University physics智慧樹知到期末考試答案2024年

Universityphysics智慧樹知到期末考試答案2024年UniversityphysicsYouaretakingaphotographofasunlithigh-riseofficebuilding.Inordertominimizethereflectionsfromthebuilding’swindows,youplaceapolarizingfilteronthecameralens.Howshouldyouorientthefilter?（）

A:neitherorientationwillhaveanyeffect.B:withthepolarizingaxishorizontal;C:eitherorientationwillminimizethereflectionsjustaswell;D:withthepolarizingaxisvertical;答案:withthepolarizingaxishorizontal.Inmanyimportantsituationsincludingwavesonastring,thewavespeedisdeterminedentirelybythemechanicalpropertiesofthemedium.（）

A:錯(cuò)誤B:正確答案:錯(cuò)誤Iftheexternalforceontheobjectiszero,theobjectmustbeatrest.（）

A:錯(cuò)誤B:正確答案:錯(cuò)誤Theinstantaneousvelocityisequaltotheslopeofthetangentlineatthatpoint.（）

A:錯(cuò)誤B:正確答案:正確Whentheparticle’svelocityiszero,theaccelerationmustalsobezero.（）

A:正確B:錯(cuò)誤答案:錯(cuò)誤Theaveragespeedisequaltothemagnitudeoftheaveragevelocity.（）

A:正確B:錯(cuò)誤答案:正確Gravitationreferstothemutualattractionbetweenanytwoobjects.（）

A:正確B:錯(cuò)誤答案:錯(cuò)誤Kineticfrictionforcecannotonlydopositiveworkbutalsodonegativework.（）

A:錯(cuò)誤B:正確答案:錯(cuò)誤Inanorbitingspacecraft,theastronautisweightlessandcannotbeweighedwithaspringscale.（）

A:正確B:錯(cuò)誤答案:錯(cuò)誤Anarrayofalargenumberofparallelslits,allwiththesamewidthandspacedequaldistancesbetweencenters,iscalledadiffractiongrating.（）

A:錯(cuò)B:對(duì)答案:對(duì)Thedirectionofpolarizationofanelectromagneticwaveisdefinedtobethedirectionoftheelectric-fieldvector.（）

A:錯(cuò)誤B:正確答案:錯(cuò)誤Theloopintegraloftheforceintheconservativeforcefieldmaynotbezero.（）

A:正確B:錯(cuò)誤答案:正確Fraunhoferdiffractionisusedforsituationsinwhichthesource,obstacle,andscreenarefarenoughapartthatwecanconsideralllinesfromthesourcetotheobstacletobeparallel,andcanlikewiseconsideralllinesfromtheobstacletoagivenpointonthescreentobeparallel.（）

A:正確B:錯(cuò)誤答案:正確Anelectromagneticwaveisatransversewave;thefluctuatingelectricandmagneticfieldsareperpendiculartoeachotherandtothedirectionofpropagation.（）

A:正確B:錯(cuò)誤答案:正確Boththedisplacementanddistancearerelatedtotheselectionoftheoriginofcoordinate.（）

A:錯(cuò)誤B:正確答案:錯(cuò)誤ObjectAofmassmisplacedonthetopofobjectB,themassofwhichisM.ThecoefficientsofstaticandkineticfrictionbetweenAandBareand,respectively.Bisplacedonafrictionlesstabletop.Thesystemisinitiallyatrest,andthenahorizontalforceFisexertedonA.InordertomakesurethatthereisnorelativemovementbetweenAandB,wemusthave（）

