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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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The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point
Masja [62]

Answer:

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

Explanation:

Given a point in rectangular form, that is P(x,y) = (x,y), its polar form is defined by:

P(x,y) = (r,\theta) (1)

Where:

r - Norm, measured in meters.

\theta - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

r = \sqrt{x^{2}+y^{2}} (2)

And direction is calculated by following trigonometric relation:

\theta = \tan^{-1} \frac{y}{x} (3)

If we know that x = -3.50\,m and y = -2.50\,m, then the components of coordinates in polar form is:

r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}

r \approx 4.301\,m

Since x < 0\,m and y < 0\,m, direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)

\theta \approx 215.538^{\circ}

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

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Explanation:

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An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
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Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

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E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

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So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

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