CCEA GCSE Physics - Hodder Education



WORKBOOK ANSWERSCCEA GCSE PhysicsUnit 1Unit 2This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the Workbook. They are not exhaustive and other answers may be acceptable – they are intended as a guide to give teachers and students feedback.Exam-style question answersThe answers given for exam-style questions of an explanatory, or evaluative, type set out what is called ‘indicative content guidance’. Just showing the examiner you are familiar with some or most of the content will not suffice. You need to demonstrate that you understand it and are willing and able to use it in a way that directly addresses the question. The indicative content shown for each question is not exhaustive. Questions may be approached in a number of different ways. The choice of approach is yours. Just make sure your approach answers the question.Assessing whether or not a question has been answered is where ‘levels marking’ comes in. For questions with maximum marks of 6, 12 or 15, three levels of attainment are recognised. Where the maximum mark is 20, four levels are defined. The table below gives an idea of the descriptors that are used to determine the quality of an answer and the mark to be awarded.LevelDescriptorMark (depending on maximum)1Demonstrates isolated elements of knowledge and understanding.Question not addressed.1–2 (max. 6)1–4 (max. 12)1–5 (max. 15)2Demonstrates knowledge and understanding — not always relevant or accurate.Shows some awareness of the question.3–4 (max. 6)5–8 (max. 12)6–10 (max. 15)3Demonstrates knowledge and understanding that is mostly relevant and accurate.Addresses the question directly.5–6 (max. 6)9–12 (max. 12)11–15 (max. 15)Unit 1MotionThere are often several ways to answer these questions. Students doing GCSE Further Maths will be aware of Newton’s equations of motion, which can make the work quicker. The answers here use only the equations required by your specification.1aVectors have magnitude and direction, scalars have magnitude only.bVelocity and acceleration are vectors (V). Speed, distance and rate of change of speed are scalars (S).2aDistance = 150 + 105 + 150 + 105 = 510 mbAverage speed = cAverage velocity = Kevin’s displacement is zero, so his average velocity is 0.3aAverage speed = Average speed = Now , so v = 4?m/sbcThe acceleration down the ramp is uniform; OR friction is constant.dThe reaction time when using a stopwatch is so long that it is unsatisfactory when measuring short times. It would be better to use a light gate and electronic timer.4abc A sloped deck means that the aircraft is already rising when it leaves the carrier, so it gains height quicker. It can also mean that the aircraft experiences higher resistance on landing, so it comes to rest quicker.Note that questions requiring you to ‘suggest’ a reason mean that some creative thinking is required.5Indicative content:Hypothesis: average speed increases with height of runway, but in a non-linear way (1)Apparatus: runway, trolley (or ball bearing etc.), stopwatch (or timer with light gate), ruler (or tape), pencil (2)Description to include (3):A runway is raised to create a rampVertical height of the ramp is measuredHorizontal pencil lines at start and end of runwayMeasure the distance between the linesAllow the trolley etc. to move from upper line from restTime from the start to the ball (or trolley wheel) crossing finishing lineCarry out repetitions for reliabilityCarry out repetitions for various ramp heightsCalculation: average speed = (1)To confirm the hypothesis, plot a graph of average speed against runway height. A rising graph confirms the hypothesis; a decreasing graph refutes it. (1)QWC: to obtain this mark your spelling, punctuation and grammar must be accurate and you present your work logically using appropriate scientific terms. (1)Tip: Always give as much detail as you can in QWC questions.Exam-style questions1aiAverage speed = iiAverage speed = initial speed = 2 × 12 = 24?cm/siiiiv24?cm/sbiAverage speed = iiAverage velocity = 2aThe graph should have:the axes labelled exactly as in the tablea scale chosen to use as much of the grid as possible, consistent with ease of usethe five points plotted accurately (to within a small square)a ruled line of best fit passing through the pointsbThe initial speed is the intercept on the vertical axis, 0.3?m/scGradient = So (say) gradient = 3aFinal speed v = u + at = 0 + (10 × 2) = 20?m/sAverage speed = ?