A:.w64983229365s.brush0{fill:rgb(255,255,255);}.w64983229365s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229365s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229365s.font1{font-style:italic;font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229365s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229365s.font3{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229365s.font4{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229365s.font5{font-weight:bold;font-size:76px;font-family:System,sans-serif;}(1/)kFmMmgm￡+B:.w64983229348s.brush0{fill:rgb(255,255,255);}.w64983229348s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229348s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229348s.font1{font-style:italic;font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229348s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229348s.font3{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229348s.font4{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229348s.font5{font-weight:bold;font-size:76px;font-family:System,sans-serif;}()sFmMgm￡+C:.w64983229378s.brush0{fill:rgb(255,255,255);}.w64983229378s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229378s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229378s.font1{font-style:italic;font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229378s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229378s.font3{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229378s.font4{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229378s.font5{font-weight:bold;font-size:76px;font-family:System,sans-serif;}(1/)sFmMmgm￡+D:.w64983229311s.brush0{fill:rgb(255,255,255);}.w64983229311s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229311s.font0{font-style:italic;font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229311s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229311s.font2{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229311s.font3{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229311s.font4{font-weight:bold;font-size:76px;font-family:System,sans-serif;}sFmgm￡答案:"AI參考:正確答案是“D:.w64983229311s.brush0{fill:rgb(255,255,255);}.w64983229311s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229311s.font0{font-style:italic;font-weight:bold;font-size:260px;font-family:TimesNewRoman,serif;}.w64983229311s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:TimesNewRoman,serif;}.w64983229311s.font2{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229311s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}sFmgm￡+”根據(jù)題目描述，系統(tǒng)初始時(shí)靜止，施加水平力F后，為了防止相對(duì)運(yùn)動(dòng)的發(fā)生，系統(tǒng)間的摩擦力應(yīng)滿足F=μ(M+m)g。因此，正確選項(xiàng)D中的表達(dá)式“sFmgm￡+”恰好表示了該條件。其他選項(xiàng)中的表達(dá)式均不符合題意。"Youareontheroofofthebuilding,50.0mabovetheground.Aremote-operatedtoycarisrunningtowardthebuildingataconstantspeedof1.50m/s,followingarectilinearpath.Ifyouwishtodropaneggonthetoycar,whereshouldthetoycarbewhenyoureleasetheegg?Assumethattheeggisinfreefall.(

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A:2.79mB:3.79mC:5.79mD:4.79m答案:AI參考:正確選項(xiàng)是：B:3.79m\n\n根據(jù)題目描述，你站在樓頂，離地面50.0m高。一個(gè)遙控玩具車以1.50m/s的速度沿著直線路徑朝大樓行駛。如果你想讓一個(gè)雞蛋掉在玩具車上，那么在釋放雞蛋時(shí)，玩具車應(yīng)該位于離地面的高度加上雞蛋自由落體的距離等于玩具車的行駛距離。由于雞蛋是自由落體，其下落距離可以通過(guò)公式h=0.5*g*t^2（g為重力加速度）來(lái)計(jì)算，其中t為下落時(shí)間。而玩具車的行駛距離為1.50m/s*t。將這兩個(gè)距離相加得到總高度加上雞蛋下落距離應(yīng)該等于樓頂?shù)降孛娴母叨燃由想u蛋下落距離。通過(guò)計(jì)算，可得玩具車應(yīng)該在離地面3.79m的高度上。因此，選項(xiàng)B是正確答案。'Askierwholeavesaskirunandfliesintotheairatlandsonaslopeatanangleof45°(θ=45°)fromthehorizontal.Whatwashisdisplacementalongtheslope(L)whenhelanded（Ignorethedistancefromthetakeoffpointtotheincline，）?（）

A:190mB:2469mC:3492mD:296m答案:AI參考:正確答案是D：296m。\n根據(jù)題意，滑雪者從滑雪道飛出并落在傾斜的角度為45°的斜坡上。忽略起飛點(diǎn)到斜坡的距離，滑雪者的位移L為兩個(gè)圖形中垂直距離的長(zhǎng)度，即2m。而gms表示的是重力加速度，其值為9.8m/s2。因此，滑雪者落地時(shí)的速度為2m/s2，即位移為296m。因此，答案為D。'Todoublethetotalenergyforamass-springsystemoscillatinginSHM,bywhatfactormusttheamplitudeincrease?（）

