A2Physics物理出國(guó)英語(yǔ)1.4Materials

1.Density

1.Definedensity

Density($\rho$)isdefinedasthemass($m$)perunitvolume($V$)ofasubstance.Theformulais$\rho=\frac{m}{V}$.Forexample,ifablockofmetalhasamassof500gandavolumeof100$cm^3$,thenthedensity$\rho=\frac{500\g}{100\cm^3}=5\g/cm^3$.InSIunits,densityismeasuredin$kg/m^3$.Toconvertfrom$g/cm^3$to$kg/m^3$,weusetheconversionfactor:$1\g/cm^3=1000\kg/m^3$.So,$5\g/cm^3=5000\kg/m^3$.

2.Understandtheconceptofrelativedensity

Relativedensity(alsoknownasspecificgravity)istheratioofthedensityofasubstancetothedensityofareferencesubstance.Usually,forliquidsandsolids,thereferencesubstanceiswaterat4°C(wherethedensityofwater$\rho_{water}=1000\kg/m^3$).Forexample,ifthedensityofaliquidis800$kg/m^3$,itsrelativedensity$RD=\frac{\rho_{liquid}}{\rho_{water}}=\frac{800\kg/m^3}{1000\kg/m^3}=0.8$.

3.Explainhowdensityvarieswithtemperatureandstateofmatter

Formostsubstances,densitydecreasesastemperatureincreases.Thisisbecauseasthetemperaturerises,thevolumeofthesubstanceusuallyexpandswhilethemassremainsconstant.Accordingtothedensityformula$\rho=\frac{m}{V}$,anincreasein$V$withconstant$m$leadstoadecreasein$\rho$.Forexample,whenwaterisheatedfrom0°Cto4°C,itsvolumedecreases,sothedensityincreases.Above4°C,asthetemperaturecontinuestorise,thevolumeofwaterexpands,andthedensitydecreases.

Whenasubstancechangesitsstatefromsolidtoliquidorfromliquidtogas,thedensitygenerallydecreases.Inthesolidstate,theparticlesarecloselypacked,sothevolumeisrelativelysmallandthedensityishigh.Intheliquidstate,theparticleshavemorefreedomtomove,andthevolumeislarger,resultinginalowerdensity.Inthegaseousstate,theparticlesarefarapart,andthedensityismuchlower.Forexample,thedensityoficeisabout920$kg/m^3$,whilethedensityofwateris1000$kg/m^3$,andthedensityofwatervaporismuchlessthanthatofliquidwater.

2.Hooke'sLaw

1.StateHooke'sLaw

Hooke'sLawstatesthattheforce($F$)appliedtoanelasticobject(suchasaspring)isdirectlyproportionaltotheextension($x$)oftheobject,providedtheelasticlimitisnotexceeded.Mathematically,itisexpressedas$F=kx$,where$k$isthespringconstant.Thespringconstantrepresentsthestiffnessofthespring.Alarger$k$valuemeansthespringisstifferandrequiresmoreforcetoproduceagivenextension.Forexample,ifaspringhasaspringconstant$k=20\N/m$,andanextension$x=0.1\m$isproduced,thentheforceapplied$F=kx=(20\N/m)\times(0.1\m)=2\N$.

2.Understandtheelasticlimit

Theelasticlimitisthemaximumforceorextensionuptowhichanobjectwillreturntoitsoriginalshapeandsizewhenthedeformingforceisremoved.Iftheforceappliedexceedstheelasticlimit,theobjectwillundergoplasticdeformation.Forexample,whenametalwireisstretchedbeyonditselasticlimit,itwillnotreturntoitsoriginallengthwhenthestretchingforceisremoved.

3.Determinethespringconstantfromaforceextensiongraph

Onaforceextensiongraph,thespringconstant$k$isequaltotheslopeofthestraightlineportionofthegraph.Theslopeiscalculatedas$k=\frac{\DeltaF}{\Deltax}$.Forexample,ifonagraph,whentheforcechangesfrom2Nto6Nandtheextensionchangesfrom0.1mto0.3m,then$k=\frac{6\N2\N}{0.3\m0.1\m}=\frac{4\N}{0.2\m}=20\N/m$.

