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
The rubber band has elastic potential energy
Answer:
1) force = 8.4 N
2) acceleration = 1 m/s^-2
Explanation:
I've attached the working
10 m/s2 for a distance of 65.2 ft
Answer:
radius of curvature is 1.2 m
Explanation:
given data
distance u = 30 cm
to find out
radius of curvature
solution
we know radius of curvature is double of focal length
so we find focal length
lens formula for focal length
1/f = 1/u + 1/v ..............................1
so here from magnification
M = - u /v
2 = - u /v )
v = -2u = -2(30)
v =- 60 cm
so from equation 1
1/f = 1/u + 1/v
1/f = 1/30 + 1/-60
f = 60.02 cm
so R = 2f
R = 2( 60.02)
R = 120.04 cm
so radius of curvature is 1.2 m