The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.
Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
Answer:
c.) sewage and industrial waste
atoms of lithium, sodium,
and potassium almost never found alone in nature because t<span>hey have only one electron in their outer most
shell which makes them very reactive and they react with other atoms thats why.
I hope this can help you</span>