【温州2019届高三适应性测试】2018年8月温州市普通高中选考适应性测试技术试题(含答案)
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2023届浙江省温州市普通高中高三下学期选考适应性考试(三模)全真演练物理试题一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题如图所示,一根弹簧一端固定在左侧竖直墙壁上,另一端连着小球A,同时水平细线一端连着A球,另一端固定在右侧竖直墙上,弹簧与竖直方向的夹角是60°,A、B两小球分别连在另一根竖直弹簧两端。
开始时A、B两球都静止不动,A、B两小球的质量相等,重力加速度为g,若不计弹簧质量,在水平细线被剪断瞬间,A、B两球的加速度分别为( )A .B.,C.,D.,第(2)题战绳训练中,运动员抖动战绳一端,使其上下振动,运动状态可视为简谐振动。
如图所示,足够长的战绳两端,两位运动员均以的频率、相同的起振方向同时上下抖动战绳,在战绳上传播的波速为,下列说法正确的是( )A.战绳上每个部分振幅都相同B.战绳上每个部分振动频率都为C.战绳上相邻的振动加强区相距为D.战绳上相邻的振动减弱区相距为第(3)题有一种瓜子破壳器其简化截面如图所示,将瓜子放入两圆柱体所夹的凹槽之间,按压瓜子即可破开瓜子壳。
瓜子的剖面可视作顶角为θ的扇形,将其竖直放入两完全相同的水平等高圆柱体A、B之间,并用竖直向下的恒力F按压瓜子且保持静止,若此时瓜子壳未破开,忽略瓜子重力,不考虑瓜子的形状改变,不计摩擦,若保持A、B距离不变,则( )A.圆柱体A、B对瓜子压力的合力为零B.顶角θ越大,圆柱体A对瓜子的压力越小C.顶角θ越大,圆柱体A对瓜子的压力越大D.圆柱体A对瓜子的压力大小与顶角θ无关第(4)题《史记》中对日晕有“日有晕,谓之日轮”的描述.如图(a)所示,日晕是日光通过卷层云时,受到冰晶的折射或反射而形成的。
图(b)为太阳光射到六边形冰晶上发生两次折射的光路图,对于图(b)中出射的单色光a,b,下列说法正确的是()A.单色光a的折射率比单色光b的折射率大B.在冰晶中,单色光a的传播速度比单色光b的传播速度大C.单色光a的频率比单色光b的频率大D.单色光a的单个光子能量比单色光b的单个光子能量大第(5)题如图所示,一绝热容器被隔板K分为A、B两部分,一定量的空气处于A中,而B中真空。
2024届浙江省温州市普通高中高三第二次适应性考试英语试题学校:___________姓名:___________班级:___________考号:___________一、阅读理解Art Gallery of NSW(New South Wales) ExhibitionLouise Bourgeois: Has the Day Invaded the Night or Has the Night Invaded the Day?25 November 2023-28 April 2024Day and night, love and rage, calm and chaos. Enter a world of emotional extremes in this exhibition of the art of Louise Bourgeois, one of the most influential artists of the past century. Born in Paris in 1911 and living and working in New York until her death in 2010, Bourgeois is well-known for her fearless exploration of human relationships across a seven-decade career.Louise Bourgeois: Has the Day Invaded the Night or Has the Night Invaded the Day? reveals the extraordinary reach and intensity of Bourgeois’ art, from unforgettable sculptures of the 1940s to her tough yet tender weaving works of the 1990s and 2000s. It also reveals the psychological tensions that powered her search, through a dramatic presentation in two contrasting exhibition spaces. Moving from the well-lit rooms of “Day” to the darkened area of “Night”, viewers will encounter more than 120 works, including many never seen before in Australia.Tickets can be booked online via the exhibition or event page on our website, or in person at the welcome desk at the Art Gallery. Tickets cannot be exchanged, but if something unexpected happens that prevents you from attending, you can change the date of your reservation in your confirmation email.1.What do we know about Louise Bourgeois?A.Her art is conservative.B.She was an emotional artist.C.She was raised in Paris.D.Her art explores human relationships. 2.What does the exhibition feature?A.Various themes.B.Contrasting layout.C.Intensive colors.D.Extraordinary paintings.3.What can you do if you can’t attend the exhibition?A.Reschedule the date.B.Cancel the booking.C.Claim the money back.D.Exchange the event.Do I think the sky is falling? Sort of.My husband and I were recently in Egypt, where the temperature was a bit warm for my tiny princess self. So, we left Egypt. Back home, my dearest friends struggled with health stuff, with family craziness...The game of life is hard, and a lot of us are playing hurt.I ache for the world but naturally I’m mostly watching the Me Movie, where balance and strength are beginning to fail. What can we do as the creaking elevator of age slowly arrive? The main solution is to get outside every day, ideally with friends. Old friends — even thoughts of them — are my comfort.Recently I was walking along a beach with Neshama. We go back 50 years. She is 84, short and strong. Every so often, she bent down somewhat tentatively (踌躇地) and picked up small items into a small cloth bag.