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horsena [70]
3 years ago
5

What are the effects of moon rotation and revolution​

Physics
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

The effects of the Moon's rotation includes;

1) The Moon rotation and revolution gives the appearance of a perfectly still Moon to observers of the Moon from the Earth

2) The Moon has two sides, the near side that continuously faces the Earth and the "back" or far side, which is also known as the dark side of the Moon

The effect of the Moon's revolution

1) The tides in the ocean and water bodies, due to the gravitational forces from the Moon

2) The changes in the observed shape of the Moon due to the amount of Sunlight that is able to be reflected from the Moon as a result of the relative position of the Moon, the Earth and the Sun, at a given point in time

3) Lunar and Solar eclipse, when the Earth and the Moon blocks the light coming from the Sun respectively, due to their combined revolution

Explanation:

The duration of the Moon's orbit round the Earth = 27.322 days

The time it takes the Moon to rotate round its axis = 27 days

The Moon is the closest cosmic body to the Earth.

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Rearrange the formula for mechanical energy to solve for height:
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Explanation:

Given formula:

            ME=\frac{1}{2}mv²+mgh

  To make height the subject of the formula, follow the following procedures;

     Subtract \frac{1}{2}mv² from both side of equation

 M.E - \frac{1}{2}mv² = \frac{1}{2}mv² - \frac{1}{2}mv²+mgh

                  This gives:

                        M.E - \frac{1}{2}mv² = mgh

Multiply both sides of the expression by \frac{1}{mg}

  ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg} =  \frac{1}{mg} x mgh

       h = ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg}

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

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3 years ago
What is the hardest fruit to open??
Airida [17]

Answer:

pizza

Explanation:

3 0
2 years ago
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A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

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Answer:

Two estimates

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There are mainly two estimates used in the calculation of depreciation such as the useful life and the salvage value of an asset. The salvage value is defined as the predicted amount that will be obtained by a company from an asset when it is disposed at the end of the useful life of the particular asset. On the other hand, the useful life commonly refers to the estimation of how long the asset is useful for the company. This is different from the lifespan of the asset.  

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3 years ago
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Which two statements correctly describe the role of a semiconductor in a
KIM [24]

Answer:

It's A and C

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3 years ago
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