1.a. Squaring the equation v = at gives: v2 = a2t2 equation 1 The equation s = 1/2(at2) can be rearranged to give: t2 = 2s/a which can be substituted into equation 1 v2 = a2(2s/a) =>v2 = 2as b.i. Funbalanced = Fthrust - Ffriction Funbalanced = 3150 - 450 Funbalanced = 2700N a = Funbalanced/m a = 2700/1000 a = 2.7m/s/s b.ii. To calculate the length of the track use : v2 = u2 + 2as s = (v2 - u2)/2a s = (332- 02)/2x2.7 s = 201.67m c. Use the cosine rule to calculate the resultant velocity,or, draw an accurate scale diagram. a2 = b2 + c2 -2bcCosA a2 = 362 + 122 -2x36x12Cos140 a2 = 1296 + 144 - 864Cos140 a2 = 1296 + 144 - 864x-0.766 a2 = 2101.9 a = 45.8m/s Use the sine rule to calculate angle C. a/sinA = c/sinC sinC = csinA/a sinC = 12sin140o/45.8 sinC = 0.168 angle C = 9.7o Resultant velocity = 45.8m/s with a bearing of 350.3o 2.a.i. Vhorizontal = Vcos36o Vhorizontal = 41.7cos36o Vhorizontal = 33.74m/s a.ii. Vvertical = Vsin36o Vvertical = 41.7sin36o Vvertical = 24.5m/s b. The time for the ball to reach Q(ttotal) can be calculated by summing the time to travel from O to P(t1) and the time to travel from P to Q(t2). ttotal = t1 + t2 When P is reached the vertical component of the velocity is 0m/s. This velocity is the final velocity(v) of the first part of the journey. The initial component of the vertical velocity(u) for this part of the journey is 24.5m/s. v = 0m/s v = u +at1 u = 24.5m/s t1 = (v-u)/a a = -9.8m/s/s t1 = (0-24.5)/-9.8 t1 = ? t1 = 2.5s When falling from P to Q take the initial velocity(u) as 0m/s for this part of the journey. u = 0m/s s = ut2 + 1/2(at22) s = -19.6m s = 1/2(at22) a = -9.8m/s t22 = 2s/a t2 = ? t22 = 2x-19.6/-9.8 t22 = 4 t2 = 2s ttotal = t1 + t2 ttotal = 2.5 + 2 ttotal = 4.5s (As required) c. Shorizontal = Vhorizontalxttotal Shorizontal = 33.74x4.5 Shorizontal = 151.8m/s 3.a. Use the law of conservation of momentum to solve this problem. Pbefore = Pafter Before collision Pbefore = mPuP + mQuQ Pbefore = 0.2x0.5 + 0.3x0 Pbefore = 0.1kgm/s After collision Pafter = mPvP + mQvQ 0.1 = 0.2vP + 0.3x0.4 vP = (0.1 - 0.3x0.4)/0.2 vP = -0.1m/s The negative sign indicates the direction is to the left. b. DPp = Pp(after) - Pp(before) Pp(after) = mPvP Pp(after) = 0.2x-0.1 Pp(after) = -0.02kgm/s Pp(before) = mPuP Pp(before) = 0.2x0.5 Pp(before) = 0.1kgm/s DPp = -0.02-0.1 DPp = -0.12kgm/s c.i. Faveraget = mP(uP - vP) Faverage = mP(uP - vP)/t Faverage = 0.2[(0.5-(-0.1)]/0.06 Faverage = 2N Note: the direction of this force is to the left. c.ii. 4.a. The assumption in this experiment is that the gas in the container is the same as the water temperature. To facilitate this the can must be fully immersed to allow the gas and water temperature to equilibrate. b. P1 = 100kPa T1 = (17+273)K = 290K P2 = ? T2 = (75+273)K = 348K P1/T1 = P2/T2 P2 = P1T2/T1 P2 = 100x348/290 P2 = 120kPa c. P = F/A F = PA F = 120x103x0.001 F = 120N d. The mass and the volume of gas are fixed in this experiment,therefore, the density(mass/volume) must also remain constant. 5.a. A battery emf of 12V will supply 12J of energy to each coulomb of charge passing through the cell. b.i. P = V2/R R = V2/P R = 122/48 R = 3W b.ii. When S1 is closed the circuit can be treated as a simple series circuit with the internal resistor(r) in series with the headlamp resistance(Rh). Rtotal = Rh + r Rtotal = 3 + 0.05 Rtotal = 3.05W I = emf/Rtotal I = 12/3.05 I = 3.93A Vheadlamp = IRh Vheadlamp = 3.93x3 Vheadlamp = 11.8V (as required) c.i. The bulb and the starter motor are in parallel. The resistance of this network Rp is calculated as shown below: 1/Rp = 1/Rbulb + 1/Rmotor 1/Rp = 1/3 + 1/0.12 1/Rp = 8.66... Rp = 0.1154W Rtotal = Rp + r Rtotal = 0.115 + 0.05 Rtotal = 0.165W c.ii. I = emf/Rtotal I = 12/0.165 I = 72.7A 6.a.i. Gain = Voutput/Vinput Gain = -9x30mv/5x2mV Gain = -270mV/10mv Gain = 27 a.ii. Vpeak = 1.414Vrms Vrms = Vpeak/1.414 Vrms = 190.9mV a.iii.Gain = Voutput/Vinput =-Rf/R1 To produce a gain of magnitude 27 choose: Rf = 270kW R1 = 10kW Other values are acceptable as long as the ratio of Rf to R1 is 27. b.i. The bridge is balanced when the digital voltmeter reads 0V. This means the potential at X and Y are equal and there is no potential difference between these points. b.ii. Voutput = (-Rf/R-)(V--V+) (V--V+) = -Voutput(R-/Rf) (V--V+) = -(-0.18)(100x103/1x106) (V--V+) = 