“What are you doing?”“I’m picking up micro litter. I try to help where I can.”I reminded her of an old story. A great warhorse comes upon a tiny sparrow (麻雀) lying on its back with its feet in the air, eyes tightly shut with effort. The horse asks it what it’s doing.“I’m trying to help hold back the darkness.”The horse laughs loudly, “That is so funny. What do you weigh?”And the sparrow replies, “One does what one can.”This is what older age means. We do what we can.We continued our walk. Neshama bent tentatively to pick up bits of litter and started to slip, but I caught her and we laughed. We are so physically vulnerable in older age, but friendship makes it all a rowing machine for the soul. We can take it, as long as we feel and give love, and laugh gently at ourselves as we fall apart. We know by a certain age the great lie in our life — if you do or achieve this or that, you will be happy and rich. No. Love and service make us rich.4.What does the author say about her present life?A.She leads a balanced life.B.She enjoys meeting old friends.C.She is really into movies.D.She struggles with family crises.5.Why did the author mention the sparrow?A.To confirm Neshama’s fear.B.To offer her comfort.C.To change Neshama’s mind.D.To show her approval.6.What does the underlined “it” in the last paragraph refer to?A.Physical weakness.B.The great lie.C.The rowing machine.D.The broken soul.7.What largely determines happiness in older age according to the author?A.Achieving important life goals.B.Enjoying life as you can.C.Living a life of love and service.D.Loving what is being done.Bonobos often form friendly relationships with other bonobo s in separate social groups — the first time this has been seen in non-human primates (灵长类). This is in line with humans, but in contrast to chimpanzees, another primate, which frequently kill chimps in other groups. The findings challenge the idea that humans evolved (进化) from violent apes, says Surbeck at Harvard University. “This potential to form cooperative links between different groups is not uniquely human and it might have occurred earlier than we thought,” he says.Many animals cooperate, but they seem to do so only with those within their social circle, or in-group. Hostile (敌对的) interactions between groups are common among animals, including chimpanzees, so scientists have often assumed that hostility towards other social groups in humans is natural, says Samuni, also at Harvard. However, humans also often cooperate with people in different social circles, for example, by trading or teaching.Bonobos are one of our closest living relatives. They are less studied than chimpanzees, but are known to be more peaceful, says Surbeck. To learn more about interactions between groups, Surbeck and Samuni observed 31 adult bonobo s from two social groups in Congo over a two-year period. The pair documented 95 encounters between the groups, which represented about 20% of their total observation time. Unlike chimpanzees observed in previous studies, they showed cooperation with out-group members. In fact,10% of all mutual grooming (梳毛) and 6% of all food sharing occurred among