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No expanded working has been attached for this question yet. The verified answer is Option A.

F = \(\frac{GM_1M_}{r^2}\) = \(\6.67 \times 10^{-11} \times 70 \times 60}{1^2}\) = 2.8014 x 10-7 N

F = qvB = 10 x 10-6 x 1.0 x 105 x 10-4 = 5.0 x 10-4N

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option C.

Quality Dimensions a) Impulse b) Acceleration c) Work MLT\(^{-1}\) LT\(^{-2}\) ML\(^2\)T\(^{-2}\)

The components of velocity at maximum height are: Vy = 0 Vx = UCos \(\theta\) = 20 Cos 40^\circ =15.32ms\(^{-1}\)

F = evB sin \(\theta\) V = \(\frac{F}{evB sin \theta} = \frac{9.6 \times 10^{12}}{1.6 \times 10^{-19} \times 0.12 sin 90^o}\) = \(\frac{9.6 \times 10^{-12}}{1.6 x 10^{-19} \times 0.12 \times 1}\) = 5.0 x 10\(^8 ms{-1}\)

Uses rockets are: — for space travel — for space exploration — for warfare — for launching of artificial satellites into orbits — for entertainment, that is fireworks

(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body. (b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc. (c) V\(_1^2\) = \(U_1^2 + 2gh_1\) = 0 + 2 x 10 x 2 = 40 \(V_1 = \sqrt{40}\) = 6.325m/s \(v_2^2 = u_2^2 - 2gh_2\) O = \(U^2_2\) - 2 x 10 x 1.8 u = \(\sqrt{36}\) = 6ms\(^{-1}\) Impulse Change in momentun mv\(_1\) - (-mu\(_2\)) = m (v\(_1\) + u\(_2\)) = 0.2(6.325 + 6.000) = 2.46 Ns (d) v = \(\omega \sqrt{r^2 - y^2}\) 2\(\pi\)f\(\sqrt{r^2 - y^2}\) 2 = 2 x 3.14 x 5 x \(\sqrt{r^2 - 0.1^2}\) r = 0.12m (ii) Vmax = \(\omega\)r = 2\(\pi\)fr = 2 x 3.142 x 5 x 0.12 = 3,77 m/s (iii) P.E = \(\frac{1}{2}m \omega^2e^2 = \times \frac{1}{2}m(2 \pi f)^2r^2\) = \(\frac{1}{2} \times 0.02(2 \times 3.142 \times 5)^2 \times (0.12)^2\) = 1.42 x 10\(^{-1}J\)

(a)(i) Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them. (ii) Fundamental interval is the difference in temperature between the lower fixed point, i.e. temperature of melting ice at standard atmospheric pressure and that upper fixed point, i.e. temperature of steam at standard atmospheric pressure. (b) Uses of hydraulic press are:- -Compressing metal sheets; -Compressing soft materials into bales; -Compressing paper in printing industries; -Lifting of objects. (c) The material used to reset the steel index in the six's maximum and minimum thermometer is magnet. (d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium. M\(_mC_w\Delta \theta_m\) + C\(_a\Delta\theta_a\) 0.3 x C\(_m\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26) 13.5 C\(_m\) = 29400 + 1260 C\(_m\) = \(\frac{30660}{13.5}\) = 2271.1 Jkg\(^{-1}\)K\(^{-1}\) (ii) (\(\alpha\)) The bottle losses heat through conduction and convection. (\(\beta\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.

(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy. Mass defect = total mass of nucleus – mass of nucleons. (ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J). (b)(i) The effect of decreasing the speed of a photoelectron on: (\(\alpha\)) wavelength --> wavelength increases. (\(\beta\)) energy ---> energy decreases. (ii) \(\lambda = \frac{h}{mv}\) = \(\frac{6.6 \times 10^{-34}}{4.4 \times 10^{-2} \times 10^{-16}}\)m (c) Pairs of isotopes in the figure are; \(^{208}_{82}Pb\) and \(^{212}_{82}Pb\) or \(^{212}_{84}Po\) and \(^{216}_{84}Po\) or \(^{224}_{88}Ra\) and \(^{288}_{88}Ra\) or \(^{288}_{90}Th\) and \(^{232}_{90}Th\) (ii) The series started with isotopes \(^{232}_{90}Th\) (iii) Nuclear equation for; \(\alpha\)Alpha decay \(^{232}_{90}Th\) \(\to\) \(^{228}_{88}Ra\) + \(^{4}_{2}He\) + Energy; \(^{228}_{90}Th\) \(\to\) \(^{22}_{88}Ra\) + \(^{4}_{2}He\) + Energy; \(^{224}_{88}Ra\) \(\to\) \(^{220}_{86}Rn\) + \(^{4}_{2}He\) + Energy; \(^{212}_{84}Po\) \(\to\) \(^{208}_{82}Pb\) + \(^{4}_{2}He\) + Energy; \(^{216}_{84}Po\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy; \(^{212}_{84}Rn\) \(\to\) \(^{208}_{82}Po\) + \(^{4}_{2}He\) + Energy; \(^{220}_{84}Rn\) \(\to\) \(^{216}_{84}Po\) + \(^{4}_{2}He\) + Energy; \(^{216}_{84}Rn\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy; (\(\beta\)) Beta decay \(^{224}_{88}Ra\) \(\to\) \(^{224}_{89}Ac\) + \(^{o}_{-1}e\) + Energy; \(^{228}_{88}Ac\) \(\to\) \(^{228}_{90}Ac\) + \(^{o}_{-1}e\) + Energy; \(^{212}_{82}Pb\) \(\to\) \(^{212}_{83}Bi\) + \(^{o}_{-1}e\) + Energy; \(^{212}_{83}Ra\) \(\to\) \(^{212}_{84}Ac\) + \(^{o}_{-1}e\) + Energy;

