湖北省黄石市第三中学上册期末精选综合测试卷(word含答案)
湖北省黄石市第三中学高一上学期物理期末考试及答案
湖北省黄石市第三中学高一上学期物理期末考试及答案一、选择题1.生活中常使用贴在墙上的吸盘挂钩来挂一些物品,吸盘受到拉力不易脱落的原因是()A.大气压力大于重力B.大气压力与重力平衡C.吸盘所受摩擦力大于所挂物体的重力D.吸盘所受摩擦力与所挂物体的重力平衡2.如图所示,两块相互垂直的光滑挡板OP、OQ,OP竖直放置,小球A、B固定在轻杆的两端。
现用水平力F将B向左缓慢推动一小段距离,则此过程中A.杆对A的弹力变小B.挡板OP对A作用力变大C.水平力F变大D.挡板OQ对B支持力变大3.通过理想斜面实验得出“力不是维持物体运动的原因"的科学家是A.亚里士多德B.伽利略C.阿基米德D.牛顿4.为提高生产效率,工厂用智能机械手代替工人搬运货物。
如图所示,一机械铁夹竖直夹起一个金属小球,小球在空中处于静止状态,铁夹与球接触面保持竖直,则()A.铁夹受到的摩擦力方向竖直向上B.小球受到的摩擦力大小与小球受到的重力大小相等C.小球没有掉下,是因为摩擦力大于小球的重力D.增大铁夹对小球的压力,小球受到的摩擦力变大5.第26届国际计量大会通过“修订国际单位制(SI)”的决议,将千克、安培、开尔文和摩尔4个国际单位制的基本单位改由物理常数重新定义,从而提高计量单位的稳定性和精确度。
基本单位“千克”所对应的物理量是()A.力B.时间C.长度D.质量6.如图所示,吊篮P悬挂在天花板上,与吊篮质量相等的物体Q被固定在吊篮中的轻弹簧托起,当悬挂吊篮的细绳被剪断的瞬间,吊篮P和物体Q的加速度为()A.g,g B.2g,g C.g,0 D.2g,07.在同一水平直线上的两位置分别沿同方向水平抛出两小球A和B,其运动轨迹如图所示,不计空气阻力,要使两球在空中相遇,则必须()A.先抛出A球B.先抛出B球C.A球的初速度小于B球的初速度D.A球的初速度大于B球的初速度8.一个物体自由下落6s后落地,g取 10m/s2,则最后 2s内通过的位移为()A.20m B.80m C.100m D.180m9.将一小球以5m/s的速度水平抛出,经过1s小球落地,不计空气阻力g=10m/s2.关于这段时间小球的运动,下列说法正确的是()A.小球着地速度是10m/sB.小球竖直方向的位移是10mC.小球着地速度是5m/sD.小球水平方向位移是5m10.如图所示,小球用一根轻弹簧悬于天花板下,已画出重物和弹簧的受力图.关于这四个力的以下说法错误..的是A.F1与F4是一对平衡力B.F2与F3是一对作用力与反作用力C.F2的施力物体是弹簧D.F3的施力物体是小球11.质量为10kg的物体在光滑水平面上受到20N的水平拉力作用,由静止开始做匀加速直线运动,4s末撤去拉力,第5s末物体的速度和加速度分别为( )A.v=8m/s;a=0 B.v=8m/s;a=2m/s2C.v=10m/s;a=0 D.v=10m/s;a=2m/s2 12.如图所示,质量为m的物体放在水平地面上,受到力F的作用后仍静止,则()A.物体所受摩擦力大于F,所以物体没被拉动B.物体对地面的压力大小等于mgC.地面对物体的支持力等于FsinθD.物体受的合力为零13.下列物理量中,属于矢量的是()A.加速度B.质量C.路程D.时间14.下列两个实验中体现出的共同的物理思想方法是()A.极限法B.控制变量法C.放大法D.等效替代法15.如图,汽车向右沿直线做匀减速运动,原来的速度是v1,经过一小段时间之后,速度变为v2,用Δv表示速度的变化量,则()A.汽车的加速度方向与v1的方向相同B.汽车的加速度方向与v2的方向相同C.汽车的加速度方向与v1的方向相反D.汽车的加速度方向与Δv的方向相反16.一物体受到三个共面共点力F1、F2、F3的作用,三力的矢量关系如图所示(各小方格边长相等),则下列说法正确的是()A.三力的合力为F1+F2+F3,方向不确定B.三力的合力为3F2,方向与F2同向C.三力的合力为2F3,方向与F3同向D.由题给条件无法求出合力大小17.甲、乙两车沿同一平直公路行驶,两车运动的x﹣t图象如图所示,则下列表述正确的是()A.乙车做曲线运动,甲车做直线运动B.甲车先做匀减速运动,后做匀速运动C.两车相遇两次D.乙车的速度不断减小18.关于曲线运动,下列说法中正确的是()A.速度方向一定改变B.速度方向一定不变C.速度大小一定改变D.速度大小一定不变19.倾角为a、质量为M的斜面体静止在水平桌面上,质量为m的木块静止在斜面体上,下列说法正确的是()A.木块受到的摩擦力大小是mg cos a B.木块对斜面体的压力大小是mgC.桌面对斜面体的摩擦力大小是mg sinαcosαD.桌面对斜面体的支持力大小是(M+m)g 20.从牛顿第二定律知道,无论怎样小的力都可以使物体产生加速度.可是当我们用一个很小的水平力去推很重的桌子时,却推不动它,这是因为()A.牛顿第二定律不适用于静止物体B.根据a=F/m和判断,加速度很小,速度增量很小,眼睛不易觉察到C.推力小于静摩擦力,加速度是负值D.重力、地面支持力、推力和静摩擦力的合力等于零.根据牛顿第二定律加速度等于零,所以原来静止的桌子还是静止的二、多选题21.如图所示为研究木板与木块之间摩擦力大小的实验装置,将一木块和木板叠放于水平桌面上,轻质弹簧测力计一端固定,另一端用细线与木块水平相连,长木板右端连有一细绳.开始时,整个装置处于静止状态,弹簧测力计的示数为0,现在用手向右水平拉细绳,使长木板在桌面上滑动,下列说法中正确的是A.木块与木板之间的摩擦力始终是滑动摩擦力B.如果测出木块的重力,可求出木块与木板间的动摩擦因数C.木板运动过程中,弹簧秤示数先变大,后不变D.木板运动过程中,弹簧秤示数保持不变22.如图示,两物块质量为M和m,用绳连接后放在倾角为θ的斜面上,物块和斜面的动摩擦因数为μ,用沿斜面向上的恒力F 拉物块M做匀加速直线运动,则中间绳子的张力T ()A.T的大小与运动情况无关B.T 的大小与θ无关C.T的大小与μ无关D.以上说法都不对,若不计绳与滑轮的摩擦,23.如图所示,人的质量为M,物块的质量为m,且M m则当人拉着绳向右跨出一步后,人和物仍保持静止,则下列说法中正确的是()A.地面对人的摩擦力增大B.绳子上的张力不变C.人受到的合力减小D.人对地面的压力增大24.如图所示,把一质量为m的物体用一水平方向的推力F=kt(k为恒量,t为时间)压在竖直的足够高的平整墙上,物体与墙壁间动摩擦因数为μ,从t=0开始计时,静止释放此物体,则物体所受的摩擦力f、加速度a、速度v和位移x随t变化的关系正确的是图中的()A.B.C.D.25.甲、乙两玩具车(视为质点)沿同一方向做直线运动,某时刻经过同一地点,若以该时刻作为计时起点,得到两车的位移图象如图所示,图象中的OC与AB平行,CB与OA平行,则下列说法中正确的是()A.0~t2时间内两车的距离越来越远B.0~t3时间内甲车的平均速度大于乙车的平均速度C.t2~t3时间内甲车和乙车距离越来越小D.t3时刻甲车和乙车的位移相等26.A、B两车沿同一平直公路同向运动,从某时刻开始,两车的v—t图象如图所示.假设t=8s时两车恰好相遇.则在0~8s内A.A车加速度大小小于B车加速度大小B.t=6s时两车相距最远C.t=0时B车在A车前方33m处D.这段时间内A、B两车可能相遇两次三、实验题27.某次“探究加速度a跟物体所受的合外力F和质量m的关系”时,实验如下⑴.下列做法正确的是________ (填字母代号)A.调节滑轮的高度,使牵引木块的细绳与长木板保持平行B.在调节木板倾斜角度平衡木块受到的滑动摩擦力时,将装有砝码的砝码桶通过定滑轮拴在木块上C.实验时,先放开木块再接通打点计时器的电源D.通过增减木块上的砝码改变质量时,不需重新调节木板倾角⑵图甲为实验装置图。
湖北省黄石市物理初三上册期末监测试题及答案
湖北省黄石市物理初三上册期末监测试题班级:________________ 学号:________________ 姓名:______________一、单选题(每题3分)1.下列有关声现象的说法中,正确的是( )A. 声音可以在真空中传播B. 振动停止后,发声体仍能发声C. 发声体的振动频率越高,响度越大D. 戴耳罩是在人耳处减弱噪声答案:D解析:声音的传播需要介质,真空不能传声,故A错误;声音是由物体的振动产生的,振动停止,发声也停止,但声音的传播不会立即停止,故B错误;发声体的振动频率越高,音调越高,故C错误;戴耳罩是在人耳处减弱噪声,故D正确。
2.下列有关光现象的说法中,正确的是( )A. 岸边景物在水中形成倒影,属于光的折射现象B. 小孔成像是由于光的直线传播形成的C. 电影屏幕选用粗糙的布料,目的是使光发生镜面反射D. 红外线最显著的性质是使荧光物质发光答案:B解析:岸边景物在水中形成倒影,属于平面镜成像,是光的反射形成的,故A错误;小孔成像是由于光的直线传播形成的,故B正确;电影屏幕选用粗糙的布料,目的是使光发生漫反射,让各个方向的观众都能看到,故C错误;紫外线最显著的性质是使荧光物质发光,故D错误。
3.下列物态变化中,属于液化的是( )A. 冰雪消融B. 露的形成C. 霜的形成D. 冰冻的衣服变干答案:B解析:冰雪消融,是冰雪由固态变成液态,是熔化现象,故A错误;露是空气中的水蒸气遇冷液化形成的小水滴,故B正确;霜是空气中的水蒸气遇冷凝华形成的冰晶,故C错误;冰冻的衣服变干,是固态冰直接变成气态水蒸气,是升华现象,故D错误。
4.下列说法正确的是( )A. 导体在磁场中一定会受到力的作用B. 发电机是根据电磁感应现象制成的C. 通电螺线管外部的磁场与条形磁铁的磁场相似D. 电流方向改变时,通电螺线管的磁极也会改变答案:BCD解析:通电导体在磁场中受到力的作用,静止的导体在磁场中不会受到力的作用,故A错误;发电机是根据电磁感应现象制成的,故B正确;通电螺线管外部的磁场与条形磁铁的磁场相似,故C正确;通电螺线管的磁极与电流方向有关,电流方向改变时,通电螺线管的磁极也会改变,故D正确。
2024届湖北省黄石市第三中学化学高三第一学期期末综合测试试题含解析
2024届湖北省黄石市第三中学化学高三第一学期期末综合测试试题注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
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一、选择题(每题只有一个选项符合题意)1、下列实验可达到实验目的的是A.用相互滴加的方法鉴别Ca(OH)2和NaHCO3溶液B.向CH3CH2Br中滴入AgNO3溶液以检验溴元素C.用溴的四氯化碳溶液吸收SO2气体D.与NaOH的醇溶液共热制备CH3-CH=CH22、下列有关溶液的说法中,正确的是A.(NH4)2SO4溶液浓缩时温度过高可能导致生成NH4HSO4B.相同温度下,强酸溶液的导电能力一定大于弱酸溶液C.通电时,溶液中溶质粒子分别向两极移动D.蒸干Na2CO3溶液最终得到NaOH3、铅霜(醋酸铅)是一种中药,具有解毒敛疮、坠痰镇惊之功效,其制备方法为:将醋酸放入磁皿,投入氧化铅,微温使之溶化,以三层细布趁热滤去渣滓,放冷,即得醋酸铅结晶;如需精制,可将结晶溶于同等量的沸汤,滴醋酸少许,过七层细布,清液放冷,即得纯净铅霜。
制备过程中没有涉及的操作方法是A.萃取B.溶解C.过滤D.重结晶4、下列量气装置用于测量CO2体积,误差最小的是()A.B.C.D.5、下列各组离子在溶液中可以大量共存,且加入或通入试剂X后,发生反应的离子方程式也正确的是A.A B.B C.C D.D6、只用一种试剂即可区别的:NaCl、MgCl2、FeCl3、Al2(SO4)3四种溶液,这种试剂是A.AgNO3 B.NaOH C.BaCl2 D.HCl7、下列实验操作、现象及结论均正确的是选实验操作和现象实验结论项向Na2SO3溶液中加入足量的Ba(NO3)2溶液,出现白色沉淀;再加入足量BaSO3溶于稀盐酸A稀盐酸,沉淀溶解B 向KI溶液中加入CCl4,振荡后静置;液体分层,下层呈紫红色碘易溶于CCl4,难溶于水将AgCl与AgBr的饱和溶液等体积混合,再加入足量的浓AgNO3溶液,K SP(AgCl)>K SP(AgBr)C析出沉淀的物质的量AgCl多于AgBrD 室温下,同浓度的Na2A溶液的pH比NaB溶液的pH大说明酸性H2A<HB A.A B.B C.C D.D8、对下列化学用语的理解正确的是A.丙烯的最简式可表示为CH2B.电子式既可以表示羟基,也可以表示氢氧根离子C.结构简式(CH3)2CHCH3既可以表示正丁烷,也可以表示异丁烷D.比例模型既可以表示甲烷分子,也可以表示四氯化碳分子9、下列烷烃命名错误的是A.2─甲基戊烷B.3─乙基戊烷C.3,4─二甲基戊烷D.3─甲基己烷以下叙述错误的是A.NO2由反应N2+2O22NO2生成B.总反应可表示为2SO2+O2+2H2O2H2SO4C.还可能发生的反应有4NO2+O2+2H2O=4HNO3D.还可能发生的反应有4NO+3O2+2H2O=4HNO311、下列实验能达到实验目的且符合安全要求的是()A.利用排空气法收集CO2B.收集氧气C.制备并检验氢气的可燃性D.稀释浓硫酸12、地球在流浪,学习不能忘”,学好化学让生活更美好。
湖北省黄石三中2016-2017学年高一上学期期末数学试卷(word版含答案)
