辽宁省实验中学、鞍山一中、东北育才学校2017届高三上学期期末联考英语试题Word版含答案

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辽宁省重点高中协作校(沈阳二中、大连育明中学、辽师大附中、本溪高中等)2017届高三上学期期末考试英语

辽宁省重点高中协作校(沈阳二中、大连育明中学、辽师大附中、本溪高中等)2017届高三上学期期末考试英语

辽宁省重点高中协作校2017届高三上学期期末考试英语试题考生注意:1.本试卷分第I卷(选择题)和第II卷(非选择题)两部分,共150分,考试时间120分钟。

2.答题前,考生在答题考生务必用直径0.5毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚,并贴好条形码,请认真核准条形码上的准考证号、姓名和科目。

3.每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,在选涂其他答案标号,在试题卷上作答无效.........。

4.本试卷主要考试内容:高中综合。

第I卷第一部分听力(共两节,满分30分)第一节(共5小题,每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. How old is the woman?A. 16 years oldB. 19 years oldC. 20 years old2. What does the woman do?A. A teacherB. A studentC. A doctor3. What does the man do on holiday?A. Read booB. Listen to musicC. Travel with friends4. What is the woman doing?A. Using a travel websiteB. Flying to AtlantaC. Looing for a travel agent5. What is the man’s reaction to the news?A. He feels surprised at itB. He doesn’t care about itC. He is very happy at it第二节(共15小题,每小题1.5分,满分22.5分)听下面5段对话或独白。

辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校2017届高三上学期期末物理试卷Word版含解

辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校2017届高三上学期期末物理试卷Word版含解

2016-2017学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末物理试卷一、选择题1.两物体A、B按如图所示连接且处于静止状态,已知两物体质量为m A>m B,A物体和地面的动摩擦因数为μ.现在B上加一个水平力F,使物体B缓慢移动,物体A始终静止,则此过程中()A.物体A对地面的压力逐渐变小B.物体A受到的摩擦力不变C.绳的拉力逐渐变小D.地面对A的作用力不变2.一皮带传送装置如图所示,轻弹簧一端固定,另一端连接一个质量为m的滑块,已知滑块与皮带之间存在摩擦.现将滑块轻放在皮带上,弹簧恰好处于自然长度且轴线水平.若在弹簧从自然长度到第一次达最长的过程中,滑块始终未与皮带达到共速,则在此过程中滑块的速度和加速度变化情况是()A.速度增大,加速度增大B.速度增大,加速度减小C.速度先增大后减小,加速度先增大后减小D.速度先增大后减小,加速度先减小后增大3.如图所示,固定在水平桌面上的光滑金属导轨cd、eg处于方向竖直向下的匀强磁场中,金属杆ab与导轨接触良好.在两根导轨的端点d、e之间连接一电阻,其它部分电阻忽略不计.现用一水平向右的外力F1作用在金属杆ab上,使金属杆由静止开始向右沿导轨滑动,滑动中杆ab始终垂直于导轨.金属杆受到的安培力用F f表示,则关于图乙中F1与F f随时间t变化的关系图象可能的是()A.B.C.D.4.下列说法正确的是()A.太阳辐射的能量主要来自太阳内部的核聚变反应B.大量的氢原子从n=3的能级向低能级跃迁时只会辐射两种不同频率的光C.一束单色光照射到某种金属表面不能发生光电效应,是因为该束光的波长太短D.发生光电效应时,入射光的频率一定,光强越大,光电子最大初动能越大5.质量为M的木块在光滑的水平面上以速度v1向右运动,质量为m的子弹以速度v2水平向左射入木块(子弹留在木块内),要使木块停下来,必须发射子弹的数目为()A.B.C.D.6.如图所示,三个质点a、b、c质量分别为m1、m2、M,(M>>m1,M>>m2).a、b在同一平面内绕c沿逆时针方向做匀速圆周运动,它们的周期之比T a:T b=1:k.(k>1,为正整数)从图示位置开始,在b运动一周的过程中,则()A.a、b距离最近的次数为k次 B.a、b距离最近的次数为k+1次C.a、b、c共线的次数为2k次 D.a、b、c共线的次数为2k﹣2次7.一匀强电场的电场强度E随时间t变化的图象如图所示,在该匀强电场中,有一个带电粒子于t=0时刻由静止释放,若带电粒子只受电场力作用,则下列说法中正确的是()A.带电粒子只向一个方向运动B.0~2s内,电场力所做的功等于零C.4s末带电粒子回到原出发点D.2.5s~4s内,电场力的冲量等于零8.“嫦娥二号”卫星由地面发射后,进入地月转移轨道,经多次变轨最终进入距离月球表面100km,周期为118min的工作轨道,开始对月球进行探测,则()A.卫星在轨道Ⅲ上的运动速度比月球的第一宇宙速度小B.卫星在轨道Ⅲ上经过P点的速度比在轨道Ⅰ上经过P点时大C.卫星在轨道Ⅲ上运动的周期比在轨道Ⅰ上短D.卫星在轨道Ⅰ上的机械能比在轨道Ⅱ上大9.一物体静止在水平地面上,在竖直向上的拉力F的作用下开始向上运动,如图甲所示.在物体向上运动过程中,其机械能E与位移x的关系图象如图乙所示(空气阻力不计),已知曲线上点A处的切线的斜率最大,则()A.在x1处物体所受拉力最大x2过程中,物体的动能先增大后减小B.在x1~C.在x2处物体的速度最大x2过程中,物体的加速度先增大后减小D.在x1~10.如图所示,平行板电容器AB两极板水平放置,A在上方,B在下方,现将其和二极管串连接在电源上,已知A和电源正极相连,二极管具有单向导电性,一带电小球从AB间的某一固定点水平射入,打在B极板上的N点,小球的重力不能忽略,现通过上下移动A板来改变两极板AB间距(两极板始终平行),则下列说法正确的是()A.若小球带正电,当AB间距增大时,小球打在N点的右侧B.若小球带正电,当AB间距减小时,小球打在N点的左侧C.若小球带负电,当AB间距减小时,小球可能打在N点的右侧D.若小球带负电,当AB间距增大时,小球可能打在N点的左侧11.如图所示,闭合的矩形导体线圈abcd在匀强磁场中绕垂直于磁感线的对称轴OO′匀速转动,沿着OO′方向观察,线圈沿逆时针方向转动.已知匀强磁场的磁感强度为B,线圈匝数为n,ab边的边长为l1,ad边的边长为l2,线圈总电阻为R,转动的角速度为ω,则当线圈转至图示位置时()A.线圈中感应电流的方向为abcdaB.线圈中的感应电动势为2nBl2ωC.穿过线圈磁通量随时间的变化率最大D.线圈ad边所受安培力的大小为12.在光滑的水平面上方,有两个磁感应强度大小均为B,方向相反的水平匀强磁场,如图.PQ为两个磁场的边界,磁场范围足够大.一个边长为a、质量为m、电阻为R的金属正方形线框,以速度v垂直磁场方向从如图实线位置开始向右运动,当线框运动到分别有一半面积在两个磁场中时,线框的速度为,则下列说法正确的是()A.此过程中通过线框截面的电量为B.此时线框的加速度为C.此过程中回路产生的电能为mv2D.此时线框中的电功率为二、实验题13.在研究弹簧的形变与外力的关系的实验中,将弹簧水平放置测出其自然长度,然后竖直悬挂让其自然下垂,在其下端竖直向下施加外力F,实验过程是在弹簧外力的弹性限度内进行的.用记录的外力F与弹簧的形变量x,作出F﹣x图线如图所示,由图可知弹簧的劲度系数为N/m,图线不过坐标原点的原因是由于.14.用下列器材组装成一个电路,既能测量出电池组的电动势E和内阻r,又能同时描绘小灯泡的伏安特性曲线.A.电压表V1(量程6V、内阻很大)B.电压表V2(量程3V、内阻很大)C.电流表A(量程3A、内阻很小)D.滑动变阻器R(最大阻值10Ω、额定电流4A)E.小灯泡(2A、5W)F.电池组(电动势E、内阻r)G.开关一只,导线若干实验时,调节滑动变阻器的阻值,多次测量后发现:若电压表V1的示数增大,则电压表V2的示数减小.(1)请将设计的实验电路图在图甲的虚线方框中补充完整.(2)每一次操作后,同时记录电流表A、电压表V1和电压表V2的示数,组成两个坐标点(I1,U1)、(I2,U2),标到U﹣I坐标中,经过多次测量,最后描绘出两条图线,如图乙所示,则电池组的电动势E=V、内阻r=Ω.(结果保留两位有效数字)(3)在U﹣I坐标中两条图线在P点相交,此时滑动变阻器连入电路的阻值应为Ω(结果保留两位有效数字).三、计算题15.如图所示,与水平面夹角为θ=30°的倾斜传送带始终绷紧,传送带下端A点与上端B点之间的距离为L=4m,传送带以恒定的速率v=2m/s向上运动.现将一质量为1kg的物体无初速度地放于A处,已知物体与传送带间的动摩擦因数μ=,取g=10m/s2,求:(1)物体从A运动到B共需多少时间?(2)电动机因传送该物体多消耗的电能.16.排球运动是一项同学们喜欢的体育运动.为了了解排球的某些性能,某同学让排球从距地面高h1=1.8m处自由落下,测出该排球从开始下落到第一次反弹到最高点所用时间为t=1.3s,第一次反弹的高度为h2=1.25m.已知排球的质量为m=0.4kg,g取10m/s2,不计空气阻力.求:①排球与地面的作用时间.②排球对地面的平均作用力的大小.17.如图所示,与水平面成37°的倾斜轨道AC,其延长线在D点与半圆轨道相切,轨道半径R=1m,全部轨道为绝缘材料制成且位于竖直面内,整个空间存在水平向左的匀强电场,MN的右侧存在垂直于纸面向里的匀强磁场(C点在MN边界上).一质量为0.4kg的带电小球沿轨道AC下滑,至C点时速度为v0=m/s,接着沿直线CD运动到D处进入半圆轨道,进入时无动能损失,且恰好能通过F 点,在F点速度v f=4m/s,(不计空气阻力,g=10m/s2,cos37°=0.8)求:(1)小球带何种电荷;(2)小球在半圆轨道部分克服摩擦力所做的功.(3)小球从F点飞出时磁场同时消失,小球离开F点后的运动轨迹与直线AC(或延长线)的交点为G(G点未标出),求G点到D点的距离.2016-2017学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末物理试卷参考答案与试题解析一、选择题1.两物体A、B按如图所示连接且处于静止状态,已知两物体质量为m A>m B,A物体和地面的动摩擦因数为μ.现在B上加一个水平力F,使物体B缓慢移动,物体A始终静止,则此过程中()A.物体A对地面的压力逐渐变小B.物体A受到的摩擦力不变C.绳的拉力逐渐变小D.地面对A的作用力不变【考点】共点力平衡的条件及其应用;力的合成与分解的运用.【分析】先以A为研究对象,分析受力情况,分析平衡条件得到F、绳子拉力与绳子和竖直方向夹角的关系,分析F和拉力的变化情况.再对A研究,根据平衡条件分析地面的支持力和摩擦力如何变化.【解答】解:ABC、以B为研究对象,分析受力情况如图1,设绳子与竖直方向的夹角为θ,则由平衡条件得:Tsinθ=FTcosθ=m B g得:T=,F=m B gtanθ由题,θ增大,cosθ减小,tanθ增大,则绳子拉力T增大,F增大.对A研究,分析受力情况如图2所示,由平衡条件得:Tsinα+N=m A g,f=TcosαT增大,α不变,则地面对A的支持力N减小,摩擦力f增大,则由牛顿第三定律得知,物体A对地面的压力逐渐减小.故A正确,BC错误.D、地面对A的作用力是N和f的合力,根据平衡条件得知,N和f的合力与M A g 和T的合力大小相等,T增大,则知m A g和T的合力增大.故地面对A的作用力增大.故D错误.故选:A2.一皮带传送装置如图所示,轻弹簧一端固定,另一端连接一个质量为m的滑块,已知滑块与皮带之间存在摩擦.现将滑块轻放在皮带上,弹簧恰好处于自然长度且轴线水平.若在弹簧从自然长度到第一次达最长的过程中,滑块始终未与皮带达到共速,则在此过程中滑块的速度和加速度变化情况是()A.速度增大,加速度增大B.速度增大,加速度减小C.速度先增大后减小,加速度先增大后减小D.速度先增大后减小,加速度先减小后增大【考点】牛顿第二定律;匀变速直线运动的速度与时间的关系.【分析】根据滑块的受力,通过合力的变化判断加速度的变化,根据加速度的方向与速度方向的关系判断速度的变化.【解答】解:物块滑上传送带,受到向左的摩擦力,开始摩擦力大于弹簧的弹力,向左做加速运动,在此过程中,弹簧的弹力逐渐增大,根据牛顿第二定律,加速度逐渐减小,当弹簧的弹力与滑动摩擦力相等时,速度达到最大,然后弹力大于摩擦力,加速度方向与速度方向相反,物体做减速运动,弹簧弹力继续增大,根据牛顿第二定律得,加速度逐渐增大,速度逐渐减小.故D正确,A、B、C错误.故选:D3.如图所示,固定在水平桌面上的光滑金属导轨cd、eg处于方向竖直向下的匀强磁场中,金属杆ab与导轨接触良好.在两根导轨的端点d、e之间连接一电阻,其它部分电阻忽略不计.现用一水平向右的外力F1作用在金属杆ab上,使金属杆由静止开始向右沿导轨滑动,滑动中杆ab始终垂直于导轨.