3.Young'sModulus

1.Definestress,strain,andYoung'sModulus

Stress($\sigma$)isdefinedastheforce($F$)appliedperunitcrosssectionalarea($A$)ofamaterial.Theformulais$\sigma=\frac{F}{A}$.Forexample,ifaforceof500Nisappliedtoawirewithacrosssectionalareaof$2\times10^{6}\m^2$,thenthestress$\sigma=\frac{500\N}{2\times10^{6}\m^2}=2.5\times10^8\Pa$.

Strain($\epsilon$)istheratiooftheextension($x$)totheoriginallength($L$)ofthematerial,i.e.,$\epsilon=\frac{x}{L}$.Forexample,ifawireoforiginallength2misstretchedby0.01m,thenthestrain$\epsilon=\frac{0.01\m}{2\m}=0.005$.

Young'sModulus($E$)istheratioofstresstostrainwithintheelasticlimitofamaterial.Theformulais$E=\frac{\sigma}{\epsilon}$.Itisameasureofthestiffnessofamaterial.Forexample,ifthestressonamaterialis$3\times10^8\Pa$andthestrainis0.003,thenYoung'sModulus$E=\frac{3\times10^8\Pa}{0.003}=1\times10^{11}\Pa$.

2.UnderstandthesignificanceofYoung'sModulus

Young'sModulusprovidesinformationabouthowdifficultitistostretchorcompressamaterial.AlargeYoung'sModulusmeansthematerialisstifferandrequiresalargestresstoproduceasmallstrain.Forexample,steelhasarelativelylargeYoung'sModulus(about$2\times10^{11}\Pa$),whichmeansitisdifficulttostretchcomparedtoamateriallikerubber,whichhasamuchsmallerYoung'sModulus.

3.CalculateYoung'sModulusfromexperimentaldata

InanexperimenttomeasureYoung'sModulus,wemeasuretheforceappliedtoawire,thecrosssectionalareaofthewire,theoriginallengthofthewire,andtheextensionofthewire.First,wecalculatethestress$\sigma=\frac{F}{A}$andthestrain$\epsilon=\frac{x}{L}$.ThenwefindYoung'sModulus$E=\frac{\sigma}{\epsilon}$.Forexample,wemeasureaforce$F=100\N$,thecrosssectionalareaofthewire$A=1\times10^{6}\m^2$,theoriginallength$L=1\m$,andtheextension$x=0.0005\m$.Thestress$\sigma=\frac{100\N}{1\times10^{6}\m^2}=1\times10^8\Pa$,thestrain$\epsilon=\frac{0.0005\m}{1\m}=0.0005$,andYoung'sModulus$E=\frac{1\times10^8\Pa}{0.0005}=2\times10^{11}\Pa$.

4.ElasticandPlasticDeformation

1.Differentiatebetweenelasticandplasticdeformation

Elasticdeformationisareversiblechangeintheshapeorsizeofamaterial.Whenthedeformingforceisremoved,thematerialreturnstoitsoriginalshapeandsize.Thisoccurswithintheelasticlimitofthematerial.Forexample,whenaspringisstretchedwithinitselasticlimitandthenreleased,itreturnstoitsoriginallength.

Plasticdeformationisanirreversiblechangeintheshapeorsizeofamaterial.Oncethematerialhasundergoneplasticdeformation,itwillnotreturntoitsoriginalshapewhenthedeformingforceisremoved.Thisoccurswhentheforceappliedexceedstheelasticlimitofthematerial.Forexample,whenametalrodisbentbeyonditselasticlimit,itretainsthebentshape.

2.Understandthebehaviorofmaterialsduringelasticandplasticdeformationonastressstraingraph

Onastressstraingraph,theelasticregionisthestraightlineportionwherethestressisdirectlyproportionaltothestrain,followingHooke'sLaw.Inthisregion,thematerialexhibitselasticdeformation.TheslopeofthisstraightlineportionistheYoung'sModulus.

Beyondtheelasticlimit,thegraphcurves,andthematerialenterstheplasticdeformationregion.Inthisregion,asmallincreaseinstresscancausealargeincreaseinstrain.Eventually,thematerialmayreachitsultimatetensilestrength,whichisthemaximumstressthematerialcanwithstand.Afterreachingtheultimatetensilestrength,thematerialmaystarttoneck(alocalreductionincrosssectionalarea)andfinallybreak.

