2020-2021学年山西省晋城市(高平一中、阳城一中、高平实验中学)高二上学期英语试题
山西省晋城市(高平一中、阳城一中、高平实验中学)2020-2021学年高二上学期生物试题
考试时间:90 分钟;总分:100 分 第 I 卷(选择题)
一、单选题(前 20 题,每题 1 分,21-35 每题 2 分,共 50 分) 1.下列有关课本实验课题与相应方法的叙述,正确的是( )
A.研究种群数量变化规律时运用了建构物理模型的方法 B.研究分泌蛋白的合成与分泌,利用了荧光标记法 C.证明细胞膜具有流动性,利用了同位素标记法 D.研究细胞核的功能时通常采用去核、核移植等方法 2.如图为某种酶催化的化学反应在不同温度条件下反应物浓度随时间变化的曲线。下列叙述错 误的是( ) A.该酶的化学本质可能是 RNA 或蛋白质 B.实验过程中 pH 的变化不会对该实验结果产生影响 C.t2 时将温度由 45℃变为 25℃,反应底物浓度不变 D.t1 时将温度由 65℃变为 25℃,反应底物浓度不变 3.为验证食虫植物猪笼草的分泌液中含有蛋白酶,某学生设计了两组实验,如图所示。在 35℃ 水浴中保温一段时间后,甲、乙试管中分别加入适量双缩脲试剂,丙、丁试管中不加任何试 剂,可以达到实验目的的是( ) A.只有实验②可以 B.只有实验①可以 C.实验①、实验②都可以 D.实验①、实验②都不可以
14.某实验小组用等量不同浓度的 2,4-D 溶液分别浸泡绿豆种子,探究不同浓度的 2,4-D 溶液 对绿豆发芽的影响。在相同且适宜条件下培养 12h,得到下图的实验结果。下列分析正确的 是( )
16.为调查某池塘的鲫鱼数量,调查人员进行了捕捞,第一次捕获鲫鱼 108 条,第二次捕获 94 条,其中 18 条有标记。调查人员根据以上数据估算的结果和实际数量相比出现了很大偏差。 以下分析判断正确的是( ) A.两次捕获的样本太小,导致估算数据小于真实数据 B.若第二次捕获引起鱼死亡,会直接影响估算的结果 C.根据以上数据能准确判断该鲫鱼种群的数目变化趋势 D.两次捕捞所用鱼网应相同,网眼太大会降低估算数据
山西省晋城市高平一中、阳城一中、高平一中2020-2021学年高二下学期期中联考英语试卷【含答案】
2020--2021学年度山西省高平一中、阳城一中、高平一中实验校三校高二年级期中联考英语试题考生注意:1.本试卷共150分。
考试时间120分钟。
2.请将各题答案填写在答题卡上。
第一部分听力(共两节,满分30分)第二部分阅读理解(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和D四个选项中,选出最佳选项。
AOnline Bachelor's Degrees and ProgramsThe overviewColleges offer online degrees in a range of fields. Click on a degree program to explore what courses you take, how long the program takes to complete, career options and average salary. For example, to earn an online health science bachelor's degree, you would take courses in public health, health communications and epidemiology(流行病学).The benefitsPursuing an accredited(官方认可的)online bachelor's degree can help you take the next step in your career. Through your program, you'll attend online lectures, study from home manage a busy schedule and interact virtually with professors and classmates.The applicantsAn online college degree may appeal to those who want more flexibility in completing their education, or who want to take classes while working full time or parenting. Choosing where to register online will likely be challenging, but below you'll find tools, advice and other resources to make your search easier.Best online bachelor's programsU. S. News evaluated several factors to rank the best online bachelor's degree programs, including graduation rates and support services available remotely.To see the full ranking list, please click here.()21. What will you do after attending online degree programs?A. Get a degree for free.B. Attend lectures in a hall.C. Take some related coursesD. Talk with teachers face to face.()22. Which university is the last one to end the application?A. Georgia Washington University.B. University of Georgia.C. North Carolina University.D. University of Arizona.()23. Where are you likely to read the text?A. On the Internet.B. In a magazine.C. In a textbook.D. In a newspaper.BKemira Boyd had just jumped in the shower when she heard her stepmother, Tammy Boyd, knocking on the door. Kemira's 12-day-old daughter Ryleigh was choking. Kemira tried everything, but she still couldn't breathe. Kemira knew Ryleigh needed to get to the hospital fast.They had barely driven out of their neighborhood when a police car appeared behind them. Deputy Will Kimbro figured that the speeding driver was either too distracted to notice him or unconcerned. Kimbro soon found out it was a frightening combination of the two.Once she'd pulled over, Kemira handed the baby to Kimbro. He put a hand on her little chest. Ryleigh's heart was barely beating. Kimbro radioed for an ambulance. At that time Ryleigh's lips were already blue.The fact that Kimbro was there was a miracle. He had recently completed a CPR(心肺复苏)class and knew how to treat a child."Although I was shocked, my training kicked in, and I went to work to keep that child alive," said Kimbro. Then he began tapping(轻拍)Ryleigh's chest, hoping to press her heart back into action. Thanks to the CPR class, Kimbro knew the choking child didn't have a chance if there was a blockage, and he used one finger to clear her airway(气道). That was a magic touch. 20 seconds later, Ryleigh began to cry. "If she's crying like that, she's breathing," said Kimbro.But they still had five more minutes until the ambulance would arrive, and Kimbro worried that Ryleigh would choke again. He continued with delicate chest compression and clearing her airway.After transferring Ryleigh to the ambulance, Kimbro drove away. At the hospital, Ryleigh recovered quickly thanks to a determined police officer who was in the right place at the right time.