2016-2017学年湖北省黄石三中高一(上)期末数学试卷一、选择题(本大题共12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.设集合A={x|﹣3≤2x﹣1≤3},集合B为函数y=lg(x﹣1)的定义域,则A ∪B=()A.(1,2) B.[﹣1,+∞)C.(1,2]D.[1,2)2.若sinα<0且tanα>0,则α是()A.第一象限角B.第二象限角C.第三象限角D.第四象限角3.已知函数,则f(2)=()A.9 B.3 C.0 D.﹣24.已知向量,若,则=()A.1 B.2 C.3 D.45.已知tanx=﹣,则sin2x+3sinxcosx﹣1的值为()A.﹣ B.2 C.﹣2或2 D.﹣26.同时满足两个条件:(1)定义域内是减函数;(2)定义域内是奇函数的函数是()A.f(x)=﹣x|x|B.C.f(x)=tanx D.7.已知函数则f(x)在区间[0,]上的最大值与最小值分别是()A.1,﹣2 B.2,﹣1 C.1,﹣1 D.2,﹣28.若将函数y=cos 2x的图象向左平移个单位长度,则平移后图象的对称轴为()A.x=﹣(k∈Z)B.x=+(k∈Z)C.x=﹣(k∈Z)D.x=+(k∈Z)9.若sin(﹣α)=﹣,则cos(+2α)=()A.B.C.D.10.f(x)是定义在R上的奇函数,满足f(x+2)=f(x),当x∈(0,1)时,f(x)=2x﹣1,则的值等于()A.B.﹣6 C.D.﹣411.若向量,,且,若,则β﹣α的值为()A.或B.C. D.或12.函数f(x)=,若f(0)是f(x)的最小值,则a的取值范围为()A.[﹣1,2]B.[﹣1,0]C.[1,2]D.[0,2]二、填空题(本大题共4小题,每小题5分,共20分,将答案填在题中的横线上)13.已知函数g(x)=(a+1)x﹣2+1(a>0)的图象恒过定点A,且点A又在函数(x+a)的图象上.则实数a=.14.若函数f(x)=x2﹣ax﹣b的两个零点是2和3,则函数g(x)=bx2﹣ax﹣1的零点是.15.在△ABC中,M是BC的中点,AM=3,点P在AM上且满足,则=.16.已知向量,.三、解答题(本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.计算下列式子的值:(1);(2).18.已知平面向量,,.(1)求满足的实数m,n;(2)若,求实数k的值.19.已知函数f(x)=Acos(+),x∈R,且f()=.(1)求A的值;(2)设α,β∈[0,],f(4α+π)=﹣,f(4β﹣π)=,求cos(α+β)的值.20.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,)的部分图象如图所示.(1)求函数f(x)的解析式;(2)求函数f(x)的单调递增区间和对称中心.21.已知,,函数f(x)=•(x∈R)(1)求函数f(x)的周期;(2)若方程f(x)﹣t=1在内恒有两个不相等的实数解,求实数t 的取值范围.22.已知f(x)是定义在[﹣1,1]上的奇函数,且f(1)=1,若m,n∈[﹣1,1],m+n≠0 时,有.(1)求证:f(x)在[﹣1,1]上为增函数;(2)求不等式的解集;(3)若对所有恒成立,求实数t的取值范围.2016-2017学年湖北省黄石三中高一(上)期末数学试卷参考答案与试题解析一、选择题(本大题共12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.设集合A={x|﹣3≤2x﹣1≤3},集合B为函数y=lg(x﹣1)的定义域,则A ∪B=()A.(1,2) B.[﹣1,+∞)C.(1,2]D.[1,2)【考点】对数函数的定义域;并集及其运算.【分析】先化简集合A,B再根据并集的定义即可求出.【解答】解:A={x|﹣3≤2x﹣1≤3}=[﹣1,2],y=lg(x﹣1)的定义域为{x|x>1}=(1,+∞),∴A∪B=[﹣1,+∞)故选B.2.若sinα<0且tanα>0,则α是()A.第一象限角B.第二象限角C.第三象限角D.第四象限角【考点】三角函数值的符号.【分析】由正弦和正切的符号确定角的象限,当正弦值小于零时,角在第三四象限,当正切值大于零,角在第一三象限,要同时满足这两个条件,角的位置是第三象限,实际上我们解的是不等式组.【解答】解:sinα<0,α在三、四象限;tanα>0,α在一、三象限.故选:C.3.已知函数,则f(2)=()A.9 B.3 C.0 D.﹣2【考点】函数的值.【分析】可根据解析式,先计算f(2)=f(1)=f(0),按照由内到外的顺序计算即可.【解答】解:∵,∴f(2)=f(2﹣1)=f(1)=f(1﹣1)=f(0)=﹣2.故选D.4.已知向量,若,则=()A.1 B.2 C.3 D.4【考点】平面向量共线(平行)的坐标表示.【分析】根据向量的坐标运算和向量的平行的条件以及向量模的计算即可.【解答】解:∵,∴=(3,3m),∵,∴3m=﹣3m,解得m=0,∴=(2,0),∴=2,故选:B.5.已知tanx=﹣,则sin2x+3sinxcosx﹣1的值为()A.﹣ B.2 C.﹣2或2 D.﹣2【考点】三角函数的化简求值;同角三角函数间的基本关系.【分析】化tanx=﹣为=,得出,cosx=﹣2sinx.由sin2x+cos2x=1,求得sin2x=,将原式化为关于sin2x的三角式求解.【解答】解:tanx=﹣,即=,cosx=﹣2sinx.由sin2x+cos2x=1,得5sin2x=1,sin2x=所以原式=sin2x﹣6sin2x﹣1=5sin2x﹣1=﹣1﹣1=﹣2故选D6.同时满足两个条件:(1)定义域内是减函数;(2)定义域内是奇函数的函数是()A.f(x)=﹣x|x|B.C.f(x)=tanx D.【考点】奇偶性与单调性的综合.【分析】根据函数奇偶性的定义域判断出f(x)是奇函数、化简f(x)后由二次函数的单调性判断出f(x)的单调性,可判断A;由基本初等函数的单调性判断B、C,根据f(x)的定义域判断D.【解答】解:A、因为f(x)的定义域是R,且f(x)=x|﹣x|=﹣f(x),所以f(x)是奇函数,因为f(x)=﹣x|x|=,所以f(x)在定义域上是减函数,可知符合题中条件,A正确;B、函数在定义域{x|x≠0}不是单调函数,不符合题意,B不正确;C、f(x)=tanx在定义域内不是单调函数,C不正确;D、函数f(x)的定义域是(0,+∞),关于原点不对称,不是奇函数,D不正确.故选A.7.已知函数则f(x)在区间[0,]上的最大值与最小值分别是()A.1,﹣2 B.2,﹣1 C.1,﹣1 D.2,﹣2【考点】正弦函数的定义域和值域.【分析】根据正弦函数的图象和性质,即可求出函数的最大值和最小值.【解答】解:∵0≤x≤,∴≤2x≤,∴当2x=时,即sin(2x)=1时,函数取得最大值为2﹣1=1,当2x=时,即sin(2x)=时,函数取得最小值为﹣×2﹣1=﹣2,故选:A.8.若将函数y=cos 2x的图象向左平移个单位长度,则平移后图象的对称轴为()A.x=﹣(k∈Z)B.x=+(k∈Z)C.x=﹣(k∈Z)D.x=+(k∈Z)【考点】函数y=Asin(ωx+φ)的图象变换.【分析】利用函数y=Asin(ωx+φ)的图象变换规律,正弦函数、余弦函数的图象的对称性,得出结论.【解答】解:由题意,将函数y=cos 2x的图象向左平移个单位得y=cos 2(x+)=cos(2x+)的图象,令2x+=kπ,求得x=﹣,故平移后函数的对称轴为x=﹣,k∈Z,故选:C.9.若sin(﹣α)=﹣,则cos(+2α)=()A.B.C.D.【考点】两角和与差的正弦函数;两角和与差的余弦函数;二倍角的正弦;二倍角的余弦.【分析】利用诱导公式求得cos(+α)的值,再利用二倍角的余弦公式求得=2﹣1的值.【解答】解:∵=cos(+α),∴=2﹣1=﹣,故选A.10.f(x)是定义在R上的奇函数,满足f(x+2)=f(x),当x∈(0,1)时,f(x)=2x﹣1,则的值等于()A.B.﹣6 C.D.﹣4【考点】奇偶函数图象的对称性;函数的值.【分析】先确定函数的正确,再转化,利用当x∈(0,1)时,f(x)=2x﹣1,结合函数为奇函数,即可求得结论.【解答】解:∵f(x+2)=f(x),∴函数的周期为2∵>>∴﹣2>>﹣3∴0>2+>﹣1∴0<﹣2﹣<1∵当x∈(0,1)时,f(x)=2x﹣1∴f(﹣2﹣)=﹣1=∴f(﹣)=∴=﹣故选A.11.若向量,,且,若,则β﹣α的值为()A.或B.C. D.或【考点】平面向量数量积的运算;三角函数中的恒等变换应用.【分析】由与表示出﹣,由⊥(﹣),得到•(﹣)=0,利用平面向量的数量积运算法则列出关系式,由同角三角函数间的基本关系及两角和与差的余弦函数公式化简,求出β﹣α的度数即可.【解答】解:∵=(cosα,sinα),=(2cosβ,2sinβ),∴﹣=(2cosβ﹣cosα,2sinβ﹣sinα),∵⊥(﹣),∴•(﹣)=0,即cosα(2cosβ﹣cosα)+sinα(2sinβ﹣sinα)=0,整理得:2cosαcosβ﹣2cos2α+2sinαsinβ﹣2sin2α=0,即cosαcosβ+sinαsinβ=,∴cos(β﹣α)=,∵≤α<<β≤,∴0<β﹣α<,则β﹣α=.故选:B.12.函数f(x)=,若f(0)是f(x)的最小值,则a的取值范围为()A.[﹣1,2]B.[﹣1,0]C.[1,2]D.[0,2]【考点】函数的最值及其几何意义.【分析】由分段函数可得当x=0时,f(0)=a2,由于f(0)是f(x)的最小值,则(﹣∞,0]为减区间,即有a≥0,则有a2≤x++a,x>0恒成立,运用基本不等式,即可得到右边的最小值2+a,解不等式a2≤2+a,即可得到a的取值范围.【解答】解:由于f(x)=,则当x=0时,f(0)=a2,由于f(0)是f(x)的最小值,则(﹣∞,0]为减区间,即有a≥0,则有a2≤x++a,x>0恒成立,由x+≥2=2,当且仅当x=1取最小值2,则a2≤2+a,解得﹣1≤a≤2.综上,a的取值范围为[0,2].故选:D.二、填空题(本大题共4小题,每小题5分,共20分,将答案填在题中的横线上)13.已知函数g(x)=(a+1)x﹣2+1(a>0)的图象恒过定点A,且点A又在函数(x+a)的图象上.则实数a=1.【考点】函数的图象与图象变化.【分析】令x﹣2=0得x=2并求出g(2),可得定点A的坐标,由题意代入f(x)的解析式,由对数的运算性质化简后求出a的值.【解答】解:令x﹣2=0得x=2,则g(2)=(a+1)0+1=2,所以函数g(x)的图象恒过定点A(2,2),代入(x+a)得,2+a=,解得a=1,故答案为:1.14.若函数f(x)=x2﹣ax﹣b的两个零点是2和3,则函数g(x)=bx2﹣ax﹣1的零点是.【考点】函数零点的判定定理.【分析】函数f (x )=x 2﹣ax ﹣b 的两个零点是2和3,即f (2)=0,f (3)=0,得到关于a 和b 的两个方程,解方程组即可求出a 和b ,代入函数g (x )=bx 2﹣ax ﹣1中,解方程g (x )=0即可.【解答】解:由题意:,解得∴g (x )=﹣6x 2﹣5x ﹣1的零点为﹣,﹣.故答案为:15.在△ABC 中,M 是BC 的中点,AM=3,点P 在AM 上且满足,则= ﹣4 .【考点】平面向量数量积的运算.【分析】先根据AM=3,点P 在AM 上且满足,求||的值,再根据M是BC 的中点,计算,最后计算即可.【解答】解:∵AM=3,点P 在AM 上且满足,∴||=2∵M 是BC 的中点,∴=2=∴=•=﹣=﹣4故答案为﹣416.已知向量,120° .【考点】数量积表示两个向量的夹角.【分析】由知,此两向量共线,又=﹣,故与的夹角为与的夹角的补角,故求出与的夹角即可,由题设条件利用向量的夹角公式易求得与的夹角【解答】解:由题意,故有=(﹣1,﹣2)=﹣,故与的夹角为与的夹角的补角,令与的夹角为θ又,∴cosθ==,∴θ=60°故与的夹角为120°故答案为:120°三、解答题(本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.计算下列式子的值:(1);(2).【考点】三角函数的化简求值;对数的运算性质.【分析】(1)根据对数的运算法则和指数幂的运算法则计算即可.(2)利用诱导公式化简后计算即可.【解答】解:(1);原式=lg(8×125)﹣72++1=lg1000﹣49+8+1=3﹣49+8+1=﹣37(2).原式=sin(4π++cos()﹣tan()==+﹣1=018.已知平面向量,,.(1)求满足的实数m,n;(2)若,求实数k的值.【考点】平面向量数量积的运算.【分析】(1)由题意和向量的坐标运算,以及向量的相等的条件列出方程,即可求出m,n的值,(2)由题意和向量的坐标运算,以及向量的垂直的条件列出方程,即可求出k 的值【解答】解:(1)∵m=(﹣m,2m),n=(4n,n),∴m+n=(4n﹣m,2m+n)∵=m+n,∴,解得m=,n=;(2)∵+k=(3+4k,2+k),2﹣=(﹣5,2),∵,∴﹣5×(3+4k)+2×2(2+k)=0,∴k=﹣19.已知函数f(x)=Acos(+),x∈R,且f()=.(1)求A的值;(2)设α,β∈[0,],f(4α+π)=﹣,f(4β﹣π)=,求cos(α+β)的值.【考点】两角和与差的余弦函数.【分析】(1)直接利用条件求得A的值.(2)由条件根据f(4α+π)=﹣,求得sinα的值,再利用同角三角函数的基本关系求得cosα的值;由f(4β﹣π)=,求得cosβ的值,再利用同角三角函数的基本关系求得sinβ的值;从而求得cos(α+β)=cosαcosβ﹣sinαsinβ的值.【解答】解:(1)对于函数f(x)=Acos(+),x∈R,由f()=Acos=A=,可得A=2.(2)由于α,β∈[0,],f(4α+π)=2cos(+)=2cos(α+)=﹣2sinα=﹣,∴sinα=,∴cosα==.又f(4β﹣π)=2cos(+)=2cosβ=,∴cosβ=,∴sinβ==.∴cos(α+β)=cosαcosβ﹣sinαsinβ=×﹣×=.20.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,)的部分图象如图所示.(1)求函数f(x)的解析式;(2)求函数f(x)的单调递增区间和对称中心.【考点】由y=Asin(ωx+φ)的部分图象确定其解析式;正弦函数的单调性.【分析】(1)利用最值求出A,利用周期求出ω,利用特殊点,求出φ,即可求函数f(x)的解析式;(2)利用正弦函数的性质,求函数f(x)的单调递增区间和对称中心.