A - 28.4-μC charge is placed 16.4 cm from a charge q, the force between the two charges is 1240 N. What is the value of q?

1.31×10⁻⁴ C or 131 μC

Explanation:

From the question,

Applying Coulomb's Law,

F = kqq'/r² Equation 1

Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulomb's constant.

make q' the subject of the equation

q' = F×r²/(kq)

q' = Fr²/kq Equation 2

Given: F = 1240 N, r = 16.4 cm = 0.164 m, q = 28.4 μC = 2.84×10⁻⁵ C,

Constant: k = 8.98×10⁹ Nm/C²

Substitute these values into equation 2

q' = (1240×0.164²)/[(2.84×10⁻⁵)×(8.98×10⁹)

q' = (33.35104)/(25.5032×10⁴)

q' = 1.31×10⁻⁴ C

q' = 131 μC

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

no

explanation:

no i dont know how to solve this you should probably try to google the answer