()24. Why did Kemira stop her car outside her neighborhood?A. She wanted to ask for help.B. She had broken traffic rulesC. She needed to care for the child.D. She planned to talk with Kimbro()25. What was Kimbro like when he dealt with the emergency?A. Nervous.B. Frightened.C. Calm.D. Curious.()26. What made Ryleigh come back to life?A. The CPR class.B. The doctors' help.C. Kemira's reaction.D. Kimbro's first aid.()27. What may be the best title for the text?A. An encounter saved a lifeB. A CPR class is importantC. A clever and brave motherD. A policeman's experienceCChildren's average daily time spent watching television or using mobile device increased from 53 minutes at 12 months old to more than 150 minutes at 3 years old, according to an analysis by researchers at the National Institutes of Health. Children aged 7 were more likely to spend the highest amount of screen time if they had been in bad home-based childcare or were born to first-time mothers"Our results indicate that screen habits begin early," said Edwina Yeung, an investigator in National Institute of Child Health and Human Development(NICHD), "This finding suggests that interventions to reduce screen time could have a better chance of success if introduced early.”In the research, mothers of 4,000 children responded to questions on their kids' media habits when they were 12, 18, 24, 30, and 36 months of age.The American Academy of Pediatrics recommends avoiding digital media exposure for children under 18 months of age, introducing children 18 to 24 months of age to screen media slowly, and limiting screen time to an hour a day for children from 2 to 5 years of age. In the current study, researchers found that 87% of the children had screen time exceeding(超过)these recommendations. However, while screen time increased throughout infancy (幼儿期), after 8 years of age, screen time fell to under 1. 5 hours per day. The researchers believe this decreaserelates to time consumed by school-related activities.The study authors classified the children into two groups based on how much their aver- age daily screen time increased from age 1 to age 3. The first group, 73% of the total, had the lower increase, from an average of nearly 51 minutes a day to nearly an hour and 47 minutes a day. The second group, 27% of the total, had the higher increase, from nearly 37 minutes of screen time a day to about 4 hours a day. Higher levels of parental education were associated with the lower odds of inclusion in the second group.()28. Which of the following is a reason for children's addiction to the media?A. Low economic level.B. Poor family educationC. The media's attraction.D. The shortage of parents' love.()29. What's Edwina's advice?A. To stop children using the media.B. To help parents care for children wellC. To reduce children's screen time earlierD. To increase intervention to children()30. How did the author develop the main body of the text?A. By giving some examples.B. By showing some data.C. By analyzing some reasons.D. By concluding some results.()31. What does the underlined word "odds" in the last paragraph mean?A. Probability.B. Price.C. Cost.D. Income.DIn times of stress, particularly when the water gets too warm, the coral(珊瑚)erupts the algae(海藻), and the coral turns white, causing a state called coral bleaching(漂白). Just a few degrees of heat can lead to coral bleaching, putting the coral on a path to starvation and death.Driven by climate change, marine heat waves are becoming one of the greatest threats to the existence of coral, which is important to the ocean ecology. But in some rare good news researchers have discovered coral can recover from bleaching even before a heat wave ends, suggesting it has the potential to survive long heat waves. Coral was thought to survive only if a heat wave lasted just a few weeks.But no one had studied this process during a longer heat wave. Then in 2015, Julia Baum, a marine ecologist at the University of Victoria, began a survey of two common species: brain and star coral around Kiritimati in the central Pacific Ocean. They checked the condition of the coral as the heat wave struck and disappeared.Starting in May 2015, the temperature rose about 1 ℃within 2 months. As expected, coral that housed heat-sensitive algae bleached sooner than those housing the heat-tolerant kind of algae. As the water continued to warm, even