2025年中北大學(xué)框架考試題庫及答案大學(xué)物理一、選擇題1.一質(zhì)點(diǎn)作直線運(yùn)動(dòng)，某時(shí)刻的瞬時(shí)速度$v=2m/s$，瞬時(shí)加速度$a=-2m/s^2$，則一秒鐘后質(zhì)點(diǎn)的速度（）A.等于零B.等于-2m/sC.等于2m/sD.不能確定答案：D解析：因?yàn)閮H知道某時(shí)刻的瞬時(shí)速度和瞬時(shí)加速度，加速度不一定是恒定的，所以不能用勻變速直線運(yùn)動(dòng)的速度公式$v=v_0+at$來確定一秒鐘后質(zhì)點(diǎn)的速度。2.一質(zhì)量為$m$的質(zhì)點(diǎn)，在半徑為$R$的半球形容器中，由靜止開始自邊緣上的$A$點(diǎn)滑下，到達(dá)最低點(diǎn)$B$時(shí)，它對容器的正壓力為$N$。則質(zhì)點(diǎn)自$A$滑到$B$的過程中，摩擦力對其做的功為（）A.$\frac{1}{2}R(N-3mg)$B.$\frac{1}{2}R(3mg-N)$C.$\frac{1}{2}R(N-mg)$D.$\frac{1}{2}R(N-2mg)$答案：A解析：在最低點(diǎn)$B$，根據(jù)牛頓第二定律$N-mg=\frac{mv^{2}}{R}$，可得$v^{2}=\frac{R(N-mg)}{m}$。根據(jù)動(dòng)能定理$W_f+mgR=\frac{1}{2}mv^{2}$，將$v^{2}$代入可得$W_f=\frac{1}{2}R(N-3mg)$。二、填空題1.一物體作斜拋運(yùn)動(dòng)，初速度為$v_0$，拋射角為$\theta$，則物體到達(dá)最高點(diǎn)時(shí)的切向加速度$a_t=$______，法向加速度$a_n=$______。答案：$0$；$g$解析：在最高點(diǎn)，速度方向水平，加速度為重力加速度$g$，切向加速度是加速度沿速度方向的分量，此時(shí)速度水平，重力加速度垂直于速度方向，所以切向加速度$a_t=0$，法向加速度$a_n=g$。2.一彈簧原長$l_0=0.1m$，勁度系數(shù)$k=50N/m$，其一端固定在半徑為$R=0.1m$的半圓環(huán)的端點(diǎn)$A$，另一端與一套在半圓環(huán)上的小環(huán)相連。在把小環(huán)由半圓環(huán)中點(diǎn)$B$移到另一端$C$的過程中，彈簧的拉力對小環(huán)所做的功為______。答案：$-0.25J$解析：在$B$點(diǎn)，彈簧伸長量$x_1=\sqrt{2}R-R$；在$C$點(diǎn)，彈簧伸長量$x_2=2R-R$。根據(jù)彈簧彈力做功公式$W=-\frac{1}{2}k(x_2^{2}-x_1^{2})$，代入數(shù)據(jù)可得$W=-0.25J$。三、計(jì)算題1.質(zhì)量為$m$的子彈以速度$v_0$水平射入沙土中，設(shè)子彈所受阻力與速度反向，大小與速度成正比，比例系數(shù)為$k$，忽略子彈的重力，求：（1）子彈射入沙土后，速度隨時(shí)間變化的函數(shù)式；（2）子彈進(jìn)入沙土的最大深度。解：（1）根據(jù)牛頓第二定律$F=ma$，子彈所受阻力$F=-kv$，則$m\frac{dv}{dt}=-kv$，分離變量$\frac{dv}{v}=-\frac{k}{m}dt$，兩邊積分$\int_{v_0}^{v}\frac{dv}{v}=-\frac{k}{m}\int_{0}^{t}dt$，可得$\ln\frac{v}{v_0}=-\frac{k}{m}t$，即$v=v_0e^{-\frac{k}{m}t}$。（2）因?yàn)?v=\frac{dx}{dt}=v_0e^{-\frac{k}{m}t}$，分離變量$dx=v_0e^{-\frac{k}{m}t}dt$，兩邊積分求最大深度$x_{max}=\int_{0}^{\infty}v_0e^{-\frac{k}{m}t}dt$，根據(jù)積分公式$\int_{0}^{\infty}e^{-at}dt=\frac{1}{a}(a\gt0)$，這里$a=\frac{k}{m}$，所以$x_{max}=\frac{mv_0}{k}$。高等數(shù)學(xué)一、選擇題1.函數(shù)$f(x)=\frac{x-1}{x^{2}-1}$在$x=1$處（）A.連續(xù)B.有可去間斷點(diǎn)C.有跳躍間斷點(diǎn)D.