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yawa3891 [41]
3 years ago
13

two electrons (each with the charge of 1.6 x 10^ -19 c) are seperated by a distance of 5.4 x 10^ -5 c what is the electric force

pushing them apart?
Physics
1 answer:
Elan Coil [88]3 years ago
6 0

The electric force between two charges is:

F = (9 x 10⁹) Q₁ Q₂ / D²

F is the force, in Newtons

Q₁ and Q₂ are the two charges, in Coulombs

D is the distance between them, in meters

I'm going to assume that the first little 'c' in the question stands for "Coulombs", and the second little 'c' stands for "centimeters".  And now, I'll proceed to answer the question that I've just invented.

For these two electrons:

F = (9 x 10⁹) (1.6 x 10⁻¹⁹) (1.6 x 10⁻¹⁹) / (5.4 x 10⁻⁷)²

F = (9 x 1.6 x 1.6 x 10⁻²⁹) / (5.4² x 10⁻¹⁴)

F = (23.04 / 29.16) x 10⁻¹⁵

<em>F = 7.9 x 10⁻¹⁶ of a Newton</em>

<em></em>

EACH electron feels that same force, pushing it away from the other electron.

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Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles
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3 years ago
Giving brainiest to correct answer.
mixas84 [53]

Answer:

5.33\ m/s

Explanation:

We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s

Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s

5 0
3 years ago
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