heat-tolerant algae erupted.Many brain and star coral on Kiritimati recovered from bleaching while the water was still unusually warm. Baum said, "The unexpected recovery provides new hope, because it means that even under lasting heat waves, there's a path forward for some of them."An unusual feature of the recovery is that brain coral that started out with heat-sensitive algae had a higher survival rate(82%)than coral that began with heat-tolerant algae(25%). "That finding is surprising," said Baum, expecting that heat-tolerant algae would be better suited for helping coral survive a heat wave. But during a longer heat wave, it might be more advantageous to start with a heat-sensitive algae.()32. What results in coral bleaching?A. The white algae.B. The coral's death.C. An attack of waves.D. A rise in ocean temperature.()33. Why did Baum begin the survey?A. To prove that coral can stop climate change.B. To study how coral bleaching comes about.C. To figure out whether coral survives long heat wavesD. To explain why coral bleaching is a big threat to coral()34. How did Baum feel about the finding?A. Ashamed.B. ConfusedC. Worried.D. Astonished.()35. What can the finding be used to do?A. Protect the ocean environment.B. Reduce coral bleaching.C. Grow more different algae.D. Regulate the heat wave.第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
晋城市高平一中阳城一中高平实验中学高二数学上学期期末考试试题文
山西省晋城市(高平一中、阳城一中、高平实验中学)2020-2021学年高二数学上学期期末考试试题 文一、选择题(本大题共12小题,每小题5分,共60分) 1.已知点A(2,0),B(3,-3),则直线的倾斜角为 A.30° B.45° C.120° D 。
135°2。
给出下列命题:①在圆柱的上、下底面的圆周上各取一点,则这两点的连线是圆柱的母线;②存在每个面都是直角三角形的四面体;③若三棱锥的三条侧棱两两垂直,则其三个侧面也两两垂直;④棱台的上、下底面可以不相似,但侧棱长一定相等。
其中正确命题的个数是A 。
0 B.1 C 。
2 D 。
3 3.双曲线C :221916x y -=的左右焦点分别为F 1,F 2,点P 在双曲线C 上且|PF 1|=20,则|PF 2|等于A.12或28B.14或26 C 。
16或24 D 。
17或23 4。
已知直线l 1:(m +2)x +(m +3)y -5=0和l 2:6x +(2m -1)y =5互相平行,则m =A 。
4 B.-52 C 。
4,-52 D.-1,-925.一个几何体的三视图如图所示,则该几何体的体积是A.23B.13 C 。
43 D.836。
已知m ∈R ,则“m 〉3”是“方程22113x y m m -=--表示双曲线"的A.充分必要条件 B 。
充分不必要条件C.必要不充分条件 D 。
既不充分也不必要条件7。
已知向量a ,b 满足|a |=5,|b|=6,a ·b =-6,则cos 〈a ,a +b 〉=A 。
-3135B 。
-1935 C.1735 D.19358.直三棱柱ABC -A 1B 1C 1中,若∠BAC =90°,AB =AC =AA 1,则异面直线BA 1与AC 1所成的角等于A.30°B.45°C.60° D 。
90°9。
山西省晋城市(高平一中、阳城一中、高平实验中学)2020-2021学年高二历史上学期期末联考试题答案
历史参考答案1.【答案】A【详解】周天子或者诸侯在处死某个贵族之时,首先要知会对应的宗主,否则容易引发内乱,这说明宗法制度下的相关规定对周天子和宗主以及贵族都有约束,有助于维系社会稳定,A正确;等级秩序在材料中未提及,排除B;西周并不是中央集权制度,排除C;题干内容与周天子的权威无关,排除D。
2.B【详解】汉武帝让九卿和丞相不得直接奏事,与此同时,他还提拔了一批中下层官员作为自己的高级侍从和助手,替他出谋划策,发号施令,削弱了丞相为首的中枢官僚权力,有助于加强皇权,B正确;题干举措与诸侯国势力变化无关,排除A;题干涉及的是中枢体制的变化,排除C;材料举措的目的是为了加强皇权而非调整中央决策机构,排除D。
3.C【详解】材料信息“阴阳不调,日月薄蚀,年谷不登,大遭旱蝗饥馑之害,滴见天地,灾及万民”——“议可以佐百姓之急”;“今鼎至甘泉,以光润龙变……有黄白云降”——“议尊宝鼎”;“六年有日食之异——“议郡守不宜频更”体现的是政治与天命的结合,说明皇权政治具有神秘色彩,故C项正确;材料信息体现的是皇权政治,不是民间迷信,故A项错误;天人感应理论普遍流行开始于汉武帝,故B项错误;D项说法错误,材料主旨是强调皇权的神秘化。
”4.D【详解】贞观前期的朝廷重臣住房条件较为简陋,反映了唐太宗吸取隋朝灭亡的教训,身体力行“戒奢从简”所产生的示范效应,而统治阶层戒奢从简有助于减轻百姓的负担,D正确;贞观时期国力逐渐强盛,而非发生了财政危机,排除A;戒奢从简的社会风尚不利于形成贪污腐败的官场风气,排除B;题干中官员的住房条件以及君主提倡节俭与君民关系无关,排除C。
5.B【解析】中国封建农耕经济形态下,拥有土地即可依赖土地获得地租收入,以此维持佛教出家团体的生活,因此佛教传入中国后出家团体生活方式的改变适应了中国封建农耕的经济形态,故答案为B项。
材料没有反映佛教教义的改变,排除A项;C项材料不能反映,排除;D项说法绝对,“主要力量”材料不能反映,排除。
2020-2021学年山西省晋城市(高平一中、阳城一中、高平实验中学)高二上学期期末考试物理试题
绝密★启用前山西省晋城市(高平一中、阳城一中、高平实验中学)2020-2021学年高二上学期期末考试物理试题注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上第I卷(选择题)请点击修改第I卷的文字说明一、单选题1.如图所示,直线是一簇未标明方向的由点电荷产生的电场线,曲线是某一带电粒子通过电场区域时的运动轨迹,ab是轨迹上两点。
若带电粒子运动中只受到电场力作用,根据此图可判断()A.带电粒子所带电荷的正负B.带电粒子在a、b两点的受力方向C.带电粒子在a、b两点的速度b处大D.带电粒子在a、b两点的电势能a处大2.以下说法正确的是()A.将体积一定、粗细均匀的导线均匀拉长到原来长度的2倍,则其电阻变为原来的2倍B.电阻率越大的导体对电流的阻碍作用一定大C.在电源内部正电荷能从负极到达正极是因为电源内部只存在非静电力而不存在静电力。
D.在闭合电路中,当外电阻变大时,电源的效率也变大,但电源的电动势不变。
3.2020年爆发了新冠肺炎,该病毒传播能力非常强,因此研究新冠肺炎病毒株的实验室必须是全程都在高度无接触物理防护性条件下操作。
武汉病毒研究所是我国防护等级最高的P4实验室,在该实验室中有一种污水流量计,其原理可以简化为如右下图所示模型:废液内含有大量正、负离子,从直径为d的圆柱形容器右侧流入,左侧流出。
流量值Q等于单位时间通过横截面的液体的体积。
空间有垂直纸面向里的磁感应强度为B的匀强磁场,下列说法正确的是()A.带电粒子所受洛伦兹力方向由M指向NB.M点的电势高于N点的电势C.污水流量计也可以用于测量不带电的液体的流速D.只需要测量MN两点电压就能够推算废液的流量4.如图所示,等边三角形线框LMN由三根相同的导体棒连接而成,固定于匀强磁场中,线框平而与磁场方向垂直,线框顶点M、N与直流电源两端相接,已知导体棒ML受到的安培力的大小为F,则线框LMN受到的安培力的大小为()A .0B .1.5FC .2FD .3F5.如图所示,同一竖直面内的正方形导线框a 、b 的边长均为l ,电阻均为R ,质量分别为2m 和m 。
山西晋城三校2020-2021学年高二联考
2020-2021学年度晋城高平一中、阳城一中、高平一中高二年级联考化学试题可能用到的相对原子质量:H 1 C 12 N 14 O 16 Na 23 Cl 35.5 K 39 Hg 201第I卷(选择题共48分)一、选择题(本题包括16小题,每小题3分,共48分。
每小题只有一个选项符合题意)1.化学与人类的生活、生产有着紧密的联系,推动着社会的进步和人类的发展。
下列说法错误的是A.“嫦娥五号”的碳化硅特种陶瓷望远镜是一种无机非金属材料B.采用“国和一号(CAP1400)”核电机组发电可减少二氧化碳等温室气体的排放C.半纤维素酶高效生产及应用的关键技术中半纤维素酶的主要成分是蛋白质D.古代染坊常用某种“碱剂(主要成分是一种盐)”来精炼丝绸,该“碱剂”可能是火碱2.基态R原子的3p能级上只有一对成对电子,则R的最高价氧化物对应水化物的酸根离子是A.RO32-B.RO42-C.RO3-D.RO4-3.下列物质的性质与其应用对应关系正确的是4.在电池工业上,碳酸乙烯酯(EC)可作为锂电池电解液的优良溶剂,其结构为,熔点为35℃。