金属杆受到的安培力用F f表示,则关于图乙中F1与F f随时间t变化的关系图象可能的是()A.B.C.D.【考点】导体切割磁感线时的感应电动势.【分析】根据安培力大小表达式F f=,可知安培力与速率成正比,安培力均匀增大,速度均匀增大,可分析出导体棒应做匀加速运动,根据牛顿第二定律推导出外力F1与速度的关系式,再选择图象.【解答】解:根据安培力大小表达式F f=,可知安培力与速率成正比.图中安培力随时间均匀增大,则速度随时间均匀增大,说明导体棒应做匀加速运动,加速度a一定,根据牛顿第二定律得:F1﹣F f=ma,得F1=+ma,可见外力F1与速度是线性关系,速度随时间均匀增大,则外力F1也随时间均匀增大,故B正确.故选B4.下列说法正确的是()A.太阳辐射的能量主要来自太阳内部的核聚变反应B.大量的氢原子从n=3的能级向低能级跃迁时只会辐射两种不同频率的光C.一束单色光照射到某种金属表面不能发生光电效应,是因为该束光的波长太短D.发生光电效应时,入射光的频率一定,光强越大,光电子最大初动能越大【考点】玻尔模型和氢原子的能级结构;原子核衰变及半衰期、衰变速度.【分析】太阳辐射的能量是核聚变反应;从n=3的能级向低能级跃迁时只会辐射3种不同频率的光;不能发生光电效应,是因为该束光的频率太低,波长太长;光电效应的最大初动能由频率决定.【解答】解:A、太阳辐射的能量主要来自太阳内部的核聚变反应,故A正确;B、大量的氢原子从n=3的能级向低能级跃迁时只会辐射三种不同频率的光,故B错误;C、一束单色光照射到某种金属表面不能发生光电效应,是因为该单色光的频率小于这种金属的极限频率,所以不会产生光电效应,再由频率与波长关系,则有该束光的波长太长,故C错误;D、发生光电效应时,入射光的频率一定,光强越大,光子数越多,故单位时间内逸出的光电子数就越多,而依据光电效应方程:E km=hγ﹣W,可知,光电子最大初动能不变,故D错误;故选:A.5.质量为M的木块在光滑的水平面上以速度v1向右运动,质量为m的子弹以速度v2水平向左射入木块(子弹留在木块内),要使木块停下来,必须发射子弹的数目为()A.B.C.D.【考点】动量守恒定律.【分析】子弹在射木块的过程中,木块和所有子弹组成的系统动量守恒,结合动量守恒定律求出射入的子弹个数.【解答】解:对所有子弹和木块组成的系统研究,根据动量守恒定律得,Mv1﹣Nmv2=0解得N=.故C正确,A、B、D错误.故选:C.6.如图所示,三个质点a、b、c质量分别为m1、m2、M,(M>>m1,M>>m2).a、b在同一平面内绕c沿逆时针方向做匀速圆周运动,它们的周期之比T a:T b=1:k.(k>1,为正整数)从图示位置开始,在b运动一周的过程中,则()A.a、b距离最近的次数为k次 B.a、b距离最近的次数为k+1次C.a、b、c共线的次数为2k次 D.a、b、c共线的次数为2k﹣2次【考点】线速度、角速度和周期、转速.【分析】质点a、b均在c点的万有引力的作用下绕c做圆周运动,根据周期比,每多转半圈,三质点共线一次,可先求出多转半圈的时间,与总时间相比,得出三点共线次数.【解答】解:A、设每隔时间T,a、b相距最近,则(ωa﹣ωb)T=2π,所以T==故b运动一周的过程中,a、b相距最近的次数为:n==即a、b距离最近的次数为k﹣1次,故AB均错误.C、设每隔时间t,a、b共线一次,则(ωa﹣ωb)t=π,所以t==;故b运动一周的过程中,a、b、c共线的次数为:n= =.故a、b、c共线的次数为2k﹣2,故C错误,D正确.故选:D7.一匀强电场的电场强度E随时间t变化的图象如图所示,在该匀强电场中,有一个带电粒子于t=0时刻由静止释放,若带电粒子只受电场力作用,则下列说法中正确的是()A.带电粒子只向一个方向运动B.0~2s内,电场力所做的功等于零C.4s末带电粒子回到原出发点D.2.5s~4s内,电场力的冲量等于零【考点】带电粒子在匀强电场中的运动.【分析】从图中可以看出,电场变化的周期是3s,在每个周期的前一秒内,带电粒子的加速度为,后两秒的加速度为,可得a2是a1的2倍,从而绘出v﹣t图,有题意可知粒子在反向加速0.5s后开始反方向运动,可判断A错误;由图可以看出0~2s的时间内,做功不会为零;由v﹣t图象中图线与坐标轴围成的图形的面积为位移可判断C选项的正误;由动量定理可计算出2.5s~4s内的动量的变化,从而判断D的正误.【解答】解:由牛顿第二定律可知,带电粒子在第1s内的加速度,为第2s内加速度的,因此先加速1s再减小0.5s速度为零,接下来的0.5s将反向加速,v﹣﹣t图象如图所示,由对称关系可得,反向加速的距离使带电粒子刚回到减速开始的点,所以选项A错;0~2s内,带电粒子的初速度为零,但末速度不为零,由动能定理可知电场力所做的功不为零,选项B错误;由v﹣﹣t图象中图线与坐标轴围成的图形的面积为物体的位移,由对称可以看出,前4s内的位移不为零,所以带电粒子不会回到原出发点,所以C错误;2.5s~4s内,电场力的冲量为I=2qE0×0.5+(﹣qE0)×1=0,选项D正确.故选D.8.“嫦娥二号”卫星由地面发射后,进入地月转移轨道,经多次变轨最终进入距离月球表面100km,周期为118min的工作轨道,开始对月球进行探测,则()A.卫星在轨道Ⅲ上的运动速度比月球的第一宇宙速度小B.卫星在轨道Ⅲ上经过P点的速度比在轨道Ⅰ上经过P点时大C.卫星在轨道Ⅲ上运动的周期比在轨道Ⅰ上短D.卫星在轨道Ⅰ上的机械能比在轨道Ⅱ上大【考点】人造卫星的加速度、周期和轨道的关系;万有引力定律及其应用.【分析】月球的第一宇宙速度是卫星贴近月球表面做匀速圆周运动的速度,根据万有引力提供向心力,得出线速度与半径的关系,即可比较出卫星在轨道Ⅲ上的运动速度和月球的第一宇宙速度大小.卫星在轨道Ⅰ上经过P点若要进入轨道Ⅲ,需减速.比较在不同轨道上经过P点的加速度,直接比较它们所受的万有引力就可得知.卫星从轨道Ⅰ进入轨道Ⅱ,在P点需减速.【解答】解:A.月球的第一宇宙速度是卫星贴近月球表面做匀速圆周运动的速度,卫星在轨道Ⅲ上的半径大于月球半径,根据=m,得卫星的速度v=,可知卫星在轨道Ⅲ上的运动速度比月球的第一宇宙速度小.故A正确.B.卫星在轨道Ⅰ上经过P点若要进入轨道Ⅲ,需减速.即知卫星在轨道Ⅲ上经过P点的速度比在轨道Ⅰ上经过P点时小,故B错误.C.根据开普勒第三定律:,可知卫星在轨道Ⅲ上运动的周期比在轨道Ⅰ上短,故C正确.D.卫星从轨道Ⅰ进入轨道Ⅱ,在P点需减速.动能减小,而它们在各自的轨道上机械能守恒,所以卫星在轨道Ⅰ上的机械能比在轨道Ⅱ上大.故D正确.故选:ACD.9.一物体静止在水平地面上,在竖直向上的拉力F的作用下开始向上运动,如图甲所示.在物体向上运动过程中,其机械能E与位移x的关系图象如图乙所示(空气阻力不计),已知曲线上点A处的切线的斜率最大,则()A.在x1处物体所受拉力最大B.在x1x2过程中,物体的动能先增大后减小~C.在x2处物体的速度最大D.在x1x2过程中,物体的加速度先增大后减小~【考点】动能定理的应用.【分析】根据功能关系:除重力以外其它力所做的功等于机械能的增量,0﹣﹣x1过程中物体机械能在增加,知拉力在做正功,机械能与位移图线的斜率表示拉力.当机械能守恒时,拉力等于零,通过拉力的变化判断其加速度以及动能的变化.【解答】解:A、由图可知,x1处物体图象的斜率最大,则说明此时机械能变化最快,由E=Fh可知此时所受的拉力最大;故A正确;B、x1~x2过程中,图象的斜率越来越小,则说明拉力越来越小;x2时刻图象的斜率为零,则说明此时拉力为零;在这一过程中物体应先加速后减速,则说明最大速度一定不在x2处;故B正确;C错误;D、由图象可知,拉力先增大后减小,直到变为零;则物体受到的合力应先增大,后减小,减小到零后,再反向增大;故加速度应先增大后减小,再反向增大;故D错误;故选:AB.10.如图所示,平行板电容器AB两极板水平放置,A在上方,B在下方,现将其和二极管串连接在电源上,已知A和电源正极相连,二极管具有单向导电性,一带电小球从AB间的某一固定点水平射入,打在B极板上的N点,小球的重力不能忽略,现通过上下移动A板来改变两极板AB间距(两极板始终平行),则下列说法正确的是()A.若小球带正电,当AB间距增大时,小球打在N点的右侧B.若小球带正电,当AB间距减小时,小球打在N点的左侧C.若小球带负电,当AB间距减小时,小球可能打在N点的右侧D.若小球带负电,当AB间距增大时,小球可能打在N点的左侧【考点】带电粒子在匀强电场中的运动;平行板电容器的电容.【分析】因为二极管的单向导电性,只能给A充电而不能放电,使得Q只能增大或不变,不能减小.根据U=,C=,E===,Q不变时,改变d,E不变,所以E也只能增大或不变,即小球受的电场力只能增大或不变.【解答】A、若小球带正电,当d增大时,电容减小,但Q不可能减小,所以Q不变,根据E===,E不变所以电场力不变,小球仍然打在N点.故A 错误.B、若小球带正电,当d减小时,电容增大,Q增大,根据E===,所以d减小时E增大,所以电场力变大向下,小球做平抛运动竖直方向加速度增大,运动时间变短,打在N点左侧.故B正确.C、若小球带负电,当AB间距d减小时,电容增大,则Q增大,根据E===,所以d减小时E增大,所以电场力变大向上,若电场力小于重力,小球做类平抛运动竖直方向上的加速度减小,运动时间变长,小球将打在N点的右侧.故C 正确.D、若小球带负电,当AB间距d增大时,电容减小,但Q不可能减小,所以Q不变,根据E===,E不变所以电场力大小不变,小球做类平抛运动竖直方向上的加速度不变,运动时间不变,小球仍落在N点.故D错误.故选:BC.11.如图所示,闭合的矩形导体线圈abcd在匀强磁场中绕垂直于磁感线的对称轴OO′匀速转动,沿着OO′方向观察,线圈沿逆时针方向转动.已知匀强磁场的磁感强度为B,线圈匝数为n,ab边的边长为l1,ad边的边长为l2,线圈总电阻为R,转动的角速度为ω,则当线圈转至图示位置时()A.线圈中感应电流的方向为abcdaB.线圈中的感应电动势为2nBl2ωC.穿过线圈磁通量随时间的变化率最大D.线圈ad边所受安培力的大小为【考点】法拉第电磁感应定律;安培力;楞次定律.【分析】根据右手定则判断感应电流的方向.图示时刻ad、bc两边垂直切割磁感线,根据感应电动势公式求解线圈中的感应电动势.图示时刻线圈中产生的感应电动势最大,由法拉第电磁感应定律分析磁通量的变化率.由安培力公式F=BIL 求出安培力大小.【解答】解:A、图示时刻,ad速度方向向里,bc速度方向向外,根据右手定则判断出ad中感应电流方向为a→d,bc中电流方向为c→b,线圈中感应电流的方向为adcba.故A错误;B、线圈中的感应电动势为E=nBSω=nBl1l2ω.故B错误;C、图示时刻ad、bc两边垂直切割磁感线,线圈中产生的感应电动势最大,由法拉第电磁感应定律分析得知,磁通量的变化率最大.故C正确;D、线圈ad边所受安培力的大小为F=nBIl2=nB•l2=.故D错误.故选:C.12.在光滑的水平面上方,有两个磁感应强度大小均为B,方向相反的水平匀强磁场,如图.PQ为两个磁场的边界,磁场范围足够大.一个边长为a、质量为m、电阻为R的金属正方形线框,以速度v垂直磁场方向从如图实线位置开始向右运动,当线框运动到分别有一半面积在两个磁场中时,线框的速度为,则下列说法正确的是()A.此过程中通过线框截面的电量为。