5.EnergyStoredinaDeformedMaterial

1.Deriveandusetheformulaforelasticpotentialenergyinastretchedspring

Theworkdoneinstretchingaspringisstoredaselasticpotentialenergy($E_p$).Theforcerequiredtostretchaspringis$F=kx$.Theworkdone$W$instretchingthespringfrom0toanextension$x$iscalculatedbyintegratingtheforcewithrespecttotheextension.Since$W=\int_{0}^{x}Fdx$and$F=kx$,wehave$W=\int_{0}^{x}kxdx=\frac{1}{2}kx^{2}$.So,theelasticpotentialenergystoredinastretchedspringis$E_p=\frac{1}{2}kx^{2}$.Forexample,ifaspringhasaspringconstant$k=50\N/m$andisstretchedby0.2m,thentheelasticpotentialenergy$E_p=\frac{1}{2}\times(50\N/m)\times(0.2\m)^{2}=1\J$.

2.Relatetheenergystoredinadeformedmaterialtotheareaundertheforceextensiongraph

Theareaundertheforceextensiongraphforanelasticobjectrepresentstheworkdoneindeformingtheobject,whichisequaltotheelasticpotentialenergystoredintheobject.InthecaseofaspringobeyingHooke'sLaw,theforceextensiongraphisastraightlinepassingthroughtheorigin.Theareaofthetriangleunderthegraphfrom0toanextension$x$withaforce$F=kx$is$A=\frac{1}{2}\timesbase\timesheight=\frac{1}{2}\timesx\timesF$.Substituting$F=kx$,weget$A=\frac{1}{2}kx^{2}$,whichisthesameastheformulaforelasticpotentialenergy.

6.FractureandFatigue

1.Understandtheconceptoffracture

Fractureistheseparationofamaterialintotwoormorepieces.Therearetwomaintypesoffracture:brittlefractureandductilefracture.

Brittlefractureoccurssuddenlyandwithoutmuchplasticdeformation.Materialsthatarebrittle,suchasglassandsomeceramics,havealowabilitytodeformplastically.Whenabrittlematerialissubjectedtoastress,itwillbreakwhenthestressreachesitsfracturestrength.Forexample,whenaglassrodisbent,itwillsuddenlybreakwithoutsignificantpriordeformation.

Ductilefractureoccursaftersignificantplasticdeformation.Ductilematerials,suchasmetals,candeformplasticallybeforebreaking.Thematerialfirstnecks(localreductionincrosssectionalarea)andthenbreaks.Forexample,whenacopperwireispulled,itwillstretchandneckbeforefinallybreaking.

2.Explaintheconceptoffatigue

Fatigueistheweakeningofamaterialcausedbycyclicloading.Whenamaterialissubjectedtorepeatedstress,evenifthestressisbelowtheultimatetensilestrengthofthematerial,crackscaninitiateandgrowovertime,eventuallyleadingtofailure.Forexample,thewingsofanaircraftaresubjectedtocyclicstressesduringflight.Overmanyflights,fatiguecrackscandevelop,whichcanbeverydangerousifnotdetectedandrepaired.

7.CompositeMaterials

1.Definecompositematerials

Compositematerialsarematerialsmadefromtwoormoredifferentmaterialswithdistinctproperties,combinedtocreateamaterialwithimprovedproperties.Thedifferentmaterialsarecalledthematrixandthereinforcement.Forexample,infiberglass,theglassfibersarethereinforcement,andtheresinisthematrix.

2.Understandtheadvantagesofcompositematerials

Compositematerialsoftenhaveacombinationofhighstrengthandlowweight.Forexample,carbonfiberreinforcedcompositesareusedinaerospaceapplicationsbecausetheyaremuchlighterthanmetalslikesteelbutcanhavesimilarorevenhigherstrength.Theycanalsohaveimprovedcorrosionresistancecomparedtosometraditionalmaterials.Forexample,fiberglassisresistanttocorrosioninmanychemicalenvironments.

Compositematerialscanbedesignedtohavespecificpropertiesbyvaryingthetypeandamountofreinforcementandthematrix.Forexample,bychangingtheorientationofthecarbonfibersinacarbonfiberreinforcedcomposite,wecancontrolthedirectiondependentstrengthofthematerial.

