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1.C11.E21.C2.C12.D22.E3.B13.A23.A4.D14.D24.B5.B15.D25.E6.B16.C26.A7.E17.A27.A8.C18.E28.B9.C19.E29.B10.C20.C30.E31.Step one: calculate the time the arrow takes to reach the target using the fact that the horizontal velocity is constant. t = s_{hor}/v_{hor}t = 30/100 t = 0.3sStep two: calculate the vertical displacement of the ball in this time. u_{ver}= 0m/s t = 0.3s a = -9.8m/s/s s_{ver}= ? s_{ver}= ut + 1/2(at^{2}) s_{ver}= 0 + 1/2(-9.8x0.3^{2}) s_{ver}= -0.441mStep three: calculate the radius of the target. r = 1.5-0.9 r = 0.6m As the arrow falls less than the radius of the target it hits the target. 32.a.i. a.ii. F_{rope}= Tension(T) As the rope is stationary the forces acting on the buoy are balanced. F_{up}= w + T F_{up}= 50 + 1200Fb. The buoyancy force is not dependent on depth. Therefore, the buoyancy force is as calculated in part a.ii.(1250N). 33._{up}= 1250NIn circuit 1the total resistance in the circuit is equal to the sum of the internal resistor(r) and the resistance of the bulb (R_{bulb}). R_{total}= r + R_{bulb}In circuit 2the total resistance in the circuit is equal to the sum of the internal resistor(r) and the resistance of the two bulbs in parallel. R_{total}= r + R_{bulb}/2 The total resistance in circuit 2 is less than that in circuit 1. Thus the current(I) flowing through the internal resistor in circuit 2 is greater. The "lost" voltage (V_{lost}=Ir) across the internal resistor is therefore greater in circuit 2. This means the voltage across the lamps (V_{lamp}= E -Ir) is less in circuit 2 reducing the brightness of the lamps. 34.a. V_{rms}= V_{pk}/2^{1/2}V_{rms}= 12/1.414Vb. P = I_{rms}= 8.5V_{rms}^{2}I_{rms}= V_{rms}/R I_{rms}= 8.5/4 I_{rms}= 2.12A P = 2.12^{2}x4P = 18W35. dsinq = nl d = 1/2.5x10^{5}= 4x10^{-6}m l = 600x10^{-9}m n = 1 q = ? sinq = nl/d sinq = 1x600x10^{-9}/4x10^{-6}sinq = 0.15q = 8.6336.a. Possible transitions : E^{o}_{3}-> E_{2}E_{3}-> E_{1}E_{3}-> E_{0}E_{2}-> E_{1}E_{2}-> E_{0}E_{1}-> E_{0}b. A total of6lines are in the spectrum. Low frequency radiation will be produced when the transition is between energy levels with a small energy gap.The smallest energy gap is between levels Ec. Transitions between certain states will be more frequent than others. The more frequent a transition, the more intense the spectral line will be. 37. E_{3}and E_{2}._{k}= E_{photon}- E_{work function}E_{photon}= hf = hc/l E_{photon}= (6.63x10^{-34}x3x10^{8})/5.4x10^{-7}E_{photon}= 3.68x10^{-19}J E_{k}= 3.7x10^{-19}- 2.9x10^{-19}E_{k}= 0.8x10^{-19}## END OF QUESTION PAPER