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Soloha48 [4]
1 year ago
5

Someone pls review my work

Physics
1 answer:
Darya [45]1 year ago
6 0

In astronomy, Johannes Kepler published his three laws about planetary motion. It is one of the most important things in astronomy. So Kepler gives three laws about planetary motion.

<h3>What is motion?</h3>

Motion is a physical term in physics. If a particle of mass m and affected by some force F then it change its position in many different way. That is the motion of the object. It is a vector quantity.

<h3>What is Kepler's three laws?</h3>

In astronomy, Johannes Kepler published his three laws about planetary motion between 1609 and 1619. This shows about motion describe the orbits of planets around the Sun. So the three motions are shown following,

<u>First law</u>: Every planet in solar system that moves in a elliptical orbits where the sun always in the center of the motion.

<u>Second law</u>: Every planet covers the same amount of distance in a constant time no matter where the orbit of the planet placed. That means the velocity of every planet is not same. It varies along with the orbit. But every time the change of area is constant.

<u>Third law</u>: The orbital period of the planet is proportional with the cube of the semi major axis of the planet. It can be shown mathematically,

p²∝a³

Where we know,

p= The orbital period of the planet.

a= the semi major axis of the planet.

From the discussion we can easily shown that there are three laws of Kepler about planetary motion.

Learn more about motion:

brainly.com/question/453639

#SPJ13

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At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

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3 years ago
A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .
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Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

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The nucleus of any atom requires a strong force to hold it together. This strong force is required because
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Answer:

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Explanation:

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The physical phenomenon that MRI is based on does not depend on ionizing radiation, but on other properties of atoms instead. Wh
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A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i
Taya2010 [7]

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

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