(u + v) = ?(0 + 20) = 10?m/sDistance = average speed × time = 10?m/s × 2?s = 20?mbOver the three seconds of the motion:Final speed v = u + at = 0 + (10 × 3) = 30?m/sAverage speed = ?(u + v) = ?(0 + 30) = 15?m/sDistance = average speed × time = 15?m/s × 3?s = 45?mDistance travelled in third second = distance travelled in 3?s ? distance travelled in 2?s= 45?m ? 20?m = 25?mMotion graphs1aGnasher is at rest (not moving)bGnasher is moving at a steady speed between A and BSpeed = cStraight line from (0?s, 20?m) to (100?s, 120?m)dThe rabbits are at same position where the graphs cross: at 20?s, 40?s and 100?s.2aDistance = area under graph up to 20?s = (15 × 3) + ?(3 + 1.4) × 5 = 56?mbAcceleration = gradient = cTotal distance travelled = 56?m (up to 20?s) + 14?m (from 20?s to 30?s)Average speed = Exam-style questions1aRate of change of speed = gradient = bHorizontal line from (0?s, 15?m/s) to (60?s, 15?m/s)cDistance = area under the graphFirst bus travels ?(60 + 45) × 20 = 1050?mSecond bus travels 15 × 60 = 900?mdNo. The buses may not be travelling in same direction; nor may they have started from the same position2a10?s (between t = 80?s and t = 90?s only)bBetween t = 70?s and t = 80?s (the graph has a negative slope and is the steepest)cRetardation = change in velocitytime taken = -12.4 m/s10 s = ?1.24 m/s2dIts velocity is always positiveeDistance = area under the graph up to 50?s = m = 556 mNewton’s first and second laws1aEvery object remains at rest or moves with a constant velocity unless compelled to do otherwise by a forcebi0?Nii650?N2aFriction (or drag)bResultant force = 55 ? 15 = 40?NcdMass of bicycle = combined mass ? mass of cyclist = 80?kg ? 60?kg = 20?kg3Resultant force = ma = 0.125?kg × 25?m/s2 = 3.125?NUpward force = weight of rocket + resultant force = 1.25?N + 3.125?N = 4.375?N4aFinal speed after 4?s = 2 × average speed = 2 × 16?m/s = 32?m/sa = Resultant force = ma = 60?kg × 8?m/s2 = 480?NbDrag force = weight ? resultant force = 600?N ? 480?N = 120?NExam-style questions1aa = bIncrease in speed = a × t = 30 m/s2 × 130?s = 3900?m/scAs the burn continues, fuel is used up, so the mass being accelerated decreases2There are two ways to solve this problemFirst: The additional 500?N causes an increase of 1?m/s2 in the accelerationUsing F = ma gives:500 = m × 1 and m = 500?kgThen using Fengine ? friction = ma gives:700 ? friction = 500 × 1, which leads to friction = 200?NSecond: This requires solving simultaneous equations. Using F = ma twice gives:1200 ? friction = 2m700 ? friction = 1mSubtracting gives:500 = m (the same as in the first way above) So as above m = 500?kg and friction = 200?NMass, weight and Hooke’s law1Mass is the amount of matter in an object; weight is the force due to the pull of gravity on an object2EarthMoonMarsValue of g / m/s2101.63.8Weight of object / N50819Mass of object / kg5553aThe extension of a helical spring is directly proportional to the applied force, provided the limit of proportionality is not exceededbF = ke (or F = kx)4a6?N stretches the spring 9?cm2?N stretches the spring 3?cmSo, the total length with 2?N = 9?cm + 3?cm = 12?cmbcThe spring has been stretched beyond its elastic limit (or proportionality limit) and is now permanently deformed5Hypothesis: up to a certain limit the extension is directly proportional to the stretching forceApparatus: iron stand, boss head, clamp, helical spring, mass carrier, selection of 100?g masses, rulerDependent variable: the extension