浦东新区2014学年第二学期高三教学质量检测(理)(教研员版)
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题一、单选题 (共7题)第(1)题如图所示,一列沿轴正方向传播的简谐横波,实线是时的部分波形图,虚线是时的部分波形图,其他数据图中已标出,下列说法正确的是( )A.这列波的波长是3mB.这列波的周期可能是0.16sC.在0.6s时,处的质点振动方向沿y轴正方向D.这列波的波速可能是20m/s第(2)题图是科学史上一张著名的实验照片,显示一个带电粒子在云室中穿过某种金属板运动的径迹。
云室放置在匀强磁场中,磁场方向垂直照片向里,云室中横放的金属板对粒子的运动起阻碍作用,分析此径迹可知粒子()A.带正电,由下往上运动B.带正电,由上往下运动C.带负电,由上往下运动D.带负电,由下往上运动第(3)题如图所示,有一块半径为R的半圆形玻璃砖,OO'是对称轴。
现有平行单色光垂直照射到AB面,玻璃砖对该单色光的折射率为。
已知,不考虑二次反射,则( )A.玻璃砖的弧面上有光射出的区域弧长为B.若在纸面内将玻璃砖绕圆心逆时针旋转,有光射出的区域弧长不变C.所有从射出的光线都将汇于一点D.入射光线距OO'越远,出射光线与OO'的交点离AB面越远第(4)题如图所示为我国航天员在天宫授课时的情形,航天员授课时处于悬浮状态,下列判断正确的是( )A.航天员受到的重力近似为零,与他在太空舱中受到的空气浮力平衡B.由于太空舱内的失重环境,加速度为零C.航天员随太空舱做匀速圆周运动所需要的向心力等于其所受万有引力D.航天员出舱后若脱离飞船将会做匀速直线运动第(5)题沿x轴传播的一列简谐横波在时刻的波动图像如图甲所示,平衡位置在处的质点Q的振动图像如图乙所示,下列说法正确的是( )A.该列波的传播方向沿x轴正方向B.该列波的传播速度为6.25m/sC.经过0.4s时间,质点Q沿波的传播方向运动了3mD.时,质点Q的加速度方向沿y轴正方向第(6)题北斗三号全球卫星导航系统由24颗中圆轨道(轨道半径约)卫星、3颗地球静止同步轨道卫星和3颗倾斜地球同步轨道卫星(两种卫星轨道半径相等,均约为)组成,则( )A.倾斜地球同步轨道卫星和静止同步轨道卫星周期不相等B.北斗三号导航系统所有卫星绕地球运行的线速度均小于C.倾斜地球同步轨道卫星能定点北京上空并与北京保持相对静止D.中圆轨道卫星线速度约为地球静止同步卫星线速度的1.5倍第(7)题甲、乙两辆汽车在平直路面上同向运动,当两车同时经过同一路标开始计时,两车在时间内速度随时间的变化图像如图所示,下列说法正确的是( )A.时间内,甲车做匀加速直线运动B.时间内,甲车的平均速度大于C.时刻,两车再次同时经过同一路标D.时刻,两车再次同时经过同一路标二、多选题 (共3题)第(1)题在可调转速的电动机的转动轴上固定一根垂直于转动轴的细杆,杆的一端装一个小塑料球,转动轴轴心与塑料球球心距离为R,在小塑料球和墙壁之间放一个竖直方向上下振动的弹簧振子,振子质量为m,如图所示。
上海市浦东新区高三物理4月教学质量检测试题(含解析)
浦东新区2014-2015学年第二学期高三教学质量检测物理试卷说明:1.本试卷考试时间120分钟,满分150分,共33题。
第30、31、32、33题要求写出必要的文字说明、方程式和重要的演算步骤。
只写出最后答案,而未写出主要演算过程的,不能得分。
2.选择题答案必须全部涂写在答题卡上。
考生应将代表正确答案的小方格用2B铅笔涂黑。
注意试题题号与答题卡上的编号一一对应,不能错位。
答案需更改时,必须将原选项用橡皮擦去,重新选择。
写在试卷上的答案一律不给分。
一.单项选择题.(共16分,每小題2分,每小题只有一个正确选项,答案涂写在答题卡上.)1.【题文】从理论上预言光是一种电磁波的物理学家是A.爱因斯坦B.麦克斯韦C.安培D.卢瑟福【答案】B【解析】本题主要考查物理学史;光的特点很好的符合了麦克斯韦的电磁场理论,因此麦克斯韦预言光是一种电磁波,故选项B正确。
【题型】单选题【备注】【结束】2.【题文】物理学家通过对α粒子散射实验结果的分析A.提出了原子的葡萄干蛋糕模型B.确认在原子内部存在α粒子C.认定α粒子发生大角度散射是受到电子的作用D.认识到原子的全部正电荷和几乎全部的质量都集中在一个很小的核内【答案】D【解析】本题主要考查α粒子散射实验;α粒子散射实验中大多数粒子偏转角度较小,只有个别粒子发生大角度偏转,甚至返回,这说明原子的全部正电荷和几乎全部的质量都集中在一个很小的核内,故选项D正确。
【题型】单选题【备注】【结束】3.【题文】物理学中用到了许多科学方法,下列概念的建立有三个用到了“等效替代”的方法,有一个不属于...这种方法,这个概念是A.质点B.平均速度C.合力D.总电阻【答案】A【解析】本题主要考查物理学研究方法;质点是忽略大小和形状这个次要矛盾,抓住质量这个主要矛盾,并不是等效替代,故选项A正确。
【题型】单选题【备注】 【结束】4.【题文】关于声波和光波,下列说法正确的是 A.它们都是机械波B.它们都能在真空中传播C.它们都能发生干涉和衍射D.它们的传播速度都与介质无关【答案】C【解析】本题主要考查电磁波与机械波的联系与区别;声波是机械波,光波是电磁波,故选项A 错误,机械波传播需要介质,故选项B 错误,声波和光波都属于波,都可以发生干涉和衍射,故选项C 正确,二者的波速都与介质有关,故选项D 错误; 本题正确选项为C 。
【VIP专享】浦东新区2014年高三英语二模试卷
浦东新区2014年高考预测英语试卷2014年4月15日下午考生注意:1. 考试时间120分钟, 试卷满分150分。
2. 本试卷设试卷和答题卷两部分。
试卷分为第I卷和第II卷。
所有答案必须涂(选择题)或写(非选择题)在答题卷上,做在试卷上一律不得分。
3. 答题前,务必在答题卷纸上填写准考证号和姓名,并将核对后的条形码贴在指定位置上。
第I卷(103分)I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. A hotel B. A shopping center C. A traffic light D. A bus stop2. A. 5:00 B. 4:45 C. 5:15 D. 4:103. A. Cook B. Shop assistant C. Saleswoman D. Waitress4. A. Parent and child B. Policeman and witnessC. Bus driver and passengerD. Receptionist and guest5. A. She is not interested in movies. B. She thinks it is good news.C. She is too busy to go to the cinema.D. She has no idea about the news.6. A. It’s good for health to have some idea cream.B. He can’t eat any snacks because of his toothache.C. He doesn’t believe in what the doctor says.D. He can’t eat ice cream though he feels hot.7. A. Nervous B. Surprised C. Calm D. Happy8. A. Trying to draw a map. B. Painting the dining room.C. Discussing a house plan.D. Cleaning the kitchen.9. A. He has an ear problem. B. He never listens.C. He has never missed a meeting.D. He has something important to do.10.A. She can’t say much about her travel. B. She didn’t see the advertisement.C. She speaks highly of the advertisement.D. She doesn’t like her travel verymuch.Section BDirections:In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Because he wanted to stay connected with nature.B. Because he thought farming was a promising job.C. Because he was tired of being a chef.D. Because he found farming interesting.12. A. Giving some financial support. B. Offering specialized businesstraining.C. Promoting farm foods.D. Providing the link with thelandowners.13. A. Many Americans have developed a taste for fresh local foods.B. More people in America tend to choose farming as a job than before.C. Local governments in America encourage people to take up farming as a job.D. The United States is among the world’s leading agricultural nations.Questions 14 through 16 are based on the following passage.14. A. Famous creative individuals. B. A major scientific discovery.C. The mysteriousness of creativity.D. Creativity as shown in arts.15. A. Creative imagination. B. Logical reasoning.C. Natural curiosity.D. Critical thinking.16. A. It is beyond ordinary people. B. It is part of everyday life.C. It is yet to be fully understood.D. It is a unique human nature. Section CDirections:In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each blank.A Police RecordWitness’ name: __17__Robbery scene: A __18__ storeInformation about the robberHeight: __19__ feetHair color: DarkAge: Around 30Clothes: A dark __20__ and a light shirt.Blanks 21 through 24 are based on the following conversation.Complete the form. Write NO MORE THAN THREE WORDS for each blank.What is the name of the course?__21__What problem does the woman have?She __22__ the reference books.What is the reasonable excuse for extension?Extensions are usually given to studentswho __23__.What is the Professor’s final decision?The woman is allowed another __24__to prepare her assignment.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)One night I decided to spend some time building a happier and closer relationship with my daughter. For several weeks she __25__ (ask) me to play chess with her, I suggested a game and she eagerly accepted. It was a school night, however, and at nine o’clock my daughter asked if I __26__ hurry my moves, because she needed to go to bed; she had to get up at six in the morning.I knew she had strict sleeping habits, __27__ I thought she ought to be able to give up some of this strictness. I said to her, “Come on,you can stay up late for once. We’re having fun. ”We played on for __28__ fifteen minutes, during which time she looked anxious. Finally she said, “Please, Daddy, do it quickly.” “No,” I replied. “__29__ you want to play it well, you’re going to play it slowly.” And so we continued until suddenly my daughter burst into tears, and admitted __30__ (beat).Clearly I had made a mistake. I had started the evening wanting to have a happy time with my daughter but had allowed my desire to win to become more important than my relationship with my daughter. When I was a child, my desire __31__(win) served me well. As a parent, I realized that it got in my way. So I had to change.