2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末数学试卷(

2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末数学试卷(

2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末数学试卷(文科)一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.(5分)已知全集U={1,2,3,4},集合A={1,2},∁U B={2},则集合A∩B=()A.{1}B.{2}C.{1,2}D.{1,3,4}2.(5分)若复数,其中i为虚数单位,是z的共轭复数,则=()A.2+i B.2﹣i C.i D.﹣i3.(5分)双曲线的渐近线方程为()A.B.C.y=±2x D.4.(5分)设平面向量,则=()A.(0,0) B.C.0 D.﹣25.(5分)若,且α为第二象限角,则tanα=()A.B.C.D.6.(5分)执行如图的框图,则输出的s是()A.9 B.10 C.132 D.13207.(5分)等差数列{a n}中,a1+a5=10,a4=7,则数列{a n}的公差为()A.1 B.2 C.3 D.48.(5分)若变量x,y满足约束条件,则z=x﹣y的最小值等于()A.0 B.﹣1 C.D.9.(5分)为了得到函数y=sin2x的图象,可以将函数()A.向左平移个单位长度B.向右平移个单位长度C.向左平移个单位长度D.向右平移个单位长度10.(5分)一个空间几何体的三视图如图所示,则该几何体的外接球的表面积为()A.5πB.6πC.D.7π11.(5分)某班有三个小组,甲、乙、丙三人分属不同的小组.某次数学考试成绩公布情况如下:甲和三人中的第3小组那位不一样,丙比三人中第1小组的那位的成绩低,三人中第3小组的那位比乙分数高.若甲、乙、丙三人按数学成绩由高到低排列,正确的是()A.甲、乙、丙B.甲、丙、乙C.乙、甲、丙D.丙、甲、乙12.(5分)①“两条直线没有公共点,是两条直线异面”的必要不充分条件;②若过点P(2,1)作圆C:x2+y2﹣ax+2ay+2a+1=0的切线有两条,则a∈(﹣3,+∞);③若,则;④若函数在上存在单调递增区间,则;以上结论正确的个数为()A.1 B.2 C.3 D.4二、填空题(每题5分,满分20分,将答案填在答题纸上)13.(5分)设f(x)=,则=.14.(5分)已知圆x2+y2﹣6y﹣7=0与抛物线x2=2py(p>0)的准线相切,则p=.15.(5分)设数列{a n}的前n项和为S n,且a1=1,a n+1=3S n,n∈N+,则a n=.16.(5分)已知函数y=f(x)(x∈R)d的导函数为f′(x),若f(x)﹣f(﹣x)=2x3,且当x≥0时,f′(x)>3x2,则不等式f(x)﹣f(x﹣1)>3x2﹣3x+1的解集是.三、解答题(本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17.(12分)在△ABC中,a,b,c分别是角A,B,C所对的边,且满足.(1)求角B的大小;(2)设y=sinC﹣sinA,求y的取值范围.18.(12分)如图,在棱长为2的正方体ABCD﹣A1B1C1D1中,E、F分别为DD1,BD的中点.(1)求证:EF∥平面ABC1D1;(2)求证:EF⊥B1C;(3)求三棱锥E﹣FBC1的体积.19.(12分)随机抽取100名学生,测得他们的身高(单位:cm),按照区间[160,165),[165,170),[170,175),[175,180),[180,185]分组,得到样本身高的频率分布直方图(如图).(1)求频率分布直方图中x的值及身高在170cm以上的学生人数;(2)将身高在[170,175],[175,180),[180,185]内的学生依次记为A,B,C三个组,用分层抽样的方法从这三个组中抽取6人,求从这三个组分别抽取的学生人数;(3)在(2)的条件下,要从6名学生中抽取2人,用列举法计算B组中至少有1人被抽中的概率.20.(12分)在直角坐标系xOy中,设椭圆的上下两个焦点分别为F2,F1,过上焦点F2且与y轴垂直的直线l与椭圆C相交,其中一个交点为.(1)求椭圆C的方程;(2)设椭圆C的一个顶点为B(b,0),直线BF2交椭圆C于另一个点N,求△F1BN 的面积.21.(12分)已知函数.(1)当a=1时,求曲线y=f(x)在(e,f(e))处的切线方程;(2)当x>0且x≠1,不等式恒成立,求实数a的值.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分. 22.(10分)已知平面直角坐标系xOy中,直线l的参数方程为(t为参数,0≤α<π且),以原点O为极点,x轴正半轴为极轴建立极坐标系,曲线C 的极坐标方程为.已知直线l与曲线C交于A、B两点,且.(1)求α的大小;(2)过A、B分别作l的垂线与x轴交于M,N两点,求|MN|.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣3a|(a∈R).(1)当a=1时,解不等式f(x)>5﹣|x﹣1|;(2)若存在x0∈R,使f(x0)>5+|x0﹣1|成立,求a的取值范围.2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高三(上)期末数学试卷(文科)参考答案与试题解析一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.(5分)已知全集U={1,2,3,4},集合A={1,2},∁U B={2},则集合A∩B=()A.{1}B.{2}C.{1,2}D.{1,3,4}【解答】解:∵全集U={1,2,3,4},集合A={1,2},∁U B={2},∴B={1,3,4},∴集合A∩B={1}.故选:A.2.(5分)若复数,其中i为虚数单位,是z的共轭复数,则=()A.2+i B.2﹣i C.i D.﹣i【解答】解:∵=,∴,则=2﹣i.故选:B.3.(5分)双曲线的渐近线方程为()A.B.C.y=±2x D.【解答】解:双曲线,其渐近线方程,整理得y=±x.故选:A.4.(5分)设平面向量,则=()A.(0,0) B.C.0 D.﹣2【解答】解:平面向量,则=﹣1×0+2×2=0.故选:C.5.(5分)若,且α为第二象限角,则t anα=()A.B.C.D.【解答】解:∵,且α为第二象限角,∴sinα=,则tanα=.故选:B.6.(5分)执行如图的框图,则输出的s是()A.9 B.10 C.132 D.1320【解答】解:模拟程序的运行,可得i=12,S=1满足条件i>10,执行循环体,S=12,i=11满足条件i>10,执行循环体,S=132,i=10不满足条件i>10,退出循环,输出S的值为132.故选:C.7.(5分)等差数列{a n}中,a1+a5=10,a4=7,则数列{a n}的公差为()A.1 B.2 C.3 D.4【解答】解:设数列{a n}的公差为d,则由a1+a5=10,a4=7,可得2a1+4d=10,a1+3d=7,解得d=2,故选B.8.(5分)若变量x,y满足约束条件,则z=x﹣y的最小值等于()A.0 B.﹣1 C.D.【解答】解:由z=x﹣y得y=x﹣z,作出变量x,y满足约束条件对应的平面区域如图(阴影部分),平移直线y=x﹣z,由图象可知当直线y=x﹣z,过点A点,由,可得A(,)时,直线y=x﹣z的截距最大,此时z最小,∴目标函数z=x﹣y的最小值是﹣﹣=﹣.故选:D.9.(5分)为了得到函数y=sin2x的图象,可以将函数()A.向左平移个单位长度B.向右平移个单位长度C.向左平移个单位长度D.向右平移个单位长度【解答】解:为了得到函数y=sin2x的图象,可以将函数向左平移个单位长度,故选:C.10.(5分)一个空间几何体的三视图如图所示,则该几何体的外接球的表面积为()A.5πB.6πC.D.7π【解答】解:由几何体的三视图得到该几何体是如图所示的三棱锥P﹣ABC,其中,PC⊥底面ABC,AC⊥BC,AC=PC=,BC=1,以CA、CB、CP为三条棱构造长方体,则该几何体的外接球即长方体的外接球,∴该几何体的外接球的半径R==,∴该几何体的外接球的表面积:S=4πR2=4π×()2=7π.故选:D.11.(5分)某班有三个小组,甲、乙、丙三人分属不同的小组.某次数学考试成绩公布情况如下:甲和三人中的第3小组那位不一样,丙比三人中第1小组的那位的成绩低,三人中第3小组的那位比乙分数高.若甲、乙、丙三人按数学成绩由高到低排列,正确的是()A.甲、乙、丙B.甲、丙、乙C.乙、甲、丙D.丙、甲、乙【解答】解:甲和三人中的第3小组那位不一样,说明甲不在第三组,三人中第3小组的那位比乙分数高,说明乙不在第三组,则丙在第三组,第三组比第1小组的那位的成绩低,大于乙,这时可得乙为第二组,甲为第一组,甲、乙、丙三人按数学成绩由高到低排列,甲、丙、乙,故选B.12.