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3 years ago

The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?

Physics

2 answers:

Aloiza [94]3 years ago

6 0

Answer:

The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

$F = K\frac{q1q2}{r^{2}}$

Solve for r

$r = \sqrt{\frac{kq1q2}{F}}$

Substitution

$r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}$

Simplification

r = $\sqrt{\frac{2.3 x 10^{-28}}{2}}$

r = $\sqrt{1.15 x 10^{-24}}$

Result

r = 1.07 x 10⁻¹⁴ m

Serhud [2]3 years ago

5 0

Answer:

Explanation:

Using F = kq1q2/r^2

r^2 = kq1q2/F

r = square root of (kq1q2/F)

r = square root of(8.99 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19/2)

r = square root of(1.15072 x 10^-28)

r = 1. 07 x 10^-14m

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