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## Section A

1.C11.B2.D12.E3.D13.E4.C14.C5.B15.ESQA say D? 6.B16.D7.A17.A8.A18.D9.E19.B10.E20.C## Section B

21.a. Speed is a scalar with magnitude only. Velocity is a vector with magnitude and velocity. b.i.Distance(1) = Speed x time 10.0km/h x 0.5h 5km Distance(2) = 8.0km/h x 1.5 12km Distance(Total) = Distance(1)+Distance(2) Distance(Total) = 5km + 12kmDistance(Total) = 17kmb.ii.Draw an accurate scale diagram and take careful measurements

or use trigonometry. Use the cosine rule to calculate the resultant displacement. s^{2}=x^{2}+ y^{2}-2xycosX s^{2}=12^{2}+ 5^{2}-2x12x5cos110^{o}s^{2}=169 + 41.04 s = SQRT(210.4)s =14.49kmUse the sine rule to calculate angle X and hence the bearing. s/sinS = x/sinX sinX = (xsinS)/s sinX = (12sin110^{o})/14.49 sinX = (11.28)/14.49 sinX = 0.778 X = 51.1^{o}bearing 321.1b.iii.velocity = displacement/time s = 14.49km v = 14.49/2 t = 2 h^{o}v = 7.24km/hc. Time for Leeuvin to sail directly: t = s/v t = 14.49/7.5 t = 1.932h = 1hour 55.9minutes Time for Leeuvin to reach Y after the Mir passes X 1h 55.9min + 15min = 2h 10.9min This is longer than the 2h journey of the Mir.Leeuvin reaches Y 10.9 minutes after Mir.22.a.i.u = 60m/s v^{2}= u^{2}+ 2as v = 0m/s s = (v^{2}- u^{2})/2a t = 40s s = ? a = (v-u)/t a = (0 - 60)/40 = -1.5m/s/s s = (0^{2}- 60^{2})/2x-1.5 s = -3600/-3s = 1200ma.ii.F_{avg}= m(v-u)/t m = 7.5x10^{5}kg F_{avg}= 7.5x10^{5}(0-60)/40 u = 60m/sFv = 0m/s The force acts in the opposite direction t = 40s to the motion. Kinetic energy equated to the work done by the braking force also produces the above answer. b. I_{avg}= -11.25x10^{5}N_{rms}= 2.5x10^{3}A P_{rms}= I_{rms}x V_{rms}P = 8.5x10^{6}(J/s) or (W) V_{rms}= P_{rms}/I_{rms}V_{rms}= 8.5x10^{6}/2.5x10^{3}V23.a. As the manifold is stationary the upward forces balance the downward forces. Tension = Weight w = mg w = 5.0x10_{rms}= 3400V^{4}kg x 9.8N/kg w = 490000NT = 490000N (upwards)b.i. b.ii.Again as the manifold is stationary the upward forces balance the downward forces. T + B = W 2.5x10^{5}+ B = 490000N B = 490000 - 250000 B = 240000N The buoyancy force is a result of the pressure difference between the upper and lower surfaces. B = (P_{lower}-P_{upper})A DP = B/A DP = 240000N/8.0m^{2}DP = 30000Pa As requiredc. There is no change in the pressure difference. P_{liquid}(top) = rgh P_{liquid}(bottom) = rg(h+Dh) DP = P_{liquid}(bottom) - P_{liquid}(top) DP = rgDh Only the difference in the depth(Dh) of the top and bottom surfaces affects the pressure difference. 24.a.i.r = 2.0W emf = 9.0V (open circuit voltage) V_{tpd}= 7.8V V_{lost}= emf - V_{tpd}V_{lost}= 9.0 - 7.8 V_{lost}= 1.2V V_{lost}= Ir I = V_{lost}/r I = 1.2/2.0 I = 0.6A V_{tpd}= V_{R}V_{R}= IR R = V_{R}/I R = 7.8/0.6R = 13W24.a.ii.When S_{1}is closed there is a current through the internal resistor. The voltage drop across this resistor is "lost" and produces the decreased reading on the voltmeter. b. The 30W resistor in parallel with the original load resistor decreases the effective resistance of the load. More current therefore flows through the internal resistor resulting in more "lost volts". This means thevoltmeter reading decreases. 25.a. When fully charged the current in the circuit falls to zero and the voltage across the capacitor is equal to the supply voltage. 