members of different social groups.While bonobo s that groomed others usually got an immediate benefit, food sharing rarely resulted in a gift in return. This suggests that their actions were “not just motivated byselfish interests or immediate rewards”, Surbeck and Samuni report.Otten, a researcher from the Netherlands, finds the study “exciting”, especially as it “challenges the idea of human exceptionalism” with regard to out-group cooperation. Otten says the bonobo s that were most cooperative within their groups were the same ones that cooperated more with out-group members. This agrees with findings from humans. “Scholars used to believe that in-group ‘love’ goes together with out-group ‘hate’, but recent research suggests that often in-group cooperators are also out-group cooperators,” he says.8.What is the focus of the study on bonobos?A.Their social behavior.B.Their survival skills.C.Their evolutionary process.D.Their intelligence level.9.What can be learnt about the bonobos?A.They are humans’ closest relatives.B.They can be taught to cooperate.C.They interact friendly beyond groups.D.They share food for immediate rewards. 10.How was the study conducted?A.By comparing different primates.B.By observing bonobos’ interactions.C.By listing group members’ motivations.D.By analysing statistics of previousstudies.11.How does Otten find the study?A.Forward-looking.B.Groundbreaking.C.Controversial.D.One-sided.Studies have shown the mere exposure effect, also referred to as the familiarity principle, inspires our decisions. It is a helpful psychological mechanism that helps us sustain our energy and focus our attention on other things. Getting used to new things takes effort and it can be exhausting. So unless we have a terrible experience, we are likely to buy from companies we’ve got used to. That is why companies spend so much money on advertising and marketing and why insurance companies openly charge existing customers more than new ones.It’s not the case that we only desire things we already know. Some studies suggest when invited to share our preferences, we sometimes see less familiar options as more desirable. But when acting on that preference, we fall back to what we know. This might explain why sometimes the things we want and the things we do don’t quite match up. We might evenreturn to companies that treated us poorly in the past or stay in bad relationships.It’s easy to paint the familiarity principle as an enemy or something to battle as if it is something that holds us back from living our dreams. But this attitude might be overwhelming because it tends to encourage us toward big-picture thinking. Where we imagine that change requires a substantial dramatic swing that we don’t feel ready for. Some articles suggest the solution to familiarity frustration is complete exposure to novelty. While this can appear effective in the short run, we may only end up replacing one problem with another. It also risks overwhelm and burnout.So what if we can work with the familiarity principle instead? Familiarity is something we can learn to play with and enjoy. It is a setting for creativity and a pathway to expansion. We can broaden the zone of familiarity bit by bit. If we think of familiarity as something that can expand, we can consider changing the conditions in and around our lives to make more space for our preferences to take root and grow gently. From here, we will start to make