(B)While income worry is a rather common problem of the aged, loneliness is another problem that aged parents may face. Of all the reasons __32__ explain their loneliness, a large geographical distance between parents and their children is the major one. This phenomenon is commonly known as “Empty Nest Syndrome”(空巢综合症).In order to seek __33__(good) chances outside their countries, many young people have gone abroad, __34__(leave) their parents behind with no clear idea of when they will return home. Their parents spend countless lonely days and nights, taking care of themselves, in the hope that someday their children will come back to stay with them. The fact __35__ most of these young people have gone to Europeanized or Americanized societies makes it unlikely that they will hold as tightly to the value of duty __36__ they would have if they had not left their countries. Whatever the case, it has been noted that the values they hold do not necessarily match __37__ they actually do. This geographical and cultural distance also prevents the grown-up children from providing timely response __38__ the needs of their aged parents.The situation in which grown-up children live far away from their aged parents __39__ (describe) as “distant parent phenomenon”, __40__ is common both in developed countries and in developing countries. Our society has not yet been well prepared for “Empty Nest Syndrome”.The fact that most of these young people have gone to Europeanized or Americanized societies makes it unlikely that they will hold as tightly to the value of duty as they would have if they had not left their countries. Whatever the case, it has been noted that the values they hold do not necessarily match what they actually do.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.A. rejectedB. eventuallyC. variousD. readyE. commercializeF.prospectG. delivered H. employed I. samples J. transplants K. inevitablySince its appearance in 2007, researchers at San Diego-based Organovo have experimented with printing a wide variety of tissues, including bits of lung, kidney and heart muscle. Now the world's first publicly traded 3D bioprinting company is getting __41__ for production. In January slices of human liver tissue were __42__ to an outside laboratory for testing. These __43__ take about 30 minutes to produce, says Keith Murphy, the firm’s chief executive. Later this year Organovo aims to begin commercial sales.The invention of 3D printing provided a technology now __44__ to manufacture everything from aircraft parts to body parts. But the __45__ of 3D bioprinting is even brighter; to create human tissues for research, drug development and testing, and __46__ as replacement organs, such as a kidney, for patients desperately in need of __47__. Bioprinted organs could be made from patients’ own cells and thus would not be __48__ by their immune systems. They could also be manufactured on demand.At present only a few of companies are trying to __49__ the production of bioprinted tissues. But Thomas Boland, an early pioneer in the field, says that plenty of others are interested. He also estimates that about 80 teams at research institutions around the world are now trying to print __50__ small pieces of tissues such as skin, blood vessels, liver, lung and heart. “It’s a wonderful technology to build three-dimensional biological structures.” says Gabor Forgaces, who co-founded Organovo in 2007.III. Reading ComprehensionSection A Directions:For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.It is officially known as the Swiss Re Tower, or 30 St Mary Axe.As a typical example of green architecture, what is most remarkableabout the building is its energy-efficiency. __51__ its artful designand some fancy technology, it is expected to consume up to 50% lessenergy than a comparable conventional office building. Greenarchitecture is __52__ the way buildings are designed, built and run.Supporters of green architecture argue that the approach has many __53__. In the case of large office, for example, the __54__ of green design techniques and clever technology can not only reduce energy consumption and environmental impact, but also reduce running costs, create a more __55__ working environment, improve employee’s health andproductivity, reduce legal liability, and __56__ property values and rentalreturns.Going green saves money by reducing long-term energy costs: a surveyof 99 green buildings in America found that on average, they use 30% lessenergy than conventional buildings. So any additional building costs can be __57__ quickly. The traditional approach of trying to minimize construction costs, __58__, can lead to higher energy bills and wasted materials.Green buildings can also have less obvious __59__ benefits. The use of natural daylight in office buildings, for example, besides reducing energy costs, also seems to make workers more productive. Lockheed Martin, an aerospace firm, found that absenteeism(缺勤) __60__ by 15 after it moved 2,500 employees into a new green building in Sunnyvale, California. __61__, the use of daylight in shopping complexes appears to increase __62__. It also found that students in naturally lit classrooms performed up to 20% better. The __63__ in productivity paid for the building’s higher construction costs within a year. Despite its benefits and its growing popularity, green architecture is still not as popular as expected. The main __64__ is co-ordination(协调), for green buildings require much more planning by architects, engineers, builders and developers than traditional buildings. But, without doubt, green architecture will __65__ to reshape the construction industry over the next five years, with ever more innovative, energy-efficient and environmentally friendly buildings. “No one is doing this for fun,” he says. “There’s too much at risk.”51. A. In place of B. Thanks to C. In spite of D. In addition to52. A. giving B. discovering C. changing D. paving53. A. benefits B. factors C. techniques D. impacts54. A. contrast B. completion C. manufacture D. combination55. A. tense B. pleasant