(1) (1) (1) (1) (1) Use of energy = power time Energy = J. Example of calculation E = E = J

Transcription

1 (a) Volt is a Joule coulomb - or V = J C - or V = W/Q (not rearranged) Amp is a Coulomb sec - or A = C s - or I = Q/t (not rearranged) Show units/symbols cancelling and equating to a watt. This mark can only be given if both the other marks scored. Method must be clear, do not allow let t =. (b)(i) Use of energy = power time Energy = J E = E = J (b)(ii) QWC spelling of technical terms must be correct and the answer must be organised in a logical sequence See internal resistance / r Current will be less Less energy/power is lost in internal resistance wasted energy/power is reduced reduced lost volts it is more efficient Total for question 8

2 2 (a) Add to the diagram to show the water flow at A 2 and B 2. Laminar at A 2 minimum 2 lines, approximately straight and parallel, lines mustn t cross Turbulent at B 2 indicated by lines crossing / change in direction > 90 / chaotic lines 2 (b) Name and describe the type of water flow at A 2 and at B 2. A - laminar flow / streamline flow no abrupt change in (direction or speed of) flow/ flows in straight lines / velocity at any point constant / no mixing of layers [no eddies is not sufficient; smooth is not sufficient; no disruption of lines not sufficient] B - turbulent flow mixing of layers / eddies / sudden change in (direction or speed of) flow / velocity at a point not constant [NB - All independent marks] Total for question 6 2 4

3 3(a) (i) Show that the power available to the turbine is about 40 kw. Use of density = m/v Use of gpe = mgh Correct answer [ W] [no ue] 3 volume in s = 0.3 m 3 mass = density x V = 000 kg m -3 x 0.3 m 3 = 30 kg gpe lost = mgh = 30 kg x 9.8 N kg - x 30 m = J in one second, so power = W 3(a) (ii) 3(b) (i) [000 kg m -3 x 0.3 m 3 x 9.8 N kg - x 30 m = W gets 3 marks] Suggest a reason for output only 6 kw friction e.g. in turbine, in fluid / flow rate lower / heat due to friction [accept (electrical) resistance in turbine] Calculate maximum output of solar system for 6 hours Use of energy = power x time Correct answer [26 MJ] 2 3(b) (ii) Energy = power x time = W x 6 x 60 x 60 s = 2.6 x 0 8 J [ J, 26 MJ, kj] Discuss suitability of output of diesel generators Renewables = 46 kw [accept 40 kw], vs diesel 60 kw Backup must be enough to replace whole of renewable amount / diesel power greater than or approximately equal to renewable Total for question 8 2

4 4(a) 4(b) Diode / LED (Any type of recognised diode scores the mark but if diode is included in a list of other components the mark cannot be gained.) Infinite / infinity / Or Very high Or very large 4(c) 4(d) Use of R=V/I Correct value of R for their current in range 0.40 A to 0.43 A (Any valid pair of values for first mark. Use of tangent or gradient scores no marks) R= 0.70 V / 0.4 A R=.7 Ω Any One from, e.g. To protect components / circuits Rectification Restricts current / flow (of charge) to one direction AC to DC Produce DC supply Power indicator light Light source, e.g. Christmas tree light, torch Regulate voltage 2 (Accept any reasonable practical use for diode or LED) Total for question 5

5 5(a) (sound waves travel as) longitudinal waves Or (Air molecules) vibrate parallel to direction of travel of wave (sound waves travel as) a series of compressions and rarefactions Or (sound waves travel as) areas of high and low pressure The idea that these vibrations create a pressure/force/stress/strain on the film Or The idea that these compressions/rarefactions create a pressure/force/stress/strain on the film 5(b) 5(c) 5(d) This pressure/force/stress/strain generates a potential difference (accept idea that vibration/pressure/force/stress/strain causes redistribution of charge within crystal) Thin film is flexible / lightweight The idea that there is not much energy in sound Large area gathers more sound (energy) Or Large area generates more power/current/pd Use of P=E/t with any time, energy in J or kj Conversion of kj J and correct time in s (36000 s) P =0.56 W (accept J s ) Power = J / s Power = 0.56 W ONE Disadvantage Expensive, Not washable Only works with (loud) noise Long time to charge a phone Low output power ONE Advantage Free source of energy Lower/zero running cost Portable Can be used when away from mains electricity [Credit should be given for any reasonable correct physics point but not for generalised comments such as good for the planet environmentally friendly ] Total for question 2

6 6 Indication that 500 W is 5% of incident radiation / Apply 5% efficiency to incident flux of 20 W m -2 (i.e. find useful input) Use of radiation flux is power per unit area = 6 m 2 3 [2.38 m 2 is the answer without applying 5% mark] Input power = (500 00)/5 = 3300 W Area = input power/radiation flux = 3300 W / 20 W m -2 = 6 m 2 (5/00) x 20 W m -2 = 3.5 W m -2 Area = 500 W / 3.5 W m -2 = 6 m 2 Total for question 3 7(a) 7(b) 7(c) Use of W = QV Energy of electron = (J) Energy = J Energy of electron = J Use of energy = power time Energy = (J) Energy = Energy = J 2 ev is very much smaller than Joule kw h is very much bigger than Joule in these units, answers more easily obtained from information available answers can be found without conversions 2 2 Max 2 Total for question 6

7 8(a) (b) Current (through a conductor) is (directly) proportional to the potential difference/voltage (across it) providing the temperature of conductor remains constant external conditions remain constant. Ohmic conductor; fixed resitor horizontal straight line Filament lamp; graph showing increasing resistance (straight line or curve) from a non zero resistance start (conditional on 2nd mark) (c) Filament lamps work at high temperatures as temp of lamp increases as lamp heats up. Resistance of conductor changes the ions vibrate more. Total for question 7

8 9(a)(i) e.m.f./total resistance (accept E/(R+r) (ii) = D4*A4 or p.d. = current load resistance (B4*A4)/(A4+C4) or p.d. = e.m.f. (internal resistance x current) F4/D4 or p.d.=power/current (iii) 9.2 (W) R and r in series (b) (c)(i) R and r in series Potential divider r constant so as R increases p.d. also increases V = E - Ir E and r constant Identifies that the term Ir decreases E = I(R+r) E and r constant As R increases I decreases (power) increases and then decreases Maximum power when load equals 0.8 Ω (ii) Similar pattern of power increasing to a max and decreasing Maximum power when load resistance is.6 Ω Total for question 0 0(a) A coulomb is an Amp sec or As Do not credit current time (b) I = 0 ma I 2 = 5 ma I 3 = 30 ma 3 Total for question 4