decisions, drawing from an ever-deepening pool of valuable options.12.What allows insurance companies to charge old customers more?A.The improved service.B.The advertising cost.C.The familiarity principle.D.The law of the market.13.What can be learned from paragraph 2?A.Our preferences affect our decisions.B.Familiarity tends to generate disrespect.C.The familiarity principle is a double-edged sword.D.There can be a mismatch between desires and actions.14.What is the author’s attitude towards the solution in some articles?A.Disapproving.B.Tolerant.C.Objective.D.Reserved. 15.Which of the following is the best title for the text?A.Step Out Of Your Familiarity Zone B.Spare A Thought For Your Preference C.Gently Expand Your Familiarity Zone D.Give Priority To The Mere Exposure EffectHow to Plot a Short StoryA great short story drops the reader into its world swiftly and holds their attention all the way through. 16 It can be as simple as knowing a few key moments you want towork your way toward. You’ll wind up with things you never imagined at the start. Follow these steps to plot your next story.Brainstorm. You don’t need to have multiple short story ideas ready to go at a moment’s notice. 17 When that idea comes to you, sit down and flesh it out. Make note of any characters, settings, or bits of dialogue that you see.18 The foundations of your main conflict or theme often form a short story’s rising action. To create tension and movement, you must know exactly what your character wants and what would prevent them from getting it. Conflicts can be internal or external, so imagine at what stage the reader will be meeting your character.Create a brief outline. Sketch out (草拟) the flow of events your short story will contain, including interactions between characters and key moments. Write down identifying characteristics. But when it comes to drafting, pick your moments of backstory carefully.19Pick a point of view. Many short stories work well in first-person because of their brevity (简洁). 20 If your story needs to be told in second-person or third-person, that works, too. Regardless of which POV you choose, it’s usually best to center that narrative around one main character to ensure a consistent read on the situation at hand.A.Select the right character.B.Write out the central conflict.C.All you need is one solid concept.D.But there’s no hard and fast rule saying yours must.E.Short stories allow the freedom to experiment because of their freestyle.F.Plotting a short story doesn’t have to involve a detailed list of plot points.G.To make the cut, a piece of information must contribute to the story’s central events.二、完形填空What I will be hosting, to be exact, is a series of meditation retreats (冥想静修) to be held this spring. During each retreat, about a hundred 21 will come here from all over the world for a period of a week to ten days to deepen their meditation practices. Some of them are 22 , but many have never been to India. My role is to take care of these people during their 23 here. For most of the retreat, the participants will be in24 . For some of them, it will be the first time they’ve 25 silence as a devotional practice, and it can be intense. However, I will be the one person they are 26 to talk to if something is going wrong.I can help them. I am so 27 to help. All the listening skills I learned as a (n)28 bartender, all the antennas (直觉) I’ve ever developed throughout my lifetime