of the springIndependent variable; the stretching force applied to the springWhat to do and measure:Attach the spring and mass carrier to the clamp, boss and iron standMeasure the original length of the spring using the rulerAttach a 100?g mass (1?N force) to the carrier and measure the new length of the springRepeat for additional 100?g masses up to about 600?gConfirming the hypothesis:Subtract the original length of the spring from each of the spring length measurements to work out the extension for each loadPlot a graph of load against extensionIf the line of best fit is straight (up to a limit) and goes through the (0,?0) origin, the hypothesis is confirmed; otherwise it is rejectedExam-style questions 1aThe graph should be labelled as in the table, with a linear scale on axes covering at least half of the paper available. The graph curves after (19?cm, 12?N).b13?cm; the intercept on the total length axiscd12?N (approximately)eA straight-line graph passing through (9?cm, 0?N) and (17?cm, 12?N)Pressure1Pascal (Pa), N/cm2, N/mm223Weight = m × g = 180 × 10 = 1800?NArea = 0.6?m × 0.5?m = 0.3?m2Pressure 4aiThe raft has a large area, so the total weight of the house exerts only a small pressure on the ground; this means the house is less likely to sink into the soft earthiiThe pin’s point has a very small area so even a small force (from the thumb) allows it to exert a huge pressure on the notice boardExam-style questions1aA = 0.30?m × 0.25?m = 0.075?m2bF = P × A = 70?000 × 0.075 = 5250?NcForce from outside = 5250?N ? 3750?N = 1500?NPressure = 20?kPa2aW = mg = 0.5 × 10 = 5?NbF = P × A = 100?000 × 0.005 = 500?NcThe upward force (500?N) is greater than the force due to the cup and water on the card plus the weight of card itself (6?N); so the card is likely to remain in placeMoments180?N?cm2aThe weight on the pan plus weight of pan itself must be 2?N (because the weights are same distance from pivot)So, the weights in pan must add to 1.8?N; and there must be nine weights (each of 0.2?N)bThe clockwise moment from 3?N is now greater than the anticlockwise moment from the pan’s weightsSo the clockwise moment must be increased; so move the pivot to the leftcSuppose the distance moved by the pivot to the left is d. Then, applying the principle of moments:3 × (50 ? d) = 2 × (50 + d)150 ? 3d = 100 + 2dSo 5d = 50; and d = 10?cm3aW acts vertically downwards from the centre of the beam (100?cm from each end)F acts vertically upwards through the pivotbAnticlockwise moment = clockwise moment30?kN × 40?cm = W × 60?cmW = 20?kNForces upwards = forces downwards;so F = 30?kN + 20?kN = 50?kNExam-style questions1aAt A the anticlockwise moment = the clockwise moment600 × (3 ? 1) = RB × 5RB = 240?NbAt B the anticlockwise moment = the clockwise moment600 × 3 = RA × 5RA = 360?NcThe sum of the upward reactions is 600?N; which is the weight of the plankCentre of gravity1The centre of gravity of an object is the point through which the entire weight of the body can be thought to act.23Tumbler A is likely to be less stable than tumbler B because it has a higher centre of gravity; both tumblers are in stable equilibrium; a stationary ball on the table is likely to be in neutral equilibriumExam-style questions1It has a wide, heavy base; which gives it a low centre of massIt is in stable equilibrium, so it requires considerable force to topple2aGravitybThe tractor topples only if the weight vector is outside the base, producing a resultant moment; making the area between the four wheels as big as possible increases the area of the base, making this less likelyDensity1D = M/V and kg/m32ab3V = 76 ? 65 = 11?cm3The material is probably lead because its density is closest to 11?g/cm3Exam-style questions1Indicative content:Use a ruler (or digital calipers) to measure the length, breadth and height of the solid in cmThe product of these three numbers is the solid’s volume in cm3Measure the mass of the solid, in grams, on a top-pan balanceDivide the mass by the volume to find the density in g/cm32Repeat for reliability and average the values calculated to obtain the most reliable value for the density3aMass of salt water = density × volume = 1.06?g/cm3 × 500?cm3 = 530?gMass of water = 500?gMass of salt = 530 ? 