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 ($C_1 = 2 \mu F$, $C_2 = 3 \mu F$).

$$\frac{1}{C_{s}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_{s}} = \frac{1}{2 \mu F} + \frac{1}{3 \mu F}$$ $$\frac{1}{C_{s}} = \frac{3}{6 \mu F} + \frac{2}{6 \mu F} = \frac{5}{6 \mu F}$$ $$C_{s} = \frac{6}{5} \mu F = 1.2 \mu F$$

Step 2: Calculate the total equivalent capacitance ($C_{eq}$) by adding the series capacitance ($C_s$) to the parallel capacitor ($C_3 = 4 \mu F$).

$$C_{eq} = C_s + C_3$$ $$C_{eq} = 1.2 \mu F + 4 \mu F = 5.2 \mu F$$

Convert to Farads:

$$C_{eq} = 5.2 \times 10^{-6} F$$

Step 3: Calculate the total charge ($Q$) using the total equivalent capacitance and the supply voltage ($V = 200 V$).

$$Q = C_{eq} \times V$$ $$Q = (5.2 \times 10^{-6} F) \times (200 V)$$ $$Q = 1040 \times 10^{-6} C$$ $$Q = 1.04 \times 10^{-3} C$$

The total charge on the capacitor combination is $\boxed{\text{1.04 mC}}$.

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 = $3 \times 1000 \text{ kg} = 3000 \text{ kg}$
Velocity ($v$) = 5 M/Minute = $\frac{5 \text{ m}}{60 \text{ s}} = \frac{1}{12} \text{ m/s}$
Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$

a) Power input

Step 1: Use the formula for electrical power input.

$$P_{in} = V \times I$$ $$P_{in} = 240 \text{ V} \times 30 \text{ A}$$ $$P_{in} = 7200 \text{ W}$$

The power input is $\boxed{\text{7200 W}}$.

b) Power output

Step 1: Calculate the force required to lift the load (weight).

$$F = m \times g$$ $$F = 3000 \text{ kg} \times 9.8 \text{ m/s}^2 = 29400 \text{ N}$$

Step 2: Calculate the power output using the force and velocity.

$$P_{out} = F \times v$$ $$P_{out} = 29400 \text{ N} \times \frac{1}{12} \text{ m/s}$$ $$P_{out} = 2450 \text{ W}$$

The power output is $\boxed{\text{2450 W}}$.

c) Overall efficiency

Step 1: Use the formula for efficiency.

$$\eta = \frac{P_{out}}{P_{in}} \times 100\%$$ $$\eta = \frac{2450 \text{ W}}{7200 \text{ W}} \times 100\%$$ $$\eta \approx 0.34027 \times 100\%$$ $$\eta \approx 34.03\%$$

The overall efficiency is $\boxed{\text{34.03\%}}$.

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 ($L_f$) = $3.36 \times 10^5$ J/kg
Specific latent heat of vaporization of steam ($L_v$) = $2.26 \times 10^6$ J/kg

The process involves four stages:

Step 1: Heat required to raise the temperature of ice from -10°C to 0°C ($Q_1$).

$$Q_1 = m \times c_{ice} \times \Delta T_{ice}$$ $$Q_1 = 0.50 \text{ kg} \times 2100 \text{ J/kg/K} \times (0 - (-10)) \text{ K}$$ $$Q_1 = 0.50 \times 2100 \times 10 \text{ J} = 10500 \text{ J}$$

Step 2: Heat required to melt the ice at 0°C into water at 0°C ($Q_2$).

$$Q_2 = m \times L_f$$ $$$$