taterbuglee2003
Answered

the repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?
A. 1.07 x 10^-14 m
B. 1.28 x 10^-38 m
C. 7.19 x 10^-10 m
D. 2.68 x 10^-4 m

Answer :

joseaaronlara

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

                 [tex]F = K\frac{q1q2}{r^{2}}[/tex]

Solve for r

                [tex]r = \sqrt{\frac{kq1q2}{F}}[/tex]

Substitution

                [tex]r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}[/tex]

Simplification

                r = [tex]\sqrt{\frac{2.3 x 10^{-28}}{2}}[/tex]

                r = [tex]\sqrt{1.15 x 10^{-24}}[/tex]

Result

                r = 1.07 x 10⁻¹⁴ m

jimohfawwazajibola25

Answer:

A

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

Other Questions