C. fierce D. temporary56. A. involved B. enhanced C. shared D. showed57. A. recovered B. gained C. counted D. valued58. A. in return B. for instance C. by contrast D. in general59. A. environmental B. psychological C. academic D. economic60. A. multiplied B. estimated C. recorded D. dropped61. A. Similarly B. Contrarily C. Consequently D. Necessarily62. A. visits B. relations C. sales D. satisfaction63. A. performance B. confidence C. increase D. equal64. A. interest B. progress C. solution D. problem65. A. deserve B. help C. work D. affordSection BDirections:Read the following passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)The person who set the course of my life was a school teacher named Marjorie Hurd. When I was stepped off a ship in New York Harbor in 1949, I was a nine-year-old war refugee, who had lost his mother and was coming to live with the father he did not know. My mother, Eleni Gatzoyiannis, had been imprisoned and shot for sending my sisters and me to freedom.I was thirteen years old when I entered Chandler Junior High. Shortly after I arrived, I wastold to select a hobby to pursue during “club hours.” The idea of hobbies and clubs made no sense to my immigrant ears, but I decided to follow the prettiest girl in my class. She led me into the presence of Miss Hurd, the school newspaper adviser and English teacher.A tough woman with salt-and-pepper hair and determined eyes, Miss Hurd had no patience with lazy bones. She drilled us in grammar, assigned stories for us to read and discuss, and eventually taught us how to put out a newspaper. Her introduction to the literary wealth of Greece gave me a new perspective on my war-torn homeland, making me proud of my origins. Her efforts inspired me to understand the logic and structure of the English language. Owing to her inspiration, during my next twenty-five years, I became a journalist by profession.Miss Hurd retired at the age of 62. By then, she had taught for a total of 41 years. Even after her retirement, she continually made a project of unwilling students in whom she spied a spark of potential. The students were mainly from the most troubled homes, yet she alternately bullied and charmed them with her own special brand of tough love, until the spark caught fire.Miss Hurd was the one who directed my grief and pain into writing. But for Miss Hurd, I wouldn’t have become a reporter. She was the catalyst that sent me into journalism and indirectly caused all the good things that came after.66. What does the underlined sentence in Paragraph Two most probably mean?A. Hobbies and clubs did not interest the author.B. The author turned a deaf ear to joining clubs.C. Hobbies and clubs were inaccessible to immigrants like the author.D. The author had no idea what hobbies and clubs were all about.67. Which of the following caused the author to think of his homeland differently?A. Stepping on the American soil for the first time.B. Her mother’s miserable death.C. Being exposed to Greek literary works.D. Following the prettiest girl in his class.68. It can be inferred from Paragraph Four that ___________.A. Miss Hurd’s contribution was recognized across the nation.B. Students from troubled homes preferred Miss Hurd’s teaching style.C. The students Miss Hurd taught were all finally fired.D. Miss Hurd employed a unique way to handle these students.69. The passage is mainly concerned with ___________.A. how the author became a journalist.B. the importance of inspiration in one’s life.C. the teacher who shaped the author’s life.D. factors contributing to a successful career.(B)About PISAThe Program for International Student Assessment(PISA) is a triennial international survey which aims to evaluate education systems worldwide by testing the skills and knowledge of 15-year-old students. To date, students representing more than 70 economies have participated in the assessment.What makes PISA differentPISA is unique because it develops tests which are not directly linked to the school curriculum. The tests are designed to assess to what extent students at the end of compulsory education, can apply their knowledge to real-life situations and be equipped for full participation in society. The information collected through background questionnaires also provides context which can help analysts interpret the results.What the assessment involvesSince the year 2000, every three years, fifteen-year-old students from randomly selected schools worldwide take tests in the key subjects: reading, mathematics and science, with a focus on one subject in each year of assessment. The students take a test that lasts 2 hours. The tests are a mixture of open-ended and multiple-choice questions that are organized in groups based on a passage setting out a real-life situation. A total of about 390 minutes of test items are covered. Students take different combinations of different tests.Additional PISA initiativesPISA-based Test for Schools(PTS)As interest in PISA has grown, school and local educators have been wanting to know how their individual schools compare with students and schools in education systems worldwide. To address this need, the OECD(The Organization for Economic Co-operation and Development) has developed the PISA-based test for schools. It is currently available in the United States and the OECD is in discussions with governments to make the test available in other countries such as England and Spain.70. PISA is different from other programmes because __________.A. its test is closely related to the school curriculum.B. its test aims to assess whether students can solve real-life problems.C. its test can equip students for full participation in school.D. test scores directly determine the analysis of the test.71. Which of the following statements is true according to the passage?A. Test-takers are carefully selected.B. Test-takers answer the same questions.C. Test-takers are tested on three key subjects.D. Test-takers spend about 390 minutes on the test.72. What can we infer from the last paragraph?A. Students of all ages will be able to take PTS in the future.B. More countries are likely to have PTS in the future.C. School and local educators show little interest in PISA at present.D. PISA provides evaluation of education system within a certain country.73. Where can we most probably find the