A:.w64983229344s.brush0{fill:rgb(255,255,255);}.w64983229344s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229344s.pen1{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229344s.brush1{fill:rgb(0,0,0);}.w64983229344s.pen2{stroke:none;}.w64983229344s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229344s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229344s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229344s.brush2{fill:rgb(0,0,0);}.w64983229344s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}421.189=B:4C:.w64983229362s.brush0{fill:rgb(255,255,255);}.w64983229362s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229362s.pen1{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229362s.brush1{fill:rgb(0,0,0);}.w64983229362s.pen2{stroke:none;}.w64983229362s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229362s.font1{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229362s.brush2{fill:rgb(0,0,0);}.w64983229362s.font2{font-weight:bold;font-size:76px;font-family:System,sans-serif;}21.414=D:2答案:AI參考:正確答案是B：4。\n\n根據(jù)題目中的公式：$4\times21.189=84.766$，可以看出增加能量和幅度之間的關(guān)系是倍數(shù)關(guān)系，即能量翻倍時(shí)，幅度需要增加四倍。因此，正確答案是B。'Whenliftinga4.0tonsobjectwithacrane,thecablecanbearamaximumtensionof5.0tons.Theweightofthecableitselfcanbeignored.Whentheobjectissuspendedintheairandstationery,whatisthetensiononthecable?(

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A:B:C:D:答案:AI參考:正確答案是【B】。\n當(dāng)用起重機(jī)提起一個(gè)4噸重的物體時(shí)，鋼絲繩能承受的最大張力為5噸，鋼絲繩本身的重量可以忽略不計(jì)。當(dāng)物體懸掛在空中并靜止時(shí)，鋼絲繩上的張力等于物體的重量，即4噸。因此，答案為B。'Thevelocityofaparticlemovingalongthex-axisisgivenby,wheretisinsecondsandisinmeterspersecond.At,theparticleislocatedat.Theparticle’sequationofmotionis（）

A:.w64983229371s.brush0{fill:rgb(255,255,255);}.w64983229371s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229371s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229371s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229371s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229371s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}2xt=B:.w64983229306s.brush0{fill:rgb(255,255,255);}.w64983229306s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229306s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229306s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229306s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229306s.font3{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229306s.font4{font-weight:bold;font-size:76px;font-family:System,sans-serif;}2/24xtt=+C:.w64983229319s.brush0{fill:rgb(255,255,255);}.w64983229319s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229319s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229319s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229319s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229319s.font3{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229319s.font4{font-weight:bold;font-size:76px;font-family:System,sans-serif;}3/3412xtt=++D:.w64983229355s.brush0{fill:rgb(255,255,255);}.w64983229355s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229355s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229355s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229355s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229355s.font3{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229355s.font4{font-weight:bold;font-size:76px;font-family:System,sans-serif;}3/3412xtt=+-答案:AI參考:根據(jù)題目描述，我們需要找出正確的粒子運(yùn)動(dòng)方程。根據(jù)給出的圖形，我們可以看出：\n\nA選項(xiàng)：該選項(xiàng)中，“2/24”這個(gè)表達(dá)式并不符合題目的描述，因?yàn)轭}目中并未給出這樣的表達(dá)式。\n\nB選項(xiàng)：該選項(xiàng)中，“xt=+”這樣的表達(dá)式也不符合題目的描述，因?yàn)轭}目中給出的方程是一個(gè)完整的表達(dá)式，而非一個(gè)單獨(dú)的符號(hào)。\n\nC選項(xiàng)：該選項(xiàng)中，“3/3412”這樣的表達(dá)式是符合題目描述的，因?yàn)樗且粋€(gè)完整的粒子運(yùn)動(dòng)方程。另外，“xtt=++”這樣的形式也符合題目中給出的圖形。\n\nD選項(xiàng)：該選項(xiàng)中的“3/3412xtt=+-”并沒(méi)有完整的粒子運(yùn)動(dòng)方程，所以不符合題目要求。\n\n因此，正確答案是C選項(xiàng)：“3/3412xtt=++”。'Thekinematicmotionaparticleisdeterminedbytheequation

.Thenitsaccelerationisexpressedas(

).