500 = 30?gbThe mass of salt per unit volume is halved; so the technician must add 500?cm3 of pure water to 500?cm3 of salt waterEnergy forms and resources1HEat, ELectricity, SOund, LIght, MAgnetism, STrain, KInetic, GRavitational2aThe law of conservation of energy states that energy can be neither created nor destroyed, but it can change its form.bY, 600?J (the useful output energy can never be greater than the input energy)Exam-style questions1OilWindNuclearSolarHydroelectricCoalTidalRenewable (R) or Non-renewable (N)NRNRRNR2aiGas, peatiiThey burn to produce carbon dioxide, a polluting greenhouse gasbiNucleariiThey are very radioactiveEnergy flow and efficiency1Input energyDeviceUseful output energySoundMicrophoneelectricalChemicalCar enginekineticElectricalToasterheatElectricalLoudspeakersoundChemicalBatteryelectrical2aKinetic energy (useful form) = 1250 ? (740 + 260) = 1250 ? 1000 = 250?kJUseful fraction = b0.2 (the same as part a)3Efficiency = Exam-style questions1Input energy = 2Total input energy = total output energy = 750?kJ + 1350?kJ + 900?kJ = 3000?kJUseful output energy = 900?kJ (kinetic) + (? × 1350?kJ) = 900 + 450 = 1350?kJEfficiency = Work and power1Energy = power × time = 24?kW × (3 × 60?s) = 4320?kJ = 4.32?MJ2aThe weightlifter was applying an upward force on the mass; and it was rising upwardsbGravity3Weight lifted = 6?N; Work done = 9?J; Power = 300?mW (0.3?W)Exam-style questions1aW = force × distance = 120?N × 60?m = 7200?JbPower = 2Force = 3Measure the weight, W, of the student in newtons using bathroom scalesUse a ruler to measure the heights of about five risers in a staircaseHence work out the average height of a riserCount the number of stairs in the staircaseMultiply the number of stairs by the average height of a riser to find the vertical height, H, of the staircaseUse a stopwatch to measure the time taken, t, for the student to run from the bottom of the staircase to the topCalculate the power, P, of the student using the equation For greater reliability, repeat the experiment three times and calculate the student’s average power4a650?NbW = Fd = 650?N × (120 × 0.45)m = 35?100?JKinetic energy and gravitational potential energy1KE staying the same; GPE decreasing2aGPE = mgh = 0.050?kg × 10?N/kg × 280?m = 140?Jb140?J3Height above ground / mGPE / JKE / JTotal energy / JSpeed / m/s2.5100010002.080201003.21.664361004.20.936641005.7001001007.1Exam-style questions1aEnergy lost = 15?000 ? 7200 = 7800?JbKE = ?mv27200 = ? × 64 × v2v = 15?m/scWork done against friction = friction × distance = energy lostfriction × 195 = 7800?JSo friction = 2KE = original energy ? GPE ? energy lost as heat and sound = 120?J ? 72?J ? 28?J = 20?J3Useful work = KE of car = ? × 800 × 42 = 6400?JWork done by pushers = F × d = (2 × 250) × 20 = 10?000?JWork done against friction = 10?000 ? 6400 = 3600?JFriction = Atomic structure1ParticleRelative massRelative chargeLocationProton1+1Inside nucleusElectron1/1840?1Orbiting nucleusNeutron10Inside nucleus2Atomic number — Number of protons in the nucleus — Symbol, ZMass number — Number of particles in the nucleus — Symbol, A3aNeutrons 143, protons 92, electrons 92 bcIsotopes are different forms of the same element having the same atomic number, but different mass number. This means their nuclei have the same number of protons, but different numbers of neutrons4aRutherford–Bohr modelbPlum pudding modelcPlum pudding modelRutherford–Bohr modelElectrons are distributed throughout the atom like currents in a bunDiscrete electrons orbit a central nucleus in circular pathsPositive charge is smeared throughout the atom, like dough in a bunPositive charge exists as discrete particles within a central