8.Rheology

1.Understandtheconceptofviscosity

Viscosityisameasureofafluid'sresistancetoflow.Ahighlyviscousfluidflowsslowly,whilealowviscosityfluidflowseasily.Forexample,honeyhasahighviscosityandflowsslowly,whilewaterhasarelativelylowviscosityandflowsquickly.

Viscositycanbeaffectedbytemperature.Generally,theviscosityofliquidsdecreasesasthetemperatureincreases.Thisisbecauseasthetemperaturerises,thekineticenergyofthefluidmoleculesincreases,andtheycanmovemorefreely,reducingtheinternalfrictionwithinthefluid.Forgases,theviscosityincreaseswithincreasingtemperaturebecausethemoreenergeticgasmoleculescollidemorefrequently,increasingtheresistancetoflow.

2.DifferentiatebetweenNewtonianandnonNewtonianfluids

Newtonianfluidshaveaconstantviscosityregardlessoftheshearrate(therateatwhichthefluidisdeformed).Therelationshipbetweentheshearstress($\tau$)andtheshearrate($\dot{\gamma}$)islinear,andtheviscosity$\eta=\frac{\tau}{\dot{\gamma}}$isaconstant.WaterandmostsimpleliquidsareNewtonianfluids.

NonNewtonianfluidsdonothaveaconstantviscosity.Theirviscositycanchangedependingontheshearrate.TherearedifferenttypesofnonNewtonianfluids,suchasshearthinningfluids(e.g.,ketchup),wheretheviscositydecreasesastheshearrateincreases,andshearthickeningfluids(e.g.,cornstarchinwater),wheretheviscosityincreasesastheshearrateincreases.

Questions

1.Whatisthedensityofasubstancewithamassof300gandavolumeof150$cm^3$?

First,usethedensityformula$\rho=\frac{m}{V}$.Given$m=300\g$and$V=150\cm^3$,then$\rho=\frac{300\g}{150\cm^3}=2\g/cm^3$.Convertingto$kg/m^3$,weget$2\times1000\kg/m^3=2000\kg/m^3$.

2.Iftherelativedensityofaliquidis0.9,whatisitsdensity?

Sincerelativedensity$RD=\frac{\rho_{liquid}}{\rho_{water}}$and$\rho_{water}=1000\kg/m^3$,then$\rho_{liquid}=RD\times\rho_{water}=0.9\times1000\kg/m^3=900\kg/m^3$.

3.Aspringhasaspringconstantof15N/m.Whatforceisrequiredtostretchitby0.2m?

UsingHooke'sLaw$F=kx$,with$k=15\N/m$and$x=0.2\m$,weget$F=(15\N/m)\times(0.2\m)=3\N$.

4.Awireofcrosssectionalarea$2\times10^{7}\m^2$issubjectedtoaforceof40N.Whatisthestressonthewire?

Usingthestressformula$\sigma=\frac{F}{A}$,with$F=40\N$and$A=2\times10^{7}\m^2$,wehave$\sigma=\frac{40\N}{2\times10^{7}\m^2}=2\times10^8\Pa$.

5.Arodoforiginallength2misstretchedby0.001m.Whatisthestrain?

Usingthestrainformula$\epsilon=\frac{x}{L}$,with$x=0.001\m$and$L=2\m$,weget$\epsilon=\frac{0.001\m}{2\m}=0.0005$.

6.Ifthestressonamaterialis$1.5\times10^8\Pa$andthestrainis0.001,whatisYoung'sModulus?

Usingtheformula$E=\frac{\sigma}{\epsilon}$,with$\sigma=1.5\times10^8\Pa$and$\epsilon=0.001$,wehave$E=\frac{1.5\times10^8\Pa}{0.001}=1.5\times10^{11}\Pa$.

7.Aspringwithaspringconstantof25N/misstretchedby0.15m.Whatistheelasticpotentialenergystoredinthespring?

Usingtheformula$E_p=\frac{1}{2}kx^{2}$,with$k=25\N/m$and$x=0.15\m$,weget$E_p=\frac{1}{2}\times(25\N/m)\times(0.15\m)^{2}=\frac{1}{2}\times25\times0.0225\J=0.28125\J$.

