S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
Answer:
(a) the force is 8.876 N
(b) the magnitude of each charge is 4.085 μC
Explanation:
Part (a)
Given;
coulomb's constant, K = 8.99 x 10⁹ N.m²/C²
distance between two charges, r = 10 cm = 0.1 m
force between the two charges, F = 15 N
when the distance between the charges changes to 13 cm (0.13 m)
force between the two charges, F = ?
Apply Coulomb's law;

Part (b)
the magnitude of each charge, if they have equal magnitude

where;
F is the force between the charges
K is Coulomb's constant
Q is the charge
r is the distance between the charges

Answer:

Explanation:
Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(
) and inversely proportional to the square of the distance(d) that separates them.
Replacing the given values, where k is the Coulomb constant:

less mass is more mass but less energy in more mass. less mass has more energy