passage?A. On the InternetB. In a newspaperC. In a magazineD. In an advertisement(C)YANG YUANQING, Lenovo’s boss, hardly spoke a word of English until he was about 40: he grew up in rural poverty and read engineering at university. But when Lenovo bought IBM’s personal-computer division in 2005 he decided to immerse himself in English: he moved his family to North Carolina, hired a language tutor and -- the ultimate sacrifice -- spent hours watching cable-TV news.Lenovo is one of a growing number of multinationals from the non-Anglophone world that have made English their official language. The fashion began in places with small populations but global ambitions such as Singapore and Switzerland.Corporate English is now invading more difficult territory, such as Japan. Rakuten, a cross between Amazon and eBay, and Fast Retailing, which operates the Uniqlo fashion chain, were among the first to switch. Now they are being joined by old-economy companies such as Honda, a carmaker, and Bridgestone, a tyremaker. Chinese firms are proving harder to handle/crack: they have a huge internal market and are struggling to enroll/recruit competent managers of any description, let alone English-speakers. But some are following Lenovo’s lead. Huawei has introduced English as a second language and encourages high-flyers(ambitious employees) to become fluent.There are some obvious reasons why multinational companies want a lingua franca(共同语). Adopting English makes it easier to recruit global stars (including board members), reach global markets and assemble global production teams. Such steps are especially important to companies in Japan, where the population is shrinking.Tsedal Neeley of Harvard Business School says that “Englishnisation” can stir up a hornet’s nest of emotions. Ms. Neeley argues that companies must think carefully about implementing a policy that touches on so many emotions. Senior managers should explain to employees why switching to English is so important, provide them with classes and conversation groups, and offer them incentives(刺激) to improve their fluency, such as foreign postings. Those who are already proficient in English should speak more slowly and try not to dominate conversations. And managers must act as referees and enforcers, resolving conflicts and discouraging staff from returning to their native tongues.Intergovernmental bodies like the European Union are obliged to pretend that there is no predominant global tongue. But businesses worldwide are facing up to the reality that English is the language on which the sun never sets.74. Lenovo’s boss made all the efforts to familiarize himself with English except __________.A. hiring a language tutorB. resettling in an English-speakingenvironmentC. expanding the business overseasD. exposing himself to English Cable-TV news.75. What can we infer from the passage?A. Most Chinese firms would like to introduce corporate English.B. Chinese firms are in great need of English majors as their managers.C. Huawei followed Lenovo as the second largest multinational in China.D. Adopting corporate English is more difficult in places with a large population.76. Which of the following is true according to the passage?A. The decrease in population pushes the Japanese to learn English well.B. Neither the governmental bodies nor businesses will regard English as a global tongue.C. Companies should handle employees’ emotions carefully during the switch.D. Those good at English should be encouraged to speak more in the company.77. Which of the following might be the best title of the passage?A. English-Global Tongue in Business.B. English-Chinese Business Leaders’ NewFashion.C. English-The Best Tool in Communication.D. English-Dominating Factor ofSuccessful Business.Section C Directions: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.Now many people strive to be a follower of the LOHAS movement. LOHAS means “lifestyles of health and sustainability.” This term was coined in 2000 by two American scholars.Lohasians believe in leading a healthy lifestyle that is actively involved in preserving the earth’s environment and resources. According to Lohasians, respect for one’ own mental and physical health should exist in parallel with care for the earth’s ecology. They believe their actions, in this way, can have a positive effect on our global environment, and might be able to minimize the negative effects of people’s mindless and selfish consumption.Take organic foods for example. Lohasians prefer them, not only because they are chemical-free and good for the human body, but also because they are cultivated using natural fertilizers, which do not harm the soil. Even more Lohasians turn to locally grown produce, the transportation of which consumes far less than that of imported goods. As global warming has become a universal concern, Lohasians are anxious to find ways to cut down on energy consumption.Indeed, Lohasians are always considering the long-term impact of their behavior on the planet. As more consumers are adopting LOHAS values, this growing trend has dawned on the corporate world and they begin to practice responsible capitalism, which means providing goods and services using environmentally friendly and economically sustainable business practices. For instance, Coca-Cola’s efforts in the area of sustainable packaging focus mainly on “using and reduce its impact on the environment. As a result, the company saved 89,000 metric tons of glass in 2007 alone, and, therefore, reduced carbon dioxide emissions to a level equivalent to that of the planting of more than 13,000 acres of trees.Clearly, LOHAS values have become a significant trend in the world today. Individual or corporate “cultural creative” are promoting these values by challenging old traditions and habits, and building new lifestyles. Although whether these practices will bring immediate confident that these practices will bring immediate benefits to the environment and the health of people today remains unknown, Lohasians are confident that these practices will benefit their children and future generations. All individuals should evolve into Lohasians and take action to save the planet, before it is too late.(Note:Answer the questions or complete the statements in NO MORE THAN TWELVE WORDS. )78. Lohasians are convinced that through their responsible actions, ________________ might bereduced to a minimum.79. Why is locally grown produce favoured by Lohasians?80. Consumers’ growing trend of LOHAS values has inspired companies to _________________.81. In terms of their practices, Lohasians are not sure of _______________________.第 II 卷(共 47 分)I. TranslationDirections:Translate the following sentences into English, using the words given in the brackets. 1. 当地村民的善良感动了我们。