【解答】解:(1)显然A=2,又图象过(0,1)点,∴f(0)=1,∴sin φ=,∵|φ|<,∴φ=;由图象结合“五点法”可知ω•+=2π,得ω=2.所以所求的函数的解析式为:f(x)=2sin(2x+).(2)﹣+2kπ≤2x+≤+2kπ,可得函数f(x)的单调递增区间[﹣+kπ,+kπ](k∈Z);令,,对称中心.21.已知,,函数f(x)=•(x∈R)(1)求函数f(x)的周期;(2)若方程f(x)﹣t=1在内恒有两个不相等的实数解,求实数t 的取值范围.【考点】平面向量数量积的运算;三角函数的周期性及其求法.【分析】(1)首先利用三角函数的恒等变换,变形成正弦型函数进一步利用函数周期的定义.(2)把求方程的解得问题转化成求函数的交点问题,进一步利用函数的性质求参数的取值范围【解答】解:(1)==,∴周期T=π;(2)依题意:由=t+1,得,即函数y=t与的图象在有两个交点,∵,∴.当时,,y∈[1,2]当时,,y∈[﹣1,2]故由正弦图象得:1≤t<222.已知f(x)是定义在[﹣1,1]上的奇函数,且f(1)=1,若m,n∈[﹣1,1],m+n≠0 时,有.(1)求证:f(x)在[﹣1,1]上为增函数;(2)求不等式的解集;(3)若对所有恒成立,求实数t的取值范围.【考点】奇偶性与单调性的综合;函数的最值及其几何意义;函数奇偶性的性质.【分析】(1)由条件利用增函数的定义证得结论.(2)根据函数的奇偶性和单调性,把要解的不等式等价转化为一个不等式组,求得此不等式的解集即可.(3)根据函数的单调性求得f(x)的最大值,可得t2+t≥g(α)=+2tanα+2对的恒成立,再求得g(α)的最大值,从而求得t的范围.【解答】解:(1)证明:任取x1,x2∈[﹣1,1]且x1<x2,则,∴f(x2)>f(x1),∴f(x)为增函数.(2),等价于,求得0≤x<,即不等式的解集为.(3)由于f(x)为增函数,∴f(x)的最大值为对恒成立对的恒成立,设,则.又==1+tan2α+2tanα+2=(tanα+1)2+2,∵α∈[﹣,],∴tanα∈[﹣,1],故当tanα=1时,.∴t2+t≥6,求得t≤﹣3 t≥2,即为所求的实数t的取值范围.2017年3月8日。
湖北省黄石市第三中学高一上学期物理期末考试及答案
湖北省黄石市第三中学高一上学期物理期末考试及答案一、选择题1.一质点始终向着一个方向做直线运动,在前23t时间内平均速度为v/2,后3t时间内平均速度为 2v,则物体在t 时间内的平均速度大小是( )A.3v/4 B.3v/2 C.v D.2v/32.翻滚过山车是大型游乐园里比较刺激的一种娱乐项目.如图所示,过山车(可看成质点)从高处冲下,过M点时速度方向如图所示.在圆形轨道内经过A、B、C三点,下列说法正确的是A.过山车运动到最高点B时,速度和加速度的大小均为零B.过山车受到重力、弹力、摩擦力和向心力C.过山车经过A、C两点时的速度相同D.过山车做变加速运动3.在中国海军护舰编队“巢湖”“千岛湖”舰护送下,“河北锦绣”“银河”等13艘货轮顺利抵达亚丁湾西部预定海域,此次护航总航程4500海里.若所有船只运动速度相同,则下列说法正确的是()A.“4500海里”指的是护航舰艇的位移B.用GPS定位系统研究“千岛湖”舰位置时,可将“千岛湖”舰看作质点C.以“千岛湖”舰为参考系,“巢湖”舰一定是运动的D.根据本题给出的条件可以求出护舰编队此次航行过程中的平均速度4.如图所示,一定质量的物体用轻绳AB悬挂于天花板上,用水平向左的力F缓慢拉动绳的中点O,用T表示绳OA段拉力的大小,在O点向左移动的过程中A.F逐渐变大,T逐渐变大B.F逐渐变大,T不变C.F逐渐变小,T不变D.F逐渐变小,T逐渐变小5.A、B两个物体在同一直线上作匀变速直线运动,它们的速度图象如图所示,则()A .A 、B 两物体运动方向相反B .A 物体的加速度比B 物体的加速度大C .前4s 内A 、B 两物体的位移相同D .t =4s 时,A 、B 两物体的速度相同6.一汽车通过拱形桥顶点时速度为10 m/s ,车对桥顶的压力为车重的3/4,如果要使汽车在桥顶对桥面没有压力,车速至少为;A .15 m/sB .20 m/sC .25 m/sD .30 m/s7.竖直升空的火箭,其v-t 图像如图所示,由图可知以下说法中正确的是( )A .火箭上升的最大高度为16000mB .火箭上升的最大高度为48000mC .火箭经过120s 落回地面D .火箭上升过程中的加速度始终是20m/s 28.单位制是由基本单位和导出单位所组成的一系列完整的单位体制。
2024学年湖北省黄石市第三中学化学高二上期末统考试题含解析
2024学年湖北省黄石市第三中学化学高二上期末统考试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项)1、金属原子一般具有的结构特点是A.有金属光泽,能导电,有延展性B.核外电子个数少,容易失去C.熔点和沸点较低D.最外层电子个数少,容易失去2、常温下向pH=3的醋酸中加入下列物质后,溶液中水电离出的c(H+)减小的是A.水B.NaHSO4固体C.CH3COONa固体D.NaCl固体3、于固定体积的密闭容器中进行的气体反应A(g)+B(g) C(s)+2D(g),可以说明在恒温下已达到平衡状态的是①反应容器中压强不随时间变化而变化②A气体和B气体的生成速率相等③混合气体的平均摩尔质量不随时间变化而变化④反应混合气体的密度不随时间变化而变化A.③④ B.②③ C.①③ D.①④4、下列物质属于纯净物的是()A.生铁 B.绿矾 C.不锈钢D.铝热剂5、石墨晶体是层状结构,在每一层内,每一个碳原子都跟其他3个碳原子相结合,下图是其晶体结构的俯视图,则图中7个六元环完全占有的碳原子数是A.10个B.18个C.24个D.14个6、下列仪器名称为“分液漏斗”的是A.A B.B.C.C D.D7、下列有关化学实验说法正确的是( )A.加入过量金属钠并分液,可以除去乙醇中的水B.苯中混有的己烯不能通过加入溴水除杂C.除去乙烷中混有的乙烯,可以在一定条件下通入氢气,使乙烯转化为乙烷D.将乙炔与硫化氢的混合气体通过氢氧化钠溶液能够得到纯净的乙炔气体8、a、b、c、d、e 分别是 Cu、Ag、Fe、Al、Mg 五种金属中的一种。
湖北省黄石市初三物理上册期末素质检查试题及答案
湖北省黄石市初三物理上册期末素质检查试题班级:________________ 学号:________________ 姓名:______________一、单选题(每题3分)1.下列关于声现象的说法中正确的是( )A. 声音在真空中传播得最快B. 发出强声波的喇叭能使它前面的烛焰“跳舞”,说明声波具有能量C. “闻其声知其人”主要是根据声音的音调来判断的D. 街道上的噪声监测仪是减弱噪声的专用设备答案:B解析:A、声音的传播需要介质,真空不能传声,故A错误;B、烛焰在喇叭前的气流中跳动,说明声音可以传递能量,故B正确;C、“闻其声而知其人”主要是根据声音的音色来判断的,故C错误;D、街道上的噪声监测仪只是监测声音的大小,不能减弱噪声,故D错误。
2.下列现象中,能用光的直线传播规律解释的是( )A. 雨后彩虹B. 水中倒影C. 日食现象D. 钢笔“错位”答案:C解析:A、雨后彩虹是由于光的折射形成的,不符合题意;B、水中倒影是平面镜成像,其实质是光的反射形成的,不符合题意;C、日食现象是由光的直线传播形成的,符合题意;D、钢笔“错位”是光由水中斜射入空气中时发生折射形成的,不符合题意。
3.下列关于物态变化的说法中,正确的是( )A. 春天,河里冰雪消融,是升华现象B. 夏天,冰糕周围冒“白气”,是汽化现象C. 秋天,早晨花草上出现的小露珠是熔化现象D. 冬天,温暖的车内窗玻璃会变模糊,是因为车内水蒸气液化的缘故答案:D解析:A、春天,河里冰雪消融,属于熔化,故A错误;B、夏天,冰糕周围冒“白气”,是空气中的水蒸气遇到温度较低的冰糕液化成小水珠形成的,是液化现象,故B错误;C、秋天,草叶上结有小露珠,是水蒸气液化形成的,这不是熔化现象,故C错误;D、冬天,温暖的车内窗玻璃会变模糊,是因为车内水蒸气遇冷液化形成小水珠,附着在玻璃的内表面,故D正确。
4.下列关于平面镜成像的说法中,正确的是( )A. 物体离平面镜越近,所成的像越大B. 物体越大,所成的像越大C. 物体离平面镜越近,像离平面镜也越近D. 平面镜所成的像是实像答案:B解析:A、平面镜所成的像与物体大小相同,与物体到镜面的距离无关,故A错误;B、平面镜所成的像与物体大小相同,物体越大,像也越大,故B正确;C、根据平面镜成像特点可知,物体到平面镜的距离与像到平面镜的距离相等,物体离平面镜越近,像离平面镜也越近,故C错误;D、平面镜所成的像是虚像,故D错误。
2025届湖北省黄石市黄石港区数学三上期末统考试题含解析
2025届湖北省黄石市黄石港区数学三上期末统考试题请考生注意:1.请用2B铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。
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小军走一步大约是5________ 做10道口算题大约需要20________和谐号动车每小时大约行220________ 一头蓝鲸的质量可达195________2.8个534的和是(________);179的4倍是(________)。
3.2个17加上3个17是(________),再加上(________)个17就是1。
4.在括号里填上“>”“<”或“=”。
715-152-138(____)715-(152-138)4×(9-6)(____)72÷9+2 263+(900-600)(____)465-(325+75)67-29-12(____)67-(29+12)5.有2瓶同样的矿泉水,其中一瓶丽丽喝了后还剩17,另一瓶丁丁喝了后还剩16,________喝的多。
6.47是由(______)个17组成的,6个18是(______),1里面有(______)个111.7.(1)从正面看到的有_____.(2)从正面看到的有_____.(3)从侧面看到的有_____.8.2020年,新冠肺炎席卷了全球,个人卫生习惯得到了人们的高度重视,消毒洗手液也成为了畅销产品。
某品牌的消毒洗手液现价12.8元,比原来便宜1.2元,这种洗手液原价为(________)元。
二、我是小法官。
(对的打√,错的打×)9.在乘法里,两个因数都乘1,积也要乘1._____10.学校举行庆祝活动,需要1500把椅子,原来有九百多把椅子,再租来六百把椅子,肯定够。
2023-2024学年湖北省黄石市第三中学高一化学第一学期期末监测模拟试题含解析
2023-2024学年湖北省黄石市第三中学高一化学第一学期期末监测模拟试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题(每题只有一个选项符合题意)1、下列有关0.2mol·L-1 BaCl2溶液的说法不正确的是( )A.500mL溶液中Cl-离子浓度为0.2mol·L-1B.500mL溶液中Ba2+离子浓度为0.2mol·L-1C.500mL溶液中Cl-离子总数为0.2N AD.500mL溶液中Ba2+和Cl-离子总数为0.3N A2、设N A为阿伏加德罗常数,下列说法正确的是A.18gNH4+离子中所含的电子总数为12N AB.标准状况下,11.2L水含有的分子数为0.5N AC.0.3mol/LNa2SO4溶液中,含有Na+和SO42-总数为0.9N AD.H2SO4的摩尔质量是98g/mol3、下列化合物的类别完全正确的是()A.Na2CO3,碱,电解质B.MgO,金属氧化物,不是电解质C.H2SO4,酸,电解质D.CO2,酸性氧化物,电解质4、下列物质中,属于电解质的是()A.稀盐酸B.Cu丝C.NaCl D.蔗糖5、下列气体中既可用浓硫酸干燥,又可用固体氢氧化钠干燥的是()A.Cl2B.O2C.SO2D.NH36、将等物质的量的Cl2和SO2混合后通入品红溶液中,观察到的现象是()A.迅速退色B.不退色C.先退色后又恢复红色D.慢慢退色7、用N A表示阿伏德罗常数,下列叙述正确的是()A.标准状况下,22.4LH2O含有的分子数为N AB.常温常压下,1.06g Na2CO3含有的Na+离子数为0.02N AC.通常状况下,N A个CO2分子占有的体积为22.4LD.在0.5mol/L的MgSO4溶液中,Mg2+数目为0.5N A8、已知N A为阿伏加德罗常数的值,下列说法中正确的是( )A.7.8 g过氧化钠与足量水反应转移的电子数为0.2N AB.标准状况下,22.4 L H2O含有N A个水分子C.2 g H2的分子数为N AD.1 mol/L Mg(NO3)2溶液中含有NO3—的数目为2N A9、下列变化属于氧化-还原反应的是A.CaCO3CaO+CO2↑B.2KI+Br2→2KBr+I2C.NaCl+H2SO4NaHSO4+HCl↑D.NH4Cl NH3↑+HCl↑10、将烧碱溶液和盐酸依次滴入某溶液A中,其沉淀(y轴)和加入溶液的体积(x轴)的关系可能出现两种图示形式,则A溶液不可能是()A.三氯化铝溶液B.偏铝酸钠溶液C.明矾溶液D.硝酸铝溶液11、下列方法中,不能用于实验室里制取氨气的是A.在试管中将熟石灰和氯化铵混合后加热B.加热试管中的氯化铵固体C.将烧瓶中的浓氨水加热D.将分液漏斗中的浓氨水滴入装有生石灰的烧瓶中12、下列化合物中,能用金属与酸溶液反应直接制取的是()A.AgCl B.FeSO4C.FeCl3D.CuSO413、能用H++OH-=H2O来表示的化学反应是A.氢氧化镁和稀盐酸反应B.Ba(OH)2溶液滴入稀硫酸中C.澄清石灰水和稀硝酸反应D.二氧化碳通入澄清石灰水中14、下列碳原子的物质的量为1 mol的是( )A.0.012 kg C B.0.012 kg12CC.0.012 kg13C D.0.012 kg12C和13C15、将铁屑溶于过量盐酸后,再加下列物质,会有三价铁生成的是()A.稀硫酸B.氯气C.氯化锌D.氯化铜16、当我们外出旅游的时候,在车站会看到这样的告示:严禁携带易燃、易爆、剧毒品、易腐蚀品等上车。
