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alexandr402 [8]
2 years ago
6

Newton’s law of gravitation says that gravity is a mutually attractive force. Explain the following observation: A small object

is dropped on Earth and we see it fall toward Earth. However, we do not observe Earth moving toward the object.
Physics
2 answers:
bulgar [2K]2 years ago
5 0

Answer:

Explanation:

The explanation involves another Newton's law: his second law of motion says force is a product of mass and acceleration.

When a small object is dropped on Earth, gravity pulls the object down towards Earth. The same mutually attractive gravity force also pulls Earth towards the object.  But as mass of Earth is much larger than that of the small object, the acceleration of Earth and hence its movement toward the object will be much smaller as to be not noticeable.

Airida [17]2 years ago
4 0

Answer: See below

Explanation:

The Earth attracts the falling object with the same intensity of gravity as the object attracts the Earth, according to Newton's law of gravitation. The displacement of the two bodies, however, is inversely proportional to their respective masses.

Example: The Earth attracts a ball that falls 3 metres from the ground, even though the ball's mass is insignificant in comparison to the Earth's. Similarly, the ball draws the Earth with the same power, but the Earth's mass is enormously more than the ball's. As a result, the Earth collides with a billionth of a millimetre ball (or even less). Restart the Earth's descent on the ball you'll never see again.

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3 years ago
At what point does the external energy enter the system?
Phoenix [80]
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If a car go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/
Taya2010 [7]

Answer:

v = 87.57 m/s

Explanation:

Given,

The initial velocity of the car, u = 0

The final velocity of the car, v = 60 mi/hr

The time period of car, t = 8 s

                                         = 0.00222 hr

The acceleration of the car is given by the formula,

                                       a = (v -u) / t

                                           = 60 / 0.00222

                                            = 27027 mi/hr²

If the car has initial velocity, u = 50 mi/hr

The time period of the car, t = 5.0 s

                                         = 0.00139 hr

Using first equations of motion

                                      <em> v = u + at</em>

                                          = 50 + (0.00139 x 27027)

                                          = 87.57 mi/hr

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3 years ago
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Answer:

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\Delta P = \frac{gH}{2}(\rho_A + \rho_B)

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

F = mg

so we have

\Delta P \pi R^2 = mg

so we have

\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g

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An astronaut goes out for a "space-walk" at a distance above the earth equal to twice the radius of the earth. What is her accel
Drupady [299]

Answer:

Acceleration due to gravity will be a=2.45 m/s^{2}.

Explanation:

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The F is equal to the weight of the astronaut, so we will have:

ma=G\frac{mM_{e}}{R^{2}}

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But the distance between the astronaut and the center of the earth is 2R, then we have:

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Therefore the acceleration due to gravity will be a=2.45 m/s^{2}.

I hope it helps you!

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