上海市浦东新区2014学年第二学期高三教学质量检测地理试卷(含详细答案)
上海市浦东新区2014学年第二学期高三教学质量检测地理试卷(考试时间120分钟满分150分)考生注意:请先在答题纸上填写本人信息,请将答案写在答题纸上。
一、选择题(共60分,每小题2分,每小题只有一个正确答案)(一)位于浦东新区的上海迪斯尼乐园的美丽轮廓正越来越清晰。
上海,我来了1.上海建设迪斯尼乐园的主要区位优势是①广阔的消费市场②海陆空交通便捷③强大的科研力量④丰富的劳动力A.①②B.②③C.①③D.①④2.上海迪斯尼乐园建成后,其最突出的功能是A.吸引外商进行工业的投资B.促进长江三角洲经济转型C.拉动相关产业和行业发展D.刺激中外文化的融合发展(二)洞庭湖是我国第二大淡水湖,现有面积约为1950年时期的一半。
3.引起湖泊面积变化的主要原因是①周边植被遭破坏②全球气候变暖③南水北调工程④围湖造田A.①②B.②③C.③④D.①④4.湖泊面积缩小对湖区周边环境带来的影响是A.湿地功能丧失殆尽B.气温日较差变小C.湖泊调蓄能力下降D.降水量季节变化变小5.为准确及时监测洪水期间湖区的淹没范围,采用的现代化技术是①遥感技术②全球定位系统③地理信息系统④雷达系统A.①②B.②③C.③④D.①③(三)中国探月工程“绕”、“落”、“回”是我国迈出航天深空探测的第一步。
6.与卫星绕地飞行相比,绕月飞行A .受到的引力更大B .受到的阻力更小C .观测月球气象更复杂D .地面监控卫星更容易 7.通过探月工程可以验证A .月球微重力、强辐射环境B .地球潮汐主要受月球引力影响C .朔望月的周期为29.53日D .月球上昼夜变化周期为29.53日(四)强热带气旋是发生在热带、亚热带地区海面上的一种气旋性环流。
下图中,圆圈表示强热带气旋中心,直线表示北半球甲地所在的纬线。
8.若强热带气旋由实线位置Ⅰ向西运动到虚线位置Ⅱ,甲地风向可能发生的变化是 A .由东风转为西风 B .由西风转为东风 C .由东南风转为西北风 D .由西北风转为东南风 9.南半球发生强热带气旋次数明显少于北半球的主要原因是 A .南半球海洋面积大 B .南半球海水温度低 C .南半球洋流类型少 D .南半球海岸线平直(五)自然环境中各因素相互影响、相互制约。
上海市浦东新区高三数学4月教学质量检测试题 理(扫描版)
杨浦区2014学年度第二学期高三年级学业质量调研数学学科试卷(理科考生注意:1.答卷前,考生务必在答题纸写上姓名、考号,并将核对后的条形码贴在指定位置上2.本试卷共有23道题,满分150分,考试时间120分钟一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分1.函数的定义域是.2.若集合,则的元素个数为.3.若,则的值是.4.的展开式中的常数项的值是.5.某射击选手连续射击5枪命中环数分别为:,则这组数据的方差为.6.对数不等式的解集是,则实数的值为.7.极坐标方程所表示的曲线围成的图形面积为.8.如图,根据该程序框图,若输出的为,则输入的的值为.9.若正数满足,则的取值范围是.10.已知是不平行的向量,设,则与共线的充要条件是实数等于.11.已知方程的两根为,若,则实数的值为.12.已知从上海飞往拉萨的航班每天有5班,现有甲、乙、丙三人选在同一天从上海出发去拉萨,则他们之中正好有两个人选择同一航班的概率为.13.已知,在坐标平面中有斜率为的直线与圆相切,且交轴的正半轴于点,交轴于点,则的值为.14.对于自然数的每一个非空子集,我们定义“交替和”如下:把子集中的元素从大到小的顺序排列,然后从最大的数开始交替地加减各数,例如的交替和是;则集合的所有非空子集的交替和的总和为.二、选择题(本大题满分20分)本大题共有4题,每题有且只有一个正确答案,考生应在答题纸的相应编号上,填上正确的答案,选对得5分,否则一律得零分.15.“”是“函数只有一个零点”的()A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分又非必要条件16.在复平面中,满足等式的所对应点的轨迹是()A.双曲线B.双曲线的一支C.一条射线D.两条射线17.设反比例函数与二次函数的图像有且仅有两个不同的公共点,且,则()A.2或B.或C.2或D.或18.如图,设店是单位圆上的一个定点,动点从点出发,在圆上按逆时针方向旋转一周,点所旋转过的弧的长为,弦的长为,则函数的图像大致是()A. B. C. D.三.解答题(本大题满分74)本大题共5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19.(本题满分12分)如图,一条东西走向的大江,其河岸处有人要渡江到对岸处,江面上有一座大桥,已知在的西南方向,在的南偏西,公里.现有两种渡江方案:方案一:开车从大桥渡江到处,然后再到处;方案二:直接坐船从处渡江到对岸处.若车速为每小时60公里,船速为每小时45公里(不考虑水流速度),为了尽快到达处,应选择哪个方案?说明理由.20.(本题满分14分,其中第一小题7分,第二小题7分)在棱长为1的正方体中,点是棱的中点,点是棱上的动点. (1)试确定点的位置,使得平面;(2)当平面时,求二面角的大小(结果用反三角函数表示).21.(本题满分14分,其中第一小题4分,第二小题5分,第三小题5分)已知函数是奇函数.(1)求的值;(2)求的反函数;(3)对于任意的,解不等式:.22.(本题满分16分,其中第一小题5分,第二小题5分,第三小题6分)数列满足,(),令,是公比为的等比数列,设.(1)求证:;(2)设的前项和为,求的值;(3)设前项积为,当时,的最大值在和的时候取到,求为何值时,取到最小值.23.(本题满分18分,其中第一小题6分,第二小题6分,第三小题6分)已知抛物线的焦点,线段为抛物线的一条弦.(1)若弦过焦点,求证:为定值;(2)求证:轴的正半轴上存在定点,对过点的任意弦,都有为定值;(3)对于(2)中的点及弦,设,点在轴的负半轴上,且满足,求点坐标.。
浦东新区第二学期高三教学质量检测高三物理质量阐明
浦东新区 2014 学年第二学期高三教课质量检测高三物理质量剖析高三物理命题组张伟平执笔2015/4/27《浦东新区 2014 学年第二学期高三教课质量检测》是第二学期期中考试时期的一次全区性质量测试。
检测力争有益于发现高三第一轮及第二轮复习中的成绩和不足,为下一阶段一个多月时间里的高三物理复习冲刺供给依照。
一、试题命制的依照:《上海市高中物理课程标准》及《2015 年上海市一般高考考试说明》。
物理科考试旨在考察考生对高中物理基础知识和基本技术的掌握程度。
在此基础上,侧重对考生进行能力考察。
详细的能力目标以下:1.基础知识与基本技术1.1 能理解物理观点、规律、公式的基本含义并知道其发展历程。
1.2 能用简单的数学运算办理问题。
1.3 能理解用图像描绘的物理状态、物理过程和物理规律。
2.物理思想能力2.1 能依据物理原理进行剖析、判断、推理。
2.2 能用图像进行分折、判断、推理。
2.3 能应用简单的数学技术办理问题。
2.4 能用科学的思想方法进行剖析、判断、推理。
2.5 能阅读理解简单的新物理知识,并以此为依照进行判断、推理。
3.物理实验能力3.1 能剖析实验现象,做出合理的解说。
3.2 能对实验装置、实验操作、实验过程进行剖析、判断。
3.3 能对实验所得数据进行剖析、办理,得出结论。
4.综合应用能力4.1 能把复杂问题分解成若干简单的问题进行办理。
4.2 能从实质问题中提炼出合理的物理模型。
4.3 能针对详细问题中的各样可能性进行议论并做出判断。
4.4 能对详细问题的解决过程进行剖析,做出评论。
4.5 能发现问题,对存在的问题或给出的现象进行研究。
二、试题命制的愿景:力热电光全覆盖;难易适中有划分;能力立意有新题;陈题新用有变化;识图能力要考察;数理联合考能力;理论实质相联系;学科特点要表现。
三、测试结果剖析:全区均匀分为,与命题时的预期值()高度一致。
此中一类学校均匀分,二类学校均匀分为,三类学校均匀分为,四类学校均匀分为。
理科二模上海市浦东区高三数学
2014年上海市浦东新区高三年级二模试卷——数学(理科)2014年4月(考试时间120分钟,满分150分)一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分.1. 已知全集{}U=1,2,3,4,5,若集合{}A=2,3,则U A ð=_____2. 双曲线221916x y -=的渐近线方程为 . 3.函数()31cos 4sin x x x f =的最大值为_______4.已知直线1:210l ax y a -++=和()()2:2130l x a y a R --+=∈,若12l l ⊥,则a = .5.函数()y f x =的反函数为()1y f x -=,如果函数()y f x =的图像过点()2,2-,那么函数()11y f x -=+的图像一定过点______.6. 已知数列{}n a 为等差数列,若134a a +=,2410a a +=,则{}n a 的前n 项的和n S =_____.7.π,则球的体积为 ____ .8.(理) 一名工人维护甲、乙两台独立的机床,在一小时内,甲、乙需要维护的概率分别为、,则一小时内有机床需要维护的概率为_____9.设a R ∈,8(1)ax -的二项展开式中含3x 项的系数为7,则2lim()nn a a a →∞+++=L ____.10.(理)在平面直角坐标系xoy 中,若直线:x t l y t a =⎧⎨=-⎩(t 为参数)过椭圆3cos C :2sin x y ϕϕ=⎧⎨=⎩(ϕ为参数)的右顶点,则常数a =___.11.(理)已知随机变量ξ的分布列如右表,若3E ξ=,则D ξ=__ .12.在ABC ∆中, 角B 所对的边长6b =,ABC ∆的面积为15,外接圆半径R 5=,则ABC ∆的周长为_______13.抛物线24(0)y mx m =>的焦点为F ,点P 为该抛物线上的动点,又点A(,0)m -,则PFPA的最小值为 .14.(理)已知函数()f x 的定义域为{}1,2,3,值域为集合{}1,2,3,4的非空真子集,设点()A 1,(1)f ,()B 2,(2)f ,()C 3,(3)f ,ABC ∆的外接圆圆心为M ,且MA MC MB()R λλ+=∈u u u r u u u r u u u r,则满足条件的函数()f x 有__个.二、选择题(本大题满分20分)本大题共有4题,每题有且只有一个正确答案,考生应在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分. 15. “1x >”是“11x<”的( ) (A )充分而不必要条件 (B )必要而不充分条件 (C )充分必要条件 (D )既不充分也不必要条件 16. (理)已知z x yi =+,,x y R ∈, i 是虚数单位.若复数+1zi i+是实数,则z 的最小值为( ) (A )0 (B )52(C ) 5 (D 2 17.能够把椭圆2214x y +=的周长和面积同时分为相等的两部分的函数称为椭圆的“可分函数”,下列函数不是..椭圆的“可分函数”为( ) (A )3()4f x x x =+(B )5()ln 5x f x x -=+(C )()arctan 4x f x =(D )()x xf x e e -=+ 18. (理)方程27lg(100)(||200)(||202)2x x x -=---的解的个数为( ) (A )2 (B )4 (C )6 (D )8三、解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号规定的区域内写出必要的步骤.19.(本题满分12分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分6分. (理)如图,在直三棱柱111ABC A B C -中,AB AC ⊥,11AA AB AC ===,4ABC π∠=,D 、M 、N 分别是1CC 、11A B 、BC 的中点.(1)求异面直线MN 与AC 所成角的大小; (2)求点M 到平面ADN 之间的距离.20.(本题满分14分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分8分.如图,ABCD 是边长为10海里的正方形海域.现有一架飞机在该海域失事,两艘海事搜救船在A 处同时出发,沿直线AP 、AQ 向前联合搜索,且4PAQ π∠=(其中点P 、Q 分别在边BC 、CD 上),搜索区域为平面四边形APCQ 围成的海平面.设PAB θ∠=,搜索区域的面积为S .(1)试建立S 与tan θ的关系式,并指出θ的取值范围; (2)求S 的最大值,并求此时θ的值.21. (本题满分14分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分8分.(理)已知定义在R 上的函数)(x f ,对任意实数21,x x 都有1212()1()()f x x f x f x +=++,且(1)1f =. (1)若对任意正整数n ,有112n n a f ⎛⎫=+ ⎪⎝⎭,求1a 、2a 的值,并证明{}n a 为等比数列; (2)设对任意正整数n ,有1()n b f n =.若不等式 12226log (1)35n n n b b b x +++++>+L 对任意不小于2的正整数n 都成立,求实数x 的取值范围.BCP22.(本题满分16分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分6分.(理)已知中心在原点O ,左焦点为1(1,0)F -的椭圆1C 的左顶点为A ,上顶点为B ,1F 到直线AB|OB . (1) 求椭圆1C 的方程;(2) 过点(3,0)P 作直线l ,使其交椭圆1C 于R 、S 两点,交直线1x =于Q 点. 