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Water Flow Q = 0.7854*0.25^2 dm^2 * 18 dm/sec = 0.8836 dm^3/sec  time t = V/Q = 4400/0.8836 = 4980 sec 

Without showing all of your work, it' s difficult to know where errors occurred. Here' s how I...

... approach (separation = d₂) A will be instantaneously stationary, so it’ s kinetic energy = 0.  The kinetic energy lost will equal the electrical potential energy...

ΔEPE = 0.0610J - 0.0410J = 0.0200 J, so a) ΔVxy = 0.0200J / 2.10e-3C = 9.52 V b) E = ΔVxy / d = 9.52V / 0.25m = 38.1 V/m assuming the motion was parallel to the field.

43.91Khz                                   .

We know ω = 5.4 = √(k/m) and ω = 2πf = 2π/T so T = 2π/5.4 A = 0.1 Vmax = ωA = 5.4*0.1 = 0.54 We don't know k or m, just their ratio: 5.4² = k/m A. and D.

same, mass has no effect on period. (C)

B = µ₀*I / 2πr ... where µ₀ = 4πe-7 T·m·s/C 6e-5 T = 4πe-7T·m·s/C * 6A / 2πr solves to r = 0.02 m = 2 cm