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shepuryov [24]
3 years ago
9

The earth and the moon exert forces on each other which forces is greater? explain

Physics
1 answer:
Helen [10]3 years ago
8 0

Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.

An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

acentr=(4 pi2 r)/T2

Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.

This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.


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Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

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Where:

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F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

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A) We want to find the work function of the potassium. Apply this equation:

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B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

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c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

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