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Dominik [7]
2 years ago
12

In two or more complete sentences, explain how you can prove that the number of degrees that the Moon rotates around the Earth e

ach day is approximately 12.2⁰. Write your response in the essay box below.
Physics
2 answers:
Julli [10]2 years ago
7 0

Answer:

According to this solution, the Moon has two main motions, revolution and rotation. The Moon moves around the world in the movement called a revolution, in which an average of around 13.2 ° per day, or 92 degrees per week, or once in 27.3 days, is the annual movement of the moon around the Earth.

PolarNik [594]2 years ago
3 0

Answer:

Follows are the explanation to this question:

Explanation:

In this solution, it is defined that there are two principal motions for the moon, which are its revolution as well as rotation. In such a movement called revolution, its Moon is relocating around the Earth, in which the approximate movement of the moon from around earth has an average movement of about 13.2° per day, or 92 degrees every week, that's once in 27.3 days.

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Sinisimulan ko<br> Kaya Ko<br> Gagawin ko<br> Nalaman ko<br> Natututo Ako
ozzi

Answer:

what!!!!!!

Explanation:

didn't got the point

7 0
2 years ago
Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
3 years ago
What is an interaction pair?
Alchen [17]
A set of two forces that are in opposite directions, have equal magnitudes and act on different objects
6 0
3 years ago
Which atomic particle gives an atom gives it it’s chemical properties? Question 3 options: A.Isotopes B.Neutrons C.Electrons D.P
bulgar [2K]
C. is your answer! Hope it helps!
7 0
3 years ago
Read 2 more answers
Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a
Ahat [919]

Answer:

r_{min}=\frac{kq^2}{5m_ev^2}

Explanation:

The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:

E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}

me: mass of the electron

q: charge of the electron

k: Coulomb's constant

you take into account that v2=3v1=3v and do rmin the subject of the formula:

\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}

4 0
3 years ago
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