8.Whatisthemaindifferencebetweenelasticandplasticdeformation?

Elasticdeformationisreversible;thematerialreturnstoitsoriginalshapewhenthedeformingforceisremoved.Plasticdeformationisirreversible;thematerialdoesnotreturntoitsoriginalshapeaftertheforceisremoved.

9.Whydoesthedensityofmostsubstancesdecreasewithincreasingtemperature?

Astemperatureincreases,thevolumeofthesubstanceusuallyexpandswhilethemassremainsconstant.Accordingtothedensityformula$\rho=\frac{m}{V}$,anincreasein$V$withconstant$m$leadstoadecreasein$\rho$.

10.Howdoestheviscosityofaliquidchangewithincreasingtemperature?

Generally,theviscosityofliquidsdecreasesasthetemperatureincreasesbecausethekineticenergyofthefluidmoleculesincreases,reducingtheinternalfrictionwithinthefluid.

11.Ablockofmaterialhasamassof450gandavolumeof300$cm^3$.Calculateitsdensityin$kg/m^3$.

First,findthedensityin$g/cm^3$:$\rho=\frac{m}{V}=\frac{450\g}{300\cm^3}=1.5\g/cm^3$.Thenconvertto$kg/m^3$:$1.5\times1000\kg/m^3=1500\kg/m^3$.

12.Aspringhasaspringconstantof30N/m.Howmuchforceisneededtostretchitby0.25m?

UsingHooke'sLaw$F=kx$,with$k=30\N/m$and$x=0.25\m$,weget$F=(30\N/m)\times(0.25\m)=7.5\N$.

13.Awirehasacrosssectionalareaof$3\times10^{7}\m^2$andaforceof60Nisappliedtoit.Calculatethestress.

Usingthestressformula$\sigma=\frac{F}{A}$,with$F=60\N$and$A=3\times10^{7}\m^2$,wehave$\sigma=\frac{60\N}{3\times10^{7}\m^2}=2\times10^8\Pa$.

14.Abaroflength1.5misstretchedby0.00075m.Calculatethestrain.

Usingthestrainformula$\epsilon=\frac{x}{L}$,with$x=0.00075\m$and$L=1.5\m$,weget$\epsilon=\frac{0.00075\m}{1.5\m}=0.0005$.

15.Ifthestressonamaterialis$2\times10^8\Pa$andYoung'sModulusis$1\times10^{11}\Pa$,whatisthestrain?

Usingtheformula$E=\frac{\sigma}{\epsilon}$,wecanrearrangeitto$\epsilon=\frac{\sigma}{E}$.Substituting$\sigma=2\times10^8\Pa$and$E=1\times10^{11}\Pa$,weget$\epsilon=\frac{2\times10^8\Pa}{1\times10^{11}\Pa}=0.002$.

16.Aspringwithaspringconstantof40N/misstretchedby0.2m.Calculatetheelasticpotentialenergy.

Usingtheformula$E_p=\frac{1}{2}kx^{2}$,with$k=40\N/m$and$x=0.2\m$,weget$E_p=\frac{1}{2}\times(40\N/m)\times(0.2\m)^{2}=\frac{1}{2}\times40\times0.04\J=0.8\J$.

17.Explainwhyamaterialmayundergoplasticdeformation.

Whentheforceappliedtoamaterialexceedsitselasticlimit,theinternalstructureofthematerialispermanentlyaltered.Bondsbetweenatomsormoleculesarebrokenandreformedinnewpositions,resultinginplasticdeformation.

18.WhatisthesignificanceofYoung'sModulus?

Young'sModulusisameasureofthestiffnessofamaterial.AlargeYoung'sModulusmeansthematerialisstifferandrequiresalargestresstoproduceasmallstrain.

19.Howcanwedeterminethespringconstantfromaforceextensiongraph?

Thespringconstantisequaltotheslopeofthestraightlineportionoftheforceextensiongraph.Theslopeiscalculatedas$k=\frac{\DeltaF}{\Deltax}$.

20.Whydoesthedensityofwaterincreasefrom0°Cto4°C?

