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IRINA_888 [86]
2 years ago
10

Place the following list in order of occurrence from oldest to most recent:

Physics
1 answer:
Andre45 [30]2 years ago
4 0

Answer:

Explanation:

2.Big bang, 3.contraction of the solar nebula, 5.stellar ignition in our sun, 4.outgassing of earths secondary atmosphere, 1.appearance of the first ocean on earth ,7.evolution of photosynthesis, 6.build- up of oxygen in earths atmosphere.

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Caves most commonly form in which rock type
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Which equation correctly relates mechanical energy, thermal energy, and total energy when there is friction present in a system?
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((HURRY PLLSS)) According to Einstein, nothing in the universe can move faster than which approximate speed?
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2 years ago
Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)
Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - \frac{G.Me.m}{r} + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - \frac{G.Me.m}{Re}

The potencial energy of the object in Case 1 is

U1 = - \frac{G.Me.m}{2Re}

For the Case 2:

U2 = - \frac{G.Me.m}{3Re}

The potencial change in Case 1:

ΔU1 = - G.Me.m.(\frac{1}{Re}-\frac{1}{2Re}) = - \frac{1}{2}\frac{G.Me.m}{Re}

For Case 2:

ΔU2 = - G.Me.m(\frac{1}{Re}-\frac{1}{3Re}) = - \frac{2}{3}\frac{G.Me.m}{Re}

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ\frac{U2}{U1} = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

5 0
3 years ago
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