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2 years ago

What are the effects of moon rotation and revolution

Physics

1 answer:

polet [3.4K]2 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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if the frequency of a sound wave in air remains constant, its energy can be varied by changing its what

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Amplitude, is the answer to the question

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2 years ago

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A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin

Goshia [24]

Answer:

(A) 0.63 J

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + m $h^{2}$

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = $(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}$ )^{2}[/tex]

I = $(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}$ )^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5I $ω^{2}$

= 0.5 x 0.07875 x $4^{2}$ = 0.63 J 0.15 m

(B) from the conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

Ki + Ui = Kf + Uf

at the maximum height velocity = 0 therefore final kinetic energy = 0

Ki + Ui = Uf

Ki = Uf - Ui

Ki = mg(H-h)

where (H-h) = rise in the center of mass

0.63 = 0.42 x 9.8 x (H-h)

(H-h) = 0.15 m

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3 years ago

Assuming the bar has no weight where does the fulcrum (the top point of the tringle) need to be positioned for the two sides to

Inessa05 [86]

Fulcrum need to be positioned balanced with weight on both the sides following law of lever.

What is the physical law of the lever?

It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.

Learn more about the Fulcrum with the help of the given link:

brainly.com/question/16422662

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1 year ago

A force of 6600 N is exerted on a piston that has an area of 0.010 m2

sveticcg [70]

Answer:

Choice A: approximately $0.015\; \rm m^2$ , assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

$\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .

By Pascal's Principle, because the first piston exerted a pressure of $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ . In other words:

$P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .

To achieve a force of $9.900 \times 10^3\; \rm N$ , the surface area of the second piston should be:

$\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}$ .

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3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

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3 years ago

What are the effects of moon rotation and revolution​

What are the effects of moon rotation and revolution