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Free_Kalibri [48]
3 years ago
10

Newton theorized that the apple would experience the accelerating effect of the force of gravity no matter how far from the Eart

h the apple was. He expanded this idea to all objects near the Earth. What was Newton's explanation for why the moon doesn't crash to the Earth due to gravitational force? Do we use this principle for any application today? Give an example, if so.
Physics
1 answer:
Rudiy273 years ago
5 0

Answer:

The moon also falls towards the Earth, but having a tangential speed, the acceleration it has is used in the change of direction of its speed,

Explanation:

Newton explained that all objects attract each other, as the Earth is much greater than the apple, it is this that falls towards the Earth.

The moon also falls towards the Earth, but having a tangential speed, the acceleration it has is used in the change of direction of its speed, in general this acceleration is called centripetal

     a_{c} = g

This principle is used in all circular motion devices, for example the merry-go-round in parks, the radius of the circle is proportional to the speed of the turn and therefore to the centripetal acceleration of the device.

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A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue b
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Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

5 0
3 years ago
At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ &gt; V₀. Find W, the work done by the gas on the
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Answer:

W=\frac{p_0V_0-p_1V_1}{\gamma-1}

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(using pV^{\gamma}=k )

=\frac{p_0V_0-p_1V_1}{\gamma-1}

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