2024-2025学年黄石市下陆区三上数学期末教学质量检测模拟试题含解析
2024-2025学年黄石市下陆区三上数学期末教学质量检测模拟试题注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
用2B铅笔将试卷类型(B)填涂在答题卡相应位置上。
将条形码粘贴在答题卡右上角"条形码粘贴处"。
2.作答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
答案不能答在试题卷上。
3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。
不按以上要求作答无效。
4.考生必须保证答题卡的整洁。
考试结束后,请将本试卷和答题卡一并交回。
一、认真计算。
1.口算:68÷2= 24×4= 280÷7= 9.3元+6.6元=0×500= 90÷5= 0÷999= 9元-5.9元=2.用竖式计算,带*的题要验算.89×4=580×5=609÷7=8×706=953÷9=*843÷6=3.用递等式计算.880÷8-106 12×4×5 100-720÷8 (57+46)×4二、我会判断。
(对的画√,错的画×)4.一个西红柿大约重220千克。
(______)5.小红爸爸的身高有170米.(______)6.8的9倍是72,9的8倍是72。
(________)7.3500×6的积的末尾有2个0。
(________)8.如果□234÷34的商是两位数,□里能填的数有4个.(______)三、精挑细选。
(把正确答案的序号填在括号里)9.下列现象中,是平移的是()。
A.司机开车时,方向盘的运动B.翻开课本C.推拉抽屉10.2008年奥运会在北京举行,这一年正好是闰年,如图的年份中有()个闰年.A.1 B.2 C.3 D.411.如图所示,比一比.最长的铅笔是()A.A B.B C.C12.73×5的积是()A.345 B.355 C.365 D.375 13.用4个边长为1厘米的正方形摆成的图形中,()的周长最短。
2020年湖北省黄石市第三中学高三语文上学期期末考试试卷及答案
2020年湖北省黄石市第三中学高三语文上学期期末考试试卷及答案一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文字,完成下面小题。
汉字字序为什么从上至下汉字最初的行款应该是自由的,字序、行序并没有明确的规则。
如江苏吴县澄湖良渚古井出土的黑陶贯耳壶,腹部的4个字呈横斜状排列;而江苏高邮龙虬发现的刻文陶片,可以分为两组,4个字的那组应是竖排。
不过,这种自由式的行款在面临字数增多,或是有其他更高功能需求的记录时,不足就非常明显,新型的行款必然出现。
对新行款真正起关键作用的,应是新需求。
根据早期汉字的应用情况,我们不难猜测,它就是甲骨文占卜、竹简记录等活动。
所以,能对行款样式起主要作用的主要是两个方面:一是甲骨和竹简本身,二是活动性质的影响。
第一个方面的甲骨常被略去,因为甲骨本身对字序、行序没有多少约束;而竹简常受关注,学者们常从持简习惯、书写习惯等方面入手,认定竹简所起到的决定性作用。
然而,对于自上而下的字序,尤其是单简书写时,无论是“方便说”,或是“持简习惯说”,都略显牵强。
“方便说”认为,汉字字形便于竖写,所用的竹简宜于竖写,这是汉字书写的决定性条件。
但事实上,尚没有真正讲究书写顺序的初期汉字,是无所谓竖写横写的,这时候的书写是为字形服务,而非字形为书写服务。
而就竹简而言,客观上也没有明确的适宜横写或竖写,同样构不成竖写字序的决定性因素。
“持简习惯说”认为.古人应是席地而坐,左手持简,置于胸前,右手执笔而书。
正是持简的习惯性动作决定了字序的下行。
这个说法的前提条件是古人必须有竖向持简的习惯,而这其实是很难确定的。
正如北京大学教授李零所说:“我们最容易犯的错误,不是别的,就是‘以今人之心度古人之腹’。
”比如,甲骨是种不易写也不易刻的“不方便”材料,却被选为早期一个重要的书写载体,可见古人有时候对书写的目的更为重视,方便性反而是其次的。
这样,就剩下第二点,即活动本身。
它具有两个先天优势。
首先,地位优势。
2020年湖北省黄石市第三中学高三英语上学期期末考试试题及参考答案
2020年湖北省黄石市第三中学高三英语上学期期末考试试题及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项ASome of the world’s most talented musicians have played concerts atLincolnCenter. On September 10, the center hosted a unique class of star musicians: kids and teens. The young players are part of the World Peace Orchestra, or WPO. The group held its first Music for Peace concert inNew York City, with 134 students from more than 60 countries taking part. Musicians were chosen from all over the world.The World Peace Orchestra began in 2013. The nonprofit group brings kids together using the common language of music. To be considered for the orchestra, students first had to be recommended by their teachers and then audition online for a group of judges. Once chosen, professional musicians and teachers then worked with the players to improve their skills.Some of the young musicians played classical instruments, such as violins and flutes. Others played instruments unique to their home country. For example, Amold Mugo, 16, fromKenya, played an instrument called the djembe. The drum is originally fromWest Africa. Mugo said he was shocked when he learned he was chosen for the orchestra. “I can’t express how I felt. I was rolling on the floor. It’s a once in a lifetime opportunity,” he said.Adomas Hendrixson, 13, fromLithuania, played piano for the WPO. Before theNew York Cityperformance, Hendrixson talked about what he hoped to take away from the event. “Fun and joy-- people smiling and clapping,” he said.“This is one of the only times in your life this could happen and I'm very excited.”Mugo says his favorite part of the WPO is making new friends from around the world. “I hope when I go home, I just take a little bit of every friend that I made here-- Brazil, Portugal, Queens,” Mugo said, “I hope that I take part of their culture home with me so I understand them more.”1. Who is most likely to have played for the WPO according to this passage?A. Li Ming, 16, a student fromChina.B. Daniel, 44, a judge fromAmerica.C. Edward, 21, a clerk fromAustralia.D. Catherine, 32, a teacher fromEngland.2. We can learn from the last paragraph that Mugo ________.A.performed best for the WPO.B. has learned some foreign culture.C. was eager to be chosen for a second time.D. received a great award for his performance.3. What would be the best title for this passage?A. Kids were good at playing music.B. Mugo liked making foreign friends.C. Concerts were played atLincolncentre.D. Young musicians played for peace.BCycling has long been a sport where a competitive edge is pursued, often at any cost. Cycling shorts, however, were an area that technology had yet to enter. Little was added to them apart from, perhaps, some padding to protect a rider.Now, however, new cycling shorts are being equipped with sensors to help riders defeat opponents easily. A British start-up has created a pair of shorts that uses electric sensors and artificial intelligence to shape and strengthen cyclists' muscles while they are moving. It plans to also adapt this product next year for joggers and people at the gym.The shorts, created by Devon Lewis, a PhD student at theUniversityofSouthampton, are able to send out tiny electric currents into a user's muscles to improve their cycling technique. Mr Lewis, 27, came up with the idea after studying a connection between the brain and muscles that sends chemical signals from motor neurons to muscle fibres to make the muscle contract and strengthen.The shorts, created by Mr Lewis's start-up bypass this connection using electric cur rents to build muscle strength directly, without relying on the brain to send out signals. “We have a limited ability to control our muscles naturally,” Mr Lewis said. “You can control them more exactly, get more from your muscles and contract them more strongly, if you stimulate them directly with electricity.” Electric wires in the shorts send out currents that activate the muscle. “For those worried about getting a shock while cycling,” Mr Lewis said, “It's sort of the same thing as when you have spicy food and you get that little rush. It feels strange