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ryzh [129]
3 years ago
8

Compare an Earth year to a cosmic year

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0
Compared with an Earth year, a galactic year represents time on a grand scale but its not a consistent measurement across the galaxy
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The concept of electron transitions between energy levels in the atom can be used in the emission of
iogann1982 [59]
I would say light. Hope this helps

8 0
4 years ago
The fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that
kirill115 [55]

Answer:

B. they must be intrinsically far more luminous than the brightest galaxies.

Explanation:

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

6 0
3 years ago
Suppose you had 10 identical molecules enclosed by a box. At a given instant, one molecule has an energy of 100 Joules, and the
boyakko [2]

Answer:

A)   K_average = 1/20 m v², B) low entropy , C) entropy increases.

Explanation:

A) The kinetic energy of each molecule

         K = ½ m v²

The average kinetic energy is the sum of each kinetic energy among the number of them

        K_average = (½ m v² + 0 + 0 +) 10

        K_average = 1/20 m v²

B) Entropy is the sum of the states of each molecule, in this configuration there are only two states one with energy and the other with zero energy, so it is a system with low entropy.

       S = k ln W

  Where S is the entropy, k the Bolztmann constant, W the amount of state present in the system in this case is 2

C) Let's start by analyzing the entropy, as time goes by the molecule that is moving collides with the other molecules and transfers them some energy, so the other molecules move to a different state, after a little In time all the molecules will have the same energy, each one in a different state or volume, so the number of possible state increases to 10, so the entropy increases.

Now let's analyze what happens with the energy of the system, for this case we have two possibilities

- The system is isolated, therefore as it cannot exchange energy with the environment, the total energy remains constant even when the energy of each molecule can fluctuate.

- If the system is not isolated, it can exchange energy with the environment, therefore the total energy changes, depending on the difference in energy between the molecules and the environment

4 0
4 years ago
The force that opposes the start of motion is referred to as _____.
Xelga [282]
The force that opposes the start of motion is referred to as static friction.

Friction is a force that opposes motion in general, and static friction opposes initial movement. This is because, when two objects are stopped near one another, weak bonds begin to form between the atoms on their surfaces. These bonds must be broken in order for one object to move away from the other. This resistance caused by the bonds between motionless objects is static friction.
7 0
3 years ago
A sealed balloon filled with air that has a volume of 6 cubic inches at 99 feet will have a volume of _____ cubic inches at 33 f
SpyIntel [72]
Let's assume that the gas is behaving ideally. Hence, we can use the ideal gas equation: PV = nRT. We only know the value of the volume. Since it was not mentioned, let's assume that the temperature at those altitude do not have a significant difference. So, we can assume temperature, as well as the number of moles is constant, because it is a closed system. The other parameter, pressure, can be calculated in terms of height:

P = ρgh, where ρ is the air density equal to 0.0765 lb/ft³, g is the acceleration due to gravity equal to 32.2 ft/s², and h is the height in feet. Let's solve for the pressure, P₁.

P₁ = ( 0.0765 lb/ft³)(32.2 ft/s²)(99 feet) = 243.8667 lb/ft-s²
For consistency, let's convert V₁ to ft³ using the conversion 12 inches=1 foot.
V₁ = (6 in³)(1 foot/12 inches)³ = 0.003472 ft³

Then, let's use this to find nRT.

P₁V₁ = nRT
nRT = (243.8667 lb/ft-s²)(0.003472 ft³)
nRT = 0.8467 lb-ft²/s²

Now, let's do the same procedure for P₂:
P₂ = ( 0.0765 lb/ft³)(32.2 ft/s²)(33 feet) = 81.2889 lb/ft-s²
Then, we use the ideal gas equation knowing that nRT=0.8467

P₂V₂ = nRT = 0.8467 lb-ft²/s²
81.2889 lb/ft-s²(V₂) = 0.8467 lb-ft²/s²
V₂ = 0.010416 ft³

Finally, let's convert this to in³:

V₂ = 0.010416 ft³ (12 in/1ft)³
V₂ = 18 ft³ 
3 0
3 years ago
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