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a i. A pivot, often referred to as a fulcrum, is the point or axis around which a simple machine, such as a lever or a seesaw, rotates. ii. A load is the object or weight that is being moved, lifted, or supported by the machine. iii. Efficiency refers to how effectively a machine can perform its intended task while minimizing energy loss b. i. Given: F = 25.2N, m = 1200kg, u = 0m/s , v = 60km/h = 16.67m/s Workdone = \DeltaK.E = \(K.E_2 - K. E_1\) Workdone = \(\frac{1}{2}K.E_2 - \frac{1}{2}K.E_1\) Workdone = \(\frac{1}{2}m(v^2 - u^2)\) Workdonk = \(\frac{1}{2}\times 1200 \times 16.67^2\) ( since u = 0) Workdone = 600 x 277.89 = 166733.34J Therefore, Workdone = \(1.67 \times 10^5\)J ii. F = ma → a = \(\frac{F}{m} = \frac{25.2}{1.2 \times10^3} = 0.021ms^2\) recall, \(v^2 = u^2 +2aS\) S = \(\frac{v^2 - u^2}{ 2a} = \frac{16.67^2 - 0^2}{ 2\times 0.021}\) S = 6616.4m = 6.62km. c. Absolute zero temperature is the lowest possible temperature that theoretically represents the complete absence of thermal energy while the ice point is the temperature at which water coexists with ice in thermal equilibrium. - Absolute zero temperature is defined as 0 Kelvin (0 K) or approximately -273.15^\circC while ice point is typically defined as 0^\circC at sea level d. \(\frac{0 - ( - 6)}{ 2 - L} = \frac{ 100 - 0 }{ 15 - 2}\) \(\frac{ 6}{ 2 - L} = \frac{100 }{ 13}\) 6 x 13 = 100( 2 - L ) 78 = 200 - 100L 100L = 200 - 78 100L = 122 L = \(\frac{122}{100}\) = 1.22cm

ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. ii. Process of Charging: Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object. Contact: Requires direct physical contact with a charged object to transfer charges. - Nature of Final Charge: Induction: Results in a charge of opposite polarity to the charged object initially brought close. Contact: Results in a charge of the same polarity as the charged object with which contact is made. b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents. - Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan. - Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings. - Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire. c. \(Z^2 = R^2 + X{_L}^2\) \(Z^2 = R^2 + (2\pif_L)^2 \(Z^2 = R^2 + 4\pi^2f^2L^2\) \(Z^2 = 4\pi^2f^2L^2 + R^2\) ∴ \(Z = \sqrt{4\pi^2L^2f^2 + R^2}\) ii. from the equation \(Z^2 = 4\pi^2f^2L^2 + R^2\) using y = mx + c = y = \(Z^2\), x = \(f^2\) m = \(4\pi^2L^2\) = slope of the grape m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\) \(4\pi^2L^2 = \frac{ 500}{ 660}\) \(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4\pi^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(\pi^2\) = 10) \(L^2 = 0.0189\) ∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH

ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated. ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube. iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube. bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy. ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target. iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment. ci. \(^{235}U_{92} + ^1_0n → ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\) ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy. iii. E = ∆\(mc^2\) where ∆m is the mass defect. Total mass on the LHS: 235.044 u + 1.009 u = 236.053 u Total mass on the RHS: 91.898 u + 140.914 u + 2(1.009 u) = 234.83 u Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u ∆m = 1.223 \times 1.66 \times \(10^{-27}\) = 2.03 \times \(10^{-27}\) kg So, E = 2.03 \times \(10^{-27}\) \times (3 \times \(10^8)^2\) = 1.827 \times \(10^{-10}\) J

It is a graph that represents the force that acts on a body in a certain amount of time. Force on the object is represented by the y-axis and time taken by the object is represented by the y axis. The area under the curve indicates the impulse, the change in momentum that the object undergoes.

The kinetic-molecular theory of gases assumes that ideal gas molecules: (1) are constantly moving; (2) have negligible volume; (3) have negligible intermolecular forces; (4) undergo perfectly elastic collisions; and (5) have an average kinetic energy proportional to the ideal gas's absolute temperature.

If a body moves with uniform retardation, the velocity-time graph is a straight line.

Dimensions of the angular momentum, K = \(\frac{[K]^{2}}{[m_{1}][r]^{2}} = \frac{[G][m_{1}][m_{2}]}{[r]}\) \(\frac{[k]^{2}}{ML^{2}} = \frac{M^{-1} L^{3} T^{-2}M^{2}}{L}\) [K] = (M\(^{2}\)L\(^4\)T\(^{-2}\))½ \(\therefore\) [k] = ML\(^{2}T^{-1}\) (NOTE: the dimension of E = dimension of gravitational force.)

