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snow_tiger [21]
3 years ago
12

By how much does the earth-Sun distance change? 300,000 miles 500,000 miles 3,000,000 miles 3,500 miles

Physics
1 answer:
iogann1982 [59]3 years ago
4 0

Answer:

3,000,000 miles

Explanation:

The Earth's perihelion distance is about 147,000,000 kilometers and its aphelion distance is about 152,000,000 kilometers. So, the change in distance is about 5,000,000 kilometers which translates to about 3,000,000 in Miles.

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Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between
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Sodium chloride, NaCl, is formed when a sodium atom transfers its electron to a chlorine atom. The difference in charge between the two atoms creates a(n) electrostatic attraction that bonds them together.

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A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
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F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
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If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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