that have taught me how to 29 what people are feeling — they were all accumulated 30 I could help ease these good people into the difficult task they’ve 31 . I am so consumed by 32 at their bravery. These people have 33 their families and lives for a few weeks to go into silent retreat with a crowd of perfect 34 in India. Not everybody does this in their 35 .21.A.employees B.retirees C.interviewees D.devotees 22.A.passers-by B.old-times C.trouble-makers D.peace-lovers 23.A.work B.stay C.tour D.lecture 24.A.doubt B.memory C.silence D.surprise 25.A.treated B.interpreted C.described D.experienced 26.A.commanded B.reminded C.forced D.allowed 27.A.equipped B.moved C.wanted D.motivated 28.A.talkative B.sympathetic C.aggressive D.innocent 29.A.read B.record C.express D.reveal 30.A.so that B.now that C.in case D.even if 31.A.put aside B.taken on C.turned down D.got over 32.A.guilt B.satisfaction C.wonder D.embarrassment 33.A.contributed to B.relied on C.left behind D.prayed for 34.A.strangers B.opponents C.friends D.roommates 35.A.job B.adventure C.dream D.lifetime三、语法填空阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。
2019年下半年温州市高中选考第一次适应性测试地理试题一、选择题(本大题共25小题,每小题2分,共50分。
每小题列出的四个备选项中只有一个是符合题目要求的,不选、多选、错选均不得分)宜居带是指一颗恒星周围适宜生命存在的理想区域,下图为太阳系的宜居带分布图。
完成1题。
1.科学家研究发现太阳系宜居带有外移趋势,其产生的原因与影响可能是A.地球体积变大,地球将不再适宜人类居住B.公转轨道变大,火星将逐渐适宜生命存在C.太阳辐射增强,木星表面气温将逐渐升高D.太阳活动减弱,金星将逐渐适宜人类居住下图为2008年我国四大地区与主要出口贸易伙伴的出口额统计图。
完成2、3题。
2.工业结构以重型工业为主的是A.东部地区和东北地区B.西部地区和中部地区C.东北地区和西部地区D.东部地区和中部地区3.据图分析可知A.东部地区出口额最高的主要影响因素是面积B.影响中部地区出口地分布的主要因素是位置C.东北地区出口中国香港少是因为经济水平低D.较长的国界线促进了西部地区与东盟的贸易下图为世界某区域简图,甲乙为不同性质的锋面。
完成4、5题。
4.图示区域大陆上冰川地貌广布,主要原因是A.纬度高气候寒冷湿润,冰川厂泛分布B.离北冰洋近,受北冰洋冰川运动影响C.地质时期曾经气候寒冷冰川广泛分布D.洋流带来北冰洋冰川,塑造沿海地形5.对甲、乙两个锋面判断正确的是A.甲为冷锋正向西北移动B.甲为暖锋正向东南移动 C.乙为冷锋正向西南移动D.乙为暖锋正向东北移动下图为反映某环境问题的漫画,完成6题。
6.该漫画反映的环境问题及主要原因是A.土地荒漠化一植被破坏 B.资源枯竭一掠夺性开采C.环境污染一不合理的生产方式 D.生物多样性受损一栖息地丧失某地居民发现任何物品受当地井水浸泡几个月后就会变成“岩石”,下图为该地地质剖面图。
完成7、8题。
7.从成因上看,物品受井水浸泡变成的“岩石”类似于A.玄武岩 B.花岗岩 C.石灰岩 D.大理岩8.下列说法正确的是A.残积物上发育的土壤一般含石块少B.坡积物上发育的土壤质地分层明显C.洪积物主要来自风化基岩②的风化D.坡积物主要来自风化基岩①的风化图1为西北大西洋冬季浮冰(漂浮的冰块)南界分布图,图2为格陵兰岛冰川上广泛出现的因深色尘埃而形成的圆形孔洞。
温州市普通高中2024届高三第二次适应性考试数学试题卷2024.3本试卷共4页,19小题,满分150分.考试用时120分钟.注意事项:1.答卷前,考生务必用黑色字迹钢笔或签字笔将自己的姓名、准考证号填写在答题卷上.将条形码横贴在答题卷右上角“条形码粘贴处”.2.作答选择题时,选出每小题答案后,用2B 铅笔把答题卷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钢笔或签字笔作答,答案必须写在答题卷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,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.4.考生必须保持答题卷的整洁,不要折叠、不要弄破.选择题部分(共58分)一、选择题:本大题共8小题,每小题5分,共40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.已知z C ∈,则“2R z ∈”是“R z ∈”的()A.充分条件但不是必要条件B.必要条件但不是充分条件C.充要条件D.既不是充分条件也不是必要条件2.已知集合{{,M x y N y y ====,则M N ⋂=()A.∅B.RC.MD.N3.在正三棱台111ABC A B C -中,下列结论正确的是()A.1111113ABC A B C A BB C V V --=B.1AA ⊥平面11AB CC.11A B B C⊥ D.1AA BC⊥4.已知0.50.3sin0.5,3,log 0.5a b c ===,则,,a b c 的大小关系是()A.a b c<< B.a c b<< C.c a b<< D.c b a<<5.在()()531x x --展开式中,x 的奇数次幂的项的系数和为()A.64- B.64C.32- D.326.已知等差数列{}n a 的前n 项和为n S ,公差为d ,且{}n S 单调递增.若55a =,则d ∈()A.50,3⎡⎫⎪⎢⎣⎭B.100,7⎡⎫⎪⎢⎣⎭C.50,3⎛⎫ ⎪⎝⎭D.100,7⎛⎫⎪⎝⎭7.若关于x 的方程22112x mx x mx mx +++-+=的整数根有且仅有两个,则实数m 的取值范围是()A.52,2⎡⎫⎪⎢⎣⎭B.52,2⎛⎫ ⎪⎝⎭C.55,22,22⎛⎤⎡⎫-- ⎪⎥⎢⎝⎦⎣⎭D.55,22,22⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭8.已知定义在()0,1上的函数()()1,,1,m x m n f x n n x ⎧⎪=⎨⎪⎩是有理数是互质的正整数是无理数,则下列结论正确的是()A.()f x 的图象关于12x =对称 B.()f x 的图象关于11,22⎛⎫ ⎪⎝⎭对称C.()f x 在()0,1单调递增D.()f x 有最小值二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知角α的顶点为坐标原点,始边与x 轴的非负半轴重合,()3,4P -为其终边上一点,若角β的终边与角2α的终边关于直线y x =-对称,则()A .