nucleusThere is no central nucleusEvery atom has a tiny particle at its centre, called a nucleus, which contains almost all of the atom’s mass5, , Exam-style questions1Another element211 protons, 12 neutrons, 0 electrons (in nucleus!)3A and C are isotopes. Each has 11 protons. A has 12 neutrons, B has 13 neutrons4Indicative content:It was carried out in a highly evacuated chamberAlpha-particles were directed at a very thin gold foilThe scattered alpha-particles hit a zinc sulfide screen mounted on a microscope lensThese impacts caused the emission of flashes of light (scintillations)The light flashes allow the experimenter to ‘see’ the alpha-particle impactsThese flashes were observed using a moveable microscope5aAtoms are mostly empty space; alpha-particles were deflected when they passed close to a nucleusbThe alpha particles were back-scattered only when they came very close to a nucleus; providing evidence that a nucleus is small, relatively massive and positively chargedRadioactivity1Radioactivity is the random, spontaneous disintegration of an unstable nucleus by the emission of an alpha particle, a beta particle or a gamma wave2AlphaBetaGammaRelative charge+2?10Comes from a disintegratingNucleusnucleusnucleusRelative mass41/18400NatureparticleparticleEM waveRange2–4?cm in air0.5?cm in aluminiummany cm in leadIonisation ability (high/medium/low)highmediumlowConsists of2 protons and 2 neutronsfast electronhigh-energy EM waves3Half-life is the time taken for the activity of a radioactive source to fall to half of its original activity.4aThe time to fall from (say) 80?Bq to 40?Bq = one half-life = 2 hoursb4 hours = 2 half-lives2 hours before the experiment, the activity was 2 × 80 = 160?Bq4 hours before the experiment, the activity = 2 × 160 = 320?BqExam-style questions1a8192 → 4096 → 2048 → 1024; so 3 half-lives are involved3 half-lives = 12 hours; 1 half-life = 12 ÷ 3 = 4 hoursb24 hours represents a further 3 half-lives; 1024 → 512 → 256 → 128So the activity is 128?Bq2The radiation must penetrate paper but be partially absorbed; so it must be betaThe source must have a long half-life to avoid frequent replacement/recalibration; so the source must be D.Nuclear fission and nuclear fusion1Nuclear fission is the splitting of a uranium nucleus by a slow neutron into two highly radioactive nuclei and two or three neutronsNuclear fusion is the joining together of light hydrogen nuclei to make a much heavier helium nucleus23Achieving a temperature of about 1?000?000°C and maintaining it long enough to initiate fusion;Containment of the reactants and products at this enormous temperature4In stars (such as our Sun)5Indicative content:Uranium (nuclei) in the fuel rods in the reactor…… absorbs a slow neutron, initiating the fission processThe nucleus splits (fissions) into two lighter, radioactive nuclei; such as barium and krypton …… with the emission of two or three more neutrons …… and a considerable amount of energyThese fission neutrons go on to cause further fission; producing a chain reactionExam-style questions1Fission does not produce carbon dioxide, a major greenhouse gas that causes climate changeA fission reactor would provide many highly paid jobs for local people2aInternational Thermonuclear Experimental ReactorbControlled thermonuclear fusion; in which there is a net production of useful energyUnit 2Waves1In a longitudinal wave, the particles vibrate parallel to the direction in which the wave is movingIn a transverse wave, the particles vibrate perpendicular to the direction in which the wave is moving2The wavelength of a transverse wave is the distance between consecutive crests.The frequency of a wave is the number of waves that pass a fixed point in one second.A frequency of 1?kHz means 60?000 waves pass a fixed point every minute.The amplitude of a wave is its maximum displacement from its undisturbed position.3v = fλ = 300?Hz × 4?cm = 1200?cm/s4Gamma waves, X-rays, ultraviolet light, visible