The argon molecules become solid when the temperature decrease from 20C to -190 C.
<h3>What changes occur in argon?</h3>
The changes in arrangement and movement of the particles of the argon occurs as the temperature of the air decreased because the argon gas begins to freeze. The freezing point of argon is -189 Celsius so when the temperature decrease from 20C to -190 C, the argon molecules becomes solid.
So we can conclude that the argon molecules become solid when the temperature decrease from 20C to -190 C.
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Well,
The sun's energy is produced mainly in the core, which has sufficient temperature to initiate nuclear fusion.
Hydrogen --> Deuterium --> Tritium --> Helium --> Beryllium? --> Carbon --> ? --> Silicon --> Iron/Nickel = Most massive stars<span />
Answer:
C)Earth's orbital speed is greater when it is closer to the Sun than when it is farther from the Sun.
Explanation:
We can use Kepler's 2nd law or the law of area to answer this question.
The law states that the rate of area swept out by a planet's orbit is same throughout the orbit. For the farthest point since the distance is large as compared to to the nearest point, the possibility that area swept is large. Hence, to compensate the extra swept area. the orbital speed has to decrease at the largest point.
Hence, planet's speed is greater when it is closer to sun than the speed when it is farther.
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
+ ρ g y₁ = + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