Calculation of workdone by the force to scratch the string W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\) W = \(\frac{50^{2}}{2 \times 100}\) W = 12.5 J OR e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m w = ½ = ke\(^{2}\) w = ½ x 100 x 0.5\(^{2}\) W = 12.5 J

(a) (i) The Law of Inertia states that: A body continues in its state of rest or uniform linear motion unless acted upon by an external force to act otherwise. (ii) Explanation of wearing safety: The safety belt acts as an external force that prevents a passenger from an otherwise forward motion after impact. (b)(i)Mmoment of a force: The product of a force and the perpendicular distance from a reference point to the line of action of the force. (b)(ii) Force diagram: (ii) Determination of the tension, T, in the string; Taking moment about the knife edge Sum of clockwise moment = sum of anti-clockwise moment. T x 1.5 = 54 x 0.5 \(\therefore\) T = 18 N (ii) Determination of the force, F, acting the knife edge; At equilibrium, the sum of upward forces = the sum of forces acting downward T + F = 54 T = 18 N 18 + F = 54 F = 36 N Solid friction Viscosity It occurs on solid surfaces It occurs within fluids It is independent of areas of surfaces on It depends on the area of the surfaces in It depends on the speed of motion of the surface in contact It is independent of the speed of motion of the surfaces in contact It is independent of temperature It depends on the temperature It depends on the normal reaction. It is independent of normal reaction (d)Increasing the number of pulleys

(a)(i) Coulomb's law of electrostatics The electrostatic force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. (ii) Calculation of the magnitude of the electrostatic force F = \(\frac{kq_{1}q_{2}}{d^{2}}\) F = \(\frac{9.0 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(5.2 \times 10^{-11})^{2}}\) = 8.52 x 10\(^{-8}\) N (b)(I) Proof of V\(_{out}\) Let the current through the circuit be I V\(_{in}\) = I(\(R_1\) + \(R_2\)) OR I = \(\frac{V_{in}}{R_{1} + R_{2}}\) V\(_{out}\) = I\(R_2\) II. Calculation of \(R_1\); V\(_{out}\) = V\(_{in}\)(\(\frac{R_{1}}{R_{1} + R_{2}}\)) OR \(\frac{V_{out}}{V_{in}} = \frac{R_{1}}{R_{1} + R_{2}}\) \(\frac{1}{2.5} = \frac{R_{1}}{R_{1} + R_{30}}\) \(R_1\) = 20\(\Omega\) (ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points, (c) Why wood is not suitable for use as the Core of transformers Wood is a non-magnetic material and would not provide magnetic coupling between the coils for electromagnetic induction. (d) Applications of cathode ray tube i. For studying alternating current waveform ii. For measuring frequency/voltage/ reactance/power factor and resistance of an electronic device iii. Used to monitor signals/check FM/AM characteristics

(a) Conditions necessary for: Constructive interference of waves i. lt must involve at least two waves. ii. Waves must vibrate in phase. iii. The distance between the sources of the wave must be of the order of the wavelength. Total internal reflection i. The waves must travel from a denser medium to a less dense medium. ii. The angle of incidence (in the denser medium) must be greater than the critical angle. iii. It must involve two media of different optical densities. Production of beats 1. The amplitude of the two interfering waves should be the same. ii. The difference between the frequencies of interfering waves should be small. iii. The two sources of waves must be sounded together. (b) Calculation of velocity of sound in air V = f\(\lambda\) V = 2f(\(l_2\) - \(l_1\)) = 2 x 256(100 - 35) = 33280 cm s\(^{-1}\) = 332.80 m s\(^{-1}\) (c)(i) Classification of musical instruments i. Wind ii. Stringed iii. Percussion (ii) Examples of classified instruments saxophone I. Wind: flute, trumpet, pipe, organ, iii. Stringed: piano, guitar, violin iii. Percussion: piano, drum, bell, gong. xylophone (d) Calculation of critical angle \(\eta\) = \(\frac{1}{\text {Sin C}}\) 1.5 = \(\frac{1}{\text {Sin C}}\) Sin C = 0.667 C = 41.80º (e) Calculation of temperature \(\frac{V_{1}}{\sqrt T_{1}} = \frac{V_{2}}{\sqrt T_{2}}\) \(\frac{240} {\sqrt 375} = \frac {310} {\sqrt T_{2}}\) 240 \(\sqrt T_{2}\) = 310 x 19.36 \(\sqrt T_{2}\) = 625 K = 352 ºC