问:是否存在这样的直线l ,使||PQ 是||PR 、||PS 的等比中项?若存在,求出直线l 的方程;若不存在,说明理由.(3) 若椭圆1C 方程为:22221x y m n +=(0m n >>),椭圆2C 方程为:2222x y m nλ+=(0λ>,且1λ≠),则称椭圆2C 是椭圆1C 的λ倍相似椭圆.已知2C 是椭圆1C 的3倍相似椭圆,若直线y kx b =+与两椭圆1C 、2C 交于四点(依次为P 、Q 、R 、S ),且2PS RS QS +=u u u r u u u r u u u r,试研究动点(,)E k b 的轨迹方程.23.(本题满分18分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分8分.(理)定义区间),(d c ,),[d c ,],(d c ,],[d c 的长度均为c d -,其中c d >.(1)已知函数21xy =-的定义域为[],a b ,值域为10,2⎡⎤⎢⎥⎣⎦,写出区间[],a b 长度的最大值与最小值.(2)已知函数()M f x 的定义域为实数集[2,2]D =-,满足(),,M x x Mf x x x M ∈⎧=⎨-∉⎩(M 是D 的非空真子集) . 集合[]1,2A =,[]2,1B =-- ,求()()()()3A B A B f x F x f x f x =++U 的值域所在区间长度的总和.(3)定义函数1234()11234f x x x x x =+++-----,判断函数()f x 在区间(2,3)上是否有零点,并求不等式()0f x >解集区间的长度总和.参考答案1. _{}1,4,5___2. 43y x =±. 3. _5_____ 4. a =13. 5. __(2,3)-___.6. _23522n n -___. 7. __323π__ .8.(理)(文) _115__ 9. _13-__.10.(理) _3__. (文) 511.(理) __1 . (文) ____2 _.12. 6+__13.2. 14.(理)_12_个. (文)_20_个. 二、选择题 15. A 16. (理) ( D ) (文)( A )17.( D ) 18. (理)( B )(文)( C )三、解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号规定的区域内写出必要的步骤.19.(本题满分12分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分6分. 12EN AC =,解:(1)设AB 的中点为E ,连接EN ,则//EN AC ,且所以MNE ∠或其补角即为异面直线MN 与AC 所成的角。
浦东新区—学年第二学期高三教学质量检测.docx
高中化学学习材料唐玲出品浦东新区2014—2015学年第二学期高三教学质量检测化学试卷考生注意:1. 本试卷满分150分,考试时间120分钟。
2. 本考试设试卷和答题纸两部分,试卷包括试题与答题要求;所有答题必须涂或写在答题纸上;做在试卷上一律不得分。
3. 答题前,考生务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号,并将核对后的条形码贴在指定位置上,在答题纸反面清楚地填写姓名。
4. 答题纸与试卷在试题编号上是一一对应的,答题时应特别注意,不能错位。
相对原子质量:H-1 C-12 N-14 O-16 Na-23 S-32 Cl-35.5 Fe-56 Ba-137一、选择题(本题共10分,每小题2分,只有一个正确选项) 1.下列用途中没有用到H 2SO 4酸性的是A .实验室制氢气B .制硫酸铵C .干燥氧气D .除铁锈2.下列有机物的名称与结构简式相符的是A .石炭酸:COOHB .甘油:C .蚁酸:CH 3COOHD .木精:CH 3CH 2OH3.下列有关化学用语的表达正确的是A .CO 2的比例模型:B .NC .Cl 原子的结构示意图:D .Al 原子最外层电子排布式:3s 23p 14.下列化工生产过程中的主要反应,不涉及氧化还原反应的是A .制纯碱B .制烧碱C .制漂白粉D .制硝酸5.我国自主研发的生物航空煤油在今年三月首次商业载客飞行成功,该生物航空煤油是以“地沟油”为原料,进行裂化、加氢、去氧处理后得到的。
下列说法错误的是 A .有机化学中加氢和去氧都属于还原反应 B .裂化是化学变化,裂化产物是饱和烃2p ↑ ↓ ↓ 2s↑↓ 2OHCHOHCH 2OHC .“地沟油”的分子中含有不饱和键,加氢能改善其稳定性D .“地沟油”去氧时,氧主要以H 2O 和CO 2等形式脱去 二、选择题(本题共36分,每小题3分,只有一个正确选项) 6.下列关于物质结构的叙述中正确的是A .分子晶体中一定含有共价键B .离子晶体中一定含有离子键C .含有极性键的分子一定是极性分子D .含有非极性键的分子一定是非极性分子7.下列变化中:①合成氨,②闪电固氮,③二氧化氮与水反应;按氮元素被氧化、被还原、既被氧化又被还原的顺序排列正确的是 A .①②③B .②①③C .①③②D .③①②8.实验室用右图所示装置电解氯化铜溶液,实验中观察到碳(I )电极质量增加,碳(II )电极有气体放出。
2014理科二模-上海市浦东区高三数学
2014理科二模-上海市浦东区高三数学D的步骤.19.(本题满分12分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分6分.(理)如图,在直三棱柱111ABC A B C -中,AB AC⊥,11AA AB AC ===,4ABC π∠=,D 、M 、N 分别是1CC 、11A B 、BC 的中点.(1)求异面直线MN 与AC 所成角的大小; (2)求点M 到平面ADN 之间的距离.20.(本题满分14分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分8分.如图,ABCD 是边长为10海里的正方形海域.现有一架飞机在该海域失事,D C Q两艘海事搜救船在A 处同时出发,沿直线AP 、AQ 向前联合搜索,且4PAQ π∠=(其中点P 、Q 分别在边BC 、CD 上),搜索区域为平面四边形APCQ 围成的海平面.设PAB θ∠=,搜索区域的面积为S .(1)试建立S 与tan θ的关系式,并指出θ的取值范围; (2)求S 的最大值,并求此时θ的值.21. (本题满分14分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分8分.(理)已知定义在R 上的函数)(x f ,对任意实数21,x x 都有1212()1()()f x x f x f x +=++,且(1)1f =.(1)若对任意正整数n ,有112nna f ⎛⎫=+ ⎪⎝⎭,求1a 、2a 的值,并证明{}na 为等比数列;(2)设对任意正整数n ,有1()nbf n =.若不等式12226log (1)35n n n b b b x +++++>+对任意不小于2的正整数n 都成立,求实数x 的取值范围.22. (本题满分16分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分6分.(理)已知中心在原点O ,左焦点为1(1,0)F -的椭圆1C 的左顶点为A ,上顶点为B ,1F 到直线AB 的距离为||7OB .(1) 求椭圆1C 的方程; (2) 过点(3,0)P 作直线l ,使其交椭圆1C 于R 、S 两点,交直线1x =于Q 点. 问:是否存在这样的直线l ,使||PQ 是||PR 、||PS 的等比中项?若存在,求出直线l的方程;若不存在,说明理由.(3) 若椭圆1C 方程为:22221x y m n+=(0m n >>),椭圆2C 方程为:2222x y m n λ+=(0λ>,且1λ≠),则称椭圆2C 是椭圆1C 的λ倍相似椭圆.已知2C 是椭圆1C 的3倍相似椭圆,若直线y kx b =+与两椭圆1C 、2C 交于四点(依次为P 、Q 、R 、S ),且2PS RS QS +=,试研究动点(,)E k b 的轨迹方程.23.(本题满分18分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分8分.(理)定义区间),(d c ,),[d c ,],(d c ,],[d c 的长度均为c d -,其中c d >.(1)已知函数21xy =-的定义域为[],a b ,值域为10,2⎡⎤⎢⎥⎣⎦,写出区间[],a b 长度的最大值与最小值.(2)已知函数()Mfx 的定义域为实数集[2,2]D =-,满足(),,M x x Mf x x x M∈⎧=⎨-∉⎩ (M 是D 的非空真子集) . 集合[]1,2A =,[]2,1B =-- ,求()()()()3A B A B f x F x fx f x =++的值域所在区间长度的总和.(3)定义函数1234()11234f x x x x x =+++-----,判断函数()f x 在区间(2,3)上是否有零点,并求不等式()0f x >解集区间的长度总和.参考答案1. _{}1,4,5___2.43y x=± .3. _5_____4.a =13.5. __(2,3)-___.6. _23522nn -___.7. __323π__ . 8.(理) 0.98(文) _115__9. _13-__. 10.(理) _3__. (文) 511.(理) __1 . (文) ____2 _. 12.6+13.2.14.(理)_12_个. (文)_20_个. 二、选择题 15. A 16. (理) ( D ) (文)( A )17.( D )18. (理)( B ) (文)( C )三、解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号规定的区域内写出必要的步骤. 19.(本题满分12分)本题共有2个小题,第(1)小题满分6分,第(2)小题满分6分.解:(1)设AB 的中点为E ,连接EN ,则//EN AC ,且12EN AC =,所以MNE ∠或其补角即为异面直线MN 与AC 所成的角。
2024届上海市浦东新区高三下学期教学质量检测全真演练物理试题
2024届上海市浦东新区高三下学期教学质量检测全真演练物理试题一、单选题 (共7题)第(1)题空间P、Q两点处固定电荷量绝对值相等的点电荷,其中Q点处为正电荷,P、Q两点附近电场的等势线分布如图所示,a、b、c、d、e为电场中的5个点,设无穷远处电势为0,则( )A.e点的电势大于0B.a点和b点的电场强度相同C.b点的电势低于d点的电势D.负电荷从a点移动到c点时电势能增加第(2)题如图甲所示,在三维直角坐标系O—xyz的xOy平面内,两波源S1、S2分别位于x1=0.2m、x2=1.2m处。
两波源垂直xOy平面振动,振动图像分别如图乙、丙所示。
M为xOy平面内一点且。
空间有均匀分布的介质且介质中波速为v=2m/s,则( )A.x=0.2m处质点开始振动方向沿z轴负方向B.两列波叠加区域内x=0.6m处质点振动减弱C.两列波叠加区域内x=0.5m处质点振动加强D.两波相遇后M点振动方程为第(3)题宇宙双星系统是由两颗相距较近的恒星组成的系统,它们在相互引力作用下,围绕着共同的圆心运动。
它们为天文学家研究恒星的演化提供了很好的素材。
已知某双星之间的距离为,相互绕行周期为,引力常量为,可以估算出( )A.双星的质量之和B.双星的质量之积C.双星的速率之比D.双星的加速度之比第(4)题足球比赛中,足球以从球员身边直线滚过,在运动方向上离边界还有,该球员立即由静止开始同向直线追赶,球员和足球的速度时间图像如图所示,则( )A.球员的加速度小于足球的加速度B.内,球员的平均速度大于足球的平均速度C.时,球员刚好追上足球D.若球员不追赶足球,足球会滚出边界第(5)题如图所示,一位潜水爱好者在水下活动时,利用激光器向岸上救援人员发射激光信号,激光束与竖直方向的夹角为α。
当α大于37°以后,激光束无法射出水面。
已知光在真空中的传播速度大小为,,,则( )A.激光束由水中射向空气发生全反射的临界角为53°B.激光束在水中的折射率为C.激光束在水中传播的速度大小为D.当时,射出水面的激光束方向与水面的夹角大于60°第(6)题如图甲所示,在竖直方向的匀强磁场中,水平放置一圆形导体环,磁场的磁感应强度B随时间t变化的关系如图乙所示。
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题一、单选题 (共6题)第(1)题水面上漂浮一半径为的圆形荷叶,一条小蝌蚪从距水面的图示位置处沿水平方向以速度匀速穿过荷叶,其运动的轨迹与荷叶径向平行,已知水的折射率为,则在小蝌蚪穿过荷叶过程中,在水面之上看不到小蝌蚪的时间为( )A.3s B.4s C.5s D.6s第(2)题在如图所示的电路中,定值电阻,,电容器的电容,电源的电动势,内阻不计,当开关闭合电流达到稳定时,处在电容器中间带电量的油滴恰好保持静止,当开关闭合后,则以下判断正确的是( )A.电容器上极板是高电势点B.带电油滴加速向上运动C.两点的电势差D.通过的电量第(3)题如图所示,将一张重力不计的白纸夹在重力为G的课本与水平桌面之间,现用一水平拉力F将白纸从课本下向右抽出(书本始终未从桌面上滑落),已知所有接触面间的动摩擦因数均为μ,则在白纸被抽出的过程中( )A.课本对白纸的滑动摩擦力方向向右B.课本与桌面接触前,白纸所受滑动摩擦力大小为2μGC.白纸被抽出越快,课本动量的变化越大D.白纸被抽出越快,课本所受滑动摩擦力的冲量越大第(4)题如图所示,质量为的足够长的木板静止在粗糙水平地面上,在长木板上方右侧有质量为的物块,竖直墙面在长木板的右端,物块与木板、木板与地面间的动摩擦因数均为,某时刻对木板施加水平向右、大小的恒定拉力,作用1s后撤去,物块和木板始终未与竖直墙面碰撞,重力加速度,设最大静摩擦力等于滑动摩擦力。
下列说法正确的是( )A.外力F做的功为4JB.整个运动过程用时C.整个运动过程摩擦生热8JD.初始时,木板与墙的距离至少为第(5)题某种光电式火灾报警器的原理如图所示,由红外光源发射的光束经烟尘粒子散射后照射到光敏电阻上,光敏电阻接收的光强与烟雾的浓度成正比,其阻值随光强的增大而减小。
闭合开关,当烟雾浓度达到一定值时,干簧管中的两个簧片被磁化而接通,触发蜂鸣器报警。
2024届上海市浦东新区高三下学期教学质量检测全真演练物理试题
2024届上海市浦东新区高三下学期教学质量检测全真演练物理试题一、单项选择题:本题共8小题,每小题3分,共24分,在每小题给出的答案中,只有一个符合题目要求。