S/N A/cm B/cm 1.0 10.0 54.00 2.0 15.0 55.00 3.0 20.0 56.00 4.0 25.0 57.00 5.0 30.0 57.00 6.0 35.0 58.00 Slope (s) = \(\frac{\bigtriangleup {B}}{\bigtriangleup {A}}\) = \(\frac{B_{2} - B_{1}}{A_{2} - A_{1}}\) = \(\frac{59 - 53}{30 - 0} = \frac{6}{30}\) = \(\frac{1}{6}\) = 0.2 Intercept C on the vertical axis C = 53 Evaluate (a) \(k_1\) = [\(\frac{1 - 2S}{S}\)]100 = (\(\frac{1 - 2 \times 0.2}{0.2}\)) 100 = (\(\frac{1 - 0.4}{0.2}\))100 \(k_1\) = \(\frac{0.6}{0.2}\) 100 = 300 \(k_2\) = \(\frac{2C}{S}\) - 160 = \(\frac{2 \times 52}{0.2}\) - 160 = \(\frac{106}{0.2}= \frac{1060}{2}\) = 530 - 160 = 370 Precautions; - Avoided draught - Ensured mass did not touch/rest on the table. - Avoided error due to parallax on the metre rule. - Noted zero error on metre rule. - Repeated readings (shown on the table). (b)i. The moment of a force about a point is defined as the product of the force and the perpendicular distance from the point to the line of action of the force. ii. Conditions of rigid bodies in equilibrium; 1. The resultant of the forces in a given direction must be zero/the algebraic sum of resolved components in a given direction must be zero/the vector sum of all the points must = 0 2. The algebraic sum of the moments about any point is zero. 3. The forces must be concurrent or meet at a point.

S/N E/v I/A V/V \(\frac{1}{V}\)/V\(^{1}\) l\(^{-1}\)/A\(^{-1}\) 1 1.5 0.30 0.80 1.250 3.333 2 3.0 0.50 0.40 0.717 2.000 3 4.5 0.70 2.00 0.500 1.429 4 6.0 0.90 2.50 0.400 1.111 5 7.5 1.10 3.00 0.333 0.909 Slope = \(\frac{\bigtriangleup {V}^{-1}}{\bigtriangleup {I}^{-1}}\) = \(\frac{V_{2}^{-1} - V_{1}^{-1}}{I_{2}^{-1} - I_{1}^{-1}}\) = \(\frac{3.333 - 0.800}{1.24 - 0.26}\) = \(\frac{2.53}{0.98}\) = 2.6 Intercept C = 0.04 Precaution; - Ensured clean terminals/tight connections - Key removed after taking each reading/key removed - when reading was not being taken. - Avoided parallax error on Voltmeter/Ammeter Noted/corrected zero error on voltmeter/ammeter - Repeated readings (shown on the table) (b) Production of electric current e.g. use of - Cells (accumulators/simple cells) fuel cells (chemical energy) - Generators/dynamos/electromagnetic induction (mechanical) - Solar cells/photocells - Windmills - Tidal waves - Geothemal/thermal - Gas turbines - Hydro power - Nuclear power - Thermocouple/thermopile/themionic (heat energy) - Piezo electric generators - Electrostatic generator (e.g Van de graff generator) ii. \(\frac{1}{Re}\) = \(\frac{1}{R1} + \frac{1}{R2}\) = \(\frac{1}{R} + \frac{1}{R} = \frac{2}{R}\) Re = \(\frac{R}{2}\) RT = \(\frac{R}{2}\) + 2 = 4 \(\frac{R}{2}\) = 4 - 2 = 2 R = 4\(\Omega\)

Using the general gas laws: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) P\(_1\) = 800mmHg, V\(_1\) = 76cm\(^3\), T\(_1\) = 273 + 27 = 300k P\(_2\) = 760mmHg, V\(_2\) = ? T\(_2\) = 273 + 0 = 273 (@ STP) \(\frac{800 \times 76}{300} = \frac{760 \times V_2}{273}\) V\(_2 = \frac{800 \times 76 \times 273}{300 \times 760}\) = 72.8cm\(^3\) Therefore, the volume of gas at S.T.P = 72.8 cm\(^3\)