Answer:
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Explanation:
Energy conservation law: In isolated system the amount of total energy remains constant.
The types of energy are
- Kinetic energy.
- Potential energy.
Kinetic energy 
Potential energy =
Here, q₁= +5.00×10⁻⁴C
q₂=-3.00×10⁻⁴C
d= distance = 4.00 m
V = velocity = 800 m/s
Total energy(E) =Kinetic energy+Potential energy
+ 

=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E
+ 




m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Answer:
I do belive that it is B hrs cn I an gn
Some elements that have a symbol entirely different from the spelling are..
1.Fe. Iron
2.Na.Sodium
3.K. Pottasium
4.Ag.Silver
5.Sn.Tin
6.Sb. Antimony
7,Pb.Lead