下列有关说法错误的是A.EC由固态变成液态破坏了分子间的作用力B.一个分子中有10个σ键C.分子中至少有4个原子共平面D.EC分子间能形成氢键5.常温下,已知:一元弱酸(HR)的K a=2.95×10-8,二元弱酸(H2M)的K a1=4.3×10-7、K a2=5.6×10-11。
下列说法正确的是A.0.1 mol·L-1的NaHM溶液显碱性B.等浓度的NaR溶液与NaHM溶液相比,前者pH小C.向NaR溶液中加入少量H2M,发生反应的离子方程式为2R-+H2M=2HR+M2-D.pH相等的HR溶液和H2M溶液分别与相同的NaOH溶液反应,消耗的NaOH溶液体积相等6.下列关于能层与能级的说法中不正确的是A.原子核外每一个能层最多可容纳的电子数为2n 2B.任一能层的能级总是从s 能级开始,而且能级数等于该能层序数C.不同能层中s 电子的原子轨道半径相同D.相同能层中p 电子的原子轨道能量相同7.常温下,下列各组微粒在指定的溶液中一定能大量共存的是A.使紫色石蕊溶液变蓝的溶液:Cl -、HCO 3-、H 2PO 4-、Ca 2+B.加入金属铝粉有气体生成的溶液:F -、CH 3COO -、S 2-、Na +C.W K c(OH )-=10-1的溶液:NO 3-、SO 42-、MnO 4-、K + D.使无色酚酞不变色的溶液:ClO -、CO 32-、SO 32-、NH 4+8.用下列实验装置完成对应的实验,不能达到相应实验目的的是9.下列氢化物分子内共价键的极性由强到弱的顺序正确的是A.HF 、H 2O 、CH 4B.H 2O 、CH 4、HFC.CH 4、HF 、H 2OD.CH 4、H 2O 、HF10.N A 是阿伏加德罗常数的值。
2020-2021学年山西省晋城市(高平一中、阳城一中、高平实验中学)高二英语上学期期末考试英语试题
阅读理解:A篇: 1. A 2. C 3. BB篇:4. D 5. A 6. D 7. DC篇:8. B 9. D 10. C 11. CD篇:12. A 13. B 14. B 15. C七选五:16.A 17.F 18.B 19.G 20.D完形填空:21. B 22. D 23. A 24. A 25. C26. C 27. D 28. A 29. B 30. B31. C 32. D 33. B 34. A 35. C36. D 37. B 38. A 39. C 40. D语法填空41.have seen 42.Where 43.to learn 44.practising 45.optional46.is favored 47.to 48.rules 49.seriously 50.it短文改错:1:dated改为dating 2:Obvious改为Obviously3: a 改成an 4:去掉a good many of 中的of5:had改为have 6:appreciate改为appreciation7:them改为whom 8:but改成and \so9:It served 中间加 is 10:guest改为guests书面表达One possible version:Dear Martin,I learned from your email that you are in need of masks. I’m writing to tell you that I have bought some for you and that I will mail them tomorrow.Here are some tips for you to protect yourself from the virus. First of all, you are supposed to avoid going outside, especially crowded places. When you’re out, be sure to wear a mask. In addition, remember to wash your hands frequentlywith soap under running water. What’s more, doing some sports and drinking more water will be helpful.I hope you and your family are doing fine there. If there is anything I can do for you, please don’t hesitate to tell me.All the best!Yours,Li Hua。
2020-2021学年山西省高平一中、阳城一中、高平一中实验学校高一(下)期中数学试卷(附答案详解)
2020-2021学年山西省高平一中、阳城一中、高平一中实验学校高一(下)期中数学试卷一、单选题(本大题共12小题,共60.0分)1.下面四个几何体中,是棱台的是()A. B. C. D.2.已知α,β为不同的平面,a,b,c为不同的直线,则下列说法正确的是()A. 若a,b,c两两相交,则a,b,c在同一个平面内B. 若a⊂α,b⊂β,且α∩β=c,则a,b与c都有交点C. 若a//α,a//β,b//α,b//β,则α//βD. 若α⊥β,α∩β=a,b⊂α,且b⊥a,则b⊥β3.下列结论正确的是()A. 如果两个平面有三个公共点,那么这两个平面重合B. 若一条直线与一个平面内的两条直线都垂直,则该直线与此平面垂直C. 过平面外一点与平面内一点的直线,和平面内不经过该点的直线是异面直线D. 若两条直线和一个平面所成的角相等,则这两条直线一定平行4.已知一个长方体的长、宽、高分别为2dm,√3dm,3√2dm,则其外接球的体积为()A. 125π8dm3 B. 125π6dm3 C. 32πdm3 D. 64πdm35.有以下命题:①以直角梯形的一腰为轴旋转所得的几何体是圆台;②棱台的两个底面一定是相似多边形;③连接圆柱的上、下底面圆周上任意两点的线段是圆柱的母线;④用平行于底面的平面截去一个小圆锥后剩余的部分是圆台.其中的正确命题的个数为()A. 1B. 2C. 3D. 46.已知一个圆锥的母线长为2√2,其母线与底面所成的角为45°,则这个圆锥的体积为()A. 4πB. 8πC. 4π3D. 8π37.如图所示的是一个四边形用斜二测法画出的直观图,它是一个底角为45°,腰和上底边长都为4的等腰梯形,则原四边形的周长为()A. 6+2√2+2√6B. 8+2√2+2√6C. 12+4√2+2√6D. 16+4√2+4√68.如图,在长方体ABCD−A1B1C1D1中,AB=2AD=2AA1,则异面直线A1B与B1C所成角的余弦值是()A. √105B. 15C. √1010D. 459.《九章算术》中,称底面为矩形且有一侧棱垂直于底面的四棱锥为阳马,若阳马以该正八棱柱的顶点为顶点、以正八棱柱的侧棱为垂直于四棱锥底面的侧棱,则这样的阳马的个数是()A. 48B. 32C. 24D. 810.如图,网格纸上小正方形的边长为1,粗实线画出的是某四棱锥的三视图,则该几何体的表面积为()A. 18+3√2B. 18+9√2C. 16+2√2D. 16+3√211.已知a,b是两条相交直线,直线c分别与直线a,b异面,直线a上取4个不同的点,直线b上取3个不同的点,直线c上取2个不同的点,由这9个不同点所能确定的不同平面个数最多是()A. 36B. 24C. 12D. 1112.正方体ABCD−A1B1C1D1的棱长为2,AB,A1D1的中点分别是P,Q,直线PQ与正方体的外接球O相交于M,N两点,点G是球O上的动点,则△GMN面积的最大值为()A. 3√2+√62B. 2√2+√32C. √5+√302D. √624二、单空题(本大题共4小题,共20.0分)13.一个棱柱至少有______条棱,一个棱锥至少有______个顶点.14.已知正四棱锥P−ABCD的侧棱长为4√3,且∠APB=30°,若一只蚂蚁从点A出发沿着该四棱锥的侧面爬行一周回到点A,则蚂蚁爬行的最短距离为______.15.在棱长为6的正方体ABCD−A1B1C1D1中,E是棱AB的中点,过E,D,C1作正方体的截面,则该截面的面积是______.16.一个圆锥的轴截面是一个等边三角形,△ABC是底面圆的一个内接正三角形,V是圆锥的顶点,则二面角V−AB−C的正切值是______.三、解答题(本大题共6小题,共70.0分)17.如图,在三棱锥P−ABC中,E,F分别是PA,AB的中点,G,H分别是PC,BC上的点,且CGGP =CHHB=12.(1)证明:E,F,G,H四点共面;(2)证明:三条直线EG,FH,AC交于一点.18.如图,已知多面体ABCDEF中的四边形ABCD是正方形,△BCF是以BC为斜边的等腰直角三角形,EF//AB,△ADE是等边三角形,M是棱BC的中点,且FM⊥AB.(1)证明:AB⊥平面BCF;(2)设AB=a,求多面体ABCDEF的体积.19.如图,在直三棱柱ABC−A1B1C1中,△ABC是斜边为AC的等腰直角三角形,AB=BB1,且M是AB的中点,N是A1C与AC1的交点.(1)证明:MN//平面BCC1B1;(2)证明:平面AMN⊥平面A1B1C.20.在平面四边形ABCD中,△ABD是边长为4的正三角形,∠BCD=30°,BD⊥CD,如图1.现将△ABD沿着BD边折起,使平面ABD⊥平面BCD,点P在线段AD上,平面BPC将三棱锥A−BCD分成等体积的两部分,如图2.(1)证明:BP⊥AC;(2)若Q为CD的中点,求Q到平面BPC的距离.21.如图,已知四棱锥S−ABCD的底面ABCD是边长为1的正方形,SD⊥平面ABCD,且SD=√3.(1)求直线SB与平面ABCD所成角的余弦值;(2)点E在棱SA上,且满足SE=2EA,在直线BE上是否存在一点M,使DM//平面SBC?若存在,求出BM的长;若不存在,说明理由.22.如图,在直三棱柱ABC−A1B1C1中,M,N分别是棱BC,CC1的EB1.中点,点E在棱A1B1上,且A1E=12(1)证明:BE//平面A1MN.(2)若AB⊥BC,AB=3,BC=2,BB1=√6,求平面A1MN与平面ABC所成锐二面角的大小.答案和解析1.【答案】B【解析】解:选项A是圆台,选项C中四条棱的延长线没有相交于一点,不是棱台,选项D是棱锥.