(共8题)第(1)题放射性同位素热电机是各种深空探测器中最理想的能量源,它不受温度及宇宙射线的影响,使用寿命可达十几年。
用大量氘核轰击时可产生放射性元素,的半衰期为87.74年,含有的化合物核电池的原理是其发生衰变时将释放的能量转化为电能,我国的火星探测车用放射性材料作为燃料,中的Pu元素就是,下列判断正确的是( )A.B.C.的比结合能大于的比结合能D.1kg化合物带经过87.74年后剩余0.5kg第(2)题物理教材中有很多经典的插图能够形象地表现出物理实验、物理现象及物理规律,下列四幅图涉及不同的物理知识或现象,下列说法正确的是( )A.图甲中,卢瑟福通过分析粒子散射实验结果,发现了质子和中子B.图乙中,在光的颜色保持不变的情况下,入射光越强,饱和光电流越大C.图丙中,射线a由电子组成,射线b为电磁波,射线c由粒子组成D.图丁中,链式反应属于轻核裂变第(3)题如图所示,一质量为m的刚性圆环套在粗糙的竖直固定细杆上,圆环的直径略大于细杆的直径,圆环分别与两个相同的轻质弹簧的一端相连,弹簧的另一端连在与圆环同一高度的墙壁上。
开始时圆环处于O点,弹簧处于原长状态。
细杆上的A、B两点到O点的距离都为h,将圆环拉至A点由静止释放,重力加速度为g。
则圆环从A点运动到B点的过程中( )A.圆环通过O点时,加速度小于gB.圆环通过O点时,速度等于C.圆环通过B点时,速度等于D.圆环通过B点时,速度小于第(4)题2024年1月18日01时46分,天舟七号货运飞船成功对接空间站天和核心舱,变轨情况如图所示。
空间站轨道可近似看成圆轨道,且距离地面的高度约为。
下列关于天舟七号的说法正确的是( )A.发射速度大于B.从对接轨道变轨到空间站轨道时,速度变大C.在对接轨道上的运行周期大于空间站的运行周期D.在空间站轨道运行的速度比赤道上的物体随地球自转的线速度小第(5)题如图所示,两点分别放置两个等量异种电荷,点为,点为。
2024届上海市浦东新区高三下学期教学质量检测物理试题
2024届上海市浦东新区高三下学期教学质量检测物理试题一、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)题关于匀速圆周运动的描述正确的是()A.是匀速运动B.是匀变速运动C.合力不一定时刻指向圆心D.是加速度变化的曲线运动第(2)题铀核裂变的产物是多样的,一种典型的铀核裂变生成钡和氪,同时放出3个X粒子,核反应方程是,其中铀核、中子、钡核和氪核的质量分别为m1、m2、m3和m4。
下列说法正确的是( )A.X是B.X的质量为C.铀核的结合能大于氪核的结合能D.铀核的比结合能大于氪核的比结合能第(3)题如图所示,边界OM与ON之间分布有垂直纸面向里的匀强磁场,边界ON上有一粒子源S.某一时刻,从离子源S沿平行于纸面,向各个方向发射出大量带正电的同种粒子(不计粒子的重力及粒子间的相互作用),所有粒子的初速度大小相等,经过一段时间有大量粒子从边界OM射出磁场.已知∠MON=30°,从边界OM射出的粒子在磁场中运动的最长时间等于T(T为粒子在磁场中运动的周期),则从边界OM射出的粒子在磁场中运动的最短时间为( )A .T B.T C.T D.T第(4)题在物理学的重大发现中科学家们创造出了许多物理学研究方法,如理想实验法、控制变量法、极限思想法、微元法和建立物理模型法等。
以下关于所用物理学研究方法的叙述正确的是( )A.在物体本身的大小和形状对研究的问题影响非常小时,物体可简化为一个具有质量的点,这种方法叫极限思想法B.根据功率的定义式,当时间间隔非常小时,就可以用这一间隔内的平均功率表示间隔内某一时刻的瞬时功率,这应用了控制变量法C.如果一个力的作用效果与另外两个力的作用效果相同,这个力就是那两个力的合力。
这里采用了理想实验法D.在推导弹簧弹力做功的表达式时,把整个做功过程划分成很多小段,每一小段近似看作恒力做功,然后把各小段弹力所做的功相加,这里采用了微元法第(5)题关于光现象的叙述,以下说法正确的是( )A.太阳光照射下肥皂膜呈现的彩色属于光的干涉B.雨后天空中出现的彩虹属于光的衍射C.通过捏紧的两只铅笔间的狭缝观看工作着的日光灯管,看到的彩色条纹,属于光的干涉D.阳光照射下,树影中呈现的一个个小圆形光斑,属于光的衍射现象第(6)题如图所示,abcd是由粗细均匀的绝缘线制成的正方形线框,其边长为L,O点是线框的中心,线框上均匀地分布着正电荷,现将线框左侧中点M处取下足够短的一小段,该小段带电量为q,然后将其沿OM连线向左移动的距离到N点处,线框其他部分的带电量与电荷分布保持不变,若此时在O点放一个带电量为Q的正点电荷,静电力常量为k,则该点电荷受到的电场力大小为( )A.B.C.D.第(7)题2020年11月24日凌晨,搭载嫦娥五号探测器的长征五号遥五运载火箭从文昌航天发射场顺利升空,12月17日“嫦娥五号”返回器携带月球样品在预定区域安全着陆,在落地之前,它在大气层打个“水漂”。
2024届上海市浦东新区高三下学期教学质量检测物理试题
2024届上海市浦东新区高三下学期教学质量检测物理试题一、单选题 (共6题)第(1)题校运动会中,甲、乙两位运动员从同一起跑线同时起跑,v-t图像如图所示,已知图中阴影Ⅰ的面积大于阴影Ⅱ的面积,且t2时刻甲恰好到达50米处。
由此可知( )A.甲运动员的加速度先增大后不变B.在t1时刻,甲、乙两运动员相遇C.在t2时刻,乙运动员已经跑过了50米处D.0~t1时间内,可能存在甲、乙运动员加速度相同的时刻第(2)题如图所示圆形区域内存在一匀强磁场,磁感应强度大小为B,方向垂直于纸面向里,一带正电荷的粒子沿图中直线以速率v0从圆周上的a点射入圆形区域,从圆周上b点射出(b点图中未画出)磁场时速度方向与射入时的夹角为60°,已知圆心O到直线的距离为横截面半径的一半。
现将磁场换为平行于纸面且垂直于直线的匀强电场,同一粒子以同样速度沿直线从a点射入圆形区域,也从b点离开该区域,若不计重力,则匀强电场的场强大小为( )A.B.C.D.第(3)题如图所示,理想变压器原、副线圈的匝数比为,b是原线圈的中心抽头,电压表和电流表均为理想电表,,的最大值为2R,从某时刻开始在原线圈c、d两端加上交变电压,其瞬时值表达式为u 1=220sin100πt(V),则( )A.当单刀双掷开关与a连接时,电压表的示数为110 VB.单刀双掷开关与a连接,当滑动变阻器触头P在正中间时,消耗的功率最大C.当单刀双掷开关由a扳向b时,当滑动变阻器触头P从正中间向下移动的过程中,消耗的功率增大D.单刀双掷开关与a连接,在滑动变阻器触头P向上移动的过程中,电压表和电流表的示数均变小第(4)题如图所示,用绝缘支架将带电荷量为+Q的小球a固定在O点,一粗糙绝缘直杆与水平方向的夹角θ=30°,直杆与小球a位于同一竖直面内,杆上有A、B、C三点,C与O两点位于同一水平线上,B为AC的中点,OA=OC=L。
小球b质量为m,带电荷量为–q,套在直杆上,从A点由静止开始下滑,第一次经过B点时速度为v,运动到C点时速度为0。
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)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)题下列有关高中物理实验的描述中,正确的是()A.在“验证力的平行四边形定则”的实验中,拉橡皮筋的细绳要稍长,并且实验时要使弹簧测力计与木板平面平行,同时保证弹簧的轴线与细绳在同一直线上B.在“验证机械能守恒定律”的实验中,必须要用天平测出下落物体的质量C.在“探究弹力和弹簧伸长的关系”的实验中,弹簧必须水平使用,以克服弹簧所受重力对实验的影响D.在“研究平抛物体运动”的实验中,斜槽应尽可能光滑第(2)题下列四幅图所涉及的物理知识,论述正确的是( )A.图甲表明晶体熔化过程中分子平均动能变大B.图乙水黾可以在水面自由活动,说明它所受的浮力大于重力C.图丙是显微镜下三颗小炭粒的运动位置连线图,连线表示小炭粒的运动轨迹D.图丁中A是浸润现象,B是不浸润现象第(3)题科学研究发现,钚(Pu)是一种具有放射性的超铀元素,其衰变方程为,该衰变过程中产生的γ光子照射到逸出功为的金属上,逸出光电子的最大初动能为。
已知普朗克常量为h,下列说法正确的是( )A.X原子核中含有92个中子B.衰变产生的γ光子具有很强的电离作用C.衰变产生的γ光子的频率为D.钚(Pu)核的比结合能大于X原子核的比结合能第(4)题核电站利用核反应堆工作时释放出的热能使水汽化以推动汽轮发电机发电。
反应堆中的核反应方程式为,已知铀核的质量m U,中子的质量m n,锶(Sr)核的质量m Sr,氙(Xe)核的质量m Xe,光在真空中的传播速度为c。
则( )A.Z=38,A=136B.Z=48,A=146C.一个裂变时的质量亏损D.一个裂变时释放的能量为第(5)题如图所示,在平面直角坐标系xOy内,仅在第二象限内有垂直于纸面向外的磁感应强度为B的匀强磁场,一质量为m、电荷量为+q的粒子在纸面所在平面内从A点射入磁场。
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)全真演练物理试题
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)全真演练物理试题一、单项选择题:本题共8小题,每小题3分,共24分,在每小题给出的答案中,只有一个符合题目要求。
(共8题)第(1)题长直导线周围产生的磁感应强度大小(k为常数,I为导线中电流的大小,r为到导线的距离)。
如图所示,在等边三角形PMN的三个顶点处,各有一根长直导线垂直于纸面固定放置。
三根导线均通有电流I,且电流方向垂直纸面向外,已知三根导线在三角形中心O处产生的磁感应强度大小均为B0,若将P处导线的电流变为2I,且电流方向变为垂直纸面向里,则三角形中心O处磁感应强度的大小为()A.B 0B.B0C.2B0D.3B0第(2)题金星、地球和火星绕太阳的公转均可视为匀速圆周运动,它们的向心加速度大小分别为a金、a地、a火,它们沿轨道运行的速率分别为v金、v地、v火.已知它们的轨道半径R金<R地<R火,由此可以判定A.a金>a地>a火B.a火>a地>a金C.v地>v火>v金D.v火>v地>v金第(3)题手机软件中运动步数的测量是通过手机内电容式加速度传感器实现的,如图所示为其工作原理的简化示意图。
质量块左侧连接轻质弹簧,右侧连接电介质,弹簧与电容器固定在外框上,质量块可带动电介质相对于外框无摩擦左右移动(不能上下移动)以改变电容器的电容。
下列说法正确的是( )A.传感器匀速向左做直线运动时,电容器两极板所带电荷量将多于静止状态时B.传感器匀减速向右做直线运动时,电流表中有由b向a的电流C.传感器运动时向右的加速度逐渐增大,则电流表中有由b向a的电流D.传感器运动时向左的加速度逐渐减小,则电流表中有由a向b的电流第(4)题一架无人机在水平地面由静止开始匀加速滑行1600m后起飞离地,离地时速度为80m/s。
若无人机的加速过程可视为匀加速直线运动,则无人机在起飞离地前最后1s内的位移为( )A.79m B.78m C.77m D.76m第(5)题已知半径为、质量为的球形天体的自转周期为,式中G为引力常量,则该天体的同步卫星距该天体表面的高度为( )A.B.C.D.第(6)题在贵州台盘村的“村超”篮球比赛中,山西星动力队队员罚球时,立定在罚球线后将篮球向斜上方抛出,投进一个空心球。
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题
2024届上海市浦东新区高三下学期等级考模拟质量调研(二模)物理试题一、单项选择题:本题共8小题,每小题3分,共24分,在每小题给出的答案中,只有一个符合题目要求。
(共8题)第(1)题近代物理学的发展,催生了一大批新技术,深刻地改变了人们的生活方式和社会形态,有关近代物理学发展的相关叙述错误的是( )A.普朗克通过对黑体辐射规律的研究,提出“能量子”概念,把物理学带入了量子世界B.α粒子散射实验,揭示了原子的“枣糕”结构模型C.丹麦物理学家玻尔提出了自己的原子结构假说,解释了氢原子的光谱D.世界上第一座核反应堆装置的建立,标志着人类首次通过可控制的链式反应实现了核能的释放第(2)题某小球质量为M,现让它在空气中由静止开始竖直下落,下落过程中所受空气阻力与速率的关系满足(k为定值),当下落时间为t时,小球开始匀速下落,已知重力加速度为g,则小球在t时间内下降的高度h为( )A.B.C.D.第(3)题19世纪末,科学家们发现了电子,从而认识到:原子是可以分割的,是由更小的微粒组成的,开启了人们对微观世界探索的大门。
有关原子物理,下列说法正确的是( )A.个碘经过一个半衰期后还剩个B.发现质子的核反应方程是C.汤姆孙在粒子散射实验的基础上提出了原子的核式结构模型D.一群氢原子从的激发态向基态跃迁时,最多能放出种不同频率的光子第(4)题如图所示,矩形线圈abcd放置在垂直纸面向里的匀强磁场中,绕垂直于磁场的中心转轴以300r/min的转速匀速转动,磁场分布在的右侧,磁感应强度。
线圈匝数匝,面积。
理想变压器的原、副线圈匝数比,定值电阻,C为电容器,电流表A为理想交流电表,不计矩形线圈的电阻,下列说法正确的是( )A.矩形线圈产生的电动势最大值为B.电流表示数为5AC.变压器输出功率为200WD.增大电容器的电容,电流表示数增大第(5)题如图,低电位报警器由两个基本门电路与蜂鸣器组成,该报警器只有当输入电压过低时蜂鸣器才会发出警报.其中()A.甲是“与门”,乙是“非门”B.甲是“或门”,乙是“非门”C.甲是“与门”,乙是“或门”D.甲是“或门”,乙是“与门”第(6)题如图,静止在匀强磁场中的原子核X发生一次衰变后放出的射线粒子和新生成的反冲核均垂直于磁感线方向运动。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