light, infrared light, microwaves, radio waves5Audible sound ranges in frequency from 20?Hz to 20?kHz; ultrasound has a frequency above 20?kHz6Distance from trawler to sea-bed = (time for echo to reach trawler) × (speed of ultrasound); = (? × 0.4?s) × 1500?m/s; = 300?m7Sonar uses ultrasound; radar uses microwaves, which are electromagnetic wavesExam-style questions1Wavelength = 10?m; frequency = 0.25?Hz; speed = 2.5?m/s23Distance = (time for wave to reach Y) × (speed of radar wave)= (? × 0.0005?s) × (3 × 108?m/s) = 75?000?m = 75?km4Medicine: for example ultrasound scan of foetus to check development in womb;Industry: for example checking for cracks in railway linesReflection and refraction of waves123Water waves reflecting (increases / decreases / stays the same)Water waves refracting (increases / decreases / stays the same)Frequencystays the samestays the sameSpeedstays the samedecreaseExam-style questions1a, b, dc90°2abReflection and refraction of light1The image is same size as the object, laterally inverted, erect and virtual.2a40°b50°34Angle of reflection at A = 65°; the angle between the reflected ray at A and mirror A = 25°;The angle between the incident ray at B and mirror B = 25°; the angle of reflection at B = 65°Exam-style questions1Angle of incidence = 45°; angle of refraction = 34°2a0ob 3abDispersioncThe different colours in white light travel at different speeds in glass; so they refract by slightly different amountsCritical angle and total internal reflection1The critical angle for some glass is 42°. This means that when light, travelling in glass towards air, has an angle of incidence at the glass/air boundary of 42 degrees, the angle of refraction in the air is 90 degrees.2The light must be travelling in a dense medium towards a boundary with a rarer medium; and the angle of incidence in the rare medium must be larger than the critical angle.3aFor example endoscopy, keyhole surgerybFor example optical fibres4Exam-style questions1?Put a semicircular glass block on sheet of white paperDraw round the outline of the block with a pencilMark the centre, X, of the straight diameterDirect a ray of light towards X with light entering the glass from the curved sideMove the ray box to increase the angle of incidence until the refracted ray at X just emerges along the diameterDraw two dots on the incident ray and join them to point XDraw the normal at XMeasure the critical angleRepeat and find an average value for the critical angle2If the optical fibre is bent too much, the angle of incidence in the core falls below the critical angle and light refracts out of the core.3Slower; total internal reflection can only occur if the light is slower in the coreLenses1If the lens is broader at the centre than at the edges it is converging; if the lens is narrower at the centre than at the edges it is converging.2a, b, cd12?cmeFor example a projectorf2Exam-style questions1aA virtual image is one which cannot be projected on a screenbi, ii, iii2Real images in a convex lens are always inverted3aiLong sight (hypermetropia)iiConvexiiibiShort sight (myopia)iiConcave lensiiiConductors and insulators1Conductors have free electrons; insulators do not2a1.4?Vb4.2?Vc Exam-style questions1Q = It = 1.60?A × 60?s = 96?CNumber of electrons = Ohm’s law and resistance123ab4aXbYcZExam-style questions1Resistance of wire Y is higher by 3?Ω2Q = It = 2?A × 15?s = 30?C3Except when actually recording measurements or changing the current, ensure that the power supply is switched offThe filament lamp1a, bcdAs the current in the lamp increases, the resistance of the lamp increases.Exam-style questions1aR = V/I, but when V = 0, I = 0 and division by 0 is not possible mathematically; However, the filament does have a resistance even when no current flows through it; this can be found using a different techniquebVoltage / V0.00.20.81.83.25.07.2Current / A0.00.20.40.60.81.01.2Resistance / Ω1.02.03.04.05.06.0cQuadrupling the voltage does not cause the resistance to quadrupledThe atoms are vibrating with greater amplitude as they absorb more kinetic energy from the drifting electronsSeries and parallel resistors124?