0.18(0.1) (V--V+) = 0.018V VY-VX = 0.018V b.iii.At 20oC the potential at Y(VY) falls to a value less than it was when the bridge was balanced at 23oC. As VY is less than VX => VY-VX is negative. => Voutput=(-Rf/R-)(VY-VX) Voutput= positive value. Note: If the resistance of the thermistor varies linearly with temperature in range 20oC to 26oC the output voltage at 20oC will be +0.18V. 7.a.i. All the supply voltage is across the resistor at the instant the switch is closed. At this instant Vsupply = Vresistor. I = 100x10-6 R = 150,000W Vresistor = ? Vresistor = IRresistor Vresistor = 100x10-6x150000 Vresistor = 15V Vsupply = Vresistor = 15V a.ii. The current in the circuit must first be calculated when the voltage across the resistor is 6V. I = ? Vresistor = 6V R = 150,000W I = V/R I = 6/150000 I = 40x10-6A = 40mA From the graph read the time when the current is 40mA. Time = 40s a.iii.Vsupply = Vresistor + Vcapacitor Vcapacitor = Vsupply - Vresistor Vcapacitor = 15 - 6 Vcapacitor = 9V b.i.A Q = CV C = 16mF = 16x10-6 V = 6kV = 6000V (This is the voltage across the capacitor when fully charged) Q = ? Q = 16x10-6 x 6000 Q = 0.096C b.i.B E = 1/2(QV) E = 1/2(0.096X6000) E = 288J b.ii. I = Q/t I = 0.096/2x10-3 I = 48A 8.a.i. A maxima occurs when two waves interfere constructively. This happens when waves are in phase. A minima occurs when two waves interfere destructively. This happens when the two waves are 180o out of phase. a.ii. The minima is produced when the path difference YQ-XQ is equal to 3/2l. YQ-XQ = (5.2-4.0)m = 1.2m YQ-XQ = 3/2l l=(2x1.2)/3 l = 0.8m b.i. The speed of sound is constant in air, therefore increasing the frequency will decrease the wavelength. If the frequency is increased by a factor of 5 the wavelength will decrease by a factor of 5. Wavelength at 1000Hz = 1.2/5 = 0.24m b.ii. If the path difference is fixed at 1.2m this will represent a whole number of wavelengths for certain frequencies and produce maxima, however, for other frequencies this will represent a whole number of half wavelengths and produce minima. For this reason a series of maxima and minima are produced. 9.a. nplastic = sinqair/sinqplastic nplastic = sin15o/sin10o (Note that it is the angle between the ray and the normal that must be used.) nplastic = 1.49 b.i. qcrit = sin-1(1/nglass) qcrit = sin-1(1/1.44) qcrit = sin-1(0.694) qcrit = 44.0o b.ii. Ray Q will be refracted and partially reflected. Ray P will be totally internally reflected because the incident angle of 45o is greater than the critical angle. 10.a.i. a.ii. Individual electrons in the metal atom absorb the energy of a single photon, the energy of which is dependent on the frequency of the photon. The photon energy can be calculated using the equation E = hf. Below a certain frequency the energy absorbed by an electronin the metal, from the photon, is below that required to escape from the metal atom, thus the current in the circuit will be zero. b.i. The threshold frequency will provide electrons with just enough energy to overcome the work function of the metal and eject electrons with no excess kinetic energy. Thus to estimate the work function from the graph the line must be extrapolated until it cuts the x-axis of the graph. X-axis intercept = 6.7x1014Hz This is the threshold frequency. b.ii. The work function is calculated using: Ework function = hfthreshold Ework function = 6.63x10-34x6.6x1014 Ework function = 4.4x10-19J This is closest to the work function of the metal calcium 11.a. Alpha particles have a positive charge. It was therefore concluded that electrostatic repulsion from a like charge, of very large mass, was required to produce a large angle deflection. However, as large angle deflections were very rare, and most alpha particles were undeviated from their original path, it led Rutherford to conclude that most of the atom was empty space with the large positive mass concentrated at the centre of the atom. b.i. The total dose equivalent received is calculated using the equation: H =QD. Hgamma = QgammaDgamma Hgamma = 1x200 Hgamma = 200mSv Hneutrons = QneutronsDneutrons Hneutrons = 3x100 Hneutrons = 300mSv Htotal = Hgamma + Hneutrons Htotal = 200 + 300 Htotal = 500mSv b.ii. Three half value thicknesses are required. 3t1/2 = 3x8 = 24mm
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