Two identical, small spheres of mass 2.00 g are fastened to the ends of a 0.600 m
long light, flexible, insulating string. the string is suspended by a hook in the ceiling at
its exact centre. the spheres are each given an identical electric charge. they are in
static equilibrium, with an angle of 30.0° between the strings, as shown. calculate the
magnitude of the charge on each sphere.

1.19×10⁻⁷ C

Explanation:

Draw a free body diagram.  Each sphere has three forces acting on it: weight downwards, tension along the string, and electrostatic force horizontally.

Sum of the forces in the y direction:

∑F = ma

T cos 15° − mg = 0

T = mg / cos 15°

Sum of the forces in the x direction:

∑F = ma

T sin 15° − k q² / r² = 0

Substituting and solving for q:

mg tan 15° − k q² / r² = 0

k q² / r² = mg tan 15°

q = r √(mg tan 15° / k)

r, the distance between the spheres, is:

sin 15° = (r/2) / (L/2)

sin 15° = r / L

r = L sin 15°

Given L = 0.600 m, m = 2.00×10⁻³ kg, g = 9.81 m/s², and k = 9×10⁹ Nm²/C²:

q = (0.600 sin 15°) √(2.00×10⁻³ × 9.81 × tan 15° / (9×10⁹))

q = 1.19×10⁻⁷ C

answer: a ball

explanation: because it will bounce

waves moving from a fast medium to a slow medium will experience a reflection as light travels more slowly than red light. total internal reflection occurs when the critical angle is.