有無窮間斷點(diǎn)答案：B解析：$f(x)=\frac{x-1}{x^{2}-1}=\frac{x-1}{(x-1)(x+1)}=\frac{1}{x+1}(x\neq1)$，$\lim\limits_{x\rightarrow1}f(x)=\frac{1}{2}$，但函數(shù)在$x=1$處無定義，所以$x=1$是可去間斷點(diǎn)。2.設(shè)函數(shù)$y=f(x)$在點(diǎn)$x_0$處可導(dǎo)，且$f^\prime(x_0)\gt0$，則曲線$y=f(x)$在點(diǎn)$(x_0,f(x_0))$處的切線傾斜角$\alpha$的取值范圍是（）A.$(0,\frac{\pi}{2})$B.$[0,\frac{\pi}{2})$C.$(\frac{\pi}{2},\pi)$D.$[\frac{\pi}{2},\pi)$答案：A解析：因?yàn)榍芯€斜率$k=f^\prime(x_0)\gt0$，而$k=\tan\alpha$，且$\alpha\in[0,\pi)$，所以$\alpha\in(0,\frac{\pi}{2})$。二、填空題1.設(shè)函數(shù)$y=\ln(1+x^{2})$，則$y^\prime|_{x=1}=$______。答案：$1$解析：根據(jù)復(fù)合函數(shù)求導(dǎo)法則，$y^\prime=\frac{2x}{1+x^{2}}$，將$x=1$代入可得$y^\prime|_{x=1}=1$。2.曲線$y=x^{3}-3x^{2}+2x+1$的拐點(diǎn)坐標(biāo)為______。答案：$(1,1)$解析：先求$y^\prime=3x^{2}-6x+2$，再求$y^{\prime\prime}=6x-6$，令$y^{\prime\prime}=0$，得$x=1$，當(dāng)$x\lt1$時(shí)，$y^{\prime\prime}\lt0$；當(dāng)$x\gt1$時(shí)，$y^{\prime\prime}\gt0$，將$x=1$代入原函數(shù)得$y=1$，所以拐點(diǎn)坐標(biāo)為$(1,1)$。三、計(jì)算題1.求不定積分$\int\frac{x+1}{x^{2}+2x+3}dx$。解：令$u=x^{2}+2x+3$，則$du=(2x+2)dx=2(x+1)dx$，所以$\int\frac{x+1}{x^{2}+2x+3}dx=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|x^{2}+2x+3|+C$。2.求由曲線$y=x^{2}$與直線$y=x+2$所圍成的平面圖形的面積。解：先求交點(diǎn)，聯(lián)立方程$\begin{cases}y=x^{2}\\y=x+2\end{cases}$，即$x^{2}-x-2=0$，解得$x=-1$或$x=2$。根據(jù)定積分求面積公式$S=\int_{-1}^{2}[(x+2)-x^{2}]dx=\left(\frac{1}{2}x^{2}+2x-\frac{1}{3}x^{3}\right)\big|_{-1}^{2}=\frac{9}{2}$。大學(xué)英語一、詞匯與語法選擇題1.Theprofessor,aswellashisstudents,______interestedinthenewresearch.A.areB.isC.havebeenD.were答案：B解析：當(dāng)主語后面跟有aswellas連接的詞組時(shí)，謂語動(dòng)詞的數(shù)要與前面的主語保持一致，這里前面的主語是Theprofessor，是第三人稱單數(shù)，所以用is。2.Ican'tfindmykeys.I______theminthecar.A.mightleaveB.mighthaveleftC.mayleaveD.mayhavebeenleft答案：B解析：“mighthavedone”表示對過去事情的可能性推測，“我可能把鑰匙落在車?yán)锪恕?，這里是對過去發(fā)生的事情的推測，所以用mighthaveleft。二、閱讀理解題Passage:Inmodernsociety,peoplearemoreandmoreconcernedabouttheirhealth.Healthexpertspointoutthatitisveryimportanttoeatabalanceddiet.Abalanceddietincludesavarietyoffoodsfromdifferentfoodgroups,suchasgrains,fruits,vegetables,dairyproducts,andproteins.Grainsarethemainsourceofenergyforourbody.Theycanprovideuswithcarbohydrates,whichareessentialforourdailyactivities.Fruitsandvegetablesarerichinvitaminsandminerals.Theycanhelpuspreventmanydiseases.Dairyproductsaregoodforourbonesandteethbecausetheycontaincalcium.Proteinsarenecessaryforthegrowthandrepairofourbodytissues.However,manypeoplehavebadeatinghabits.Somepeopleliketoeatjunkfood,suchashamburgers,fries,andsoda.Thesefoodsarehighincalories,fat,andsugar,butlowinnutrients.Eatingtoomuchjunkfoodcanleadtoobesity,highbloodpressure,andotherhealthproblems.Tohaveahealthydiet,weshouldmakeagoodplan.Weshouldeatthreeregularmealsadayandavoideatingbetweenmeals.Weshouldalsocontroltheportionsizeofourmeals.Inaddition,weshoulddrinkplentyofwatereveryday.1.Whatisthemainideaofthispassage?A.Peoplearemoreconcernedabouttheirhealthinmodernsociety.B.Abalanceddietisimportantforourhealth.C.Junkfoodisbadforourhealth.D.Weshouldmakeagoodplanforourdiet.答案：B解析：文章開篇指出人們越來越關(guān)注健康，接著強(qiáng)調(diào)均衡飲食的重要性，后文分別介紹了各類食物的作用以及不良飲食習(xí)慣的危害，最后提出健康飲食的建議，所以文章主旨是均衡飲食對健康很重要。2.WhichofthefollowingisNOTmentionedasabenefitofabalanceddiet?A.Providingenergy.B.Preventingdiseases.C.Makingpeoplemorebeautiful.D.Helpingwiththegrowthandrepairofbodytissues.答案：C解析：文中提到谷物能提供能量，水果和蔬菜能預(yù)防疾病，蛋白質(zhì)有助于身體組織的生長和修復(fù)，而沒有提到均衡飲食能讓人更美麗。三、寫作題Writeacompositiononthetopic"MyHobbies".Youshouldwriteatleast120wordsaccordingtothefollowingoutline:1.Introduction:Brieflyintroduceyourhobbies.2.Body:Describeoneofyourhobbiesindetail,includingwhenyoustartedit,whyyoulikeit,andwhatbenefitsyougetfromit.3.Conclusion:Summarizeyourhobbiesandhowtheyenrichyourlife.MyHobbiesInmysparetime,Ihaveseveralhobbiesthatbringmegreatjoyandfulfillment.Amongthem,readingismyfavoriteone.IstartedreadingwhenIwasalittlekid.Myparentsoftenboughtmepicture-books,andIwasimmediatelyattractedbythecolorfulpicturesandsimplestories.AsIgrewolder,Ibegantoreadnovels,poetry,andvariouskindsofnon-fictionbooks.ThereareseveralreasonswhyIlikereading.Firstly,itisawonderfulwaytoescapefromtherealworld.WhenIopenabook,Icanenteradifferentworldfullofimaginationandadventure.Sec