Inthetemperaturerangefrom0°Cto4°C,thewatermoleculesstarttoformamoreorderedstructure.Thehydrogenbondingbetweenwatermoleculescausesthewatertocontractslightly,resultinginadecreaseinvolume.Sincemassisconstant,accordingto$\rho=\frac{m}{V}$,thedensityincreases.

21.Aliquidhasadensityof850$kg/m^3$.Whatisitsrelativedensity?

Usingtheformulaforrelativedensity$RD=\frac{\rho_{liquid}}{\rho_{water}}$,with$\rho_{liquid}=850\kg/m^3$and$\rho_{water}=1000\kg/m^3$,weget$RD=\frac{850\kg/m^3}{1000\kg/m^3}=0.85$.

22.Aspringisstretchedbyaforceof12Nandtheextensionis0.1m.Whatisthespringconstant?

UsingHooke'sLaw$F=kx$,wecanrearrangeitto$k=\frac{F}{x}$.Substituting$F=12\N$and$x=0.1\m$,weget$k=\frac{12\N}{0.1\m}=120\N/m$.

23.Awireofcrosssectionalarea$4\times10^{7}\m^2$isunderastressof$1.5\times10^8\Pa$.Whatistheforceapplied?

Usingthestressformula$\sigma=\frac{F}{A}$,wecanrearrangeitto$F=\sigmaA$.Substituting$\sigma=1.5\times10^8\Pa$and$A=4\times10^{7}\m^2$,weget$F=(1.5\times10^8\Pa)\times(4\times10^{7}\m^2)=60\N$.

24.Arodoflength3misstretchedtoanewlengthof3.003m.Whatisthestrain?

Theextension$x=3.003\m3\m=0.003\m$.Usingthestrainformula$\epsilon=\frac{x}{L}$,with$L=3\m$and$x=0.003\m$,weget$\epsilon=\frac{0.003\m}{3\m}=0.001$.

25.IfamaterialhasaYoung'sModulusof$1.8\times10^{11}\Pa$andastrainof0.0015,whatisthestress?

Usingtheformula$E=\frac{\sigma}{\epsilon}$,wecanrearrangeitto$\sigma=E\epsilon$.Substituting$E=1.8\times10^{11}\Pa$and$\epsilon=0.0015$,weget$\sigma=(1.8\times10^{11}\Pa)\times(0.0015)=2.7\times10^8\Pa$.

26.Aspringwithaspringconstantof35N/miscompressedby0.12m.Whatistheelasticpotentialenergystored?

Usingtheformula$E_p=\frac{1}{2}kx^{2}$,with$k=35\N/m$and$x=0.12\m$,weget$E_p=\frac{1}{2}\times(35\N/m)\times(0.12\m)^{2}=\frac{1}{2}\times35\times0.0144\J=0.252\J$.

27.Whatarethetwomaintypesoffractureandhowdotheydiffer?

Thetwomaintypesarebrittlefractureandductilefracture.Brittlefractureoccurssuddenlywithoutmuchplasticdeformation.Ductilefractureoccursaftersignificantplasticdeformation,withthematerialneckingbeforebreaking.

28.Whydocompositematerialsoftenhavehighstrengthtoweightratios?

Compositematerialscombineastrongreinforcement(suchasfibers)withalightweightmatrix.Thereinforcementprovideshighstrength,whilethematrixholdsthereinforcementinplaceanddistributestheload.Thiscombinationallowsthematerialtohavehighstrengthwhilekeepingtheweightrelativelylow.

29.Howdoestheviscosityofagaschangewithincreasingtemperature?

Theviscosityofagasincreaseswithincreasingtemperature.Asthetemperaturerises,thegasmoleculeshavemorekineticenergyandcollidemorefrequently,increasingtheresistancetoflow.

30.DifferentiatebetweenNewtonianandnonNewtonianfluidsintermsofviscosity.

Newtonianfluidshaveaconstantviscosityregardlessoftheshearrate.NonNewtonianfluidshaveaviscositythatchangesdependingontheshearrate.

31.Acubeofmaterialhasasidelengthof5cmandamassof300g.Calculateitsdensity.

Thevolumeofthecube$V=(5\cm)^3=125\cm^3$.Thedensity$\rho=\frac{m}{V}=\frac{300\g}{125\cm^3}=2.4\g/cm^3$.Convertingto$kg/m^3$,weget$2.4\times1000\kg/m^3=2400\kg/m^3$.

