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Nostrana [21]
3 years ago
5

A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

Physics
2 answers:
Digiron [165]3 years ago
8 0

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

Marina CMI [18]3 years ago
6 0

31.5N

I have confirmed it is the right answer.

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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
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Answer:

0.68 s

Explanation:

We are given that

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\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

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