(5分)①“两条直线没有公共点,是两条直线异面”的必要不充分条件;②若过点P(2,1)作圆C:x2+y2﹣ax+2ay+2a+1=0的切线有两条,则a∈(﹣3,+∞);③若,则;④若函数在上存在单调递增区间,则;以上结论正确的个数为()A.1 B.2 C.3 D.4【解答】解:对于①,两条直线没有公共点,则这两条直线不一定是异面直线,若两条直线是异面直线,则这两条直线没有公共点,所以是必要不充分条件,①正确;对于②,过点P(2,1)作圆C:x2+y2﹣ax+2ay+2a+1=0的切线有两条,则D2+E2﹣4F=a2+(2a)2﹣4(2a+1)>0,化简得5a2﹣8a﹣4>0,解得a>2或a<﹣;又点P代入圆的方程得22+12﹣2a+2a+2a+1>0,解得a>﹣3;所以a的取值范围是﹣3<a<﹣或a>2,②错误;对于③,若,则1+2sinxcosx=,∴2sinxcosx=﹣,∴(sinx﹣cosx)2=1﹣2sinxcosx=,∴;对于④,函数f(x)=﹣x3+x2+2ax,f′(x)=﹣x2+x+2a=﹣(x﹣)2++2a;当x∈(,+∞)时,f′(x)<f′()=2a+,令2a+≥0,解得a≥﹣,所以a的取值范围是[﹣,+∞),④正确;综上,正确的命题序是①③④,共3个.故选:C.二、填空题(每题5分,满分20分,将答案填在答题纸上)13.(5分)设f(x)=,则=.【解答】解:由分段函数的表达式得f()=ln=﹣1,则f(﹣1)=e﹣1=,故f[()]=,故答案为:14.(5分)已知圆x2+y2﹣6y﹣7=0与抛物线x2=2py(p>0)的准线相切,则p=2.【解答】解:整理圆方程得(x﹣3)2+y2=16,∴圆心坐标为(3,0),半径r=4,∵圆与抛物线的准线相切,∴圆心到抛物线准线的距离为半径,即=4,解得p=2.故答案为:2.15.(5分)设数列{a n}的前n项和为S n,且a1=1,a n+1=3S n,n∈N+,则a n=.【解答】解:a1=1,a n+1=3S n,n∈N+,当n≥2时,a n=3S n﹣1,由a n=S n﹣S n﹣1,可得a n+1﹣a n=3a n,=4a n,即为a n+1由于a2=3a1=3,则a n=a2q n﹣2=3•4n﹣2,综上可得,,故答案为:.16.(5分)已知函数y=f(x)(x∈R)d的导函数为f′(x),若f(x)﹣f(﹣x)=2x3,且当x≥0时,f′(x)>3x2,则不等式f(x)﹣f(x﹣1)>3x2﹣3x+1的解集是(,+∞).【解答】解:令F(x)=f(x)﹣x3,则由f(x)﹣f(﹣x)=2x3,可得F(﹣x)=F(x),故F(x)为偶函数,又当x≥0时,f′(x)>3x2即F′(x)>0,所以F(x)在(0,+∞)上为增函数.不等式f(x)﹣f(x﹣1)>3x2﹣3x+1化为F(x)>F(x﹣1),所以有|x|>|x﹣1|,解得x>,故答案为(,+∞).三、解答题(本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17.(12分)在△ABC中,a,b,c分别是角A,B,C所对的边,且满足.(1)求角B的大小;(2)设y=sinC﹣sinA,求y的取值范围.【解答】解:(1)由正弦定理知,,即,在△ABC中,∴即,又B∈(0,π)∴,∴,即.(2)依题知y=sinC﹣sinA=sinC﹣sin(B+C)∴=∴.由(1)知,∴,∴,即.18.(12分)如图,在棱长为2的正方体ABCD﹣A1B1C1D1中,E、F分别为DD1,BD的中点.(1)求证:EF∥平面ABC1D1;(2)求证:EF⊥B1C;(3)求三棱锥E﹣FBC1的体积.【解答】(1)证明:∵E、F分别为DD1,BD的中点,连结BD1,∴EF∥BD1,又∵EF⊄平面ABC1D1,BD1⊂平面ABC1D1,∴EF∥平面ABC1D1;(2)证明:∵B1C⊥BC1,B1C⊥D1C1,B1C∩D1C1=C1,∴B1C⊥平面BD1C1,∵BD1⊂平面BD1C1∴BD1⊥B1C,又∵EF∥BD1,∴EF⊥B1C;(3)解:∵EF∥BD1,EF⊂平面EFC1,BD1⊄平面EFC1,∴BD1∥平面EFC1,即点B、D1到平面EFC1的距离相等,∴,取CD中点M,连FM,则FM∥BC.在正方体AC1中BC⊥平面DC1,BC=2.∴FM⊥平面DC1设点F到平面ED1C1的距离为h,则,∴,即三棱锥E﹣FBC1的体积为.19.(12分)随机抽取100名学生,测得他们的身高(单位:cm),按照区间[160,165),[165,170),[170,175),[175,180),[180,185]分组,得到样本身高的频率分布直方图(如图).(1)求频率分布直方图中x的值及身高在170cm以上的学生人数;(2)将身高在[170,175],[175,180),[180,185]内的学生依次记为A,B,C三个组,用分层抽样的方法从这三个组中抽取6人,求从这三个组分别抽取的学生人数;(3)在(2)的条件下,要从6名学生中抽取2人,用列举法计算B组中至少有1人被抽中的概率.【解答】解:(1)由频率分布直方图可知5x=1﹣5×(0.07+0.04+0.02+0.01)所以.(3分)100×(0.06×5+0.04×5+0.02×5)=60(人).(5分)(2)A,B,C三组的人数分别为30人,20人,10人.因此应该从A,B,C组中每组各抽取(人),20×=4(人),10×=2(人).(8分)(3)在(2)的条件下,设A组的3位同学为A1,A2,A3,B组的2位同学为B1,B2,C组的1位同学为C1,则从6名学生中抽取2人有15种可能:(A1,A2),(A1,A3),(A1,B1),(A1,B2),(A1,C1),(A2,A3),(A2,B1),(A2,B2),(A2,C1),(A3,B1),(A3,B2),(A3,C1),(B1,B2),(B1,C1),(B2,C1).其中B组的2位学生至少有1人被抽中有9种可能;(A1,B1),(A1,B2),(A2,B1),(A2,B2),(A3,B1),(A3,B2),(B1,B2),(B1,C1),(B2,C1)所以B组中至少有1人被抽中的概率为.(13分)20.(12分)在直角坐标系xOy中,设椭圆的上下两个焦点分别为F2,F1,过上焦点F2且与y轴垂直的直线l与椭圆C相交,其中一个交点为.(1)求椭圆C的方程;(2)设椭圆C的一个顶点为B(b,0),直线BF2交椭圆C于另一个点N,求△F1BN 的面积.【解答】解:(1)椭圆的上下两个焦点分别为F2,F1,过上焦点F2且与y轴垂直的直线l与椭圆C相交,其中一个交点为.c==,,解得a2=4,b2=2,∴椭圆C的方程为:.(2)直线BF2的方程为,由,得点N的横坐标为,又,∴,综上,△F1BN的面积为.21.(12分)已知函数.(1)当a=1时,求曲线y=f(x)在(e,f(e))处的切线方程;(2)当x>0且x≠1,不等式恒成立,求实数a的值.【解答】解:(1)a=1时,f(x)=lnx﹣x+1,f(e)=2﹣e,∴切点为(e,2﹣e),,∴切线方程为即曲线y=f(x)在(e,f(e))处的切线方程(e﹣1)x+ey﹣e=0;(2)∵当x>0且x≠1时,不等式恒成立∴x=e时,∴又即对x>0且x≠1恒成立等价于x>1时f(x)<0,0<x<1时f(x)>0恒成立∵x∈(0,1)∪(1,+∞),令f'(x)=0∵a>0∴x=1或①时,即时,时,f'(x)>0,∴f(x)在单调递增,∴f(x)>f(1)=0,∴不符合题意,②当时,即时,x∈(0,1)时f'(x)<0,∴f(x)在(0,1)单调递减,∴f(x)>f(1)=0;x∈(1,+∞)时f'(x)<0,∴f(x)在(1,+∞)单调递减,∴f(x)<f(1)=0,∴符合题意.③当时,即时,时,f'(x)>0,∴f(x)在单调递增∴f(x)<f(1)=0∴不符合题意,④当时,即a>1时,x∈(0,1)时,f'(x)>0,∴f(x)在(0,1)单调递增,∴f(x)<f(1)=0,∴a>1不符合题意.综上,.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分. 22.(10分)已知平面直角坐标系xOy中,直线l的参数方程为(t为参数,0≤α<π且),以原点O为极点,x轴正半轴为极轴建立极坐标系,曲线C 的极坐标方程为.已知直线l与曲线C交于A、B两点,且.(1)求α的大小;(2)过A、B分别作l的垂线与x轴交于M,N两点,求|MN|.(1)由已知直线l的参数方程为:(t为参数,0≤α<π且),【解答】则:,∵,,∴O到直线l的距离为3,则,解之得.∵0<α<π且,∴(2)直接利用关系式,解得:.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣3a|(a∈R).(1)当a=1时,解不等式f(x)>5﹣|x﹣1|;(2)若存在x0∈R,使f(x0)>5+|x0﹣1|成立,求a的取值范围.【解答】解:(1)由已知|x﹣3|+|x﹣1|>5,当x<1时,解得,则;当1≤x≤3时,解得x∈∅,则x∈∅,当x>3时,解得,则综上:解集为或(2)∵||x﹣3a|﹣|x﹣1||≤|(x﹣3a)﹣(x﹣1)|=|3a﹣1|∴|x﹣3a|﹣|x﹣1|≤|3a﹣1|当且仅当(x﹣3a)(x﹣1)≥0且|x﹣3a|≥|x﹣1|时等成立.∴|3a﹣1|>5,解之得a>2或,∴a的取值范围为.。

