Gamma rays then x rays then UVA rays then visible light then IR then radio waves (from highest to lowest frequency).
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)
then the time would be 5x of the time as gravity is acceleration. So 3.9s*5=19.5s
As the force of gravity is less, then the acceleration of masses is also less, therefore it will take more time for the object to fall by the factor of the force of gravity difference
Answer:
Assuming it starts at 72 kmph and hits a dead stop: Divide 72 by 60 for distance per minute. So, 1.2km per minute. 1.2km is 1200m and 4 seconds is one fifteenth of a minute.
Explanation:
Answer:
B . energy cannot be created or destroyed