故选:B.根据圆台、棱台和棱锥的结构特征,即可得解.本题考查简单空间几何体的结构特征,考查空间立体感,属于基础题.2.【答案】D【解析】解:若a,b,c两两相交,且有三个不同的交点,则a,b,c在同一个平面内,故A错误;若a⊂α,b⊂β,且α∩β=c,则a,b与c可能都没有交点,即平行,故B错误;若a//α,a//β,b//α,b//β,且a,b为相交直线,则α//β,故C错误;若α⊥β,α∩β=a,b⊂α,且b⊥a,则b⊥β,故D正确.故选:D.考虑a,b,c两两相交,其交点情况可判断A;由线线的位置关系可判断B;由面面平行的判定定理可判断C;由面面垂直的性质定理可判断D.本题考查空间中线线、线面和面面的位置关系,考查转化思想和推理能力,属于基础题.3.【答案】C【解析】解:如果两个平面有三个公共点,那么这两个平面相交或重合,故A错误;若一条直线与一个平面内的两条相交直线都垂直,则该直线与此平面垂直,故B错误;过平面外一点与平面内一点的直线,和平面内不经过该点的直线是异面直线,故C正确;若两条直线和一个平面所成的角相等,则这两条直线平行、相交或异面,故D错误.故选:C.由三个公共点是否共线,可判断A;由线面垂直的判定定理可判断B;由异面直线的判定定理可判断C;由线面角的定义、以及两直线的位置关系,可判断D.本题考查空间中线线、线面和面面的位置关系,考查推理能力和空间想象能力,属于基础题.4.【答案】B【解析】解:设长方体的外接球半径为R,所以:(2R)2=22+(√3)2+(3√2)2,解得:R=52dm,故V=43⋅π⋅(52)3=125π6dm3.故选:B.直接利用长方体和外接球的关系,球的半径的求法,球的体积公式的应用求出结果.本题考查的知识要点:长方体和外接球的关系,球的半径的求法,球的体积公式,主要考查学生的运算能力和数学思维能力,属于基础题.5.【答案】B【解析】解:对于①,以直角梯形的垂直于底的一腰为轴旋转所得的几何体是圆台,故①错误;对于②,棱台的两个底面一定是相似多边形,故②正确;对于③,连接圆柱的上、下底面圆周上任意平行母线上的两点的线段是圆柱的母线,故③错误;对于④,用平行于底面的平面截去一个小圆锥后剩余的部分是圆台,故④正确.故选:B.直接利用圆锥的定义和圆台的定义和性质的应用判断①②③④的结论.本题考查的知识要点:圆锥的定义和圆台的定义和性质,主要考查学生的运算能力和数学思维能力,属于基础题.6.【答案】D【解析】解:因为圆锥的母线长为2√2,其母线与底面所成的角为45°,所以圆锥的底面半径r=2√2cos45°=2,圆锥的高ℎ=2√2sin45°=2,所以圆锥的体积为13πr2ℎ=13π×22×2=8π3.故选:D.利用边角关系求出圆锥的底面半径以及圆锥的高,由圆锥的体积公式求解可.本题考查了圆锥几何性质的应用,圆锥的体积公式的应用,考查了逻辑推理能力、空间想象能力与运算能力,属于基础题.7.【答案】D【解析】解:把直观图还原出原图形,如图所示:所以四边形ABCD是直角梯形,且AB=A′B′=4,BC=2B′C′=8,CD=C′D′=4+4√2,AD=√82+(4√2)2=4√6,所以原四边形的周长为AB+BC+CD+AD=4+8+(4+4√2)+4√6=16+4√2+4√6.故选:D.把直观图还原出原图形,结合图形求出原四边形的周长.本题考查了直观图与原平面图形的应用问题,是基础题.8.【答案】C【解析】解:如图,连接A1D,BD,因为A1D||B1C,所以∠BA1D是异面直线A1B与B1C所成角,设AD=AA1=1,则AB=2,A1B=BD=√5,A1D=√2,所以△A1BD是以A1D为底边的等腰三角形,所以cos∠BA1D=√22√5=√1010,故异面直线A1B与B1C所成角的余弦值是√1010.故选:C.连接A1D,BD,于是∠BA1D是异面直线A1B与B1C所成角,在△A1BD中即可求出cos∠BA1D,进而得出所求的答案.本题考查异面直线所成角的求法,考查空间想象能力和计算能力,属中档题.9.【答案】A【解析】解:如图,在正八边形ABCDEFGH中,矩形有ABEF,BCFG,CDGH,DEHA,ACEG,BDFH,共6个.所以以棱柱的下底面为底面的阳马有4×6=24个,同理,以棱柱的上底面为底面的阳马有4×6=24个,共有48个.故选:A.以正八边形ABCDEFGH的顶点为顶点的矩形共有6个,每个顶点对应1条侧棱,故有24个阳马.又棱柱有2个底面,故共有48个.本题以棱锥为背景考查计数原理,属于基础题.10.【答案】B【解析】解:三视图可知,该四棱锥底面为边长3的正方形,高为3,S=3×3+2×12×3×3+2×12×3×3√2=18+9√2.故选:B.根据三视图可知,该四棱锥底面为边长3的正方形,高为3,即可求解表面积.本题考查三视图求解几何体的表面积,解题时要认真审题,注意空间思维能力的培养,是中档题.11.【答案】A【解析】解:根据题意,不共线的三点确定一个平面,由此分5种情况讨论: ①直线c 上选出2个点,直线a 或b 上选出1个点,最多可以确定3+4=7个平面, ②直线a 上选2点,直线c 上选1点,可以确定2个平面, ③直线b 上选2点,直线c 上选1点,可以确定2个平面,④直线a 、b 上各选1个点,直线c 上选1个点,最多有4×3×2=24个平面, ⑤直线a 、b 确定一个平面,故最多可以确定7+2+2+24+1=36, 故选:A .根据题意,平面的基本性质,据此分5种情况讨论,由加法原理计算可得答案. 本题考查排列组合的应用,涉及分类计数原理的应用,注意平面的基本性质,属于基础题.12.【答案】C【解析】解:如图,设正方体外接球球O 的半径为r ,过球心O 作OH ⊥PQ ,垂足为H ,知H 为PQ 的中点. 因为正方体ABCD −A 1B 1C 1D 1的棱长为2, 所以OP =√2,PQ =√6,HP =√62,ON =R =√3,所以OH =√OP 2−HP 2=√22,HN =√ON 2−OH 2=√102,所以MN =√10.因为点G 是球O 上的动点,所以点G 到MN 的最大距离为ℎ=OH +r =√22+√3,故△GMN 面积的最大值为12⋅MN ⋅ℎ=√5+√302.故选:C .△GMN 的外接圆为球的大圆,利用圆的弦长公式求出MN 的值,则△GMN 的高为圆上一点到直线的距离,最大值为圆心到直线的距离加半径, 即当OG ⊥MN 时,面积最大.本题考查正方体和球的性质,圆上一点到直线距离的应用,考查直观想象能力,属于中档题.13.【答案】94【解析】解:棱长最少的棱柱为三棱柱,有9条棱;顶点最少的棱锥为三棱锥,有4个顶点.故答案为:9;4.根据三棱柱和三棱锥的结构特点,即可得解.本题考查简单空间几何体的结构特征,熟练掌握棱柱和棱锥的结构是解题的关键,考查空间立体感,属于基础题.14.【答案】12【解析】解:根据题意,正四棱锥P−ABCD的展开图为:若一只蚂蚁从点A出发沿着该四棱锥的侧面爬行一周回到点A,则最短距离为线段AA′的长度,又由PA=4√3,且∠APB=30°,则∠APA′=120°,故|AA′|2=PA2+PA′2−2PA⋅PA′cos120°=144,则|AA′|=12,故答案为:12.由题意,作出四棱锥的展开图,利用侧面展开图,分析求出|AA′|的值,即可得答案.本题考查利用侧面展开图求最短路程,涉及余弦定理的应用,属于基础题.15.【答案】812【解析】解:正方体ABCD−A1B1C1D1中,E是棱AB的中点,过E,D,C1作正方体的截面,是等腰梯形DEFC1,如图所示:其中F 是BB 1的中点,EF =12DC 1=12×6√2=3√2,DE =FC 1=√62+32=3√5; 所以梯形底面上的高为ℎ=√(3√5)2−(3√22)2=9√22,则该截面的面积是S =12×(6√2+3√2)×9√22=812.故答案为:812.过E ,D ,C 1作正方体的截面,是等腰梯形,结合图中数据,求出截面图形的面积. 本题考查了正方体截面面积的计算问题,也考查了运算求解能力,是基础题.16.【答案】2√3【解析】解:设圆锥底面圆心为O ,取AB 的中点D ,则点D 在平面VOC 内,连接VD ,OD ,VO ,轴截面如图所示,则二面角V −AB −C 的平面角为∠VDC , 设VC =2,则VO =√3,OD =12,且VO ⊥CD , 所以tan∠VDC =VOOD =2√3,则二面角V −AB −C 的正切值是2√3. 故答案为:2√3.设圆锥底面圆心为O ,取AB 的中点D ,则点D 在平面VOC 内,连接VD ,OD ,VO ,利用二面角的平面角的定义得到二面角V −AB −C 的平面角为∠VDC ,由边角关系求解即可. 本题考查了空间角的求解,圆锥的几何性质的应用,圆锥轴截面的理解与应用,二面角的平面角的应用,考查了逻辑推理能力、空间想象能力,属于中档题.17.【答案】证明:(1)根据题意,CG GP =CH HB =12,则有HG//BP ,而E ,F 分别是PA ,AB 的中点,则EF//BP ,则有HG//EF,故E,F,G,H四点共面,(2)由(1)的结论,HG//EF,且HF≠GE,则HF与GE交于一点,设HF与GE的交点为M,则M∈EG,则M∈平面APC,同理:M∈平面ABC,又由平面APC∩平面ABC=AC,则M在AC上,故三条直线EG,FH,AC交于一点.【解析】(1)根据题意,由直线平行的判断定理可得HG//BP,由中位线定理可得EF//BP,则有HG//EF,即可得结论;(2)根据题意,分析可得HF与GE交于一点,设HF与GE的交点为M,由平面的基本性质可得M在AC上,即可得结论.本题考查平面的基本性质,涉及平面的判定定理,属于基础题.18.【答案】解:(1)证明:因为四边形ABCD是正方形,所以AB⊥BC.又FM⊥AB,FM∩BC=M,所以AB⊥平面BCF.(2)如图,因为AD//BC,过AD作平行于平面BCF的平面将多面体ABCDEF补形成三棱柱ADN−BCF,由(1)知三棱柱AND−BCF是直棱柱,因为△BCF是以BC为斜边的等腰直角三角形,且AB=BC=a,所以S△BCF=a24,所以V ADN−BCF=a34,设AD的中点为G,连接EG,NG,可知NG=a2,EG=√3a2,所以EN=√EG2−NG2=√2a2,所以V E−ADN=13⋅a24⋅√22a=√2a324,故V ACCDEF=V ADN−BFF−V E−NDN=a34−√2a324=6−√224a3.