()3cos π5α+=B.()π2π22k k βα=++∈Z C.7tan 24β=D.角β的终边在第一象限10.已知圆221:6C x y +=与圆222:20C x y x a ++-=相交于,A B 两点.若122C AB C AB S S =△△,则实数a的值可以是()A.10B.2C.223D.14311.已知半径为r 球与棱长为1的正四面体的三个侧面同时相切,切点在三个侧面三角形的内部(包括边界),记球心到正四面体的四个顶点的距离之和为d ,则()A.r 有最大值,但无最小值B.r 最大时,球心在正四面体外C.r 最大时,d 同时取到最大值D.d 有最小值,但无最大值非选择题部分(共92分)三、填空题:本大题共3小题,每题5分,共15分.把答案填在题中的横线上.12.平面向量,a b满足()2,1a = ,a b ,a b ⋅= ,则b = ______.13.如图,在等腰梯形ABCD 中,12AB BC CD AD ===,点E 是AD 的中点.现将ABE 沿BE 翻折到A BE ' ,将DCE △沿CE 翻折到D CE '△,使得二面角A BE C '--等于60︒,D CE B '--等于90︒,则直线A B '与平面D CE '所成角的余弦值等于______.14.已知P ,F 分别是双曲线()22221,0x y a b a b -=>与抛物线()220y px p =>的公共点和公共焦点,直线PF 倾斜角为60 ,则双曲线的离心率为______.四、解答题:本大题共5小题,共77分.解答应写出文字说明,证明过程或演算步骤.15.记ABC 的内角,,A B C 所对的边分别为,,a b c ,已知2sin c B =.(1)求C ;(2)若tan tan tan A B C =+,2a =,求ABC 的面积.16.已知直线y kx =与椭圆22:14xC y +=交于,A B 两点,P 是椭圆C 上一动点(不同于,A B ),记,,OP PA PB k k k 分别为直线,,OP PA PB 的斜率,且满足OP PA PB k k k k ⋅=⋅.(1)求点P 的坐标(用k 表示);(2)求OP AB ⋅的取值范围.17.红旗淀粉厂2024年之前只生产食品淀粉,下表为年投入资金x (万元)与年收益y (万元)的8组数据:x1020304050607080y12.816.51920.921.521.92325.4(1)用ln y b x a =+模拟生产食品淀粉年收益y 与年投入资金x 的关系,求出回归方程;(2)为响应国家“加快调整产业结构”的号召,该企业又自主研发出一种药用淀粉,预计其收益为投入的10%.2024年该企业计划投入200万元用于生产两种淀粉,求年收益的最大值.(精确到0.1万元)附:①回归直线ˆˆˆu bv a =+中斜率和截距的最小二乘估计公式分别为:1221ˆni ii n ii v unv ubv nv ==-⋅=-∑∑,ˆˆa u bv =-⋅②81ii y=∑81ln ii x=∑821ii x=∑()128ln i i x =∑81ln iii y x=∑1612920400109603③ln20.7,ln5 1.6≈≈18.数列{}{},n n a b 满足:{}n b 是等比数列,122,5b a ==,且()()*1122238N n n n n a b a b a b a b n ++⋅⋅⋅+=-+∈.(1)求,n n a b ;(2)求集合()(){}*0,2,Ni i A x x a x b i n i =--=≤∈中所有元素的和;(3)对数列{}n c ,若存在互不相等的正整数()12,,,2j k k k j ⋅⋅⋅≥,使得12j k k k c c c ++⋅⋅⋅+也是数列{}n c 中的项,则称数列{}n c 是“和稳定数列”.试分别判断数列{}{},n n a b 是否是“和稳定数列”.若是,求出所有j 的值;若不是,说明理由.19.如图,对于曲线Γ,存在圆C 满足如下条件:①圆C 与曲线Γ有公共点A ,且圆心在曲线Γ凹的一侧;②圆C 与曲线Γ在点A 处有相同的切线;③曲线Γ的导函数在点A 处的导数(即曲线Γ的二阶导数)等于圆C 在点A 处的二阶导数(已知圆()()222x a y b r -+-=在点()00,A x y 处的二阶导数等于()230r b y -);则称圆C 为曲线Γ在A 点处的曲率圆,其半径r 称为曲率半径.(1)求抛物线2y x =在原点的曲率圆的方程;(2)求曲线1y x=的曲率半径的最小值;(3)若曲线e x y =在()11,ex x 和()()2212,e x x xx ≠处有相同的曲率半径,求证:12ln2x x +<-.。
温州市2024届普通高中高三第三次适应性考试高三数学试题卷一、选择题:本题共8小题,每小题5分,共40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.在ABC 中,三个内角,,A B C 成等差数列,则()sin A C +=()A .12B.2CD .12.平面向量()(),2,2,4a m b ==-,若()a ab - ∥,则m =()A .1-B .1C .2-D .23.设,A B 为同一试验中的两个随机事件,则“()()1P A P B +=”是“事件,A B 互为对立事件”的()A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件4.已知*m ∈N ,()21mx +和()211m x ++的展开式中二项式系数的最大值分别为a 和b ,则()A .a b <B .a b=C .a b>D .,a b 的大小关系与m 有关5.已知5πsin 4⎛⎫β+=-⎪⎝⎭()()sin 2cos cos 2sin αβαβαα---=()A .2425-B .2425C .35-D .356.已知函数()223,02,0xx x x f x x ⎧-+>=⎨≤⎩,则关于x 方程()2f x ax =+的根个数不可能是()A .0个B .1个C .2个D .3个7.已知12,F F 是椭圆2222:1(0)x y C a b a b +=>>的左右焦点,C 上两点,A B 满足:222AF F B = ,14cos 5AF B ∠=,则椭圆C 的离心率是()A .34BC .23D8.数列{}n a 的前n 项和为()*1,n n n n S S a n a +=∈N ,则5622111i i i i a a -==-∑∑可以是()A .18B .12C .9D .6二、选择题:本题共3小题,每小题6分,共18分。
在每小题给出的选项中,有多项符合题目要求。
温州市普通高中2024届高三第一次适应性考试化学试题卷参考答案2023.11一、选择题(本大题共16小题,每小题3分,共48分。
每个小题列出的四个备选项中只有一个是符合题目要求的,不选、多选、错选均不得分)题号12345678910答案B A D A D A BCCC题号111213141516答案BABDDC二、非选择题(本大题共5小题,共52分)17.(共10分)(1)(1分)【没标4s、4P,或者轨道数与电子自旋方向有错,都0分,即全对1分,但4S4P 写在上或下给分】(2)①AB (2分)(选A 或者B 项给1分,二项且正确2分,其它多选少选或选错0分)②sp 2、sp 3(1分)(写出一个且正确给1分)金刚石型BN 为共价晶体(1分),B-N 的键能大于金刚石中C-C 键能,所有硬度更大(1分)。
(指出晶体类型1分,键能或键长比较正确给1分。
分步抓重点给分)③2B n H m +6n H 2O=2n H 3BO 3+(3n+m )H 2↑(2分,需写成B n H m 的形式,不然不给分,系数为分数也可,如B n H m +3n H 2O=n H 3BO 3+()32n m +H 2↑,化学式有错0分,配平系数错扣1分,箭头不要求。
