Two point charges are separated by a distance of 40.0 cm. The numerical value of one charge is twice that of the other. Each charge exerts a force of magnitude 65.0 N on the other. 1) Find the magnitude of the charge with a smaller magnitude. (Express your answer to three significant figures.)

2.4×10⁻⁵ C

Explanation:

From coulomb's law,

F = kqq'/r² Equation 1

Where F = force on each charge, q = magnitude of the first charge, q' = magnitude of the second charge, r = distance between both charges, k = coulomb's constant.

From the question,

q' = 2q

Therefore,

F = 2kq²/r² Equation 2

make q the subject of the equation

q = √(Fr²/2k) Equation 3

Given: F = 65 N, r = 40 cm = 0.4 m.

Constant: k = 9×10⁹ Nm²/C²

Substitute this values into equation 3

q = √[(65×0.4²)/(2×9×10⁹)]

q = √(5.78×10⁻¹⁰)

q = 2.4×10⁻⁵ C

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27