2.5 miles an hour but yea hope that helps
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;

where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
= (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
= v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)