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The uses of ferromagnetic materials are: (i) Electromagnets (ii) Magnetic Storage (iii) Magnetic Sensors (iv) Magnetic Shielding (v) Permanent Magnets (vi) Inductive Components, others are transformers, electric bells, electric generators, etc.

(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas (b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force. (c) SEE THE DIAGRAM ABOVE. (ii) Taking a moment about the support: clockwise moment = anti-clockwise moment 400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above 4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i) 400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii) divide eqn i by ii \(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\) \(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\) 3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15) \(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75) \(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45 L = \(\frac{2L}{2}\) = 45 cm From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15) 2400 = m x (\(\frac{45}{2}\) - 15) 2400 = m x 7.5 m = \(\frac{2400}{7.5}\) = 320 g Therefore, I. m = 320g and II. L = 45 cm (d) Reasons for less pressure at the top of a mountain: The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level. (e)(i) Density = \(\rho\) = \(\frac{m}{v}\) m = v x \(\rho\), but v = A x h m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg (e)(ii) pressure exerted by the surface of the mercury P\(_t\) = P\(_{atm}\) + pressure of mercury P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa

(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\) (ii) F = Qv\(\beta\)sin\(\theta\) (iii) for a charged particle moving parallel to the magnetic field, \(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\) F = Qv\(\beta\)sin 0 = 0 Thus, F = 0 (b)The main parts of an electrical transformer are the coil, Soft iron core (ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law. c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path. (ii) I. radius of curved path qv\(\beta\) = \(\frac{mv^2}{r}\) r = \(\frac{mv}{q\beta}\) r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m. II. time taken by the proton to move through the magnetic field, t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s

(v) M(g) W\(_A\)(N) W\(_W\)(N) U\(_1\) = W\(_A\) - W\(_W\). W\(_L\)(N) U\(_2\) = W\(_A\) - W\(_L\) 20.0 0.20 0.16 0.04 0.18 0.02 40.0 0.40 0.33 0.07 0.36 0.04 60.0 0.60 0.49 0.11 0.54 0.06 80.0 0.80 0.66 0.14 0.72 0.08 100.0 1.00 0.82 0.18 0.90 0.10 (vi) SEE THE GRAPH ABOVE (vii) s = \(\frac{\Delta U_1}{\Delta U_1}\) = \(\frac{0.162 - 0.2}{0.09 - 0.067}\) = \(\frac{0.042}{0.023}\) = 1.826 (viii) K = \(\frac{1}{\text{s}}\) = \(\frac{1}{1.826}\) = 0.5476 (ix)(I) I avoided errors due to parallax when reading the spring balance. (II) I cleared the mass thoroughly before immersing it in liquid L b(i) Archimedes's principle states that when a body is completely or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of fluid displaced. (ii) Density has a unit whereas relative density has none. (II) The value of density is different in different systems of measurement, whereas relative density is the same in all systems of measurement.

(vii) L(cm) L\(_x\)(cm) L\(_y\)(cm) R\(_1\) = (\(\frac{L_y}{L_x})R_x\)\(\Omega\) L\(_{x2}\) L\(_{y2}\) R\(_2\) = \(\frac{L_{y2}}{L_{x2}}R_x\)\(\Omega\) 100.0 5.2 94.8 18.2 21.5 78.5 3.6 90.0 6.0 94.0 15.7 22.5 77.5 3.4 80.0 6.8 93.2 13.7 23.4 76.6 3.3 70.0 7.6 92.4 12.2 24.5 75.5 3.1 60.0 8.4 91.6 10.9 25.6 74.4 2.9 (viii) SEE THE GRAPH ABOVE (ix) Slope = \(\frac{\Delta R_2}{\Delta R_1}\) = \(\frac{3.5 - 3.3}{17.0 - 14.6}\) = \(\frac{0.2}{2.4}\) = 0.083 (x) K = \(\sqrt{0.083}\) = 0.2881 (xi) I Tight connection/clean terminal was ensured II. Key removed when readings were not being taken. b(i) I. A potentiometer can be used for measuring the internal resistance of a cell and also to compare the e.m.f. of two cells. II. A potentiometer gives the exact value of potential difference across any two points in a circuit. (ii) It refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat.

Dry Leclanché: This is a common non-rechargeable cell.