)(3)①4(1分,有错0分)②A (1分,有错0分)18.(10分)(1)H ··N ··H ····H (2分)(孤对电子有错,电子点数有错,元素符号有错,排例不符规则均0分,也不出现1分,要么0分,要么2分)(2)ACD (2个正确得1分,全对2分,有错不给分)(3)SOCl 2+4NH 3=SO(NH 2)2+2NH 4Cl (2分)(选B 或者D 项给1分,二项且正确2分,其它多选少选或选错0分)(4)S 4N 4或N 4S 4或(SN)4(2分)(其它原子数倍数的均0分)(5)取少量反应液滴加稀盐酸并加热,若产生的气体使品红溶液褪色,则有SO 32-;另取少量反应液滴加先滴加硝酸酸化,再滴加硝酸银,若产生白色沉淀,则有Cl -。
浙江省温州市普通高中2023届高三第一次适应性考试1. 已知全集R,集合,Z,则( )A. B. C. D.2. 若复数z满足,其中i为虚数单位,则复数z的虚部是( )A. B. C. 2i D. 23. 浙江大学2022年部分专业普通类平行志愿浙江录取分数线如下表所示,则这组数据的第85百分位数是( )专业名称分数线专业名称分数线人文科学试验班663工科试验班材料656新闻传播学类664工科试验班信息674外国语言文学类665工科试验班海洋651社会科学试验班668海洋科学653理科试验班类671应用生物科学农学652工科试验班664应用生物科学生工食品656A. 652B. 668C. 671D. 6744. 若,则( )A. 5B.C. 3D.5. 一个袋子中装有大小相同的5个小球,其中有3个白球,2个红球,小明从中无放回地取出3个小球,摸到一个白球记1分,摸到一个红球记2分,则小明总得分的数学期望等于( )A. 分B. 4分C. 分D. 分6. 某制药企业为了响应并落实国家污水减排政策,加装了污水过滤排放设备,在过滤过程中,污染物含量单位:与时间单位:之间的关系为:其中,k是正常数已知经过1 h,设备可以过滤掉的污染物,则过滤一半的污染物需要的时间最接近参考数据:( )A. 3 hB. 4 hC. 5 hD. 6 h7.已知P为直线上一动点,过点P作抛物线C:的两条切线,切点记为A,B,则原点到直线AB距离的最大值为( )A. 1B.C.D. 28. 在三棱锥中,平面BCD,,,则三棱锥外接球表面积的最小值为( )A. B. C. D.9. 一组样本数据,,…,的平均数为,标准差为s;另一组样本数据的平均数为,标准差为两组数据合成一组新数据,,…,,新数据的平均数为,标准差为,则( )A. B. C. D.10. 已知向量,,,其中R,则下列命题正确的是( )A. 在上的投影向量为B. 的最小值是C. 若,则D. 若,则11. 已知实数a,b满足:且,则( )A. B.C. D.12. 若函数的图象上存在两个不同的点P,Q,使得在这两点处的切线重合,则称函数为“切线重合函数”,下列函数中是“切线重合函数”的是( )A. B. C. D.13. 在函数图象与x轴的所有交点中,点离原点最近,则可以等于__________写出一个值即可14. 在棱长为1的正方体中,E是线段的中点,F为线段AB的中点,则直线CF 到平面的距离等于__________.15. 已知,是椭圆C的两个焦点,点M在C上,且的最大值是它的最小值的2倍,则椭圆的离心率为__________.16. 定义在R上的函数满足,,若,则__________,__________.17.已知数列是等差数列,,且,,成等比数列.给定N,记集合N的元素个数为求,的值;求最小自然数n的值,使得…18. 记锐角的内角A,B,C的对边分别为a,b,c,已知求证:;若,求的最大值.19.如图,线段是圆柱的母线,是圆柱下底面的内接正三角形,劣弧上是否存在点D,使得平面?若存在,求出劣弧的长度;若不存在,请说明理由.求平面和平面夹角的余弦值.20. 2021年11月10日,在英国举办的《联合国气候变化框架公约》第26次缔约方大会上,100多个国家政府、城市、州和主要企业签署了《关于零排放汽车和面包车的格拉斯哥宣言》,以在2035年前实现在主要市场、2040年前在全球范围内结束内燃机销售,电动汽车将成为汽车发展的大趋势.电动汽车生产过程主要包括动力总成系统和整车制造及总装.某企业计划为某品牌电动汽车专门制造动力总成系统.动力总成系统包括电动机系统、电池系统以及电控系统,而且这三个系统的制造互不影响.已知在生产过程中,电动机系统、电池系统以及电控系统产生次品的概率分别为,,求:在生产过程中,动力总成系统产生次品的概率;动力总成系统制造完成之后还要经过检测评估,此检测程序需先经过智能自动化检测,然后再进行人工检测,经过两轮检测恰能检测出所有次品.已知智能自动化检测的合格率为,求:在智能自动化检测为合格品的情况下,人工检测一件产品为合格品的概率.随着电动汽车市场不断扩大,该企业通过技术革新提升了动力总成系统的制造水平.现针对汽车续航能力的满意度进行用户回访.统计了100名用户的数据,如下表:产品批次合计对续航能能力是否满意技术革新之前技术革新之后满意285785不满意12315合计4060100试问是否有的把握可以认为用户对续航能力的满意度与该新款电动汽车动力总成系统的制造水平有关联?▲参考公式:,21. 已知双曲线:的左右焦点分别为,,P是直线l:上不同于原点O的一个动点,斜率为的直线与双曲线交于A,B两点,斜率为的直线与双曲线交于C,D两点.求的值;若直线OA,OB,OC,OD的斜率分别为,,,,问是否存在点P,满足,若存在,求出P点坐标;若不存在,说明理由.22. 已知,函数的最小值为2,其中,求实数a的值;,有,求的最大值.答案和解析1.【答案】B【解析】【分析】本题考查集合的交集、补集运算,一元二次不等式的求解,为基础题.【解答】解:或,,2.【答案】D【解析】【分析】本题考查复数的概念与分类、复数的模及其几何意义、复数的除法运算,属于基础题.【解答】解:,,虚部为3.【答案】C【解析】【分析】本题考查百分位数,属于基础题.【解答】解:,从小到大排倒数第二个数是4.【答案】B【解析】【分析】本题考查二项式定理,考查二项展开式的应用,为基础题.【解答】解:,的系数5.【答案】C【解析】【分析】本题考查离散型随机变量的均值,属于基础题.【解答】解:可取3,4,5,则,,,6.【答案】A【解析】【分析】本题考查对数运算的实际应用,属于中档题.【解答】解:,,,,,,,,,故接近7.【答案】B【解析】【分析】本题考查曲线的切线方程和点到直线的距离公式,为中档题.【解答】解:设,过P作抛物线的切线,切点为A,切点弦,即8.【答案】D【解析】【分析】本题考查三棱锥外接球表面积,解题关键是用一个变量表示出球的表面积,前提是选定一个参数,由已知设,其他量都用表示,并利用三角函数恒等变换,换元法,基本不等式等求得最小值.考查了学生的运算求解能力,逻辑思维能力,属于难题.设,在等腰中,求得CD,设的外心是M,外接圆半径是r,由正弦定理,设外接球球心是O,可得OMDA是直角梯形,设可得,将h也用表示,然后可表示出外接球半径,利用三角恒等变换,换元法,变形后由基本不等式求得最小值,从而得球表面积的最小值.【解答】解:设,在等腰中,,设的外心是M,外接圆半径是r,则,,设外接球球心是O,则平面BCD,平面BCD,则,同理,,又平面BCD,所以,OMDA是直角梯形,设,外接球半径为R,即,则,所以,在直角中,,,,,,,令,则,,当且仅当,时等号成立,所以的最小值是故选:9.【答案】BC【解析】【分析】本题考查平均数,标准差的计算,属于中档题.【解答】解:由题意,B正确;,同理两式相加得,,正确.10.【答案】ABD【解析】【分析】本题考查投影向量、向量的数量积运算,向量模长的求解,为中档题.【解答】解:在上投影向量为,,,A对.时取最小值10,,B对.,则,无法判断符号,C错.,则,则,D对.11.【答案】ACD【解析】【分析】本题考查利用导数比较大小,涉及基本不等式求最值、利用对数函数的图象与性质比较大小,属于中档题.【解答】解:方法一:,则,在单调递增,,,即,A对.当,时,,,此时,B错.又,,,,C对.令,令,故在单调递减,,,在单调递减,,,,D对.方法二:,由在R上单调递增,A正确对于B,例如取,,知,B错.