Key concepts. For values of physical quantities, you need to know the values of these multiples or prefixes. centiprefix c. 100 e.g.

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1 Cor Ky concpts You nd to know th SI units of physical quantitis, as wll as thir multipls and submultipls, how to conrt btwn units and how to us significant figurs and standard form. Bas SI units Drid SI units Thr ar six main bas SI units you nd to know. Thr ar nin drid SI units you nd to know. Nam lngth mass tim currnt tmpratur amount of a substanc Unit mtr kilogram scond ampr or amp klin mol Abbriation m kg s A K mol Nam frquncy forc nrgy powr prssur lctric charg lctric potntial diffrnc lctric rsistanc magntic flux dnsity Unit hrtz nwton joul watt pascal coulomb olt Abbriation Hz N J W Pa C V ohm tsla Ω T Multipls and sub-multipls For alus of physical quantitis, you nd to know th alus of ths multipls or prfixs gigaprfix G g. 5 GHz mgaprfix M g. 5 MW kiloprfix k 1000.g. 2 km cntiprfix c 100.g. 8 cm milliprfix m 1000.g. 6 ma micronanoprfix μ prfix n g. 7 μs.g. 2 nm Significant figurs Standard form Significant figurs ar ithr non-zro digits (i.. th digits 1 to 9) or zros btwn non-zro digits. Th numbr has 6 significant figurs. Th numbr 0.23 has two significant figurs. Th numbr has 3 significant figurs. Th zros bfor th 2 ar not significant figurs. Th numbr has 4 significant figurs. It is usful to writ ry larg and ry small numbrs in standard form. This is a numbr btwn 1 and 10 Th numbr roundd to 5 significant figurs is Stat th numbr of significant figurs in ach numbr. (a) 4.56 (b) (c) Conrt: (a) 12 hours to sconds This is an intgr a whol numbr that can b positi or ngati. A 10B abl to d to b n o ls a u uch Yo n units s tw b rt n co s, or s to mtr tr m ilo k as conds. hours to s (b) 64 km/h to m/s 3 Conrt 58.3 MW to watts. Gi your answr in standard form. 4 Th spd of light is mils pr scond. Conrt this to mtrs pr scond and writ it in standard form. 1 mil is qual to 1609m 1

2 Papr 1 Colour of an objct Th colour an objct appars is rlatd to th transmission, rflction and absorption of diffrnt walngths of light. Spcular rflction Diffus rflction Spcular rflction occurs whn was ar rflctd from a smooth surfac. Whn paralll rays of light ar incidnt on a smooth, plan surfac such as a mirror, th rflctd light rays will also b paralll. Th sizs of any irrgularitis on th surfac ar much smallr than th walngth of th wa. Diffus rflction occurs whn th surfac is not smooth and has rough irrgularitis. Th siz of th irrgularitis is comparabl with th walngth of th wa. Th incidnt wa is thn rflctd at many diffrnt angls and th rflctd rays will not b paralll, such as whn light is rflctd off a paintd wall. Th colour spctrum Visibl light maks up a ry small part of th lctromagntic spctrum. Th colours that w s can b split into diffrnt colours by a prism. rfraction rfraction ht lig whit prism whit spctrum R O Y G B I V Ths colours all ha a diffrnt walngth, ranging from th longst walngth at th rd nd of th spctrum to th shortst walngth at th iolt nd. Diffrntial absorption at surfacs Th colour an objct appars is basd on how th atoms at its surfac rspond to th light bing shon on thm. A matrial appars grn bcaus its atoms rflct th grn walngths and absorb all of th othrs. whit light coming in t igh l n gr grn surfac Filtrs Filtrs lt through diffrnt colours of light and absorb all grn filtr othr colours. For xampl, a grn filtr will lt through or transmit grn light and absorb all of th othr walngths. 1 Explain which typ of rflction you would associat with a gral path Explain how a filtr works to lt through rd light. Explain which typ of rflction you would associat with a plan mirror or a calm lak surfac. spcular rflction, bcaus th surfac is smooth, so paralll rays or wafronts that ar incidnt on ths surfacs will b rflctd as paralll rays or wafronts 3 Explain what you would s whn whit light trals through a rd filtr and thn a grn filtr.