(a)i c = 50.0cm (ii) M\(_R\) = 90.0 g (vii) d = 0.8m (ix) S/N h(cm) t(s) T= \(\frac{t}{20}\)s T\(^2\)s\(^2\) 1 40.0 26.12 1.306 1.7056 2 50.0 27.80 1.390 1.9321 3 60.0 30.68 1.534 2.3532 4 70.0 34.60 1.730 2.9929 5 80.0 35.80 1.790 3.2041 (x) SEE THE GRAPH ABOVE (xi) Slope, s = \(\frac{\DeltaT^2}{\Delta h}\) = \(\frac{3.0 - 1.75}{75 - 46}\) = \(\frac{1.25}{29}\) = 0.431s\(^2\)cm\(^{-1}\) (xii) K = \(\frac{\text{S}}{\text{Q}}\) where, Q = \(\frac{2}{25d^2}\) = \(\frac{2}{25 \times (0.8)^2}\) = 0.125 K = \(\frac{\text{S}}{\text{Q}}\) = \(\frac{0.431}{0.125}\) = 0.3448 b(i) A couple, as related to oscillatory motion, refers to a pair of forces that are equal in magnitude but opposite in direction, and act on opposite sides of a pivot or fulcrum, causing rotation or torsion. b(ii) I. Wrenches: When using a wrench to tighten or loosen a bolt, the force applied creates a couple that produces rotational motion. II. Steering Wheels: Turning the steering wheel of a car applies a couple that rotates the front wheels, allowing for directional change. III. Door Handles: When pushing or pulling on one side of a door handle, a couple is created that rotates the door about its hinges. IV. Bicycle Pedals: When pedaling a bicycle, the force applied to the pedals generates a couple that turns the crankshaft, propelling the bike forward.

The patient cannot see clearly an object closer than 50 cm Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens u=25cm ;v=-50cm (virtual image); p= \(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\) ⇒\(\frac{1}{f}\) =\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = f = 50cm =0.5m ⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\) p = \(\frac{1}{f}\) p = \(\frac{1}{0.5}\) p = 2 diopters ; The patient needs a converging lens with a power of 2 diopters

Mechanics is the branch of physics that deals with the motion of objects and the forces acting on them. It is the foundation of all other branches of physics, and it is used to study everything from the motion of planets to the motion of electrons.

m=1200kg, \(V_1\)= \(10ms^{-1}\) \(V_2\) = \(15ms^{-1}\), w= ? work=►K.E = \( K.E_2\) = \(K.E_2\) - \(K.E_1\) ⇒work= \(\frac{1}{2}{mv^2_2}-\frac{1}{2}{mv^2_1}\) ⇒work= \(\frac{1}{2}m({v^2_2}-{v^2_1}\)) ⇒work= \(\frac{1}{2}\times 1200\times (15^2-10^2)\) ⇒work = 600 \times (225 -100) ⇒work= 600 \times 125 ⇒work= 7.5\times\(10^4 j\)

The work done against the gravitational force is calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately \(9.8 ms^2\) on Earth), and h is the height. Substituting the given values, we get W = 3.0 kg * \(9.8 m/s^2\) * 240 m = 7.2 kJ. Therefore, the correct answer is '7.2 kJ'.

ε= 80%, L=200\times 10 = 2000N, \(d_L\)=10m \(d_E\)= 110-10 = 100m, E=? ε = \(\frac{work done on the load}{work done by the effort}\) \times 100% ⇒ work done on load = 2000 \times10 =20,000 j ⇒ work done by effort = E \times 100 = 100E ⇒ 80 = \(\frac{20,000}{100E}\) \times 100% ⇒ 80 = \(\frac{20,000}{E}\) ⇒ E = \(\frac{20,000}{80}\) ⇒ E = 250N

We first find the lorry's initial velocity a = \(4ms^-2\), s= 250m, t= 10, u=? ⇒ ut + \(\frac{1}{2}\times4\times(10^2) ⇒250= 10u + 200 ⇒ 250-200 = 10u ⇒ 50 = 10u ⇒U =\frac{50}{10}\) = \(5ms^-1\) For the next 10s, we set t=20s = s = (5 \times 20) + \(\frac{1}{2}\times4\times{20^2}\) = s = 100 + 800 ⇒ 900 m ;Distance covered:900-250 =650m

W = mg = 220 x 9.8 = 2156 N ⇒Sin 38º = \(\frac{2156}{T_1}\) ⇒ \(T_1\) = \(\frac{2156}{Sin 38}\) ⇒ \(T_1\) = 3502 N Cos 38º = \(\frac{T_2}{T_1}\) ⇒ \(T_2\) = 3502 x Cos 38º ⇒ \(T_2\) = 2760 N ; \(T_1\) = 3502 N, \(T_2\) = 2760 N.

Fx= 12cos32^\circ-16cos26^\circ = -4.20N (right is taken +ve and left is taken -ve) Fy=12sin32^\circ+16sin26^\circ-21 = -7.63N (up is taken +ve and down is taken -ve) \(R^2\)= 4.202+7.632 ⇒\(R^2\) = 75.86 ⇒ R=\sqrt75.86 ∴R = 8.71 N tan \theta=\(\frac{opp}{adj}\) ⇒tan \theta=\(\frac{7.63}{4.20}\) ⇒tan \theta =1.8095 ⇒ \theta = \(tan^-1\) (1.8095) \theta = 61^\circ ∴The angle of the resultant force with the x-axis is 61^\circ