【解析】(1)由题意利用线面垂直的判断定理,即可证证明AB⊥平面BCF;(2)首先将结合体补形为一个三棱柱,然后结合几何体的结构特征,求出多面体ABCDEF 的体积即可.本题主要考查线面垂直的判断定理,补形思想的应用,空间几何体体积的求解等知识,属于中档题.19.【答案】证明:(1)如图,连接BC1,∵M,N分别是AB,AC1的中点,∴MN//BC1.又∵MN⊄平面BCC1B1,BC1⊂平面BCC1B1,∴MN//平面BCC1B1;(2)由图可知平面AMN就是平面ABC1.∵AB=BB1,AB=BC,∴BB1=BC,又ABC−A1B1C1是直三棱柱,∴四边形BCC1B1是正方形,可得BC1⊥B1C.∵ABC−A1B1C1是直三棱柱,∴BB1⊥AB,又BC⊥AB,BB1∩BC=B,∴AB⊥平面BCC1B1,从而AB⊥B1C.又AB∩BC1=B,∴B1C⊥平面ABC1,即B1C⊥平面AMN.又B1C⊂平面A1B1C,∴平面AMN⊥平面A1B1C.【解析】(1)连接BC1,可得MN//BC1,再由直线与平面平行的判定得到MN//平面BCC1B1;(2)由图可知平面AMN就是平面ABC1.证明BC1⊥B1C,由已知可得BB1⊥AB,结合BC⊥AB,得AB⊥平面BCC1B1,从而AB⊥B1C.再由直线与平面垂直的判定可得B1C⊥平面ABC1,即B1C⊥平面AMN.进一步得到平面AMN⊥平面A1B1C.本题考查直线与平面平行、平面与平面垂直的判定,考查空间想象能力与思维能力,考查推理论证能力,是中档题.20.【答案】解:(1)证明:因为平面BPC将三棱锥A−BCD分成等体积的两部分,所以P 为线段AD的中点.因为△ABD是正三角形,所以BP⊥AD,记BD的中点为E,连接AE,则AE⊥BD.因为平面ABD⊥平面BCD,所以AE⊥平面BCD,因为CD⊂平面BCD,所以AE⊥CD.又BD⊥CD,所以CD⊥平面ABD,所以BP⊥CD,又CD∩AD=D,所以BP⊥平面ACD.因为AC⊂平面ACD,所以BP⊥AC.(2)由(1)知AE⊥平面BCD,△ABD是边长为4的正三角形,所以AE=2√3,所以P到平面BCD的距离为√3.因为Q为CD的中点,在Rt△BCD中,∠BCD=30°,BD=4,CD=43,所以Rt△BCD的面积为8√3,V P−BCQ=12V P−BCD=12×13×8√3×√3=4.由(1)知△BPC是直角三角形,S△BPC=12BP⋅CP=12×2√3×2√13=2√39.设Q到平面BPC的距离为d,则V Q−BCP=V P−BCQ=13d⋅S△BCP=2√393d=4,解得d=2√3913.【解析】(1)证明BP⊥AD,连接AE,说明AE⊥BD.利用平面ABD⊥平面BCD,推出AE⊥平面BCD,即可证明AE⊥CD,结合BD⊥CD,证明CD⊥平面ABD,然后证明BP⊥平面ACD,推出BP⊥AC.(2)通过V P−BCQ=12V P−BCD,求出S△BPC,设Q到平面BPC的距离为d,通过V Q−BCP=V P−BCQ,求解即可.本题考查直线与平面垂直的判断定理,平面与平面垂直的性质定理和等体积法的应用,考查空间点、线、面距离的求法,是中档题.21.【答案】解:(1)由题意SD ⊥平面ABCD ,连接BD ,可得SD ⊥BD ,BD ⊂平面ABCD ,那么直线SB 与平面ABCD 所成角为∠DSB ,∵△SDB 是直角三角形,SD =√3.底面ABCD 是边长为1的正方形, ∴DB =√2,那么SB =√5 ∴cos∠DSB =SD BD=√2√5=√105. (2)过A 作SB 的平行线交BE 于M ,可得M 为直线BE 上一点,证明DM//平面SBC ,如下: ∵AM//SB ,AD//BC , AM ∩AD =A ,SB ∩BC =BSB 、BC ⊂平面SBC ,AM 、AD ⊂平面ADM ∴平面ADM//平面SBC , 而MD ⊂平面ADM ∴MD//平面SBC .又∵SE =2EA ,∴EM =12BE∵底面ABCD 是边长为1的正方形,SD ⊥平面ABCD , ∴EA ⊥AB ,AB =1,AE =23,∴BE =√133,那么BM =32BE =√132.【解析】(1)SD ⊥平面ABCD ,连接BD ,可得SD ⊥BD ,BD ⊂平面ABCD ,那么直线SB 与平面ABCD 所成角为∠DSB ,即可求解其余弦值.(2)过A 作SB 的平行线交BE 于M ,可得M 为直线BE 上一点,证明DM//平面SBC 即可,再进行计算.本题考查直线与平面所成角,考查使直线平行于平面的点是否存在的判断与证明,解题时要认真审题,注意空间思维能力的培养.22.【答案】(1)证明:如图,连接BC1,C1E,延长MN,B1C1交于点F,连接A1F,由M,N分别是棱BC,CC1的中点,得BC1//MF,且C1F=MC=12B1C1,∴B1C1B1F =B1EB1A1,从而C1E//A1F,∵BC1⊄平面A1MF,MF⊂平面A1MF,C1E⊄平面A1MF,A1F⊂平面A1MF,∴BC1//平面A1MF,C1E//平面A1MF,且BC1∩C1E=C1,∴平面BC1E//平面A1MF,而BE⊂平面BC1E,∴BE//平面A1MF,即BE//平面A1MN.(2)解:由(1)知,平面A1MN与平面ABC所成锐二面角等于平面BC1E与平面A1B1C1所成的锐二面角,又BB1⊥平面A1B1C1,AB⊥BC,AB=3,BC=2,BB1=√6,取C1E的中点H,连接BH,B1H,∵B1E=B1C1=2,∴B1H⊥C1E,可证BH⊥C1E,∴平面BC1E与平面A1B1C1所成的锐二面角的平面角为∠BHB1,由已知求得B1H=√2,∵tan∠BHB1=√6√2=√3,∴∠BHB1=π3.即平面A1MN与平面ABC所成锐二面角的大小为π3.【解析】(1)连接BC1,C1E,延长MN,B1C1交于点F,连接A1F,分别证明BC1//MF,C1E//A1F,可得BC1//平面A1MF,C1E//平面A1MF,进一步得到平面BC1E//平面A1MF,从而得到BE//平面A1MN.(2)由(1)知,平面A1MN与平面ABC所成锐二面角等于平面BC1E与平面A1B1C1所成的锐二面角,取C1E的中点H,连接BH,B1H,可得平面BC1E与平面A1B1C1所成的锐二面角的平面角为∠BHB1,求解三角形得答案.本题考查平面与平面平行的判定与性质,考查空间想象能力与思维能力,训练了空间角的求法,考查运算求解能力是中档题.。
山西省晋城市高平一中、阳城一中、高平一中实验学校2020-2021学年高二政治下学期期中联考试题(P
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晋城市高平一中阳城一中高平实验中学高二地理上学期期末考试试题
山西省晋城市(高平一中、阳城一中、高平实验中学)2020-2021学年高二地理上学期期末考试试题第I卷(选择题)一、单选题(24个小题,每题2分,共48分)1.春季是我国西北干旱区的融雪季,沙尘天气频繁,沉降的沙尘会加速积雪的融化.导致我国西北干旱区春季沙尘天气多发的天气系统为下图为世界部分板块分布示意图,图中实线表示板块边界,箭头表示板块运动方向.完成下面小题.2。
甲、乙板块分别是六大板块中的A.欧亚板块、太平洋板块B.印度洋板块、欧亚板块C.欧亚板块、美洲板块D。
印度洋板块、太平洋板块3。
图中丙处附近的海底地形多是A。
裂谷 B.海岭C。
海沟 D.海岸山脉4。
新加坡地理位置的优势是①扼守马六甲海峡②连接太平洋和印度洋③位于东亚、南亚、非洲、欧洲的贸易海运线上④地处海上石油贸易航线上A。
① B.①② C.①②③D。
①②③④中国为挪威生产的深海半潜式智能养殖场被称为“超级渔场”(下左图).2017年6月,“超级渔场”经中国南海、马六甲海峡、好望角,穿越大西洋,运抵挪威海域(下右图)。
“超级渔场”规模庞大,可实现全自动监测、喂养、清洁等工作,一次可养鱼150万条,可抗12级大风。
“超级渔场”将推动渔业养殖从近海向深海、从网箱式向大型装备式、从传统人工方式向自动化智能化转变。
据此完成下面小题。
5.“超级渔场”不选择经苏伊士运河抵达欧洲的主要原因是A。
运输距离远 B.红海风浪大C。
地中海雨量大D。
苏伊士运河通航能力有限6.挪威西南部渔场形成的主要原因是A。
离岸西风引起上升流B。
寒暖流在此交汇C.大河入海口营养物质丰富D.水域开阔,鱼类数量大近年来,“认养农业”在我国蔚然兴起,它是指消费者预付生产费用,生产者为消费者提供绿色、有机食品.辽河三角洲的盘锦市大洼区曾是传统稻作区,所产盘锦大米因品质高而响誉全国。
2015年,大洼区首创“互联网+认养农业”新模式,利用互联网平台,认养客户自选水稻良种,可通过手机APP随时监测水稻的生产过程,收获后直接配送到客户家中。
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2020-2021 学年度第一学期高二期末联考英语试题第二部分:阅读理解(共两节,满分60 分)第一节(共15 小题;每小题3 分,满分45 分)阅读下列短文,从每题所给的四个选项(A、B、C 和D)中,选出最佳选项。
