Using conservation of energy, PE = KE: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 12 m) = 15.49 m/s

Potential Energy (PE) = m * g * h = 3 kg * 9.8 m/s^2 * 4 m = 117.6 J (approximately 120 J)

Kinetic Energy (KE) = 1/2 * m * v^2 = 1/2 * 3 kg * (2 m/s)^2 = 6 J

Momentum (p) = mv = 3 kg * 6 m/s = 18 kg m/s

Using F = ma, a = F/m = 12 N / 3 kg = 4 m/s².

If the object is moving with constant velocity, the net force acting on it is 0 N according to Newton's first law.

Momentum p = mv = 5 kg * 10 m/s = 50 kg m/s.

Momentum = mass × velocity = 5 kg × 4 m/s = 20 kg m/s.

Using F = ma, a = F/m = 18 N / 6 kg = 3 m/s².

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Using F = ma, a = F/m = 15 N / 5 kg = 3 m/s².

Net force = Applied force - Frictional force = 20 N - 5 N = 15 N. Acceleration = Net force / Mass = 15 N / 5 kg = 3 m/s².

When the plates of a disconnected capacitor are moved apart, the capacitance decreases, leading to an increase in potential difference.

Capacitance C is given by C = Q/V. Here, C = 24μC / 12V = 2μF.

The capacitance C is defined as C = Q/V.

Capacitance (C) = Charge (Q) / Voltage (V) = 6 µC / 12 V = 0.5 µF.

When the distance is doubled, the potential difference across the capacitor also doubles, resulting in 24V.

Once disconnected, the charge remains constant. Doubling the area increases capacitance but does not change the potential difference since the charge is fixed.

When the distance between the plates is doubled, the capacitance decreases, which causes the voltage to double since Q remains constant.

If the charge is removed, the potential difference across the capacitor becomes 0 volts.

The charge Q stored in a capacitor is given by Q = CV.

Work Done = Change in Kinetic Energy = 1/2 * m * (v^2 - u^2) = 1/2 * 1000 kg * (20 m/s)^2 = 200,000 J

Using the formula: acceleration = (final velocity - initial velocity) / time. Acceleration = (0 - 15) / 5 = -3 m/s².

Force F = qE = (1 × 10^-6 C)(1000 N/C) = 0.001 N.

Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.

Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.

Work done W = F * d = (E * q) * d = (300 N/C * 5 × 10^-6 C) * 0.2 m = 0.3 J.

The force on a negative charge is in the direction opposite to the electric field, hence towards the positive charge.

First, find the equivalent resistance R_eq = 1/(1/4 + 1/6) = 2.4Ω. Then, I = V/R_eq = 12V/2.4Ω = 5 A.

Using Ohm's law, I = V/R = 15V/5Ω = 3A.

The voltage drop across the resistor is equal to the battery voltage in a simple circuit: 9V.