25.bi. I = 20mA R = 400W V_{R}= IR V_{R}= 20x10^{-3}x400 V_{R}= 8V V_{supply}= V_{R}+ V_{C}V_{C}= V_{supply}- V_{R}V_{C}= 12 - 8Vb.ii. E = 1/2(CV_{C}= 4V^{2}) E = 0.5 x 100x10^{-6}x 4^{2}E = 0.0008J (800mJ) c. Reduce the value of the resistor in the circuit to less than 400W. d. The charging time is less. This means the capacitor must have a valueless than 400mF26.a. Differential op-amp V_{out}= (R_{f}/R_{1})(V_{2}-V_{1}) V_{out}= (120/10)(7.52-7.50) V_{out}= 12 x 0.02Vb. As the temperature increases the: Resistance of the thermistor decreases increasing V_{out}= 0.24V_{2}and V_{out}. When V_{out}increases to 0.7V or above the transistor conducts. The electromagnet in the relay gets magnitised and closes the switch to the alarm. c. V_{out}= (R_{f}/R_{1})(V_{2}-V_{1}) 0.72 = 12(V_{2}- 7.50) V_{2}= (0.72/12) + 7.50 V_{2}= 0.06 + 7.50 V_{2}= 7.56V This corresponds to atemperature of 3627.a.i.Angle air = 82^{o}C^{o}Angle liquid = 45^{o}n_{liquid(red)}= sin(air)/sin(liquid) n_{liqiud(red)}= sin82^{o}/sin45^{o}n_{liquid(red)}= 0.990/0.707na.ii.The angle of refraction for blue light is_{liquid(red)}= 1.40greater. sinq_{air}= nsinq_{liquid}q_{air}= sin^{-1}(nsinq_{liquid}) As n_{liquid(blue)}>_{liquid(red)}q_{air}must be greater. Note: q_{air}is the angle of refraction. b. q_{critical}= sin^{-1}(1/n) q_{critical}= sin^{-1}(1/1.44) q_{critical}= sin^{-1}(0.694) q_{critical}= 43.9^{o}This means that light incident at 45^{o}will be totally internally reflected. 28.a.i.A passing photon can encourage or stimulate an electron to fall from a higher energy level in an atom to a lower one. This will happen if the passing photon has the same energy as the energy gap between the two energy levels in the atom. a.ii.Amplification is produced because each photon produced by stimulated emission becomes a new stimulating photon. One becomes two, two becomes four, four becomes eight and so on. b. Grating Equation : dsinq = nl d = ?m n = 1 q = 37/2^{o}=18.5^{o}l = 633nm = 633 x10^{-9}m d = nl/sinq d = 1 x 633 x10^{-9}/sin18.5^{o}d = 1.995x10^{-6}m Lines per metre = 1/d Lines per metre = 1/1.995x10^{-6}Lines per metre = 501271c. Thewavelength has been decreased. Shorter wavelengths are diffracted less than longer ones so the maxima are closer together. 29.a. b. In forward bias the junction conducts if the applied voltage is high enough. Higher energy electrons flowing from the n-type material fall into holes, at a lower energy level, in the p-type material. The electrons give up energy in the form of photons as they do this. c.i. E = 3.68x10^{-19}J E = hf h = 6.63x10^{-34}Js f = E/h f = ? f = 3.68x10^{-19}/6.63x10^{-34}f = 5.55x10^{14}Hz l = v/f l = 3x10^{8}/5.55x10^{14}l = 540nmii. E = 3.68x10^{-19}J E = qV q = 1.60x10^{-19}C V = E/q V = V = 3.68x10^{-19}/1.60x10^{-19}V = 2.3V30.a.i. 92 is the atomic number. The number of protons in the uranium nucleus. a.ii.235 is the mass number. The number of protons plus neutrons in the uranium nucleus. b. The two neutrons produced in the fission reaction can be absorbed by two other uranium nuclei and produce two more fissions. c. Total mass before = (390.173 + 1.675)x10^{-27}= 391.848x10^{-27}kg Total mass after = (232.242 + 155.884 + 2x1.675)x10^{-27}391.476x10^{-27}kg Mass defect(Dm) = (391.848 - 391.476)x10^{-27}= 0.372x10^{-27}kg E = Dmc^{2}E = 0.372x10^{-27}x (3x10^{8})^{2}E = 33.48x10^{-12}JReturn to past paper index page. ## END OF QUESTION PAPER