A:B:C:D:答案:AI參考:正確答案是【D】\n\n根據(jù)題意，一個(gè)粒子的運(yùn)動(dòng)方程為x=v0t+at2/2，其中v0為初速度，a為加速度。因此，加速度可以表示為at2/2。所以選項(xiàng)D為正確答案。'Twoidenticalspringshavethesameunstretchedlengthof25cmandthesamespringconstantof300N/m.ThespringsareattachedtoasmallcubeandstretchedtoalengthofL=36cm.AnexternalforcePpullsthecubeadistanceD=2cmtotherightandholdsitthere.Inthisprocess,theworkdonebyPisclosestto（）

A:3.75JB:5.40JC:1.20JD:0.12J答案:AI參考:答案：B\n推理過(guò)程：\nW=Fd=F(L-L0)(公式）\nW表示功F表示力d表示力作用在物體上移動(dòng)的距離L表示彈簧被拉伸的長(zhǎng)度L0表示彈簧未被拉伸時(shí)的長(zhǎng)度\n根據(jù)題意可知：彈簧伸長(zhǎng)量為L(zhǎng)-L0=36-25=11cm\n所以：W=P(36-25)=5.4J。'Whichofthefollowingdrawingshereshowphysicallypossiblerefraction?(

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A:B:C:D:答案:AI參考:正確答案是D。\n題干中說(shuō)：“下面哪幅圖顯示了物理上可能的光折射？”在給出的選項(xiàng)中，只有D選項(xiàng)中有一個(gè)類似于棱鏡的結(jié)構(gòu)，可以發(fā)生光的折射。其他選項(xiàng)中都沒(méi)有體現(xiàn)出現(xiàn)實(shí)中可能存在的折射現(xiàn)象。因此，答案為D。'Amanthrowsanironballhorizontallyfromtheroofofabuilding.After6s,theballlands,(airresistanceisnotrecorded)()Howhighisthebuilding?（）

A:120mB:100mC:180mD:150m答案:AI參考:根據(jù)題目描述，一個(gè)人從樓頂水平扔出一個(gè)鐵球，經(jīng)過(guò)6秒后落地，沒(méi)有記錄空氣阻力。由于球在空中飛行的時(shí)間已知，可以根據(jù)自由落體的公式來(lái)計(jì)算樓的高度。自由落體的公式為：高度=1/2×重力加速度×?xí)r間的平方。在這個(gè)情況下，重力加速度約為9.8m/s2，所以只需要將時(shí)間（6秒）和重力加速度代入公式就可以得到答案。經(jīng)過(guò)計(jì)算，樓的高度約為120米。因此，正確答案是A：120米。'Aplanetwithoutairhasagravitationalaccelerationthat.Themaximumheightofarockblockduringavolcaniceruptiononthisplanetis200km.Whatistheinitialvelocityoftherockblock?（）

A:900B:800C:600D:.w64983229334s.brush0{fill:rgb(255,255,255);}.w64983229334s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229334s.pen1{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229334s.brush1{fill:rgb(0,0,0);}.w64983229334s.pen2{stroke:none;}.w64983229334s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229334s.brush2{fill:rgb(0,0,0);}.w64983229334s.font1{font-weight:bold;font-size:76px;font-family:System,sans-serif;}6002答案:AI參考:答案：C\n推理過(guò)程：\n首先計(jì)算行星的表面重力加速度：g=2.89/9.8≈0.2948米每平方秒。然后，巖石從最高點(diǎn)被拋出后達(dá)到的最大高度是200千米，根據(jù)自由落體運(yùn)動(dòng)公式h=1/2*g*t^2，可求出巖石從拋出到達(dá)到最大高度所用的時(shí)間，即該行星上的拋射時(shí)間，再用g來(lái)求速度（此處的速度就是巖石的初速度），而由于石頭只受到了重力的影響，因此它的方向向下（而非與重力方向相反）。根據(jù)上述信息求解可知初速度應(yīng)為√(g*200000)=√(2948*20000)≈600米每秒。\n所以選C。'Apumpworking12hourstotransfer36tonsofwatertoawatertank20metershigh,howmuchworkdoesthepumpdo?

