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cubic and linear expansivity are related by this equation. γ=3\alpha.

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

Given Data: L = 10, V = 50, F = 100, I = ? Inductive Reactance [X\(_L\)] = 2\(\pi\)FL → 2 * 3.14 * 100 * 10 X\(_L\) = 6,280 Current[I] = \(\frac{V}{X_L}\) → \(\frac{50}{6280}\) I = 0.0079 ≈ 0.008A

Initial Mass: 512 g, Mass Decayed: \(\frac{7}{8}\) of 512 = 448g, Remaining mass = 512 - 448 = 64g Using, N(t) = No(\(\frac{1}{2}\))\(^{\frac{T}{T1/2}}\) \(\frac{N(t)}{No}\) = (\(\frac{1}{2}\))\(^{\frac{T}{T1/2}}\) \(\frac{64}{512}\) = \(\frac{1}{8}\) = (\(\frac{1}{2}\))\(^{\frac{T}{15}}\) \(\frac{1}{8}\) = (\(\frac{1}{2}\))\(^3\) = (\(\frac{1}{2}\))\(^{\frac{T}{15}}\) equating coefficient 3 = \(\frac{T}{15}\) T = 3 x 15 = 45hours.

Area expansivity(ß) = \(\frac{ A_2 - A_1}{ A_1(\theta_2 - \theta_1})\) Area = \(\frac{\pi d^2}{4}\) \(d_1\) = 50.0cm, \(d_2\) = 50.29cm, \(\theta_1\) = 30ºC, \(\theta_2\) = ? and ß = 2\alpha \(2 \times 1.9 \times 10^{-5}\) = \(\frac{50.29^2 - 50^2}{ 50^2( \theta_2 - 30)}\) \(\theta_2 - 30\) = \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\) \(\theta_2\) = 30 + \(\frac{50.29^2 - 50^2}{ 50^2(2 \times 1.9 \times10^{-5})}\) = 335.3 ^\circC note : \(\frac{\pi}{4}\) cancels out.

Properties of intrinsic semiconductors. They; i. Are tetravalent ii. Have thermally generated charge cariers iii. Have equal number of charge carriers iv Have low electrical conductivity/high resistivity.

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) Ageostrationary satelite: is an artificial satellite whose period of revolution is equal to the period of rotation of the Earth. OR A satellite whose period of revolution is equal to 24 hrs/ 1 day. (b) Types of lasers i. Gas ii. Solid state iii. Semi-conductor iv. Liquid v. Chemical vi. Fibre

(a) S.l unite for the intensity of a blackbody radiation = Wm-² OR Js\(^{-1}\)m-² (b) Two features of the graph i. The shape of the graph is non-symmetric. ii. Peaks of the intensity have lower wavelengths/Peaks decrease with wavelength. iii. The line joining the peaks have a negative gradient.

Vernier calipers commonly used in industry provide a precision to 0.01 mm (10 micrometres), or 0.01cm or one thousandth of an inch. Usually, it is the total length of the main scale. Vernier calipers, generally, have a range of 300 mm 300 millimeters equal 30.0 centimeters (300mm = 30.0cm).

|PQ| = Range = Ux.T = UT Cos\(\theta\) = 250 x 40 Cos 42º = 7431 m (7.4 km)

Magnetic materials Non-magnetic materials They are attracted to/by magnets They are not attracted to/by magnets The domains of the magnetic fields align The domains of the magnetic fields are randomly aligned themselves in an orderly fashion. They can be magnetized They cannot be magnetized

(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