when you first start using it but then you adapt to it quite quickly.”His idea has received support from Future Worlds, a start-up accelerator atSouthamptonUniversity. The product is being shown next week at the technology conference CES, held digitally this year.4. What's the strength of the newly invented cycling shorts ?A. They help cyclists perform better.B. They last longer than normal ones.C. They can keep a balanced comfort.D. They have tools to make good figures.5. Why can the cycling shorts help the runners and joggers?A. They are made of quality cotton materials.B. They are created by world-top designers.C. They can make running or jogging less boring.D. They are made intelligently with electric sensors.6. What may be some people's worry about the shorts?A. They may spend more than they can afford.B. They get little help from the new shorts.C. They may get shocked by the electricity.D. The new shorts can't function well enough.7.What's the best title of the text?A. The Special CyclingShortsB. Waysto Be a Good RunnerC. How Shorts Benefit YouD. Some Tips about Running FastCOne day, when I was working as a psychologist(心理学家)in England, an adolescent boy showed up in my office. It was David. He kept walking up and down restlessly, his face pale, and his hands shaking slightly. His head teacher had referred him to me.“This boy has lost his family,” he wrote. “He is understandably very sad and refuses to talk to others, and I’m very worried about him. Can you help?”I looked at David and showed him to a chair. How could I help him? There are problems psychology doesn’t have the answer to, and which no words can describe. Sometimes the best thing one can do is to listen openly and sympathetically(同情)The first two times we met, David didn’t say a word. He sat there, only looking up to look at the children’s drawings on the wall behind me. I suggested we play a game of chess. He nodded. After that he played chess with me every Wednesday afternoon---in complete silence and without looking at me. It’s not easy to cheat in chess, but I admit I made sure David won once or twice.Usually, he arrived earlier than agreed, took the chess board and pieces from the shelf and began setting them up before I even got a chance to sit down. It seemed as if he enjoyed my company(陪伴). But why did he never look at me?“Perhaps he simply needs someone to share his pain with,” I thought. “Perhaps he senses that I respect hissuffering.” Some months later, when we were playing chess, he looked up at me suddenly.“It’s your tum,” he said.After that day, David started talking. He got friends in school and joined a bicycle club. He wrote to me a few times, about his biking with some friends, and about his plan to get into university. Now he had really started to live his own life.Maybe I gave David something. But I also learned that one…without any words---can reach out to another person. All it takes is a hug, a shoulder to cry on, a friendly touch, and an ear that listens.8. When he first met the author, David .A. felt a little excitedB. looked a little nervousC. walked energeticallyD. showed up with his teacher9. David enjoyed being with the author because he .A. wanted to ask the author for adviceB. liked the children’s drawings in the officeC. beat the author many times in the chess gameD. needed to share sorrow with the author10. What can be inferred about David?A. He liked biking before he lost his family.B. He recovered after months of treatment.C. He went into university soon after starting to talk.D. He got friends in school before he met the author.11. What made David change?A. The author’s silent communication with him.B. His teacher’s help.C. The author’s friendship.D. His exchange of letters with the author.DAs one of the world’smost popular cultural mediums, cinema is at the leading position at the Taihu World Cultural Forum(论坛), an annual event that aims to improve cultural exchanges. So far this year, Chinese box officehas already topped 58 billion yuan, the China Film Administration said. This indicates that China, the world’s second-largest movie market, is closing the gap with the United States, the world’s top cinema market.Cao Yin, director of the program center at China Movie Channel, said that China has 67,000 cinema screens. the most of any country, and the number is expected to increase to 80,000 by the end of next year. Saying that China has signed coproduction agreements with 22 countries (including the United States, Canada, Japan. India etc.) ,Cao added it has deepened cinematic exchanges between Chinese filmmakers and their foreign counterparts (同行).With the country’s huge native market, which produced more than 1,000 films in 2019, Hong Kong director Stanley Tong said he believes foreign filmmakers will increasingly seek cooperation with China. “International coproduction will become an important platform to send Chinese stories oversea and give us a broader vision of creation,” Tong said.Recently, over 80 percent of the world’s top 100 highest box-office titles have been action films. Tong said the films, in which plots are basically secondary to shining stunts (特技), is one of the easiest ways to appeal to foreign audiences. But it has been a decades-long struggle for Chinese filmmakers to sell their stories overseas. When asked what kinds of Chinese films would have the most global appeal. Yan Zhaozhu, chairman of the Taihu World Cultural Forum, said stories that address universal issues, such as environmental protection and climate change, are perhaps the best options.12. What is the purpose of the Taihu World Cultural Forum?A. To strengthen cultural exchanges.B. To build more cinemas in China.C. To invite more tourists to Taihu.D. To attract more international investment.13. What can we infer about Cao Yin’s opinion in Paragraph 2?A. Chinese box office has already overtaken the United States so farB. The screens of Chinese cinemas will be world-leading in the futureC. China has signed the most coproduction agreements with other countriesD. There will be more and more cooperations between Chinese filmmakers and other countries.14. How do Chinese filmmakers feel about selling their movies to the world right now?A. Easy.B. Disappointed.C. Challenged.D. Confused.15. What is the best title for the text?A. Box Office: Unclear FutureB. China: Top Second Movie MarketC. Cinema: The Leading Cultural PositionD. Chinese Filmmakers: Severe Competition Situation第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年湖北省黄石市第三中学高三英语上学期期末试题及答案
