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1.D11.D21.B2.A12.C22.A3.D13.D23.B4.D14.B24.C5.B15.B25.E6.D16.A26.A7.B17.C27.E8.E18.D28.B9.C19.B29.E10.B20.C30.E31. Calculate the magnitude of the displacement vector (

s) using Pythagoras' theorem and calculate the corresponding bearing using basic trigonometry. s^{2}= AB^{2}+BC^{2}Let angle BAC = q s^{2}= 50^{2}+150^{2}tanq=BC/BA s^{2}= 2,500+22,500 tanq=150/50 s^{2}= 25,000 tanq=3 s= (25000)^{1/2}q=TAN^{-1}3 s= 158.11m q=71.6^{O}S=158.11mat a bearing of071.632. To solve this problem use conservation of momentum. P^{O}_{before}=P_{after}P_{before}= m_{ball}u_{ball}+m_{student}u_{student}P_{before}= 0.42xu_{ball}+50x0 P_{before}= 0.42xu_{ball}P_{after}= (m_{ball}+m_{student})x v P_{after}= (0.42+50)0.10 P_{after}= 50.42x0.1 P_{after}= 5.042kgm/s Equating P_{before}to P_{after}leads to: 0.42xu_{ball}= 5.042 u_{ball}= 5.042/0.42 (m/s)u33. Step 1 :Calculate the effective resistance of the parallel network containing the 6W and 3W resistors. 1/R_{ball}= 12.00m/s_{p}=1/R_{1}+1/R_{2}1/R_{p}=1/6_{1}+1/3_{2}1/R_{p}=3/6 R_{p}=6/3 R_{p}=2W Step 2 :Add the 2W resistor to the value calculated above. R=2W+2W R=4W Step 3 :The 4W resistor is in parallel with the resistance calculated in step 2. The total resistance is therefore given by: 1/R_{total}= 1/4 + 1/4 1/R_{total}= 2/4R34.a. When the wheatstone bridge is balanced: R_{total}= 2W_{1}/R_{2}= R_{v}/R_{t}R_{t}= R_{v}xR_{2}/R_{1}R_{t}= 0.225x2.2/3.3 R_{t}= 0.15kWRb. At 80mV the temperature of the thermistor has increased by 4_{t}= 150W^{o}C. The temperature of the thermistor is therefore24. 35.a. E = hf f=v/l E = hv/l l = hv/E l = 6.63x10^{o}C^{-34}x3x10^{8}/6.9x10^{-19}l = 2.88x10b. As 4.0x10^{-7}m^{-7}m is greater than the maximum wavelength that will cause an electron to be emitted, no electrons will be emitted from the zinc surface. 36.a. When an electron in a high energy level is made to fall to a lower energy level in an atom as a result of a passing photon, with an energy equal to the energy gap between the high energy level and the lower energy level, a photon produced. This is called stimulated emission of radiation where the passing photon is called the stimulating photon. b. The stimulating and the stimulated radiation have the same frequency and wavelength. They are also travelling in the same direction and are in phase. 37. To solve this problem use the grating equationdsinq = nl. d = 1/300 mm = (1/300)x10^{-3}m = 3.33x10^{-6}m q = 24.5^{o}n = 2 l = ? l = dsinq/n l = 3.33x10^{-6}sin24.5^{o}/2l = 6.91x10^{-7}m = 691nmReturn to past paper index page. ## END OF QUESTION PAPER