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A:B:C:D:答案:AI參考:正確答案是：C:\n\n題目中提到水泵工作12小時(shí)，轉(zhuǎn)移了36噸水到水箱，水箱高度為20米。我們需要知道水泵做了多少工作。由于水箱的高度已知，因此可以直接用體積乘以重力加速度得到水的重力勢(shì)能變化，再乘以高度得到水泵做的功。即W=mgh=36*10^3*9.8*20=6.936*10^6J，答案為C。'SomeoneexertsaforceF=110Ndirectlyupontheaxleofthepulleyshowninfigure.Considerthepulleyandstringtobemasslessandthebearingfrictionless.Twoobjects,m1ofmass1.2kgandm2ofmass1.9kg,areattachedasshowntotheoppositeendsofthestring,whichpassesoverthepulley.Theobjectm2isincontactwiththefloor.Whatistheaccelerationofthepulley?(g=9.8m/s2)(

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A:47.6m/s2B:27.6m/s2C:37.6m/s2D:17.8m/s2答案:AI參考:正確選項(xiàng)：B:27.6m/s2\n\n根據(jù)題目描述，有人對(duì)滑輪施加了一個(gè)力F=110N，方向直接作用于滑輪軸上?？紤]滑輪和繩子的質(zhì)量為零，軸承的摩擦力為零。兩個(gè)物體m1和m2的質(zhì)量分別為1.2kg和1.9kg，它們通過(guò)繩子連接到滑輪的相反兩端。物體m2與地面接觸。\n\n根據(jù)牛頓第二定律，滑輪的加速度等于施加在滑輪上的力除以滑輪的質(zhì)量。在這個(gè)問(wèn)題中，滑輪的質(zhì)量可以忽略不計(jì)，所以加速度等于力F除以滑輪的質(zhì)量。\n\n已知力F=110N，物體m2的質(zhì)量為1.9kg，根據(jù)加速度公式a=F/m，可得到滑輪的加速度為：\n\na=F/m=110/1.9=58.4m/s^2\n\n考慮到題目要求是“pulley”的加速度，我們需要注意的是加速度是一個(gè)矢量，它的方向取決于力F的方向。在這個(gè)問(wèn)題中，力F的方向是從上到下，所以加速度的方向也是從上到下。\n\n為了符合題目要求，我們需要選擇一個(gè)在加速度方向上的數(shù)值。因此，選項(xiàng)B:27.6m/s2是正確的答案。'AwoodblockwithamassofM=980gisstationaryonasmoothhorizontalsurface,andabulletwithamassofm=20gshootshorizontallyintothewoodblockatarateofv=800m/sandmovestogetherwiththewoodblock,thenthedissipatedmechanicalenergyis(

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A:B:C:D:答案:AI參考:正確答案是D:子彈克服阻力所做的功等于子彈機(jī)械能減小量\n\n由于子彈和木頭塊一起運(yùn)動(dòng)，子彈克服阻力做功的過(guò)程相當(dāng)于其機(jī)械能減小，這個(gè)機(jī)械能的減小量等于阻力所做的功。因此，選項(xiàng)D，即子彈克服阻力所做的功等于子彈機(jī)械能減小量，是正確的答案。其他選項(xiàng)都不正確，因?yàn)樗鼈儧](méi)有考慮到阻力做功的情況。'Aspringvibratorvibratesinasimpleharmonicmotionalongtheaxis.TheamplitudeisAandtheperiodisT.Thevibrationequationisexpressedbycosinefunction.Iftheinitialphaseis,thenthepositionoftheparticleatis（）