2020-2021学年湖北省黄石市第三中学高三英语上学期期末试题及答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项A阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项,并在答题卡上将该项涂黑。
The opening ceremony for the 2020 Tokyo Olympic Games was held at the Olympic Stadium in Tokyo on July 23, 2021 after a year of waiting due to the pandemic. Let's take a look at the highlights which make this opening ceremony unique.No spectatorsThere are no spectators during the opening ceremony in an effort to prevent the spread of the coronavirus infections. About 950 people, including officials and journalists, will watch in the main stadium, organizers said. Fewer athletes are expected to attend the ceremony because many are not allowed to stay for the entire Games due to COVID-19 restrictions.Male and female flag-bearersThe International Olympic Committee has actively encouraged all delegations toselect two flag-bearers at the opening ceremony, including one woman, as women's participation in the Tokyo Games reaches 48.8 percent, making it the most gender-balanced Olympics in the history of the event.Oath (宣誓) session updatedOne of the most symbolic parts of the opening ceremony—the Olympic oath—has been significantly adapted in order to “highlight the importance of solidarity, inclusion, non-discrimination and equality”. The number of oath-takers is also extended from three to six—two athletes, two coaches and two judges.Simplified ceremonyIn the 30-second video showing the rehearsal of the opening ceremony released to the public, the performances are integrated with dances, scenery and other display elements used in Japanese traditional festival activities.In order to prevent and control the COVID-19 pandemic and manage the cost, the Tokyo 2020 Organizing Committee said that the performances of the ceremony will be greatly simplified, with more light and shadow technology used instead of real decorations at the scene.1.Why isn't there any spectator during the opening ceremony?A.To limit the number of people present.B.To highlight the spirit of the Olympic Games.C.To control the spread of an infectious disease.D.To make room for the athletes all over the world.2.Why is each delegation encouraged to select one woman as a flag-bearer in Tokyo Olympics?A.Women will be in a leading position in sports.B.Women athletes make up a larger percentage than ever.C.Women will play a more important part in society.D.More and more women will take part in the Olympic Games.3.Which of the following statements is True according to the passage?A.Oath-takers are made up of six athletes.B.Tokyo uses more real decorations to cut the cost.C.About 950 people watch the entire games in the Tokyo Games.D.The opening ceremony presents the traditional culture of Japan.BWhena person in the United States gets the COVID-19 vaccine (疫苗), the person receives a small piece of paper called a “COVID-19 Vaccination Record Card”.It is a piece of paper with the logo of the Centers for Disease Control and Prevention, or CDC, and the name and date of the vaccine. Because it is so simple, it could be easy to fake.Many people in theU.S.are still unsure about getting the vaccine. As universities, workplaces and other places are requiring proof of vaccination, some people are now buying fake vaccination cards without getting a shot.The Associated Press reports that students and teachers at universities around theU.S.are worried about fake cards. Sellers are using social media apps like Instagram to advertise fake vaccination cards. The prices range from $25 to $200. The AP notes that many college students seem interested in buying the cards. On the site Reddit, one person wrote, “I need one, too, for college. I refuse to be a guinea pig.”It is reported that more than 700 universities and colleges require proof of vaccination. Most schools simply ask their students to take a photo of their card and send it to a school website. Benjamin Mason Meier is a professor at theUniversityofNorth CarolinaatChapel Hill, UNC. He studies international health policy. He said, unlike some countries, theU.S.is not using a digital system to record vaccine status. He said theU.S.is depending on“aflimsypaper card”, and students have told him they knew of others who had used fake vaccination cards. Rebecca Williams also works at UNC. She is a researcher at the school’s Center for Health Promotion and Disease Prevention. She said she was not surprised that people were worried about fake vaccination cards. “This is why I think the development of a reliable national digital vaccine passport app is very important,” she said.There is a law that should prevent people from making false vaccination cards. If someone uses the CDC logo without permission, they can be lined and punished by up to five years in prison. The U.S. Department of Justice recently charged a person inCaliforniawith making fake vaccination cards.College students who already have the vaccine are criticizing those who would rather spend money to buy a fake than get a free shot. Maliha Reza is an electrical engineering student atPennsylvaniaStateUniversity, She called those students “dumb”. “I’m angry about that,” she said. “Like, there is more anger than I could describe now.”4. Why do some college students buy fake vaccination cards?A. They are easy to get.B. Many Americans are still uncertain about getting the vaccine.C. Students have an interest in the fake vaccination