3 Papr 1 Instigating rfraction Practical skills You can instigat rfraction in rctangular glass blocks in trms of th intraction of light was with mattr. Thr is mor information about th rfraction of light on pags 32 and 33. Aim To instigat th natur of how light was chang dirction whn thy mo from air into glass. Apparatus ray box and slits, 12 V powr supply, glass block, protractor, A3 papr, sharp pncil glass box Mthod r 1 Plac th rctangular block on th A3 papr and draw i around it with th sharp pncil. normal 2 Draw th normal lin, which will b at right angls to th ray box sid of th block towards which th light ray will b shon. 3 Using th protractor and pncil, mark on th papr angls of incidnc of 0 through to 80 in 10 intrals. 4 Starting with th 0 angl (th light ray tralling along th normal lin), dirct th light ray towards th block and mark its xit point from th block with a sharp pncil dot. 5 Rmo th glass block and join th dot to th point of incidnc by drawing a straight lin. Masur and rcord this angl, which is th angl of rfraction. 6 Rpat for all of th othr angls from 10 to 80. Rsults Angl of incidnc Angl of rfraction Conclusion Whn a light ray trals from air into a glass block, its dirction changs and th angl of rfraction will b lss than th angl of incidnc unlss it is tralling along th normal. 38 Whn carrying out th instigation, mak sur that you dirct a thin bam of light towards th point at which th normal maks contact with th glass block. Clarly mark th angls from 0 to 80 with a sharp pncil and rulr. Rfraction Th rfraction of a light ray inols a chang in: th dirction of th light ray th spd of th light. Light slows down whn it mos from air into glass and spds up whn it mos from glass into air. Th only tim whn th dirction dos not chang is whn th bam is tralling along th normal. Rmmbr that th angl of incidnc is masurd with rspct to th normal lin. Light will slow down mor whn it trals from air into glass than it will whn it trals from air into watr. This is bcaus th optical dnsity of glass is gratr than that of air. 1 Stat th two things that can chang whn a light ray is rfractd. 2 Draw a tabl of rsults for a light ray tralling from air into a block of watr instad of a block of glass. 3 Explain how th (a) frquncy and (b) walngth of a light ray ar affctd whn it ntrs glass from air.

4 Papr 1 Masuring radioactiity Photographic film and a Gigr Müllr tub can both b usd to masur and dtct radiation. Th Gigr Müllr tub Th Gigr Müllr (GM) tub is usd to dtct nuclar radiation. It is connctd to a countr or ratmtr which shows th amount of radiation that has bn dtctd. Th txt in rd shows what is happning insid th GM tub, but you do not nd to rmmbr this in dtail. Th GM tub contains argon gas. Whn radiation ntrs th tub, th atoms of argon ar ionisd and lctrons tral towards th thin wir. sourc Am-241 As th lctrons tral towards th thin wir, mor lctrons ar knockd from atoms and an aalanch ffct occurs. Th amount of radiation dtctd is shown by th ratmtr. ratmtr Dtcting radiation In 1896, Hnri Bcqurl discord that uranium salts would lad to th darkning of a photographic film, n if it was wrappd up so that no light could rach it. Radiation was bing mittd from th uranium nucli and this was rsponsibl for th darkning of th film. This is now mad us of in th nuclar industry as workrs will war a film badg to dtrmin if thy ar bing xposd to diffrnt forms of nuclar radiation. photographic film insid thin and thick plastic windows or aluminium stop som bta particls opn window lad btwn th plastic cas and th film stops bta and most gamma radiation This dosimtr is a film badg usd to monitor th radiation rcid by its warr. (a) Explain why thr is an opn window in th film badg. Light rays will darkn th film. Sinc th film badg is dsignd to dtct nuclar radiation, this acts as a control so that you know that th film is working. (b) Explain why th film badg has aluminium and lad shts insrtd in it. Aluminium absorbs bta particls and lad absorbs gamma-rays. If thr is darkning bhind ths shts, thn high-nrgy bta- or gamma-rays may ha got through and so th natur of th radiation can b dtrmind and th risk idntifid. 50 Th GM tub has a thin wir in th middl which is connctd to a oltag of +400 V. 1 Explain why GM tubs ar bttr at dtcting alpha particls than bta- or gamma-rays. You should us your knowldg of th pntrating powrs of th diffrnt typs of radioactiity to answr th qustion. 2 Explain how a film from a film badg would tll you if thr was a potntial risk to humans.

5 Papr 1 Wa quations Both of ths quations can b us to find th spd or locity of a wa. Spd, frquncy and walngth =f l = wa spd (mtrs pr scond, m/s) f = frquncy (hrtz, Hz) l = walngth (mtrs, m) f 3 λ Spd, distanc and tim x = t = wa spd (mtrs/scond, m/s) x = distanc (mtrs, m) t = tim (sconds, s) x 3 t A sismic wa has a frquncy of 15 Hz and trals at 4050 m/s. Calculat its walngth. l= f 4050 m/s = 15 Hz = 270 m bda. k lttr lam l is th Gr A wa on th sa is tralling at 4 m/s. Calculat how long it taks to tral along a 20 m long pir. x t= 20 m = 4 m/s = 5s Start with th quation = f λ and rarrang it to mak λ th subjct. Diid both sids of th quation by f, to gt λ= f Substitut th alus for and f to calculat th alu for th walngth, λ. Th units for ar m/s and th units for f ar Hz (or pr scond), so th units for λ ar m / s = m. 1/ s Always stat th corrct units as part of your answr. _x_ Rarrang th quation = t to gi _x_ t =. You can also find this asily by coring th lttr t in th triangl at th top of th pag. Whn diiding th units, m (m/s), you gt m s/m = s, which is th corrct unit for tim. Again, always includ th corrct unit in your answr. 1 A sound wa with a frquncy of 100 Hz has a spd of 330 m/s. Calculat its walngth. 2 A wa in th sa trals at 25 m/s. Calculat th distanc it trals in on minut. 24