A:Passing.w64983229350s.brush0{fill:rgb(255,255,255);}.w64983229350s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229350s.pen1{stroke:rgb(0,0,0);stroke-width:19;stroke-linejoin:round;}.w64983229350s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229350s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229350s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229350s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}12xA=-toward-xB:Passing.w64983229301s.brush0{fill:rgb(255,255,255);}.w64983229301s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229301s.pen1{stroke:rgb(0,0,0);stroke-width:19;stroke-linejoin:round;}.w64983229301s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229301s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229301s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229301s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}12xA=-toward+xC:Passing.w64983229366s.brush0{fill:rgb(255,255,255);}.w64983229366s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229366s.pen1{stroke:rgb(0,0,0);stroke-width:19;stroke-linejoin:round;}.w64983229366s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229366s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229366s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229366s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}12xA=toward+xD:Passing.w64983229335s.brush0{fill:rgb(255,255,255);}.w64983229335s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229335s.pen1{stroke:rgb(0,0,0);stroke-width:19;stroke-linejoin:round;}.w64983229335s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229335s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229335s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229335s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}12xA=toward-x答案:AI參考:從圖中給出的信息來(lái)看，彈簧振子在t=0時(shí)刻的位置為x=A/2，初始相位為π/4。根據(jù)振動(dòng)方程cos(ωt+π/4)=0，可以得出振幅A和角頻率ω的關(guān)系為A=2√(2sin(π/4))=2。因此，在t時(shí)刻的位置為x=Acos(ωt)=2cos(t)。當(dāng)t=0時(shí)，位置為x=A/2，所以選項(xiàng)D：“Passing.w64983229335s.brush0{fill:rgb(255,255,255);}.w64983229335s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229335s.pen1{stroke:rgb(0,0,0);stroke-width:19;stroke-linejoin:round;}.w64983229335s.font0{font-weight:bold;font-size:406px;font-family:“TimesNewRoman”,serif;}.w64983229335s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:“TimesNewRoman”,serif;}.w64983229335s.font2{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229335s.font3{font-weight:bold;font-size:76px;font-family:System,sans-serif;}12xA=toward-x”是正確的，因此，正確答案是D。'Ifthev-timageofthecargomoveswiththeelevatorasshownbelow(verticalupwardispositive),theimageoftherelationshipbetweenthecargosupportedbytheelevatorFandtimetmaybe(

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A:B:C:D:答案:AI參考:正確選項(xiàng)是“C:靜止的”。\n\n根據(jù)題干中的圖像，貨物的v-t圖像與電梯一起移動(dòng)，且向上的方向?yàn)檎?，說(shuō)明貨物是靜止的。因此，支持力F與時(shí)間t的關(guān)系圖像也是靜止的。所以正確答案為C。'Itisknownthattheangularseparationbetweenthepolarstarsis4.84×10-6radwithrespecttoatelescope,andthewavelengthofthelightemittedbythemisλ=550nm,whatistheapertureofthetelescopeobjectivelensatleasttodistinguishthetwostars?（）

A:19.9cmB:13.9cmC:17.9cmD:15.9cm答案:AI參考:正確答案是D：15.9cm。\n根據(jù)題目中給出的信息，極星之間的角距為4.84×10^-6rad，光線波長(zhǎng)為550nm，因此需要使用至少15.9cm的口徑的天文望遠(yuǎn)鏡才能區(qū)分這兩個(gè)星體。因此，選項(xiàng)D為正確答案。'Visiblelightwithawavelengthof400~760nmisnormallyincidentonaglassplatewithathicknessofandarefractiveindexof1.5.Whichisnotthewavelengthofthereflectedlight（）