cards.D. To get a vaccine shot is expensive.5. Which of the following measures can NOT be used to stop the fake cards?A. Having a law that should prevent people from making false vaccination cards.B. Developing a reliable national digital vaccine passport app.C. Using a digital system to record vaccine status.D. Having all the students take a photo of their vaccination card and send it to the school website.6. What does the underlined word “flimsy” probably mean?A. Weak.B. Effective.C. Detailed.D. Professional,7. What might be the writing purpose for the news report?A. To promote a digital system to record the vaccination shots.B. To explain why theU.S.should prevent making the fake vaccination cards.C. The stress the influence of the COVID-19 vaccination.D. To reduce theU.S.university leaders’ worry about fake vaccination cards.CIt was the first day of spring here but it didn't feel like it. The COVID-19 epidemic (流行病) was sweeping the globe and everywhere there were feelings of fear and loneliness. “Social Distancing” had become the new norm. Here the schools had been closed, the restaurant dining rooms had been shut, and people had been told to work from home whenever they could. Even the sheltered workshop where my oldest son worked had been closed until further notice. People had made a run on the stores and large areas of the shelves were bare. On the news the numbers of the sick and dead continued to rise. It felt like there was a weight on the souls of everyone in the world.My family were staying at home as much as possible and as I looked out of my window I wondered how long this crisis would last. It was then, however, that I saw something that lifted that weight off of my soul, made me smile, and made my heart feel happy again. On the street below my house there was an old friend of mine from high school who was a teacher there. With him was my younger son's former aide from the high school as well. They were delivering the school lunches door to door to the hungry children who were stuck at home. Watching them made me think of something everyone's childhood television neighbor, Mr. Roger's once said: "In the bad times, always look for the helpers."I have no doubt that this crisis will pass as all the crisis before. But it is our choice on whether it brings out the best in us or the worst in us. Let it bring out the best in you. Use it to strengthen your faith. Use it to free yourself from fear. Use it to grow kinder, more giving, and more loving. Become a helper to all those in need and you will be a happy person today and all the days to come.8. Why didn't the author like the first day of spring?A. He had to work at home instead of at office.B. There wasn't social distance between each other.C. There was an epidemic everywhere in the world.D. His own company had been closed for a long time.9. How did people feel in the face of the epidemic?A. Disappointed.B. TenseC. Confident.D. Cheerful.10. What made the author happy?A. Seeing someone delivering lunches from door to door.B. Talking with his old friend in the street.C. Watching a childhood television at home.D. Staying together with his family.11. What do Mr. Roger's words actually mean?A. It's not necessary for us to find helpers in bad times.B. In good times we needn't do much to help others.C. We should ask more people to help us in bad times.D. We are supposed to help each other in bad times.DA new look for technology, Solestrom’s new high-tech swimsuits promise to stand out all summer long. First in their new products is Solestrom’s new SmartSwimCMUV Smart Bikini featuring a smart UV meter.The bikini collects UV information through a smart fabric belt and reports the UV index to the wearer with 0.01 exactness. The electronic parts are neatly built into the removable belt, and can be worn even underwater. Next in the list is a lower cost cousin, the SmartSwimCMUV Index Detector Bikini, which has UV sensitive beads that change color1 with the level of UV intensity. The reading gives more of a range rather than an exact number, but for those who simply need to know if the UV is low, moderate or high, this bikini fits the bill.It is now available in Banana Split, more color1 s may become available later in the season. And finally, what could be better than a beach tote with built in energy source? No need to suffer dead batteries in your MP3, just plug them into your Solar Powered Beach Tote for on-the-go mobile charging. This beach tote combines fashion and functionality all in one, lightweight and roomy beach accessory. A built in solar panel charges fully most small electronics in only 2-3 hours of sunlight. Pauline Butler, Media Relations Manger at Solestrom states “the concept of blending fashion and technology is growing at amazing rate, and we are right on the leading edge. Our products are new, creative and meet the need of the young and environmentally conscious crowd.’’Solestrom’s SmartSwinirCMproducts retail from $98. 99- $189. 99,and can be found in their online store, Solestrom. com.12. What is the passage mainly about?A. Famous Solestrom and its Products.B. Fashionable and Functional Beach Tote.C. Smart swimsuits and Solar Powered Beach Tote.D. Where and How to Buy SmartSwirn and the Beach Tote.13. Why are UV sensitive beads used in the SmartSwirnCMUV Index Detector Bikini?A. To give an exact number.B. To know if the UV is low, moderate or high.C. To change color1 with the level of UV intensity.D. To improve the equality of the UV Index Detector Bikini.14. What can we know about Solar Powered Beach Tote?A. Its parts are fixed in the belt.B. It combines lightweight and functionality.C. It can tell the weaver the level of UV intensity.D. It saves people from worrying about dead batteries in their MP3.15. Who will prefer the new products mentioned in the text?A. The young people and the environmentalists.B. People who like to travel all over the world.C. People who love to buy goods online.D. People interested in the latest bag.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2021届湖北省黄石市第三中学高三生物上学期期末考试试卷及参考答案