浦东新区2014学年第二学期高三教学质量检测数学试卷(理科)注意:1. 答卷前,考生务必在答题纸上指定位置将姓名、学校、考号填写清楚. 2. 本试卷共有23道试题,满分150分,考试时间120分钟.一、填空题(本大题共有14题,满分56分);考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分. 1.不等式32x>的解为 3l o g 2x >. 2.设i 是虚数单位,复数)1)(3(i i a -+是实数,则实数a = 3 . 3.已知一个关于y x ,的二元一次方程组的增广矩阵为112012-⎛⎫⎪⎝⎭,则x y -= 2 .4.已知数列{}n a 的前n 项和n n S n +=2,则该数列的通项公式=n a n 2 .5.已知21nx x ⎛⎫- ⎪⎝⎭展开式中二项式系数之和为1024,则含2x 项的系数为 210 .6.已知直线0243=++y x 与圆()2221r y x =+-相切,则该圆的半径大小为 1 .7.在极坐标系中,已知圆θρsin 2r =(0>r )上的任意一点),(θρM 与点),2(πN 之间的最小距离为1,则=r23. 8.若对任意R x ∈,不等式0sin 22sin 2<-+m x x 恒成立,则m 的取值范围是),21(+∞+.9.已知球的表面积为64π2cm ,用一个平面截球,使截面圆的半径为2cm ,则截面与球心的距离是 .10.已知随机变量ξ分别取1、2和3,其中概率)1(=ξp 与)3(=ξp 相等,且方差13D ξ=,则概率)2(=ξp 的值为23.11.若函数223()4f x x x =+-的零点(),1,m a a a ∈+为整数.则所有满足条件a 的值为1或2-.12.若正项数列{}n a 是以q 为公比的等比数列,已知该数列的每一项k a 的值都大于从2k a +开始的各项和,则公比q 的取值范围是 . 13.等比数列{}n a 的首项1a ,公比q 是关于x 的方程2(1)2(21)0t x x t -++-=的实数解,若数列{}n a 有且只有一个,则实数t 的取值集合为130,,1,22⎧⎫⎨⎬⎩⎭.14.给定函数()f x 和()g x ,若存在实常数,k b ,使得函数()f x 和()g x 对其公共定义域D 上的任何实数x 分别满足()f x kx b ≥+和()g x kx b ≤+,则称直线:l y kx b =+为函数()f x 和()g x 的“隔离直线”. 给出下列四组函数;① x x g x f xsin )(,121)(=+=; ② x x g x x f 1)(,)(3-==; ③ x x g x x x f lg )(,1)(=+=; ④ x x g x f x=-=)(,212)(其中函数()f x 和()g x 存在“隔离直线”的序号是 ①③④ .二、选择题(本大题共有4题,满分20分); 每小题都给出四个选项,其中有且只有一个选项是正确的,考生应在答题纸相应位置上,选对得 5分,否则一律得零分. 15.已知,a b 都是实数,那么“0a b <<”是“11a b>”的 ( A ) )(A 充分不必要条件 )(B 必要不充分条件 )(C 充分必要条件)(D 既不充分也不必要条件16.平面α上存在不同的三点到平面β的距离相等且不为零,则平面α与平面β的位置关系为 ( D ))(A 平行 )(B 相交 )(C 平行或重合 )(D 平行或相交17.若直线30ax by +-=与圆223x y +=没有公共点,设点P 的坐标(,)a b ,则过点P 的一条直线与椭圆22143x y +=的公共点的个数为 ( C ) )(A 0)(B 1)(C 2 )(D 1或218.如图,若正方体12341234PP P P QQ Q Q -的棱长为1, 设j i T S Q P x ⋅=11,},{,j i j i Q P T S ∈,(}4,3,2,1{,∈j i ), 对于下列命题:①当i j i i S T PQ =时,1x =; ②当0x =时,(),i j 有12种不同取值; ③当1x =-时,(),i j 有16种不同的取值; ④x 的值仅为1,0,1-.其中正确的命题是 ( C ))(A ①②)(B ①④ )(C ①③④ )(D ①②③④三、解答题(本大题共有5题,满分74分);解答下列各题必须在答题纸的相应位置上,写出必要的步骤.19.(本题共有2个小题,满分12分);第(1)小题满分6分,第(2)小题满分6分.已知函数(),(0),af x x x a x=+>为实数. (1)当1a =-时,判断函数()y f x =在()1,+∞上的单调性,并加以证明; (2)根据实数a 的不同取值,讨论函数()y f x =的最小值.解:(1)由条件:1()f x x x=-在()1,+∞上单调递增.…………………………2分任取()12,1,x x ∈+∞且12x x <1212121212111()()()(1)f x f x x x x x x x x x -=--+=-+ ……………………4分 211x x >>,∴121210,10x x x x -<+> P 1P 2P 3P 4Q 12Q 3Q 4∴ 12()()f x f x < ∴ 结论成立 …………………………………………6分 (2)当0a =时,()y f x =的最小值不存在; …………………………………7分当0a <时,()y f x =的最小值为0;………………………………………9分当0a >时,()ay f x x x==+≥x =()y f x =的最小值为12分20.(本题共有2个小题,满分14分);共有2个小题,第(1)小题满分7分,第(2)小题满分7分.如图,在四棱锥ABCD P -中,底面正方形ABCD 的边长为2, ⊥PA 底面ABCD , E 为BC 的中点,PC 与平面PAD 所成的角为22arctan. (1) 求异面直线AE 与PD 所成角的大小(结果用反三角函数表示); (2)求点B 到平面PCD 的距离.解:方法1,(1)因为底面ABCD 为边长为2的正方形,⊥PA 底面ABCD , 则 ⊥⇒⎪⎭⎪⎬⎫=⊥⊥CD A PA AD PA CD ADCD 平面PAD ,所以CPD ∠就是CP 与平面PAD 所成的角.……………………………………………2分 在CDP Rt ∆中,由22tan ==∠PD CD CPD ,得22=PD ,…………………………3分 在PAD Rt ∆中,2=PA .分别取AD 、PA 的中点M 、N ,联结MC 、NC 、MN , 则NMC ∠异面直线AE 与PD 所成角或补角.……………4分 在MNC ∆中,2=MN,MC 3NC =,由余弦定理得,2223cos10NMC +-∠==-, 所以NMC π∠=-,…………………………6分 即异面直线AE 与PD 所成角的大小为1010arccos .……7分PA CDPA CD EMN(2)设点B 到平面PCD 的距离为h ,因为BCD P PCDB V V --=,…………………………9分 所以,11113232CD PD h BC CD PA ⨯⋅⋅=⨯⋅⋅,得h =14分 方法2,(1) 如图所示,建立空间直角坐标系,同方法1,得2=PA ,……………3分 则有关点的坐标分别为()0,0,0A ,()2,1,0E ,()0,2,0D ,()2,0,0P .………………………5分 所以()2,1,0AE =,()2,2,0-=PD .设θ为异面直线AE 与PD 所成角, 则()101085202102cos =⨯-⨯+⨯+⨯=θ, 所以,1010arccos=θ, 即异面直线AE 与PD 所成角的大小为1010arccos.…………………………………7分 (2)因为()2,2,0-=,()0,0,2=,()0,2,0=,设()w v u ,,=,则由⎩⎨⎧==⇒⎪⎩⎪⎨⎧==⋅=-=⋅w v u u CD n w v PD n 002022,………………………………………………11分 可得()1,1,0=,所以n BC d n⋅=== 14分 21.(本题共有2个小题,满分14分);第(1)小题满分6分,第(2)小题满分8分.一颗人造地球卫星在地球表面上空沿着圆形轨道匀速运行,每2将地球近似为一个球体,半径为6370道所在圆的圆心与地球球心重合.点整通过卫星跟踪站A 点的正上空A ',通过C 点.间忽略不计)y(1)求人造卫星在12:03时与卫星跟踪站A 之间的距离(精确到1千米); (2)求此时天线方向AC 与水平线的夹角(精确到1分). 解:(1)设人造卫星在12:03时位于C 点处,AOC θ∠=,33609120θ=︒⨯=︒,…2分 在ACO ∆中,222=6370+8000-263708000cos93911704.327AC ⨯⨯⨯︒=, 1977.803AC ≈(千米),……………………………………………5分 即在下午12:03时,人造卫星与卫星跟踪站相距约为1978千米.…………………6分 (2)设此时天线的瞄准方向与水平线的夹角为ϕ,则90CAO ϕ∠=+︒,s i n 9s i n (90)19788000ϕ︒+︒=,8000sin(90)sin 90.63271978ϕ+︒=︒≈,…………………9分 即cos 0.6327ϕ≈,5045'ϕ≈︒,……………………………………………………11分 即此时天线瞄准的方向与水平线的夹角约为5045'︒.………………………………12分22.(本题共有3个小题,满分16分);第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分6分.已知直线l 与圆锥曲线C 相交于,A B 两点,与x 轴、y 轴分别交于D 、E 两点,且满足AD EA 1λ=、2λ=.(1)已知直线l 的方程为42-=x y ,抛物线C 的方程为x y 42=,求21λλ+的值;(2)已知直线l :1+=my x (1>m ),椭圆C :1222=+y x ,求2111λλ+的取值范围; (3)已知双曲线C :22221(0,0)x y a b a b -=>>,22212ba =+λλ,试问D 是否为定点?若是,求出D 点坐标;若不是,说明理由. 解:(1)将42-=x y ,代入x y 42=,求得点()2,1-A ,()4,4B ,又因为()0,2D ,()4,0-E ,…………………………………………………………………………2分由AD EA 1λ= 得到,()()2,12,11λ=()112,λλ=,11=λ,同理由2λ=得,22-=λ所以21λλ+=1-.………………………………………4分 (2)联立方程组:⎩⎨⎧=-++=022122y x my x 得()012222=-++my y m , 21,22221221+-=+-=+m y y m m y y ,又点()⎪⎭⎫ ⎝⎛-m E D 1,0,0,1,由AD EA 1λ= 得到1111y m y λ-=+,⎪⎪⎭⎫⎝⎛+-=11111y m λ, 同理由BD EB 2λ= 得到2221y m y λ-=+,⎪⎪⎭⎫⎝⎛+-=22111y m λ, 21λλ+=4212)(122121-=⎪⎭⎫⎝⎛⋅+-=⎪⎪⎭⎫ ⎝⎛++-m m y y y y m ,即21λλ+4-=,……………6分2121411λλλλ-=+12144λλ+=()42421-+=λ, …………………………………………8分因为1>m ,所以点A 在椭圆上位于第三象限的部分上运动,由分点的性质可知()0,221-∈λ,所以()2,1121-∞-∈+λλ.…………………………………………10分(3)假设在x 轴上存在定点)0,(t D ,则直线l 的方程为t my x +=,代入方程12222=-by a x 得到:()()022*******=-++-b a t mty b y a m b ()22222221222221,2a m b b a t y y a m b mt b y y ---=--=+, 2221211at mty y --=+ (1) 而由AD EA 1λ=、BD EB 2λ=得到:⎪⎪⎭⎫⎝⎛++=+-2121112)(y y m t λλ (2) 22212ba =+λλ (3) ……………………………………………………………………12分由(1)(2)(3)得到:2222222ba a t mt m t -=⎪⎭⎫ ⎝⎛--+,22b a t +±=, 所以点)0,(22b a D +±,………………………………………………………………14分 当直线l 与x 轴重合时,a t a +-=1λ,a t a -=2λ,或者a t a -=1λ,at a+-=2λ, 都有222222122ba a t a =-=+λλ 也满足要求,所以在x 轴上存在定点)0,(22b a D +±.……………………………16分23.(本题共有3个小题,满分18分);第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分8分.记数列{}n a 的前n 项12,,,n a a a 的最大项为n A ,第n 项之后的各项12,,n n a a ++ 的最小项为n B ,令n n n b A B =-.(1)若数列{}n a 的通项公式为2276n a n n =-+,写出12b b 、,并求数列{}n b 的通项公式;(2)若数列{}n b 的通项公式为12n b n =-,判断{}1n n a a +-是否等差数列,若是,求出公差;若不是,请说明理由;(3)若{}n b 为公差大于零的等差数列,求证:{}1n n a a +-是等差数列. 解:(1)因为数列{}n a 从第2项起单调递增,1231,0,3a a a ===,所以112101b a a =-=-=;213132b a a =-=-=-; ………………………2分当3n ≥时,154n n n b a a n +=-=-()()2,254,13n n b n n n -=⎧⎪=⎨-=≥⎪⎩或……………………………………………………4分(2) 数列{}n b 的通项公式为12n b n =-,∴n b 递减且0n b <.由定义知,1,n n n n A a B a +≥≤……………………………………………………6分10n n n n n b A B a a +>=-≥-∴1n n a a +>,数列{}n a 递增,即121n n a a a a +<<<<< ………………8分 21112111()()()()()n n n n n n n n n n n n a a a a a a a a b b b b ++++++++---=--+-=-+=--()()12122n n =-----=⎡⎤⎣⎦…………………………………………………10分(3)①先证数列{}n a 递增,利用反证法证明如下:假设k a 是{}n a 中第一个使1n n a a -≤的项,1221k k k a a a a a --<<<<≥ ,……………………………………………………12分 111,k k k k k A A a B B ---==≤111()()k k k k k k b b A B A B ----=---()()1110k k k k k k A A B B B B ---=-+-=-≤与数列{}n b 是公差大于0的等差数列矛盾.故数列{}n a 递增.……………………………………………………………………14分② 已证数列{}n a 递增,即12n a a a <<<< ,n n A a =;1n n B a +=,………………………………………………………………16分设若{}n b 的公差为b,则2111211111()()()()()()()n n n n n n n n n n n n n n n n a a a a a a a a A B A B b b b b b++++++++++---=--+-=--+-=-+=--=-故{}1n n a a +-是等差数列.………………………………………………………18分。