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konstantin123 [22]
3 years ago
14

What determines how the different moon phases appear from earth

Physics
2 answers:
pychu [463]3 years ago
7 0

Answer:

The position of the earth and moon.

Explanation:

The position of the earth and moon are the two main factors.

Hope this helps, have great day/night!

tensa zangetsu [6.8K]3 years ago
5 0

The position of the moon in its orbit around the Earth does.

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A flame always point upwards because the flame's gas is hotter than the surrounding air and, like you said, a hot gas is always lighter or less dense than a cold gas.

Explanation:

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Look at the graph pic and answer the question correctly!
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B)

Explanation:

That the time period of which they stop.

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A pendulum of length L is suspended from the ceiling of an elevator. When the elevator is at rest the period of the pendulum is
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Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

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What is the single most important factor that determines why Earth (or almost any planet in the Solar System) experiences season
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A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

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