Ω2aR = 3 Ω + 6 Ω = 9?Ωb so R = 2?Ω3a2?VbBecause the largest resistance is 24?Ω, this has the voltage of 6?V across it; so, by proportion, the other resistors are 8?Ω and 16?Ω4SwitchResistance between X and Y / ΩCommentABOpenOpen3624?Ω in series with 2 × 24?Ω in parallelOpenClosed3224?Ω in series with 3 × 24?Ω in parallelClosedOpen12First resistor is shorted out, so only 2 × 24?Ω in parallelClosedClosed8First resistor is shorted out, so only 3 × 24?Ω in parallelExam-style questions1The current in the 2.0?Ω resistor is 1.5?A; so the voltage across the parallel combination is 3.0?V; so the resistance of the unknown resistor is 1.0?Ω2abcTotal resistance of new combination R2 = 6?ΩFactors affecting resistance1aAbIf X is (1,?6), the graph is a curve passing through (1,?6), (2,?3), (3,?2) and (6,?1)2aLength and temperaturebAs the cross-sectional area increases, the resistance decreasescResistance (vertical axis) against the reciprocal of the cross-sectional area (horizontal axis)d(say) k = RA = 12?Ω × 1?mm2 = 12?Ω?mm2eExam-style questions1aVoltage, V / V0.240.240.240.240.24Current, I / mA300200150120100Resistance, R / Ω0.81.21.62.02.4Length, L / mm100150200250300bIn every case, doubling the length causes the resistance to doublecResistance (vertical axis) against length (horizontal axis)d(say) eR = kL = (0.008?Ω/mm) × (225?mm) = 1.8?Ω2Doubling the length doubles the resistance; halving the cross-sectional area also doubles the resistance; so the resistance of the cut from the second reel is quadrupled to 2.4?Ωa0.6?Ω and 2.4?Ω connected in series gives a total resistance of 3.0?Ωb0.6?Ω and 2.4?Ω connected in parallel gives a total resistance of Electrical energy and power1P = V2R2A3P = IV = 0.015?A × 3?V = 0.045?W = 0.045?J/sThe energy produced every minute is 0.045 × 60 = 2.7?J4A kilowatt-hour is the amount of electrical energy; used in one hour; by a device that takes in 1000 joules; of electrical energy every secondExam-style questions1aCurrent in 60?W bulbCurrent in 6?W LED Excess current = 0.25?A ? 0.025?A = 0.225?AbEnergy saved = 54?J/s = 54 × 3600?J/hour = 194?400?J2aTime spent in the shower per day = (5 + 7 + 8 + 10) = 30 minutes = 0.5 hoursTime per week = 0.5 × 7 = 3.5 hoursEnergy used in 1 week = 3?kW × 3.5 hours = 10.5?kWhCost = 10.5?kWh × 16?p/kWh = 10.5 × 16?p = 168?p = ?1.68bJohn’s father is right; energy used in 1 day = 3?kW × 0.5?h = 1.5?kWh = 1500 × 60 × 60?J = 5400?000?JMagnetism and electromagnetism1ai iibExam-style questions1a(soft) ironb, cdIt is insulated; if it is not insulated, the conducting iron core would short-circuit the coil2Increase the current in the coil; increase the number of turns (per unit length) in the coilForce on a current-carrying conductor in a magnetic field1abFleming’s left-hand (motor) rule2aWhen the coil is horizontal, with A to the left of B, there is maximum turning effect on the coil. The coil turns clockwise. When the coil is vertical, there is no force on the coil, but its own momentum makes the coil overshoot the vertical. As it does this the direction of the current is reversed, so the coil continues to rotate clockwise.bIncrease the current in the coil; use a stronger magnet; increase the number of turns in the coilExam-style questions12Electromagnets can be made to be much stronger than permanent magnets; this, in turn, makes the motors more powerfulElectromagnetic induction and transmission of electricity1aThe production of a voltage across the ends of a conductor when the magnetic flux linked with that conductor has changed (or flux lines are cut)bThe taut wire is cutting through field lines; this induces a voltage across the ends of the wireBecause the wire is part of a closed circuit, a current will flow through it and it will be detected by the ammetercA current