Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.

Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.

Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².

Energy stored U = 1/2 * C * V² = 1/2 * 10 × 10^-6 F * (50 V)² = 0.0125 J.

Charge (Q) on a capacitor is given by Q = C * V. Thus, Q = 4μF * 12V = 48μC.

Energy U = 1/2 * C * V² = 1/2 * 5 × 10^-6 F * (12 V)² = 0.36 mJ.

Capacitive reactance (X_C) is given by X_C = 1/(2πfC). If the frequency (f) increases, X_C decreases.

Capacitive reactance (X_C) is given by X_C = 1 / (2πfC). Rearranging gives f = 1 / (2πX_CC). Substituting X_C = 50 ohms and C = 10 x 10^-6 F gives f = 318.31 Hz, approximately 1 kHz.

Charge Q is given by Q = CV. Here, Q = 4 µF * 12 V = 48 µC.

The energy (U) stored in a capacitor is given by U = 1/2 CV², where C is the capacitance and V is the potential difference.

Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 3 x 10^-6 F * (12 V)² = 0.36 mJ.

Energy stored U = 1/2 CV² = 1/2 * 4 x 10^-6 F * (12 V)² = 0.24 mJ.

The potential V across a capacitor is directly proportional to the charge. If the charge is doubled, the potential also doubles.

The voltage across the capacitor decreases exponentially over time when connected to a resistor due to discharge.

The time constant (τ) of an RC circuit is given by τ = RC.

When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.

When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.

Once disconnected from the battery, the charge on the capacitor remains constant as there is no path for charge to flow.

When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.

The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.

The energy stored in a capacitor is proportional to the square of the voltage (U = 1/2 CV²). If the voltage is halved, the energy becomes U/4.

The energy (U) stored in a capacitor is given by U = 1/2 CV^2, where C is the capacitance and V is the voltage.

Energy stored, U = 1/2 * C * V^2 = 1/2 * 10 × 10^-6 * (100)^2 = 0.05 J.

Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 10 × 10^-6 F * (100 V)² = 0.05 J.

Energy stored U = 1/2 * C * V² = 1/2 * 5 × 10^-6 F * (10 V)² = 0.5 mJ.

When connected in parallel, the total charge is conserved. The final voltage across both capacitors is V/2.

When connected, charge redistributes between the two capacitors, resulting in a final voltage of V/2 across each.

The energy stored in a capacitor is given by U = 1/2 CV². If the voltage is halved, the new energy becomes U' = 1/2 C(V/2)² = U/4.

When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.

The height of the water column in a capillary tube is determined by both surface tension and the density of the liquid.