辽宁省东北育才学校2017-2018学年高二上学期第一次阶段测试英语试题 Word版含答案

辽宁省东北育才学校2017-2018学年高二上学期第一次阶段测试英语试题 Word版含答案

2017-2018学年度上学期高二年级第一次阶段性考试英语科试卷本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

满分150分,考试时间120分钟。

第一部分听力(共两节,满分30分)第一节(共5小题,每小题1.5分,共7.5分)听下面5段对话,每段对话后有一个小题, 从题中所给的A,B,C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What's the time now?A. 8:30.B. 7:55.C. 8:35.2.Why does the woman look sad?A. She did badly in the exams.B. She worries about her coming tests.C. She is ill.3.What do we learn about the taxi driver?A. He drove too fast and crashed into a truck.B. He turned suddenly and ran into a tree.C. He was hit by a falling box from a truck.4.What is the man doing?A. He is asking about his letter.B. He is buying plane tickets to Europe.C. He is sending some postcard.5.Where does the conversation mostly take place?A. In a restaurant.B. In the woman's house.C. In the market.第二节(共15小题,每小题1.5分,满分22.5分)听下面5段对话或独白。

辽宁省实验中学、大连八中、某校、鞍山一中、东北育才学校高二上学期期末考试英语试题(含答案)

辽宁省实验中学、大连八中、某校、鞍山一中、东北育才学校高二上学期期末考试英语试题(含答案)