would be induced as before; but because it will flow in the opposite direction, the needle on the ammeter will flick to the other side of zero; before returning to the centredNo current is induced because no flux lines are being cut2aThe needle would deflect to one side of zero and then return to the centre againbThe needle would deflect to the other side of zero and then return to the centre againcSoft iron is an electrical conductor; if the coils were made of non-insulated wire, the iron would cause the coils to be short-circuiteddThe current is alternating; with a frequency of 1?HzExam-style questions1aStep-downbcIs × Vs = Ip × Vp so 2aiStep-upiiThe voltage increasesiiiThe current decreasesbiStep-downiiThe voltage decreasesiiiThe current increasesThe Solar System1There are eight planets in our Solar System. They all orbit our Sun in elliptical paths. The planets closest to the Sun are called rocky planets because their surfaces are solid and we can land spacecraft on them.The planets furthest from the Sun are called gas planets. There are four of them. These planets are much larger than the planets closer to the Sun.The Sun is an ordinary star. It is the source of almost all of our energy.Asteroids and comets also orbit the Sun. Asteroids are large rocks. Most asteroids are found between the planets Mars and Jupiter.2Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune3aGravitybA satellite is an object that goes around anothercA natural satellite is a heavenly body that orbits another; for example Earth is a natural satellite of the SunAn artificial satellite is put into space by humans and orbits another object; for example GAMBIT was a spy satellite that orbited the EarthdTwo from: astronomy, communications, Earth observation, weather monitoringExam-style questions1aMarsbJupitercA cloud of dust and gas (usually hydrogen)2Gravity3aIce and rocks (silicates)bSome of the ice melts as a comet approaches the Sun; comets contain water, which might contain life-forms4aSpacecraft might collide with an asteroidbThere are many more asteroids than planets so a collision with an asteroid is more likelyThe orbits of the planets are known and highly predictable; asteroids often collide with other asteroids; their orbit paths are not predictableStars and their life cycle1Asteroid, moon, planet, star, solar system, galaxy25The temperature is now high enough for nuclear fusion to begin2The pressure and density of the nebula starts to rise1Hydrogen and dust in a nebula come together due to gravity3The nebula becomes so hot that it emits light and infrared radiation4As more and more hydrogen is captured, the core temperature eventually exceeds 13 million degrees6A star is born3The outward force, called radiation pressure, has been in equilibrium with the inward force of gravity4protostar→main sequence→red giant→white dwarf→black dwarf219218228723005 Exam-style questions1A main sequence star?is one in the main phase of its life2abA neutron3Black holes are incredibly dense; and so have such enormous gravitational fields that nothing can escape from them, not even lightEvidence for Big Bang model114?000?000?000 (14 billion) years2Expanded and cooled3C4Expanding5ElectronsExam-style questions1The distance between Andromeda and the Milky Way is decreasing2MicrowaveSpace travel and life on other planets1When a planet is very close to its star, the temperature would be so high that life could not be sustained; when the planet is very far from its star, the temperature would be so low that life could not be sustainedExam-style questions1aA planet that orbits a star other than our Sun; it is not part of our Solar SystembCarbon, nitrogen and water are essential for life like our own; water can be formed from oxygen and hydrogen2aThe distance light travels in one yearbDistance = speed × time = (3 × 105) × (4.22 × 365 × 24 × 60 × 60) = 4 × 1013?kmcTime ................
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