(a) Factors affecting the rate i. Surface area of the liquid (exposed surface area) ii. Temperature of the liquid iii. Wind/ drought. iv. Humidity/ amount of water vapour v. Nature of the liquid vi. Air pressure (b) Explanation of latent heat; The energy needed to change the state of a substance without a change in temperature. (c)(i) Water in a clay pot cooler than in a rubber container; A clay pot has pores which allows evaporation to occur while a plastic container has no pores. (ii) Pressure cooker cooks faster than ordinary pot; Pressure cooker raises the pressure within it, which raises the temperature to/above the boiling point thus cooks faster. (d)(i) Calculation of latent heat of fusion of the body Ivt = mL mL = 12 x 40 x 1400 = 6.72 x 10\^{5}\)j (ii) Calculation of specific heat capacity of the body C = \(\frac{Ivt}{m \bigtriangleup \theta}\) = \(\frac{12 \times 40 \times 72}{1.5 \times 60}\) = 384 jkg\(^{-1}\) K\(^{-1}\) (e) difference between evaporation and boiling Evaporation Boiling It takes place at the surface of the liquid It occurs throughout the entire volume of the liquid It takes place across all temperatures It takes place at specific temperatures It produces a cooling effect It produces no cooling effect It is wind aided It is not wind aided

(a)() Meaning of artificial radioactivity; The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy. Emission Nature Charge Ionizing Beta (\(\beta\)) High speed electron Negative Moderately ionizing Gamma (\(\gamma\)) Electro-magnetic radiation Neutral Negligible ionizing ability Alpha particles Helium nucleus Positive Highly ionizing (b) Calculation of kinetic energy K.E = eV = 1.6 x 10\(^{-19}\) x 30 x 10\(^{3}\) = 4.8 \times 10\(^{-15}\)J (c)(i) graph of K.E versus f i. Both axes correctly distinguished ii. Reasonable scale on both axes (ii) (I) Planck's constant. h = slope of the graph Slope i. \(\bigtriangleup\)K.E and \(\bigtriangleup\)f correctly determined ii. \(\frac{\bigtriangleup K.E}{\bigtriangleup f}\) corectly determined and recorded within range. (II) Threshold frequency, \(f_0\) . The intercept on the horizontal or frequency axis. (III) Threshold frequency, \(W_o\) Negative intercept on K.E axis.

(a) OBSERVATIONS (i) Value of \(M_0\) correctly determined and recorded to at least 1 d.p in grammes (ii) Value of m correctly determined and recorded to at least 1 d.p in grammes (iii) Value of d correctly read and recorded to at least 1 d.p in cm (iv) Five values of M correctly recorded in grammes (v) Five values of Z = AK correctly read and recorded to at least1 d.p in cm and in trend. Trend: As M increases Z decreases. (vi) Five values of y = (Z - 5) correctly evaluated and recorded. (vii) Five values of x = (d - Z) correctly evaluated and recorded (vii) Five values of v = \(\frac{x}{y}\) correctly evaluated and recorded to at least 3 d.p (ix) Composite table showing at least M, Z,y, x and v. (b)(i) Condition for a body to be in stable - The algebraic sum of all external forces on the object is zero OR - The algebraic sum of all external torques from the external forces acting on the object is zero. OR - The sum of clockwise moments about a point in the object is equal to the sum of anticlockwise moment about the same point. OR - The algebraic sum of all moments about a point on the object is zero. (ii) If the distance is y, 100y = 50(6 - y) 100y = 300 - 50y 150y = 300 y = 2m

(a) OBSERVATIONS (i) Value of D correctly measure and recorded to at least 2 d.p in cm (ii) Value of A = 0.25\(\pi\)D\(^{2}\) correctly evaluated (iii) Value of \(y_0\) correctly read and recorded to at least 1 d.p in cm. (iv) Five values of m correctly recorded in grammes. (v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend. (vi) Five values of h = (y - yo) correctly evaluated. (vii) Five values of log h correctly evaluated to at least 3 d.p (viii) Five values of log m correctly evaluated to 3d. (ix) Composite table showing at least m, y, h, log h and log m. (b)(i) The Law of Flotation; A floating body displaces its own weight of fluid in which it floats. (ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6 Let y be the depth of immersion \(\therefore\) Weight of water displaced = (0.1y) x 1.26 x 1000 x 10 1260y = 453.6 y=453.6=9.36 = 1260y = 453.6 y = \(\frac{453.6}{1260}\) = 9.36