【校级联考】辽宁省实验中学、大连八中、某校、鞍山一中、东北育才学校2018-2019学年高二上学期期末考试英语试题一、阅读理解1. You can rely on Amsterdam Marketing to assist you in planning your trip to Amsterdam. They are your resources for group travel arrangements, including accommodation, excursions (短途旅游) and everything in between. Whether you’ re a professional tour organizer or arranging a trip for family and friends, Amsterdam Marketing is ready to help! Group travel in AmsterdamOur enthusiastic and experienced booking agents are specialized in advising you on unique excursions and activities for all types of groups, from family reunions and bachelor parties to company outings. We have all the inside information about Amsterdam and the surrounding area, and can offer recommendations to match your interests, needs and budget. Let us assist you with the preparations and planning, so you can focus on enjoying your trip.Accommodation for groupsAmsterdam Marketing can help you reserve accommodation for groups in and around Amsterdam. We will assist you in finding the perfect place for your requirements, ranging from youth hotels to luxury 5-star hotels. You can also combine your hotel booking with special meeting arrangements or entertaining excursions. Don’t hesitate to get in touch for assistance with your group booking!Custom advice for your tripThere is too much to see and do in Amsterdam! For custom advice to fit your interests, agenda (日程表) and budget, you can rely on the knowledge and experience of a variety of incoming tour operators, event and experience of variety of incoming tour operators, event and excursion agencies, or tour guides. You can also download the Travel Trade Manual for a list of approved partners. Get in touch with them for extended programmes, quotes (报价) and personalized suggestions.Contact usAmsterdam MarketingP.O. Box 3331 1001 AC AmsterdamTel:+31(0)20-702 6000Fax:+31(0)20-201 8850E-mail: (1)Amsterdam Marketing can offer ________.A accommodation in AmsterdamB a suitable agenda for your familyC advice about activities for family reunionsD suggestions about online ticket booking system(2)We can learn from the passage that ________.A you can contact Amsterdam Marketing in four waysB Amsterdam Marketing can help you book plane ticketsC incoming tour operators can make a budget for yourtrip D tour organizers can’t get help from Amsterdam Marketing(3)What is the passage mainly about?A Budgeting for a trip to Amsterdam.B Travelling in Amsterdam with agroup. C How to contact Amsterdam Marketing. D Accommodation advice for group travel.2. On Nov 28th 2018, police in the capital’s Shijingshan district said they caught a 43-year-old singer surnamed Chen on Monday along with a 25-year-old unemployed woman surnamed He.They were found with 7.96 grams of crystal meth and 2.14 grams of marijuana, and Chen tested positive for both drugs, the authorities said.Netizens quickly began to suspect the singer was Chen Yufan, one half of pop-rock duo (二人组) Yu Quan, who have sold12 million albums.Police declined to confirm the singer’s identity when contacted by China Daily on Wednesday afternoon. However, earlier that day, the weibo account of Beijing Public Security Bureau reposted the original information with the comment, “Illegal drugs destroyed zuimei”. Zuimei, which translates as “most beautiful”, was the title of Yu Quan’s first album.To add fuel to the fire, on Wednesday, ticket agency Tianyun Dongfang announced the band had canceled a much-anticipated (预期的) live concert on Christmas Day in Beijing due to “personal reasons.” The concert at the Worker’s Stadium was to mark the band’s 20th anniversary. Tickets for another December concert have also been canceled.Hu Haiquan, Chen’s bandmate, posted on Sina Weibo on Wedne sday to express his disappointment in an unnamed person for letting down friends who had supported them for the past 20 years.(1)When was the pop-rock duo Yu Quan founded according to the passage?A In1993.B In1998.C In2003.D In2008.(2)What is the main idea of this passage?A Members of Yu Quan took illegal drugs.B Illegal drugs destroyed manysingers. C Yu Quan’s concerts were canceled due to “personal reasons”. D Chen Yufan was caught due to illegal drugs.(3)In the fourth paragraph, “declined” probably means “________.”A were willingB refusedC agreedD felt sorry(4)Where can we probably read this passage?A In a story book.B In a newspaper.C In an official report.D In an advertisement.3. Dear strangers, I remember you.Ten months ago, when my cellphone rang, you were walking into Whole Foods, prepared to go about your food shop But I had already abandoned my cart full of groceries and I stood in the entryway of the store. My brother was telling me my father was dead. And as we hung up the phone, I started to cry and scream as my whole body trembled.Overwhelmed with emotions, I fell to the floor. and you, kind strangers, you were there You could have kept on walking, ignoring my cries, but you didn’t. You could have simply stopped and stared at my primal (原始的) display: of pain, but you didn’t. Instead you surrounded me as I yelled through my sobs.I remember in that haze (阴霾) of emotions, one of you asked for my phone and asked who you should call. I remember that I could hear your words as you tried to reach my husband for me, leaving an urgent message on his answering machine for him to call me.I recall hearing you discuss among yourselves who would drive me home in my car and who would follow that person to bring him back to the store. You didn’t know one another, but it didn’t seem to matter. You encouraged me, a stranger, in the worst moment of my life and you gathered around me with common purpose, to help.I told you that I had a friend, Pam, who worked at Whole Foods and one of you went in search of her and thankfully, she was there that morning and you brought her to me. I remember the relief I felt at seeing her face familiar and warm. She confronted and cared for me so lovingly until my husband could get to me.(1)Before the author’s brother called her, she ________.A was driving and cryingB was talking with the strangersC had filled her cart with groceriesD had paid for the groceries in her cart(2)The author develops the third paragraph mainly ________.A by processB by exampleC by comparisonD by classification(3)What can be inferred from the fourth paragraph?A One of the strangers knew the author s husband.B One stranger reached the author, s husband easily.C The strangers wanted to know the author s secrets.D The authors husband didn’t answer the phone call.(4)Which of the following can best describe the strangers?A Kind and helpful.B Clever and outgoing.C Cautious and optimistic.D Careless but kind-hearted.4.It's time to reevaluate how women handle conflict at work. Being overworked or over-committed at home and on the job will not get you where you want to be in life. It will only slow you down and hinder your career goals.Did you know women are more likely than men to feel exhausted? Nearly twice as many women than men ages 18-44 reported feeling "very tired" or "exhausted", according to a recent study.This may not be surprising given that this is the age range when women have children. It's also the age range when many women are trying to balance careers and home. One reason women may feel exhausted is that they have a hard time saying "no". Women want to be able to do it all—volunteer for school parties or cook delicious meals—and so their answer to any request is often "Yes, I can."Women struggle to say "no" in the workplace for similar reasons, including the desire to be liked by their colleagues. Unfortunately, this inability to say "no" may be hurting women's heath as well as their career.At the workplace, men use conflict as a way to position themselves, while women often avoid conflict or strive to be the peacemaker, because they don't want to be viewed as aggressive or disruptive at work. For example, there's a problem that needs to be addressedimmediately, resulting in a dispute over who should be the one to fix it. Men are more likely to face that dispute from the perspective of what benefits them most, whereas women may approach the same dispute from the perspective of what's the easiest and quickest way to resolve the problem—even if that means doing the boring work themselves.This difference in handling conflict could be the deciding factor on who gets promoted to a leadership position and who does not. Leaders have to be able to assign and manage resources wisely—including staff expertise. Shouldering more of the workload may not earn you that promotion. Instead, it may highlight your inability to assign effectively.(1)What does the author say is the problem with women?A They are often unclear about the career goals to reach.B They are usually more committed at home than on the job.C They tend to be over-optimistic about how far they could go.D They tend to push themselves beyond the limits of their ability.(2)What may hold back the future prospects of career women?A Their unwillingness to say "no".B Their desire to be considered powerful.C An underestimate of their own ability.D A lack of courage to face challenges.(3)Men and woman differ in their approach to resolving workplace conflicts in that ______.A women tend to be easily satisfiedB men are generally more persuasiveC men tend to put their personal interests firstD women are much more ready to compromise(4)What is important to a good leader?A An aggressive personality.B The ability to assign.C The courage to admit failure.D A strong sense of responsibility.二、七选五5. Whatever it means to be a good child, it doesn't mean being perfect.(1)_______Maybe one way to think of it is this: good children put themselves on the path towards being happy, successful adults. Any parent would appreciate this type of good children.Treat others like you want to be treated.(2)_______, and it really is a valuable rule to love by. For children, acting towards your parents, friends, family and other people with this guide in mind shows thoughtfulness and maturity on your part.Learn to recognize how other people are feeling.If you know how other people are feeling and are likely to react, you'll have a great advantage in deciding how you'd behave in that situation. For instance, if your parents are stressed about how they're going to pay for the bills for the month, it is probably not the best time to ask for a video game or new shoes.(3)_______, it is probably not the best time to tease him about his lack of athletic skills.(4)_______.When someone is hurting, or needs a hand, do what you can to help. The world alwaysneeds more sympathetic, helpful people.Offer gratitude to those who help you.As you become more aware of how you can help others, you should also become more aware of all the people that help you.(5)_______.A. Know when to ask for helpB. Show concern and sympathyC. Many people call this the "golden rule"D. Let them know you appreciate all they do for youE. Explain how you feel about the conflict by saying somethingF. Or, if your brother is upset about not making the baseball teamG. It does involve qualities like understanding, self﹣discipline, and appreciation,though三、完形填空6. Aged 45, I decided to learn how to surf. I surfed in Oahu-the toughest, most ____ surfing spot. Life is tough enough, but I like to make things ____ on myself, because I want to ____ my comfort zone.When I ____ in the entertainment business, I made a ____ of people I thought it was good to meet. I didn’t list those who could give me a job, ____ those who could teach me something and ____ my ideas about the world. So I started telephoning ____ in various fields. Some of them were famous ____ the world. I didn’t know any of them and ____ of them knew me. So when I called them, the ____ wasn’t always friendly. ____ they agreed to give me some time, the results weren’t always pleasant.Over the last thirty years, I’ve ____ over fifty movies. I’ve done them ____ and I’m well known in my business. I’m a guy who could ____ to the golf course tomorrow. So why do I continue? The answer is ____ : interrupting my comfort zo ne. I think it’s the best I know to keep ____ .I’m not the best surfer on the ____ , but I enjoy the challenge. All of them may be the things that we spend our time trying to ____ , But to me, they are ____ the things that in the game. (1)A comfortableB secretiveC competitiveD genuine(2)A attractiveB skillfulC difficultD necessary(3)A memorizeB interruptC ignoreD forbid(4)A brought outB got outC took outD started out(5)A listB planC nameD wall(6)A andB butC orD so(7)A challengeB praiseC supportD correct(8)A coachesB leadersC expertsD guides(9)A exceptB forC withD throughout(10)A manyB noneC neitherD each(11)A attitudeB voiceC responseD question(12)A Even whenB Even nowC Even thenD Even as(13)A recordedB watchedC starredD produced(14)A fluentlyB successfully mC happilyD crazily(15)A changeB jumpC moveD retire(16)A certainB simpleC importantD funny(17)A growingB developingC increasingD ageing(18)A seaB waveC fieldD zone(19)A beatB possessC obtainD avoid(20)A luckilyB clearlyC exactlyD naturally四、语法填空7. 阅读下面材料,在空白处填入适当的内容(1个单词)或括号内单词的正确形式。