6 Papr 2 Transformrs Transformrs can b usd to chang th siz of an altrnating oltag or currnt. Transformrs Rmmbr that th primary coil is th on connctd A transformr can b scondary to th lctricity supply. usd to chang th coil potntial diffrnc of an 5 coils 10 coils altrnating lctricity supply. V 20 V 10 V Whn an altrnating INPUT OUTPUT Th cor must b mad of a currnt flows magntic matrial so that it can channl through th th magntic lins of forc from th primary coil primary coil to th scondary coil. it producs a changing magntic fild in th iron cor. This magntic fild inducs an altrnating potntial diffrnc across th scondary coil. Th altrnating potntial diffrnc in th scondary coil lads to an altrnating currnt whn th circuit is compltd. Potntial diffrnc and turns Th rlationship btwn th potntial diffrncs in th primary and scondary coils and th turns ratio is gin by th quation: potntial diffrnc numbr of turns in across primary coil (V) primary coil = potntial diffrnc numbr of turns in across scondary coil scondary coil (V) N Vp = Np Vs s across th bcaus th potntial diffrnc potntial scondary coil is gratr than th coil ary prim diffrnc across th Transformr fficincy (a) A transformr has 20 turns on its primary coil and 240 turns on its scondary coil. Th potntial diffrnc across th primary coil is 12 V. Calculat th potntial diffrnc across th scondary coil. Vp N = p Vs Ns Vs = Vp Ns Np = 12 V = 144 V (b) What typ of transformr is this? stp-up transformr (c) Calculat th currnt in th scondary coil whn th currnt in th primary coil is 18 A. Vp Ip = Vs Is 12 V 18 A = 144 V Is Is = 12 V 18 A 144 V = 1.5 A It is assumd that transformrs ar 100% fficint. This mans that th input powr of th primary coil and th output powr of th scondary coil ar qual. Thy ar connctd by th quation: Vp Ip = Vs Is potntial potntial currnt in currnt in diffrnc across = diffrnc across scondary primary coil (A) primary coil (V) scondary coil (V) coil (A) 1 Calculat Vp whn Np = 40, Ns = 800 and Vs = 18 V. 2 A transformr with an input potntial diffrnc of 30 kv has a powr output on th scondary coil of 3 kw. Calculat th currnt in th primary coil. Stat th assumption that you mak. 3 Suggst why transformrs will not work with a d.c. input. 104

7 Papr 2 Transmitting lctricity Th National Grid is a systm of wirs that transmits lctricity from powr stations to whr th lctricity is ndd and usd. Transformr quations Th National Grid Elctricity is gnratd and transportd to our homs, hospitals and factoris by th National Grid. Th National Grid is th wirs and transformrs that transmit th lctricity; it dos not includ th powr stations and th consumrs. To nsur that transmission is fficint, and that ry littl nrgy is lost as hat, th oltag and currnt alus nd to b chosn carfully at diffrnt stags. Transformrs and th National Grid You nd to b abl to xplain th adantags of using high oltags to transmit lctrical nrgy using ths four quations: powr = nrgy tim (P = E t) powr = currnt oltag (P = I V) powr = currnt2 x rsistanc (P = I2 R) primary oltag primary currnt = scondary oltag scondary currnt (Vp Ip = Vs Is) stp-up transformrs 2 National Grid systm Fossil fuls or nuclar ful ar usd to gnrat lctrical nrgy in th 132 kv or highr powr station. A stp-up transformr stp-down incrass th oltag to transformrs 132 kv or mor. For a powr station fixd amount of lctrical 3 nrgy or powr output, homs incrasing th oltag light industry mans thr will b a dcras in currnt. 1 Elctrical nrgy is 11 kv 230 V transmittd at a high oltag and low currnt through th wirs to rduc nrgy losss as hat in th wirs. It also mans thinnr wirs can b usd, which rducs costs. Stp-down transformrs dcras th oltags from th National Grid for safr us in our homs and industry. Rducing th oltag mans thr will b an incras in currnt. Explain why lctrical powr is transmittd at a high oltag and a low currnt on th National Grid. Powr is transmittd at a high oltag and a low currnt bcaus th amount of lctrical nrgy wastd as thrmal nrgy is proportional to th squar of th currnt, from th quation P = I2R. So, if w doubl th siz of th currnt flowing, th powr wastd incrass by a factor of 22 or 4. 1 Us th Powr quations to xplain why w do not transmit lctrical nrgy through th National Grid at a high currnt. 2 Dscrib th bst way of transmitting 109 W of powr or th National Grid. Lss currnt mans thinnr wirs. Thinnr wirs mans lss wir, so lss coppr to purchas by th company. Low-rsistanc wirs also rduc powr losss. 105