**Answer:**

q = 0.0392 / V, for V= 0.1V q = 0.392 C

**Explanation:**

For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy

starting point

Em₀ = U = q V

final point

Em_f = K = ½ m v²

Em₀ = Em_f

q V = ½ m v²

q =

let's calculate

q = ½ 0.40 10⁻³ 14² / V

q = 0.0392 / V

The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is

V = 0.1 V

q = 0.0392 / 0.1

q = 0.392 C

**Answer:**

I only know one

**Explanation:**

The strength of the gravitational field.

<span> Allied Forces. they became the allies.</span>

**Answer:**

5.23 C

**Explanation:**

The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.

Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁

B₁ = 0.670 T and B₂ = 0 T

ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T

A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m

A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²

ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt

Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m

R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω

So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω

IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C

Since ΔQ = It = 5.232 C ≅ 5.23 C

So the charge is 5.23 C