对于C,,而,,C正确.对于D,令,在单调递减,,D正确.12.【答案】ABC【解析】【分析】本题考查了切线的重合问题,属于较难题.【解答】解:对于A,显然与相切,且与的切点有无穷多个,A正确;对于B,在,,,处的切线均为,B正确;对于C,,,,在R上单调递增且有无穷多个拐点,存在直线l与分别切于P,Q两点,图象类似于这种,显然存在这样的P,Q,C正确;对于D,,,在R上单调递增,当时,当时,,存在使,且在上单调递减上单调递增,大致图象如下,为上凹函数显然不存在不同两点P,Q使在这两点切线重合.13.【答案】写出中的任意一个数即可【解析】【分析】本题考查正弦函数的图象与性质,为基础题.【解答】解:,,,,解得14.【答案】【解析】【分析】本题考查空间线面间的距离,属于中档题.【解答】解:方法一:以D为坐标原点建系,,,设平面的法向量为,不妨设,则,,,平面,即求C到平面距离,方法二:易证平面,即求F到平面距离,设求F到平面距离为h,则,,,,15.【答案】【解析】【分析】本题考查椭圆的离心率问题,属于中档题.【解答】解:时,取最大值,,最小值,,16.【答案】0【解析】【分析】本题考查函数的对称性、函数的周期性及求函数值,熟练运用函数性质,属于中档题.先根据题意判定的周期为4,且,且,再根据题意求解即可.【解答】解:,,,,即的一个周期为4,又,,且,,而,,,且由,,方法二:,则关于对称,,,也是对称轴,,,又,,即,关于对称,,,,,17.【答案】解:设公差为d,由,,成等比数列,,时,中元素个数为时,中元素个数为,时,左边时,左边故最小自然数【解析】本题考查等差数列的通项公式,等比中项,分组求和,属于中档题.18.【答案】解:,,为锐角三角形,,,由知:,在中,,由正弦定理,,令,,,当且仅当时取“=”,【解析】本题考查了三角恒等变换和正弦定理的应用,属于中档题.直接利用正弦的差角公式和同角三角函数的关系即可完成证明;由正弦定理可得:又,所以,进一步利用换元法结合二次函数可求.19.【答案】解:底面圆的半径为,如图,以O为原点,OA所在直线为x轴,所在直线为z轴,在底面ABC中过O且垂直于OA的直线为x轴建立空间直角坐标系,,设平面的一个法向量取,设,若平面,又,此时,此时劣弧圆心角为劣弧长为:,,设平面的一个法向量,则,,取,平面与平面夹角余弦值为【解析】本题考查线面平行的判定及性质,考查平面与平面所成角的余弦值,为中档题.连接,过O作AB的平行线交劣弧于点D,可证平面,进而可得平面,从而可证平面平面,可得结论,从而可求的长;由题意可以O为坐标原点,分别以O连接AB的中点,过O点平行于AB的直线,所在直线为x,y,z轴建立如图所示的空间直角坐标系,求得两平面的法向量,利用向量法可求平面和平面夹角的余弦值.20.【答案】解:动力总成系统产生次品的概率;记智能自动化检测为合格品为事件A,人工检测一件产品为合格品为事件B,,,,故有的把握可以认为用户对续航能力的满意度与该新款电动汽车动力总成系统的制造水平有关联.【解析】本题考查独立事件的概率乘法公式,对立事件的概率公式,条件概率的计算及独立性检验,属于中档题.利用对立事件的概率公式即可求解;记自动化检测为合格品为事件A,人工检测为合格品为事件B,求出,代入条件概率公式即可求解;根据表格数据算出,即可求解.21.【答案】解:,,,,,设,,直线AB方程为:,设,同理CD方程为,设,仿上可联立双曲线由,,存在或符合题意.【解析】本题考查了直线与双曲线的位置关系,双曲线中的定点问题,属于较难题.22.【答案】解:,,在上递增,注意到,在上单调递减上单调递增,,由,,,,显然,令,在上单调递减上单调递增,令,当时,,在上单调递增,故只需,令当时,令且当时,,单调递减;当时,,单调递增,故只需,当且仅当且时取等号,综上:【解析】本题考查利用导数求函数最值及利用最值求参,考查分类讨论思想,为较难题.。
2024届浙江省温州市普通高中高三下学期选考适应性考试(三模)物理试题一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题如图甲所示,一物块以一定初速度沿倾角为30°的固定斜面上滑,运动过程中物块动能与路程s的关系如图乙所示。
若取出发位置所在水平面为零势能面,用表示物块的重力势能,E表示物块的机械能,重力加速度,则下列四幅图像中正确的是( )A.B.C.D.第(2)题如图所示,匀强磁场限定在一个圆形区域内,磁感应强度大小为B,一个质量为m,电荷量为q,初速度大小为v的带电粒子沿磁场区域的直径方向从P点射入磁场,从Q点沿半径方向射出磁场,粒子射出磁场时的速度方向与射入磁场时相比偏转了θ角,忽略重力及粒子间的相互作用力,下列说法错误的是( )A.粒子带正电B.粒子在磁场中运动的轨迹长度为C.粒子在磁场中运动的时间为D.圆形磁场区域的半径为第(3)题用图示装置探究气体做等温变化的规律,将一定质量的空气封闭在导热性能良好的注射器内,注射器与压强传感器相连。
实验中( )A.活塞涂润滑油可减小摩擦,便于气体压强的测量B.注射器内装入少量空气进行实验,可以减小实验误差C.0°C和20°C环境下完成实验,对实验结论没有影响D.外界大气压强发生变化,会影响实验结论第(4)题2020年12月4日,我国新一代“人造太阳”装置——中国环流器二号M装置在成都建成并成功放电(如图),为本世纪中叶实现核聚变能应用的目标打下了坚实的基础。
“人造太阳”装置中的核反应主要是一个氘核()与一个氘核()反应生成一个新核,同时放出一个中子,释放核能△E.已知真空中光速为c,下列说法正确的是( )A.核与核是两种不同元素的原子核B.一个核与一个核反应生成的新核中子数为3C.一个核与一个核反应过程中的质量亏损为D.核的比结合能为第(5)题2021年7月25日,台风“烟花”登陆上海后,“中国第一高楼”上海中心大厦上的阻尼器开始出现摆动,给大楼进行减振。
2024届浙江省温州市普通高中高三下学期选考适应性考试(三模)高效提分物理试题一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题“太极球”运动是一项较流行的健身运动,做该项运动时,健身者半马步站立,手持太极球拍,拍上放一橡胶太极球,健身者舞动球拍时,太极球却不会掉到地上,现将太极球简化成如图所示的平板和小球,熟练的健身者让小球在竖直面内始终不脱离平板且做匀速圆周运动,则( )A.小球运动过程中动量保持不变B.小球运动到B、D两处的加速度相同C.小球在B、D两处一定受到摩擦力的作用D.小球从C到A的过程中,重力的功率先增大后减小第(2)题自然界的铀中99.28%是铀238,铀238发生衰变产生新核钍234,其衰变方程为,则下列说法正确的是()A.铀238发生的是β衰变B.X比γ的穿透能力更强C.的比结合能小于的比结合能D.γ只能由原子核的衰变产生第(3)题某品牌电动汽车以额定功率在平直公路上匀速行驶,在t1时刻突发故障使汽车的功率减小一半,司机保持该功率继续行驶,到时刻汽车又开始做匀速直线运动(设汽车所受阻力不变),则在时间内( )A.汽车的加速度逐渐增大B.汽车的加速度逐渐减小C.汽车的速度先减小后增大D.汽车的速度先增大后减小第(4)题如图所示,一架执行救援任务的直升机悬停在空中,救生员抱着伤病员,缆绳正在将他们拉上飞机。
若以救生员为参考系,则处于静止状态的是( )A.伤病员B.直升机C.地面D.直升机驾驶员第(5)题“天宫课堂”第四课于2023年9月21日15时45分开课,神舟十六号航天员景海鹏、朱杨柱、桂海潮在中国空间站梦天实验舱面向全国青少年进行太空科普授课。
在奇妙“乒乓球”实验中,航天员朱杨柱用水袋做了一颗水球,桂海潮用白毛巾包好的球拍击球,水球被弹开。
对于该实验下列说法正确的是( )A.梦天实验舱内,水球体积越小其惯性越大B.击球过程中,水球对“球拍”的作用力与“球拍”对水球的作用力是一对相互作用力C.击球过程中,水球所受弹力是由于水球发生形变产生的D.梦天实验舱内可进行牛顿第一定律的实验验证第(6)题某科技小组的同学做火箭升空发射试验,火箭模型质量为1.00kg,注入燃料50g。