A 篇There are no bad authors for children. Children like and want to read and seek out, because every child is different. They can find the stories they need, and they bring themselves to stories. Here are some books children should read.The Little Prince, by Antoine de Saint-ExuperyThis book tells the story of a stranded (陷入困境的)airplane pilot, who meets an otherworldly little prince in the Sahara desert. The Little Prince is a fascinating character and the author’s watercolor illustrations are beautiful. There is an excellent pop-up(立体的)book version.At the Back of the North Wind,by George M acDonaldGeorge MacDonald has written several other lovely children’s books—The Princess and the Goblin is a must-read. But this book is perhaps his sweetest. A little boy named Diamond meets the mysterious lady North Wind one stormy night, and she brings him on many adventures through the night sky.Carry On,Mr. Bowditch, by Jean Lee LathamThis excellent book explores the world of sea travel and colonial England through its main character, Nat Bowditcli. Nat has to give up dreams of Harvard to become an indentured (受契约约束的)servant. Nonetheless, he teaches himself advanced mathematics, and grows up to become captain of his own ship.The Phantom Tollbooth,by Norman JusterIts protagonist(主角),Milo,is a bored little boy who discovers a“p hantom tollbooth”—and with it,an imaginative world in which numbers,words,music,and sounds come to life. He is given the “impossible” mission of returning two princesses to the Kingdom of Wisdom. In the process, he has learned valuable lessons, finding a love of learning. 1.Which of the following books has beautiful pictures in it?A.The Little PrinceB. At the Back of the North WindC. Carry On,Mr. BowditchD. The Phantom Tollbooth2.In Jean Lee Latham’s book, Nat Bowditch _ .A.doesn’t want to go to universityB. gives up his dream of being a captainC. learns advanced mathematics by himselfD. wants to teach mathematics in Harvard3.What does the book The Phantom Tollbooth focus on?A.Adventures through the night sky.B. Learning in a fun way.C. Exploration of the ocean world.D. Romantic stories in the desert.B 篇Recently Stiles aged 17 took her 3-year-old brother, James, to her high school. A family emergency prevented her parents from being able to pick up James. So they asked Stiles to step in.“This was a one-time event; we have never had her get him from school other than this day. But it was an emergency,” her mother said. “I figured she had gone to get him and taken him home. Later that day I saw the picture online and realized that she had taken him back to school.”Little James quickly fell asleep in his big sister’s arms at the high school.She didn’t want to be counted absent or miss any work. Stiles was studying to be a nurse. She was in her class, and the teacher gladly welcomed her little brother in. He was so tired from his school that he slept most of the class. She not only helped her family out but returned to her responsibility at s chool.Her teacher was very supportive and didn’t mind him being in the class at all. He slept most of the time and if he had become a problem she would have taken him out. She attends an amazing school that clearly sees the value in family and education.Her mom said she can now put to rest any worries she’s had that the 14-year age gap between Stiles and James would keep them from being close. As a mom, she was thankful that they had this kind of relationship, and she prayed it would always stay this way.4.Why did Stiles’ parents ask her for help?A.Her family was in trouble.B. She was close to being an adult.C. Her brother refused to attend school.D. Her parents couldn’t collect James.5.Where did Stiles want to work when she grew up?A.In a hospital.B. In a school.C. In a factory.D. In a shop.6.What was Stiles’ teacher’s attitude to her a ct?A.Opposed.B. Critical.C. Ambiguous.D. Approving.高二英语试题第 1 页(共 4 页)7.What once made Stiles’ mother concerned?A.Stiles’ bad performance i n the key school.B.Stiles’ identity of being a senior high student.C.Stiles’ terrible relationship with her c lassmates.D.The age difference between Stiles and her brother.C 篇It has always been thought that alcohol causes people to put on weight because it contains a lot of sugar, but new research suggests a glass a day cold form part of a diet. Looking at past studies they found that, while heavy drinkers do put on weight; those who drink in moderation can actually lose weight.A spokesman for the research team at Navarro University in Spain says, "Light to moderate alcohol intake, especially of wine, may be more likely to protect against, rather than promote, weight gain."The International Scientific Forum on Alcohol Research reviewed the findings and agreed with most of the conclusions, particularly that data do not clearly indicate if moderate drinking increases weight.Boston University’s Dr. Harvey Finkel found that the biologic mechanisms(生物学机制) relating alcohol to changes in body weight are not properly understood. His team pointed out the