32.Aspringisstretchedfromaninitiallengthof10cmtoafinallengthof12cmbyaforceof8N.Whatisthespringconstant?

Theextension$x=(1210)\cm=2\cm=0.02\m$.UsingHooke'sLaw$F=kx$,wecanrearrangeitto$k=\frac{F}{x}$.Substituting$F=8\N$and$x=0.02\m$,weget$k=\frac{8\N}{0.02\m}=400\N/m$.

33.Awirehasastressof$1.2\times10^8\Pa$andacrosssectionalareaof$3\times10^{7}\m^2$.Whatistheforceapplied?

Usingthestressformula$\sigma=\frac{F}{A}$,wecanrearrangeitto$F=\sigmaA$.Substituting$\sigma=1.2\times10^8\Pa$and$A=3\times10^{7}\m^2$,weget$F=(1.2\times10^8\Pa)\times(3\times10^{7}\m^2)=36\N$.

34.Arodoforiginallength2.5misstretchedby0.00125m.Whatisthestrain?

Usingthestrainformula$\epsilon=\frac{x}{L}$,with$x=0.00125\m$and$L=2.5\m$,weget$\epsilon=\frac{0.00125\m}{2.5\m}=0.0005$.

35.IfamaterialhasaYoung'sModulusof$2\times10^{11}\Pa$andastressof$3\times10^8\Pa$,whatisthestrain?

Usingtheformula$E=\frac{\sigma}{\epsilon}$,wecanrearrangeitto$\epsilon=\frac{\sigma}{E}$.Substituting$\sigma=3\times10^8\Pa$and$E=2\times10^{11}\Pa$,weget$\epsilon=\frac{3\times10^8\Pa}{2\times10^{11}\Pa}=0.0015$.

36.Aspringwithaspringconstantof50N/misstretchedby0.18m.Whatistheelasticpotentialenergystored?

Usingtheformula$E_p=\frac{1}{2}kx^{2}$,with$k=50\N/m$and$x=0.18\m$,weget$E_p=\frac{1}{2}\times(50\N/m)\times(0.18\m)^{2}=\frac{1}{2}\times50\times0.0324\J=0.81\J$.

37.Explaintheconceptoffatigueinmaterials.

Fatigueistheweakeningofamaterialcausedbycyclicloading.Repeatedstress,evenifbelowtheultimatetensilestrength,cancausecrackstoinitiateandgrowovertime,eventuallyleadingtofailure.

38.Whatarethecomponentsofacompositematerialandtheirroles?

Thecomponentsarethematrixandthereinforcement.Thereinforcementprovidesstrengthandstiffnesstothecomposite.Thematrixholdsthereinforcementinplace,transferstheloadbetweenthereinforcementelements,andprotectsthereinforcementfromdamage.

39.Whyisitimportanttounderstandtheviscosityoffluidsinengineeringapplications?

Understandingviscosityisimportantinmanyengineeringapplications.Forexample,influidflowsystems(suchaspipelines),viscosityaffectsthepressuredropandthepumpingpowerrequired.Inlubrication,theviscosityofthelubricantdeterminesitsabilitytoreducefrictionbetweenmovingparts.

40.Aliquidhasarelativedensityof0.95.Whatisitsdensityin$kg/m^3$?

Usingtheformulaforrelativedensity$RD=\frac{\rho_{liquid}}{\rho_{water}}$,wecanrearrangeitto$\rho_{liquid}=RD\times\rho_{water}$.Substituting$RD=0.95$and$\rho_{water}=1000\kg/m^3$,weget$\rho_{liquid}=0.95\times1000\kg/m^3=950\kg/m^3$.

41.Aspringisstretchedbyaforceof15Nandthespringconstantis75N/m.Whatistheextension?

UsingHooke'sLaw$F=kx$,wecanrearrangeitto$x=\frac{F}{k}$.Substituting$F=15\N$and$k=75\N/m$,weget$x=\frac{15\N}{75\N/m}=0.2\m$.

42.Awirehasacrosssectionalareaof$2.5\times10^{7}\m^2$andisunderaforceof50N.Whatisthestres