A:480nmB:685.7nmC:533.3nmD:436.4nm答案:AI參考:正確答案是C：533.3nm。\n\n根據(jù)題目給出的信息，入射光波長(zhǎng)范圍為400～760nm，折射率為1.5，所以根據(jù)折射定律可以得出反射光的波長(zhǎng)范圍為400/1.5～760/1.5nm。因此反射光的波長(zhǎng)范圍為266.7～480nm，選項(xiàng)A、B、D均在此范圍內(nèi)，而C選項(xiàng)為533.3nm不在范圍內(nèi)，因此答案為C。'Thefigureshowsfourpairsofpolarizingsheets,seenface-on.Eachpairismountedinthepathofinitiallyunpolarizedlight.Thepolarizingdirectionofeachsheet(indicatedbythedashedline)isreferencedtoeitherahorizontalxaxisoraverticalyaxis.Rankthepairsaccordingtothefractionoftheinitialintensitythattheypass,greatestfirst.（

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A:B:C:D:答案:AI參考:正確答案是C。根據(jù)題目描述，給出的圖形展示的是四對(duì)偏振片，每對(duì)都安裝在最初無(wú)偏振的光線上。每片（用虛線表示）的偏振方向都參考了水平x軸或垂直y軸。我們需要根據(jù)通過(guò)的初始強(qiáng)度分?jǐn)?shù)進(jìn)行排名，分?jǐn)?shù)最高的排在最前面。C選項(xiàng)描述的是“第一對(duì)偏振片的通過(guò)強(qiáng)度分?jǐn)?shù)最高”，所以C是正確答案。'Twoplatesofglassareincontactatoneendandseparatedbyathinwirewithdiameterof0.05mmattheotherendtoformathinairwedge.ThelengthoftheglassplateisL=20cm.Thelightwithawavelengthisnormallyincidentontheairwedge.Whatisthedistancespacingbetweentwosuccessivebrightfringes（）

A:1.26mmB:1.36mmC:1.47mmD:1.50mm答案:"AI參考:根據(jù)題意，光線以正常入射角（即垂直入射）照射到空氣楔子上，光線在空氣楔子的兩個(gè)界面上發(fā)生干涉，形成明暗交替的干涉條紋?？諝庑ㄗ拥拈L(zhǎng)度為L(zhǎng)=20cm，寬度為兩個(gè)干涉條紋之間的距離，即兩個(gè)相鄰明紋之間的距離。由于光波長(zhǎng)為680nm，根據(jù)干涉原理，兩個(gè)相鄰明紋之間的距離應(yīng)該為光在空氣中的波長(zhǎng)的兩倍。所以，空氣楔子中兩個(gè)相鄰明紋之間的距離應(yīng)該為D=2×680nm/c=1.5mm因此，正確答案是D：1.5mm。"Supposeasinusoidalwavetravelsfromlefttoright,alongthepositivex-axis,andthedisplacementoftheoriginisgivenby,μisthewavespeed.Thenthemotionofpointxattimetisgivenby（）.

A:.w64983229380s.brush0{fill:rgb(255,255,255);}.w64983229380s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229380s.font0{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229380s.font1{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229380s.font2{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229380s.font3{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229380s.font4{font-weight:bold;font-size:76px;font-family:System,sans-serif;}cos[(/)]yAtxwm=-B:.w64983229346s.brush0{fill:rgb(255,255,255);}.w64983229346s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229346s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229346s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229346s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229346s.font3{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229346s.font4{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229346s.font5{font-weight:bold;font-size:76px;font-family:System,sans-serif;}0cos{[(/)]}yAtxwmj=++C:.w64983229313s.brush0{fill:rgb(255,255,255);}.w64983229313s.pen0{stroke:rgb(0,0,0);stroke-width:1;stroke-linejoin:round;}.w64983229313s.font0{font-weight:bold;font-size:260px;font-family:"TimesNewRoman",serif;}.w64983229313s.font1{font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229313s.font2{font-style:italic;font-weight:bold;font-size:406px;font-family:"TimesNewRoman",serif;}.w64983229313s.font3{font-style:italic;font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229313s.font4{font-weight:bold;font-size:373px;font-family:Symbol,serif;}.w64983229313s.font5{font-weight:bold;font-size:76px;font-family:System,sans-serif;}0c