2021届湖北省黄石市第三中学高三生物上学期期末考试试卷及参考答案一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题目要求。
1. 下列有关实验的叙述,错误的是()A. 换成高倍镜观察叶绿体,图像不清晰要调细准焦螺旋B. 蛋白质变性后,加入双缩脲试剂仍呈现紫色反应C. 质壁分离及复原实验中先后用低倍和高倍显微镜观察三次,形成自身前后对照D. 洋葱表皮细胞浸润在一定浓度的甘油溶液中可发生质壁分离和质壁分离复原2. 科学家用含有14C的二氧化碳来追踪光合作用中的碳原子,这种碳原子的转移途径是()A.二氧化碳→叶绿素→ADPB.二氧化碳→乙醇→糖类C.二氧化碳→叶绿体→ATPD.二氧化碳→三碳化合物→糖类3. 如图表示的是一个由200个氨基酸构成的蛋白质分子。
下列叙述正确的是()A. 该分子中含有198个肽键B. 这个蛋白质中至少有200个氨基C. 合成该蛋白质时相对分子质量减少了3600D. 该蛋白质中至少含有4个游离的羧基4. 若一个学生吃的早餐是两个面包和一杯牛奶,维持整个上午的学习活动的能量直接来自()A.A TPB. 葡萄糖和脂肪C. 淀粉和蛋白质D. 面包和牛奶5. 2019年“世界水日”的宣传主题为“Leaving no one behind”(不让任何一个人掉队)。
下列有关生物体内水的叙述,错误的是()A. 细胞中的自由水可在细胞中流动,是良好的溶剂B. 水在病变细胞中以结合水和自由水的形式存在C. 细胞内核糖体在代谢过程中能够产生水D. 因为水分子间的氢键比较稳定,所以水的比热容较高6. 血友病是伴X染色体隐性遗传疾病。
患血友病的男孩,其双亲基因型不可能是A.X H X h和X h YB.X h X h和X H YC.X h X h和X h YD.X H X H和X h Y7. 脂质是细胞和生物体的重要组成成分,下列叙述错误的是()A.磷脂是构成细胞膜的重要成分B.脂肪仅由C、H、O三种元素组成C.维生素D在人体内参与血液中脂质的运输D.性激素能促进人和动物生殖器官的发育8. 伴随着人们的经济活动和国际交往,会使得一些物种由原生存地借助于人为作用或其他途径移居到另一个新的生存环境并在新的栖息地繁殖并建立稳定的种群,下图为某物种迁人一个新环境后种群的增长曲线,下列叙述中错误的是()A.t3之前种群呈“J”型增长,t3之后呈“S”型增长B.t2时(K/2)种群的出生率和死亡率的差值最大C.t3后种群的数量仍会发生变化D.t1~t3内种内斗争越来越强9. “结构与功能相适应”是生物学的基本观点之一。
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湖北省黄石市第三中学上册期末精选综合测试卷(word含答案)一、第一章运动的描述易错题培优(难)1.某班同学去参加野外游戏.该班同学分成甲、乙、丙三个小组,同时从营地A出发,沿各自的路线搜寻目标,要求同时到达营地B,如图所示为其运动轨迹,则关于他们的平均速度和平均速率的说法正确的是( )A.甲、乙、丙三组的平均速度大小相同B.甲、乙、丙三组的平均速率大小相同C.乙组的平均速度最大,甲组的平均速度最小D.乙组的平均速率最小,甲组的平均速率最大【答案】AD【解析】【详解】AC、三个质点从A到B的过程中,位移大小相等,时间相同;平均速度是位移与时间段的比值,故平均速度相同,故A正确,C错误;BD、三个质点从A到B的过程中,路程不全相同,时间相同;平均速率是路程与时间的比值,由图象知乙组的平均速率最小,甲组的平均速率最大,故C错误;D正确;故选AD.【点睛】位移是指从初位置到末位置的有向线段,路程是轨迹的长度,故从M到N过程中,三个物体的位移相同,但路程不等;平均速率是路程与时间的比值,而平均速度是位移与时间段的比值.2.若某物体做直线运动的v—t图象如图所示,则下列说法中正确的是()A.t=3s时物体运动的速度方向发生改变B.t=3s时物体运动的加速度方向发生改变C.t=3s时物体离出发点最远D.t=3s时物体的加速度为零【答案】AC【解析】【分析】解决本题要明确v —t 图象的含义:在v —t 图象中,速度的正负表示其运动方向,图象的斜率表示物体运动的加速度,图象与时间轴围成的面积为物体的位移,时间轴上方面积表示位移为正,下方表示为负.【详解】A .根据速度的正负表示速度的方向,可知t =3s 时物体运动的速度方向发生改变,故A 正确;B .在2~5s 内直线的斜率一定,说明物体的加速度恒定,则t =3s 时物体运动的加速度方向没有发生改变,故B 错误;C .物体在前3s 内沿正方向运动,3s 后沿负方向运动,则t =3s 时物体离出发点最远,故C 正确;D .根据斜率等于加速度,可知t =3s 时物体的加速度不为零,故D 错误。
故选AC 。
【点睛】图象由于具有形象直观的特点,因此在物理中广泛应用,对于图象问题要明确两坐标轴的含义,图象斜率、截距、围成面积等含义。
3.一个物体做直线运动的位移—时间图象(即x t -图象)如图所示,下列说法正确的是A .物体在1s 末运动方向改变B .物体做匀速运动C .物体运动的速度大小为5m/sD .2s 末物体回到出发点【答案】BC【解析】【分析】【详解】AB .位移时间图象的斜率表示速度,根据图象可知物体一直向负方向匀速运动,故A 错误、B 正确;C .物体运动的速度大小为5m/s ,故C 正确;D .物体的出发点在5m x =的位置,2s 末在5m x =-的位置,故2s 末物体未回到出发点,故D 错误;故选BC 。
4.心电图是现代医疗诊断的重要手段,医生在心电图上测量出相邻两波峰的时间间隔,即为心跳周期,由此可计算出1分钟内心脏跳动的次数(即心率)。
甲、乙两人在同一台心电图机上做出的心电图分别如图甲、乙所示,医生通过测量后记下甲的心率是60次/分。
该心电图机图纸的移动速度和乙的心率分别为A.20 mm/s,60次/分B.25 cm/s,70次/分C.25 mm/s,75次/分D.20 cm/s,120次/分【答案】C【解析】【详解】甲图纸中,相邻波峰的位移为25mmx=,甲的心率是60次/分,则甲心率的周期:60s1s60T==次则心电图机图纸的移动速度:25mm/s=25mm/s1xvT==乙图纸中,相邻波峰的位移为20mmx'=,乙的心率:1111.25Hz75/min20mm25mm/sfxTv=====''次故C正确,ABD错误。
故选C.5.某物体沿水平方向做直线运动,其v-t图象如图所示,规定向右为正方向,下列判断正确()A.在0~1 s内,物体做曲线运动B.在1 s~2 s内,物体向右运动,且速度大小在减小C.在1 s~3 s内,物体的加速度方向向右,大小为4 m/s2D.在3 s末,物体处于出发点左方【答案】B【解析】【详解】A .在0s ~1s 内,速度都是正值,说明物体一直向右做直线运动,不是曲线运动.故A 错误.B .在1s ~2s 内,速度都是正值,说明物体向右运动,速度大小在减小.故B 正确.C .在1s ~3s 内,物体的加速度2244m/s 4m/s 2v a t --==-= 说明物体的加速度方向向左,大小为4m/s 2.故C 错误.D .由图象的“面积”看出,前2s 内物体向右运动的位移大于第3s 内向左运动的位移,所以在3s 末,物体处于出发点右方.故D 错误.6.如图所示,从高为3m 处以某一初速度竖直向下抛出一个小球,在与地面相碰后竖直向上弹起,上升到高为2 m 处被接住,则这一过程中:( )A .小球的位移大小为1 m ,方向竖直向下,路程为5 mB .小球的位移大小为5 m ,方向竖直向上,路程为5 mC .小球的位移大小为1 m ,方向竖直向下,路程为1 mD .小球的位移大小为5 m ,方向竖直向上,路程为1 m【答案】A【解析】【分析】【详解】从高为3m 处以某一初速度竖直向下抛出一个小球,在与地面相碰后弹起,上升到高为2m 处被接住,首末位置的距离为1m ,所以位移的大小等于1m ,方向竖直向下.运动轨迹的长度为3+2=5m ,所以路程等于5m .故A 正确,BCD 错误.故选A .【点睛】此题考查了对路程和位移概念的理解;解决本题的关键知道位移是矢量,大小等于首末位置的距离,路程是标量,大小等于运动轨迹的长度.7.如图所示,气垫导轨上滑块经过光电门时,其上的遮光条将光遮住,电子计时器可自动记录遮光时间Δt .测得遮光条的宽度为x ,用x t ∆∆近似代表滑块通过光电门时的瞬时速度.为使x t∆∆更接近瞬时速度,正确的措施是_____A.换用宽度更窄的遮光条B.提高测量遮光条宽度的精确度C.使滑块的释放点更靠近光电门D.增大气垫导轨与水平面的夹角【答案】A【解析】【分析】【详解】极短时间内的平均速度表示瞬时速度;即换用宽度更窄的遮光条,通过光电门的时间更短,更接近瞬时速度,故A正确;8.甲、乙两人同时同地出发骑自行车做直线运动,前1小时内的位移—时间图象如图所示,下列表述正确的是( )A.0.2~0.5小时内,甲的加速度比乙的大B.0.2~0.5小时内,甲的速度比乙的大C.0.6~0.8小时内,甲的位移比乙的小D.0.8小时内,甲、乙骑行的路程相等【答案】B【解析】A、由图知,0.2-0.5小时内甲乙都做匀速直线运动,加速度均为零,故A错误;B、s t 图象的斜率表示速度,甲的斜率大,则甲的速度比乙的大,故B正确;C、物体的位移等于s的变化量.则知0.6-0.8小时内,甲的位移比乙的大,故C错误;D、0-0.6小时内,甲的位移比乙的大,0.6-0.8小时内,甲的位移比乙的大,所以0.8小时内,甲的路程比乙的大,故D错误.点睛:该题考查了对位移--时间图象的理解和应用,要掌握:在位移-时间图象中,图象的斜率表示质点运动的速度的大小,纵坐标的变化量表示位移.9.如图所示,两位同学从滑道最高端的同一位置先后滑下,到达低端的同一位置,对于整个下滑过程,两同学的()A.位移一定相同B.时间一定相同C.末速度一定相同D.平均速度一定相同【答案】A【解析】下滑过程中的始末位置相同,即位移相同,由于两者与滑梯间的动摩擦因数不同,所以导致运动情况不同,即运动时间,以及到达末端的速度不一定相同,A正确BC错误;位移大小相同,而所用时间不一定相同,所以两者的平均速度不一定相同,D错误.10.下列说法中正确的是()A.研究短跑运动员的起跑动作时,可将运动员看做质点B.研究汽车在上坡时有无翻倒的危险时,可将汽车看做质点C.“两岸猿声啼不住,轻舟已过万重山”是以“万重山”为参考系的D.升国旗时,观察到国旗冉冉升起,观察者是以“国旗”为参考系的【答案】C【解析】【分析】【详解】A.研究短跑运动员的起跑动作时,运动员看成质点,否则的话就没有人的动作可言了,A 错误;B.研究汽车在上坡时有无翻倒的危险时,汽车的大小和形状不能忽略,故不能看作质点,B错误;C.“两岸猿声啼不住,轻舟已过万重山“是描述轻舟相对于“万重山”在运动,是以“万重山”为参考系的,C正确;D.升国旗时,观察到国旗冉冉升起,观察者是以大地为参考系的,D错误。
【点睛】当物体的形状、大小对所研究的问题没有影响时,我们就可以把它看成质点,根据把物体看成质点的条件来判断即可;注意区分研究对象与参考系的不同。
考查对参考系概念的理解及实际判断能力。
二、第二章匀变速直线运动的研究易错题培优(难)11.甲、乙两车在同一水平路面上做直线运动,某时刻乙车在前、甲车在后,相距x=6m,从此刻开始计时,乙做匀减速运动,两车运动的v-t图象如图所示。
则在0~12s内关于两车位置关系的判断,下列说法正确的是()A.t=4s时两车相遇B.t=4s时两车间的距离为4mC.0~12s内两车有两次相遇D.0~12s内两车有三次相遇【答案】D【解析】【分析】【详解】AB.题中图像与时间轴围成的面积可表示位移,0~4s,甲车的位移为48m,乙车的位移为40m,因在t=0时,甲车在乙车后面6m,故当t=4s时,甲车会在前,乙车会在后,且相距2m,所以t=4s前两车第一次相遇,t=4s时两车间的距离为2m,故AB错误;CD.0~6s,甲的位移为60m,乙的位移为54m,两车第二次相遇,6s后,由于乙的速度大于甲的速度,乙又跑在前面,8s后,甲车的速度大于乙的速度,两车还会有第三次相遇,当t=12s时,甲的位移为84m,乙的位移为72m,甲在乙的前面,所以第三次相遇发生在t=12s之前,所以在0~12s内两车有三次相遇,故C错误,D正确。
故选D。
12.近年来,一些高级轿车的设计师在关注轿车加速性能的同时,提出了“加速度的变化率”的概念,用这一新的概念来描述轿车加速度随时间变化的快慢,轿车的加速度变化率越小,乘坐轿车的人感觉越舒适。
图示是一辆汽车在水平公路上行驶时加速度随时间变化的关系图像,取t=0时速度方向为正方向,则关于加速度变化率以及汽车的运动,下列说法正确的是()A.“加速度的变化率”的单位是m/s2B.当a t-图像是一条横线时,运动是匀速直线运动C.若加速度与速度同方向,如图所示的a t-图像,表示的是物体在做减速运动D .若加速度与速度同方向,如图所示的a t -图像,已知物体在t =0时速度为5m/s ,则2s 末的速度大小为8m/s【答案】D【解析】【分析】【详解】A .加速度的变化率为ΔΔa t ,a 的单位是m/s 2,所以“加速度的变化率”的单位应该是m/s 3,A 错误; B .当a t -图像是一条横线时,加速度不变,物体做匀加速直线运动,B 错误;C .若加速度与速度同方向,如图所示的a t -图像,物体在做加速度减小的加速运动,C 错误;D .由ΔΔv a t =,知在a t -图像中,图像与时间轴所围图形的面积表示物体速度的变化量Δv ,可得在2秒内,汽车的速度增加量为32Δm/s 3m/s 2v ⨯== 若汽车在t =0时速度为5m/s ,在2秒内汽车的速度增加3m/s ,则汽车在2秒末速度的大小为8m/s ,D 正确。