辽宁省东北育才学校2017-2018学年高一上学期第一次阶段测试英语试题 Word版含答案

辽宁省东北育才学校2017-2018学年高一上学期第一次阶段测试英语试题 Word版含答案

2017-2018学年度东北育才高中部高一年级第一次阶段测试英语科试卷第一部分:听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. At what time does the office open?A. 7:45.B. 8:00.C. 8:15.2. What did the woman do last Sunday?A. She saw a play.B. She acted in a play.C. She went to the tea house.3. What is the relationship between the speakers?A. They’re friends.B. They’re strangers.C. They’re brother and sister.4. What are the speakers talking about?A. A fine boat.B. Their friend, Tom.C. The weather.5. What will the woman do this evening?A. Meet her Mum at the airport.B. Say goodbye to her Mum at the airport.C. Fly to another city with her Mom.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题给出5秒钟的作答时间。

2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高一下学期期末考试英语卷

2017-2018学年辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校高一下学期期末考试英语卷第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man think of Beijing Opera?A. It is great.B. It is boring.C. It is terrible.2. What does the man mean?A. He tikes reading books.B. He is always organized.C. He manages his time better.3. What does the woman do?A. A receptionist.B. A cook.C. A waitress.4. Where does the conversation probably take place?A. In a factory.B. On a plane.C. At an airport.5. What will the man do near Edinburgh?A. Attend a meeting.B. Go sightseeing.C. Pick up the woman.第二节(共15题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材枓,回答第6至7题。

辽宁省实验中学、沈阳市东北育才学校等五校2016-2017学年高一下学期期末联考英语试题含答案

命题学校:大连市第八中学命题人:白艳校对人:谭璐本试卷分第I卷(选择题)和第II卷(非选择题)两部分,共150,考试时间120分钟。

第I卷第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1。

5分,满分7。

5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A。

£19。

15 B. £9。

15 C. £9。

18答案是B.1. Where does this conversation probably take place?A. In a bookstore B。

In a classroom C. In a library2. At what time will the film begin?A. 7:20B. 7:15 C。

7:00 3。

What are the two speakers mainly talking about?A. Their friend Jane B。

A weekend trip C. Aradio programme4. What will the woman probably do?A. Catch a train B。

See the man off C. Go shopping5。

Why did the woman apologize?A。

She made a late deliveryB。

She went to the wrong placeC. She couldn’t take the cake back第二节(共15小题;每小题1。

5分,满分22。

5分)听下面5段对话。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校2022届高三英语上学期期末考试答案

2022--- 2022学年(xuénián)度上学期期末考试(qī mò kǎo shì)高三年级英语(yīnɡ yǔ)科试卷答案(dáàn)听力(tīnglì)理解:1--5 BCBAC 6-10CABAC 11-15BACCB 16-20ACABA阅读理解:21-23. ABD 24-27. ACBA 28-31. DBCD 32-35. ABBD 36-40.GEACD完形填空:41-45ACCDB 46-50ABCCA 51-55BDDAB 56-60BCDAD语法填空:61.expressions 62.a 63.generally 64.to find 65.have used / have been using 66.less 67.hidden 68.that 69.heroes 70.of短文改错:1.that--when 2.deliciously--delicious 3.it--them 4.the去掉5.but--and6.goes--went7.such--so8.vitamins后加in9.healths--health 10.eat--eating书面表达 (One possible version)Dear Miss Wang,As our graduation is drawing near, I’m afraid that we have to say goodbye to each other soon. Here I’m writing to express my heartfelt gratitude for your support and guidance during the past three years.I still remember there was a time when I was struggling with my English learning and even thought of giving it up. However, it wasyou who selflessly sacrificed your spare time to guide me, cheer meup and help me with my English. Without your help, I couldn’t have made such great progress.Again, I would like to express my warm thanks to you and I wishyou good health and a happy mood every day.Yours sincerely, Li Hua内容总结可修改欢迎下载精品 Word。

辽宁省实验中学、大连八中、大连二十四中、鞍山一中、东北育才学校2019届高三上学期期末考试英语试题

2018-2019学年度上学期期末考试高三年级英语科试卷命题学校:大连八中命题人:倪春红田子毅校对人:白艳Ⅰ客观卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man want?A. A leather suit.B. A piece of leather.C. A pair of leather shoes.2. Who was absent from dinner last night?A. Robert.B. George.C. Kate.3. How often does the woman eat out?A. Five times a month.B. Four times a week.C. Five times a week.4. How much will the woman pay?A. $9.B. $6.C. $3.5. Which program does the woman want to watch?A. A movie.B. A fashion show.C. International news.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6 段材料,回答第6至7题。

6. What made the girl sick?A. The nightmares.B. The plane trip.C. Visiting the Palace.7. Where does the conversation take place?A. In London.B. In New York.C. In San Francisco.听第7段材料,回答第8至9题。

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第I 卷
第一部分:听力(共两节,满分30分)
第一节(共5小题;每小题 1.5分,满分7.5分)
听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题
和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?
A. £19.15.
B. £9.15.
C. £9.18.
答案是B。

1. What does the man like about the play?
A. The story.
B. The ending.
C. The actor.
2. Which place are the speakers trying to find?
A. A hotel.
B. A bank.
C. A restaurant.
3. At what time will the two speakers meet?
A. 5:20.
B. 5:10.
C. 4:40.
4. What will the man do?
A. Change the plan.
B. Wait for a phone call.
C. Sort things out.
5. What does the woman want to do?
A. See a film with the man.
B. Offer the man some help.
C. Listen to some music.
第二节(共15小题;每小题 1.5分,满分22.5分)
听下面5段对话或独白。

每段对话或独白后有2至4个小题,从题中做给的A、B、C 三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有5秒钟的时间阅读各个小题;听完后,各小题将给出5秒钟的作答时间。

每段对话读两遍。

听第6段材料,回答第6、7题。

6. Where is Ben?
A. In the kitchen.
B. At school.
C. In the park.
7. What will the children do in the afternoon?
A. Help set the table.
B. Have a party.
C. Do their homework.
听第7段材料,回答第8、9题。

8. What are the two speakers talking about?
A. A family holiday.
B. A business trip.
C. A travel plan.
9. Where did Rachel go?
A. Spain.
B. Italy.
C. China.
听第8段材料,回答第10至12题。

10. How did the woman get to know about third-hand smoke?
A. From young smokers.
B. From a newspaper article.
C. From some smoking patents.
11. Why does the man say that he should keep away from babies?
A. He has just become a father.
B. He wears dirty clothes.
C. He is a smoker.
12. What does the woman suggest smoking parents should do?
A. Stop smoking.
B. Smoke only outside their houses.
C. Reduce dangerous matter in cigarettes.
听第9段材料,回答第13至16题。

13. Where does Michelle Ray come from?
A. A middle-sized city.
B. A small town.
C. A big city.
14. Which place would Michelle Ray take her visitors to for shopping?
A. The Zen Garden.
B. The Highlands.
C. The Red River area.
15. What does Michelle Ray do for complete quiet?
A. Go camping.
B. Study in a library.
C. Read at home.
16. What are the speakers talking about in general?
A. Late-night shopping.
B. Asian food.
C. Louisville.
听第10段材料,回答第17至第20题。

17. Why do some people say they never have dreams according to Dr Garfield?
A. They forget about their dreams.
B. They don’t want to tell the truth.。

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