You're on a roll — For question 28: Step 1: Read the main scale reading. The main scale shows markings in millimetres. The last visible mark on the main scale before the thimble is $5 \text{ mm}$. There is also a half-millimetre mark visible after the $5 \text{ mm}$ mark. Main scale reading = $5.5 \text{ mm}$ Step 2: Read the thimble scale reading. The thimble scale shows divisions from 0 to 50. The line on the thimble scale that aligns with the main scale line is $32$. Thimble scale reading = $32 \times 0.01 \text{ mm} = 0.32 \text{ mm}$ Step 3: Calculate the total reading. Total reading = Main scale reading + Thimble scale reading Total reading = $5.5 \text{ mm} + 0.32 \text{ mm} = 5.82 \text{ mm}$ The correct option is C. $$\boxed{\text{C } 5.82 \text{ mm}}$$ For question 29: The unit of force is the newton (N). • A watt (W) is the unit of power. • A joule (J) is the unit of energy or work. • An ampere (A) is the unit of electric current. The correct option is D. $$\boxed{\text{D newton}}$$ For question 30: Step 1: Identify the given values. Load (weight of the object), $W = 900 \text{ N}$ Height raised, $h = 1 \text{ m}$ Effort applied, $E = 300 \text{ N}$ Length of the inclined plane, $L = 4 \text{ m}$ Step 2: Calculate the work output (useful work). Work output = Load $\times$ Vertical height $$W_{out} = W \times h = 900 \text{ N} \times 1 \text{ m} = 900 \text{ J}$$ Step 3: Calculate the work input (work done by effort). Work input = Effort $\times$ Distance moved along the plane $$W_{in} = E \times L = 300 \text{ N} \times 4 \text{ m} = 1200 \text{ J}$$ Step 4: Calculate the efficiency of the inclined plane. Efficiency ($\eta$) is the ratio of work output to work input, expressed as a percentage. $$\eta = \frac{W_{out}}{W_{in}} \times 100\%$$ $$\eta = \frac{900 \text{ J}}{1200 \text{ J}} \times 100\%$$ $$\eta = 0.75 \times 100\%$$ $$\eta = 75\%$$ The correct option is D. $$\boxed{\text{D } 75\%}$$ Got more? Send 'em.