused. Scientists think such a glow, if it exists, might be easier to detect at longer infrared wavelengths, but no studies with the sensitivity needed to see this faint scattered light have been performed yet.  

However, dark matter does have mass and its gravity can influence matter and light. So, dark matter does contribute to a phenomenon called gravitational lensing, in which a galaxy’s mass — including both its normal and dark matter — causes the space-time around it to curve. As light from an object in the background, such as a more distant galaxy, encounters this curved space-time, it appears to bend, which distorts and can even multiply the image of the background object. Astronomers do observe this effect, and by comparing the amount of gravity necessary to do the bending with the amount of visible matter, they have used it to confirm that galaxies are enshrouded in massive halos of dark matter.

Explanation:

TODO IS HERE BUT THAT QUESTION IS SH_T

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(Assuming acceleration is -2.5m/s2) 7.2 seconds.
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4 years ago
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Wildlife biologists are attempting to monitor the population size of the insect pest known as the gypsy moth (Lymantria dispar)
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Answer: Option A: The number of trees sampled.

The accuracy can be understood as how close is the measured value to the true value.   The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.

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4 years ago
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Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i
spayn [35]

Answer:

v = 15.56 m/s

v = 56 km/h

Explanation:

When coefficient of friction is approximately zero then we have

F_ncos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

tan\theta = \frac{v^2}{Rg}

here we know that

v = 40 km/h = 11.11 m/s

R = 30 m

tan\theta = \frac{11.11^2}{30\times 9.81}

\theta = 22.75 degree

now when friction coefficient is 0.30 then we have

F_n cos\theta = mg + F_f sin\theta

F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}

now we have

v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}

v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}

v = 15.56 m/s

v = 56 km/h

3 0
3 years ago
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion
stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0
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At what condition does a body becomes weightless at the equator?
ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0
3 years ago
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