strong protective effects of moderate drinking on the risk of getting conditions like diabetes(糖尿病), which relate to increasing obesity. Some studies suggest that even very obese people may be at lower risk of diabetes if they are moderate drinkers.The group says alcohol provides calories that are quickly absorbed into the body and are not stored in fat, and that this process could explain the differences in its effects from those of other foods. They agree that future research should be directed towards assessing the roles of different types of alcoholic drinks, taking into consideration drinking patterns and including the past tendency of participants to gain weight.For now there is little evidence that consuming small to moderate amounts of alcohol on a regular basis increases one’s risk of becoming obese. What’s more, a study three years ago suggested that resveratrol(白藜芦醇), a compound present in grapes and red wine destroys fat cells.8.The passage is mainly for those _.A.who produce wineB. who have a drinking habitC. who go on a dietD. who are eager to lose weight 9.The underlined phrase"in moderation"in the first paragraph means _ .A.ExcitedlyB. carefullyC. frequentlyD. properly10.What can we learn from the passage?A.Current data clearly show that moderate drinking increases weight.B.Resveratrol is proved to increase the risk of becoming fat.C.The research found moderate drinking has a strong protective effect.D.The specific roles of different types of alcoholic drinks are very clear.11.What is most likely to be discussed in the paragraph that follows?A.How to do some easy experiments.B.How to reduce the calories contained in wine.C.How to prove the finding mentioned above.D.How to make wine in a healthy way.D 篇Have you ever thought that your life will be perfect if something in your outside world changes? You may say, “I will live happily if I find my soul mate.” You may say, “If the perfect job comes along, I will get satisfaction. My problem is that my boss is stupid.” Or you may say, “If I have a child, I will know what it is like to be loved.” The problem is that it is not the outside world that prevents you from experiencing peace, love and joy. Instead, it is your own internal (内部的) patterns that get in your way.To prove this point, at my weekend programs, I had my students look at a flower on my table. Then I asked them to share their experiences. Some showed a state of complete joy because staring at the flower reminded them of their wedding or a great trip with their lovers. Others said the flower made them angry because it reminded them of a gift given by a former partner who turned against them. Some students even told me that the flower reminded them of their beloved grandmothers’ gardens with some shedding tears (流泪)of joy and others shedding tears of sadness. However, the flower was still just a flower.Our internal patterns decide whether each moment is filled with peace, anger, joy, love or sadness. When you don’t realize that it is your own interna l state that is creating your experience, you are likely to blame (责备)the external situations in your life. Most people aren’t taught that emotion is a choice. And they aren’t taught how to change those lower emotional patterns such as anger, anxiety, sadn ess and so on. Next time, when you meet something bad, please look at your internal state. When you accept that you are responsible for your own state, and find the courage to turn inward, you can step out of being a slave to what shows up in your external world.高二英语试题第 2 页(共 4 页)12.Why does the author mention the examples in Paragraph 1 ?A.To lead to the topic of this passage.B. To complain about imperfect life.C. To show common problems in life.D. To encourage people to change the outside w orld.13.How does the author prove his opinion?A.By giving examples.B. By doing an experiment.C. By telling a related story.D. By comparing other people’s opinions.14.What does the author advise us to do in the last paragraph?A.To step out of negative emotions.B.To pay attention to our internal world.C.To share our real emotions with others.D.To use different methods to change bad emotions.15.What does the author want to express through the passage?A.Our external world matters a lot.B.Our emotions can affect our health.C.Our internal patterns truly decide our emotions.D.Our experiences have a great influence on our emotions.第二节(共5小题;每小题3分,共15分)根据短文内容,从短文后的七个选项中选出能填入空白处的最佳选项。