Calculate the equivalent capacitance of the two capacitors in series (C1 = 2 mu F, C2 = 3 mu F).

Here are the solutions to the physics problems. 1. Calculate the charge on the capacitor shown below. Step 1: Calculate the equivalent capacitance of the two capacitors in series (C_1 = 2 F, C_2 = 3 F). (1)/(C_s) = (1)/(C_1) + (1)/(C_2) (1)/(C_s) = (1)/(2 F) + (1)/(3 F) (1)/(C_s) = (3)/(6 F) + (2)/(6 F) = (5)/(6 F) C_s = (6)/(5) F = 1.2 F Step 2: Calculate the total equivalent capacitance (C_eq) by adding the series capacitance (C_s) to the parallel capacitor (C_3 = 4 F). C_eq = C_s + C_3 C_eq = 1.2 F + 4 F = 5.2 F Convert to Farads: C_eq = 5.2 × 10^-6 F Step 3: Calculate the total charge (Q) using the total equivalent capacitance and the supply voltage (V = 200 V). Q = C_eq × V Q = (5.2 × 10^-6 F) × (200 V) Q = 1040 × 10^-6 C Q = 1.04 × 10^-3 C The total charge on the capacitor combination is 1.04 mC. 2. A motor powered by a 240V Main supply requires a current of 30A to lift a load of mass 3 Tonnes at the rate of 5M/Minute. Calculate the: Given: Voltage (V) = 240 V Current (I) = 30 A Mass (m) = 3 Tonnes = 3 × 1000 kg = 3000 kg Velocity (v) = 5 M/Minute = 5 m60 s = (1)/(12) m/s Acceleration due to gravity (g) = 9.8 m/s^2 a) Power input Step 1: Use the formula for electrical power input. P_in = V × I P_in = 240 V × 30 A P_in = 7200 W The power input is 7200 W. b) Power output Step 1: Calculate the force required to lift the load (weight). F = m × g F = 3000 kg × 9.8 m/s^2 = 29400 N Step 2: Calculate the power output using the force and velocity. P_out = F × v P_out = 29400 N × (1)/(12) m/s P_out = 2450 W The power output is 2450 W. c) Overall efficiency Step 1: Use the formula for efficiency. = P_outP_in × 100\% = 2450 W7200 W × 100\% ≈ 0.34027 × 100\% ≈ 34.03\% The overall efficiency is 34.03\%. 3. a) State 3 factors that affect the rate of evaporation. • Temperature: A higher temperature increases the kinetic energy of liquid molecules, allowing more to escape the surface. • Surface area: A larger surface area exposes more liquid molecules to the air, increasing the rate of evaporation. • Air movement (wind): Wind carries away evaporated vapor molecules, reducing the concentration of vapor above the liquid and thus increasing the evaporation rate. • Humidity: Lower humidity (drier air) allows more water vapor to escape into the atmosphere. b) State two differences and one similarity between evaporation and boiling. Differences: 1. Temperature: Evaporation occurs at all temperatures below the boiling point, while boiling occurs at a specific, constant temperature (the boiling point). 2. Location: Evaporation occurs only at the surface of the liquid, whereas boiling occurs throughout the bulk of the liquid, forming bubbles. Similarity: 1. Both evaporation and boiling are processes of phase change where a liquid turns into a gas (vaporization). c) Calculate the quantity of heat required to change 0.50 kg of ice at -10°C completely into steam at 100°C. Given: Mass (m) = 0.50 kg Specific heat capacity of ice (c_ice) = 2100 J/kg/K Specific heat capacity of water (c_water) = 4200 J/kg/K Specific latent heat of fusion of ice (L_f) = 3.36 × 10^5 J/kg Specific latent heat of vaporization of steam (L_v) = 2.26 × 10^6 J/kg The process involves four stages: Step 1: Heat required to raise the temperature of ice from -10°C to 0°C (Q_1). Q_1 = m × c_ice × T_ice Q_1 = 0.50 kg × 2100 J/kg/K × (0 - (-10)) K Q_1 = 0.50 × 2100 × 10 J = 10500 J Step 2: Heat required to melt the ice at 0°C into water at 0°C (Q_2). Q_2 = m × L_f Q_2 = 0.50 kg × (3.36 × 10^5 J/