All of those are true, except the one about the radius.
Both ends of a chord have to be on the circle, but one end
of a radius is at the center, so a radius can't be a chord.
Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
Answer:
The second system must be set in motion
seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion
seconds later