Two small spheres spaced 35cm apart have equal charge how many excess elecotrons must be present on each sphere if the magnitude of the force of repulsion between them is 2.20x10-21n?

Number of electrons, n = 395.47

Explanation:

It is given that,

Force between two spheres,

Distance between spheres, r = 35 cm = 0.35 m

A force of repulsion is acting on the spheres. It is given by :

Let n is the number of electrons on the spheres. So,

q = n e

n = 395.47

So, the the number of excess electrons on the spheres are 395.47. Hence, this is the required solution.

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer