2013学年青浦区实验中学九年级第一学期月考模拟卷

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2013年上海市青浦区中考物理一模试卷

2013年上海市青浦区中考物理一模试卷

2013年上海市青浦区中考物理一模试卷一、单项选择题(共16分)下列各题均只有一个正确选项,请将所选选项的代号用2B铅笔填涂在答题纸的相应位置上,更改答案时,用橡皮擦去,重新填涂.1.一节干电池的电压为()A.1.5伏B.24伏C.36伏D.220伏2.下列实例中,利用连通器原理工作的是()A.吸尘器B.船闸C.温度计D.订书机3.一个中学生在游泳池里潜水时,受到的浮力约为()A.5牛B.50牛C.500牛D.5000牛4.首先测出大气压值的科学家是()A.托里拆利B.阿基米德C.奥斯特D.牛顿5.关于磁场,下列描述中错误的是()A.磁体的周围一定存在磁场B.磁场具有一定的方向C.磁场的周围存在磁感线D.磁场能对磁体产生作用6.在图所示的电路中,电源电压保持不变.闭合电键S后,将滑动变阻器滑片P向右移动时,变大的是()A.电表A示数与电表A1示数的差值B.电表V示数与电表A示数的比值C.电表V示数与电表A示数的乘积D.电表V示数与电表A1示数的比值7.图(a)、(b)所示电路的电源电压相等且不变,若电流表A1、A3的示数相等,则电阻R、R1、R2的大小关系有()A.R=R1+R2B.R=R1C.R>R2D.R<R28.如图所示,底面积不同的圆柱形容器A和B分别盛有甲、乙两种液体,两液面相平且甲液体的质量大于乙液体的质量.则此时液体对各自容器底部的压强p A、p B和液体对各自容器底部的压力F A、F B的关系是()A.p A<p B,F A=F B B.p A<p B,F A>F BC.p A>p B,F A=F B D.p A>p B,F A>F BVIP二、填空题(共28分)请将结果填入答题纸的相应位置.9.教室内日光灯、饮水器等用电器的电压为伏,日光灯与饮水器是的(选填“串联”或“并联”).10.马德堡半球实验有力地证明了是很大的;实验发现了通电直导线周围存在磁场;阿基米德原理指出浸在液体中物体所受浮力的大小等于它排开液体的.11.冰的密度为0.9×103千克/米3,若1千克的冰熔化成水,其不变,变小.(均选填“质量”、“体积”或“密度”)12.滑雪运动员脚蹬滑雪板、铁轨要铺设在枕木上等实例都是通过受力面积来压强的(均选填“增大”或“减小”).潜水员要穿上潜水服才能进入较深的水域中进行作业,则是因为水的深度越深,.13.同种材料制成的导体甲、乙,它们的横截面积相同,甲长10厘米、乙长2分米,甲的电阻乙的电阻;若将它们串联在电路中,则通过甲、乙的电流之比为.14.10秒内通过某导体横截面的电荷量为3库,通过它的电流为安;当该导体两端电压为6伏时,通过它的电流为0.6安,该导体的电阻为欧;当该导体两端的电压增加9伏时,则通过它的电流增加安.15.黄浦江是一条兼有饮用水源、航运、旅游等多种利用价值的河流,其水深可达17米.在江面下10米深处,水的压强为帕;若深度每增加1米,水的压强将增加帕.16.某油轮的排水量为5×106千克,其自身质量为0.6×106千克,最多可装载原油千克,当它满载时所受的浮力为牛;若该油轮装载原油后,排开水的体积为3×103米3,它所受的浮力为牛.17.在如图所示的电路中,电源电压保持不变.电阻R1、R2可能出现了断路故障,当电键S闭合后,三个电表中示数一定等于零的电表是表,一定大于零的电表是表.18.在通过实验得知通电螺线管磁场的强弱与通过的电流、螺线管的匝数有关的基础上,某兴趣小组继续研究螺线管磁场的强弱是否与插入的金属棒有关.实验中,他们始终保持电流不变,先后两次将形状大小相同的不同种金属棒(A:铸钢;B:铸铁)插入同一通电螺线管内时,发现插入金属棒的螺线管所能吸引大头针的数量不同,实验现象如图(a)、(b)所示.①观察、比较图(a)和(b)的实验现象及相关条件可得:在螺线管匝数和通过电流不变的情况下,.②实验后将铸钢棒取出,如图c所示发现它被磁化了.在教师指导下,他们用酒精喷灯对铸钢棒进行加热,在加热过程中发现它所能吸引的大头针逐渐减少,如图(c)、(d)、(e)所示.观察、比较图(c)、(d)和(e)的实验现象及相关条件可得:.三、作图题(共9分)请在答题纸的相应位置作图,作图必须使用2B铅笔.19.在图中,重为8牛的小球静止在水面上,用力的图示法画出该球所受的浮力F浮.20.根据图中通电螺线管的N极,标出磁感线方向、小磁针的N极,并在括号内标出电源的正、负极.21.在如图中根据标出的电流方向,从电池组、电压表、电流表三个元件的符号中选出两个元件符号,分别填进电路的空缺处,填进后要求灯泡L1和L2串联,且都能发光.四、计算题(共27分)请将计算过程和答案写入答题纸的相应位置.22.重为10牛的合金块浸没在水中,合金块排开水的体积为5×10-4米3,求:①合金块的浮力F浮.②合金块重力与浮力的合力F的大小与方向.23.在如图所示的电路中,电源电压为6伏且不变,电阻R2为15欧.①求电流表A2的示数I2.②闭合电键S时,有某个电流表的示数变化了0.6安,求此时通过电阻R1的电流I1及R1的电阻.24.如图所示,两个均匀的实心正方体甲和乙放置在水平地面上,甲的边长小于乙的边长.甲的质量为5千克,边长为0.1米.①求甲的密度ρ.②求甲对水平地面的压力F和压强p.③若甲、乙各自对水平地面的压强相等,现分别在两物体上沿竖直方向截去质量相同的部分并分别放在对方剩余部分的上方,此时甲、乙剩余部分对地面的压强分别为p甲′、p乙′,则p甲′:p乙′1(选填“大于”、“等于”或“小于”).(第③题不需要计算过程,请直接在空格内填写结果)25.在图所示的电路中,电源电压为12伏且保持不变,电阻R1的阻值为10欧,滑动变阻器R2上标有“50Ω2Α”字样.闭合电键S后,电路中的电流为0.3安.①求电阻R1两端的电压U1.②求此时变阻器R2两端的电压U2.③现有阻值分别为R、2R的定值电阻,最大阻值分别为4R、10R的滑动变阻器,请选择定值电阻、变阻器各一个分别替换R1、R2,要求:在移动变阻器滑片P的过程中,定值电阻两端电压的变化量最大.选择:阻值为的定值电阻、最大阻值为的滑动变阻器.求出:定值电阻两端电压的最大变化量△U1为120V11R.(第③题不需要计算过程,请直接在空格内填写结果)五、实验题(共20分)请根据要求在答题纸的相应位置作答.26.在电学实验中,连接电路时电键应处于状态,电压表应与被测电路(选填“串联”或“并联”).图所示的仪器可用来研究.27.“探究物质质量与体积关系”实验和“测定物质的密度”实验相比较,实验目的2830。

青浦区实验中学2013年度九年级第一阶段学业模拟测试

青浦区实验中学2013年度九年级第一阶段学业模拟测试

青浦区实验中学2013学年度九年级第一阶段学业模拟测试Part 2 V ocabulary and GrammarII. Choose the best answer (共20分)31. The Double Ninth Festival reminds us to respect _____ elderly,A) a B) an C) the D) /32. I hear our head teacher will be back fr om USA _____ three weeks’ time.A) at B) of C) in D) with33. The poor girl has to depend on _____ for living because her parents are both ill in bed.A)she B) her C) hers D) herself34. _____ of the money will go to the Red Cross to help those who need help.A) a B) an C) the D) /35. Tress on _____ sides of the street are decorated beautifully to welcome the great event.B)either B) both C) neither D) all36. Miss Fang's never been to Hainan Island for holiday. ___ Mr. Lin.A) Neither is B) Neither has C) So is D) So has37. Your grandpa usually wears glasses when he reads newspaper, ________ ?A) hasn't he B) has he C) doesn't he D) does he38. Do you think which means of transport is ________ , the underground or the plane?A) safe B) safer C) safest D) the safest39. The story took place in Italy many years ago, _______ many people still remember it.A) so B) or C) but D) and40. Will you help him ____ you see an old man lying in the middle of the street?A) though B) when C) because D) unless41. —____ do you study the Chinese paper cutting?---_____ By watching TV.C)How B) Where C) When D) Why42. The famous singer ____ seven tour concerts since this May.D)gives B) gave C) is giving D) has given43. More trees ___ to make our school beautiful.E)should plant B) should be plantF)C) should be planted D) should to be planted44. Would you mind _____ another cup of coffee?A) have B) having C) to have D) had45. --- Must I finish my homework before I go to play basketball?--- Yes, you ___A) needn’t B) can’t C) must D) may46. Mum told his son _____ too much time on computer games.G)not spend B) not spending C) don't spend D) not to spend47. There ____ only three little dogs and a small cat in that small pet shop.A) am B) is C) are D) be48. Could you tell me ______?H)when will the next bullet train leave for NanjingI)when the next bullet train will leave for NanjingJ)when would the next bullet train leave for NanjingK)when the next bullet train would leave for Nanjing49. --- Are you sure you can do well in tomorrow's test, Anna?---- _____ I’ve got everything ready.L)It's hard to sayM)I’m afraid notN)I think soO)I hope not50. —Jenny, you look busy. Anything I can do to help?----______P)Yes, I’m OK B) That's all rightC) Of course D)Thanks very much, but I can manage.III. Complete the following passage with the words or phrases in the box. Each word can only be used once. (8分)Q)remind B. Think of C. comfortable D. take notesR) E. photos F. carefully G interesting H. listen to I. Be yourselfWhat to do if you don't feel popular?●Try to take part in activities. Call your friends. Plan to do something. The worst feeling is staying home alone, because it makes you feel even more lonely.●Try to make new friends. Ch oose friends 51 , not just because you think they're popular. And remember, making good friends takes time. Choosing a popular person to be friends with is okay. If they are nice.●52 . If you want to become friends with someone who's popular, don't make yourself into someone you are not just to impress that person.●Be nice. Be friendly. Be outgoing(开朗的). But don't overdo it!Talk to trusted friends if you are feeling really bad. Or you might want to write it down in a journal(期刊)or diary.●Do something special for yourself. You could take 53 of your friends, or collect their school pictures, and take a collage(拼贴画)to hang on your wall. This will 54 you, when you are feeling unpopular, that you really do have friends.●55 your own ideas. These suggestions might not work for everyone.What to do if you have a problem with your teacher?●Talk to your parents, or another adult who will56 you and perhaps can help. Talk to your friends. Maybe they have had problems with the same teacher, too.●Don't give the teacher a reason to have problems with you. Do your work, complete your task, attend the class seriously and 57 . Maybe you just need to give the teacher a chance(机会).●Try to find out what the problem is.●Talk to th e teacher if you feel 58 doing this. Don't be nasty(闹别扭的), but express your concerns(关心). Listen to the teacher, as well.●Just accept the fact that you're not going to love all your teachers.IV. Complete the sentences with the given words in their suitable forms.59. Mr Smith has recently built a ____ bridge over his pool in the garden. (wood)60. It’s clear that your drawing is much better than ______. (I)61. Every student in Grade ______ is studying hard to achieve more. (ninth)62. Among the boys, Tom is the _______ to learn new things. (quick)63. At this year’s American Music Awards. Taylor Swift took home three trophies, ________- the top honour, Artist of the Year. (include)64. The kite flew higher and higher and soon _____ in the sky(appear)65. If you feel that it hurts you, it will _______ hurt the rest too. (probable)66. Almost everyone agreed that the trick was a great ______. (succeed)V. Rewrite the following sentences as required.67. There is some food and drinks in the fridge. (改为否定句)There is ____ food _______ drinks in the fridge.68. He likes reading the book because it is a present from his father. (对划线部分提问)_______ _________ he like reading the book?69. I saw a beautiful vase on show in Shanghai Museum last week. (改为感叹句)________ _________ beautiful vase on show I saw in Shanghai Museum lase week!70. We can't smell the air, but air is around us. (保持原句意思)_______ we __________ smell the air, it is around us.71. Damin taught his grandsons how they could catch more fish. (保持句意基本不变)Damin taught his grandsons _______ ________ catch more fish.S) A big earthquake in Japan killed thousands of people in March, 2011. (改为被动语态) Thousands of people _____ ________ by a big earthquake in Japan in March, 2011.73. He did a part-time job. He wanted to support his family. (合并为一句)He did a part-time job ________ _________ he could support his family.Part 3 Reading and WritingVI. Reading comprehension.T)Choose the best answer.Elena jogged with her dad in some empty lots(小块地) at the edge of town. These out-of-the-way lots were good places to jog. But the lots were full of rocks and rubbish.One day, Elena saw a sign on the lots. It said, “New City Park.” Elena was excited. Maybe the park would have a jogging trail!Elena told her parents about the sign.“I read about the park in the newspaper,” said Elena’s father. “The city wants volunteers to help clean the area. Then the city will build a soccer field and a playground.”“Will they build a jogging trail?” Elena asked.“No, the paper didn’t mention a jogging trail.”“Well, perhaps if I help clean up the lots, they’ll put in a jogging trail,” said Elena.On Saturday Elena went to the lots. Many adults were working there. Some were picking up rubbish. Some were moving rocks.“I want to volunteer,” Elena told a man. “What can I do?”“This is not a job for children.” said the man. “You’d better go home.”Elena looked around. She could carry rubbish. She could move rocks. A lady was pulling a bag of rubbish. Elena saw a box of rubbish bags. She took an empty bag.“I’ll take this bag to the Dumpster,” said Elena. “Here’s a new bag.”“Thanks,” said the lady. “Keep those new bags coming.”Elena dragged rubbish bags. She told everyone she was glad to help build the park. Sheexplained that she hoped the park would have a jogging trail.Elena went to the area every week. Soon everyone knew her and she told everyone how wonderful a jogging trail would be.Soon the volunteers cleared the lots. The city could start building. The mayor(市长) gave a speech. He showed a drawing of the new park. He told them about the soccer field and the basketball diamond. There would be picnic tables. There would be a playground. The park would even have basketball courts. There was no sign of a jogging trail.The mayor thanked the volunteers. Then he asked if they had any suggestions. The volunteers said they wanted to build a jogging trail around the park.They told the mayor how helpful Elena had been and how she wanted a jogging trail. The mayor said, “A jogging trail won’t be hard to build. Let’s do it!”Everyone clapped and cheered. Elena cheered loudest of all.74. Where did Elena jog with her father at the beginning of the story?U)In a park B) On a jogging trailC) In some empty lots D) In center of the town75. How did Elena feel when she saw the sign on the lots?A)Happy and proud B) Excited and hopefulC) Sad and shy D) Unhappy and disappointed76. How did Elena know the city needed volunteers to help clean the lots?A)Her father told her about it B) She read it in the newspaperC) She learned it from a sign D) She was told by other volunteers.77. Why did Elena want to help clean the lots?V)She wanted to play on the playground.W)She wanted her name in the newspaperC) She wanted to work with other volunteers.D) She wanted the city to build a jogging trail.78. Why did the volunteers tell the mayor they wanted to build a jogging trail?A) They needed a park to jog around.B) They wanted to thank Elena for her help.C) They thought it was a good place for jogging.D) They hoped the mayor could jog with them every day.79. What can be the best title for the story.X)Elena and a Mayor B)Elena and New City ParkC) Elena and Ssome V olunteers D) Elena and a Jogging TrailB. Choose the words or expressions and complete the passage.Tim Becker and his neighbors are doing something to make their neighborhood a trouble-free area.When Tim Becker gets in his car to go shopping, he doesn't just drive to a store and back home. He always looks 80 up and down the streets of his neighborhood. He looks for anything unusual , such as strange cars, loud noises, 81 windows, or people gathering on street corners.Tim belonged to a neighborhood watch group in Stoneville, Indiana, USA. The neighborhood watch group meet on the third Wednesday of every month. That's 82 Tim get together with about ten of his neighbors to discuss community safety . Members of the neighborhood watch group want to help the police 83 their homes, streets, and families safe.Tim Stedman, president of 10 neighborhood watch group, agrees with Tim. "People seem to think that crime(犯罪) happens to other people but not to them. Well, it's 84 happened to me," he said, " but I don't think anyone has the right to steal from other people or to make them feel unsafe sitting in their own homes." Alex, a member of the group, said that all the neighbors look out for one another. "We watch each other's homes. We keep watch on the neighborhood at night and on weekends. Usually a group of four or five of us goes out together. If something doesn't look right, then we call the police . For example, if we notice a group of teenagers who seem to be looking for 85 , or someone destroying property, we report to the police."Alex feels the neighborhood watch group help a lot in keeping crime down. Her husband Jim agrees, "Police are good people, but they can't do everything ."80. A) carefully B) clearly C) happily D) widely81. A) dirty B) nice C) old D) broken82. A) where B) why C) why D) how83. A) keep B) hold C) let D)protect84. A) ever B) never C) often D) sometimes85. A) work B) people C) service D) troubleC. Reading the passage and fill in the blanks with proper words.As we know, museums are buildings where many valuable and important objects are kept so that people can go and see them. For example, art museums are places where people can learn about d___86___ kinds of cultures. More and more popular "design museums" that are opening today, however, perform quite a different role. Unlike most art museums, the design museum shows objects that are e 87 found in our daily life, such as fridges and washing machines. The advantage of design museums is that they are places where people feel familiar w 88 the exhibits(展品).Being different from the art museum visitors, design museum visitors s___89___ feel frightened or puzzled. This is partly because design museums clearly show how and why mass-products (批量产品) work and look as they do, and how design has i__90____ the quality of our lives. Art museum exhibits, on the other hand, would most probably fill visitors with a feeling that there is something beyond their understanding.Several new design museums have opened their doors in recent years. Each of these museums has tried to satisfy (满足) the public's growing i 91 in the field with new ideas. London's Design Museum, for example, shows a collection of mass-produced objects from electric typewriters to a group of Italian fish-tins. The choices open to design museums seem l__92___ strict than those to art museums, and visitors may also sense (感觉到) the humorous part of our society while walking around such exhibits as interesting and unusually attractive toys collected from our everyday life.D. Answer the questions.When Matty Sallin, 34, was working on a degree in art and technology at New York University, he got an interesting assignment in electronics class: Create something for the household. He decided to create an alarm clock.“To get up in time, many people have to use an alarm clock every morning. But the sound of the clock is extremely noisy and unpleasant,” he says. So before he started to design the new clock, he asked different people what they'd like to wake up in the morning. A lot of them said, " The smell of bacon(培根肉).”So Sallin and two classmates invented a new kind of alarm clock: a wooden box with a pig faceand a digital clock that uses the smell of cooking bacon rather than sound to wake someone up. He explains, "There's no danger of burning, because I built it carefully. It uses halogen light bulbs instead of a flame for cooking and turns off automatically after ten minutes." Just a few easy steps are required to set the "alarm.""What you do is put a couple of frozen strips in the night before," says Sallin. Bacon is cured, or preserved, so there is no danger of it spoiling overnight."If you set the alarm for 8:00, it will turn on at 7:50 and slow cook for ten minutes under the halogen bulbs," he says. Then the bulbs turn off and a fan blows the scent out through the nostrils of the pig."So the pleasant smell of bacon will wake you up. There will be no more noisy alarms,” says Sallin. “They you can open the door on the side and pull the bacon out and eat it.”When Sallin was a kid, he spent a lot of time making drawings of inventions. "I wanted to make an elevator in my back yard and a special tree house," he says. "But I never really thought I'd be called an inventor!"Sallin got an A in the class and went on to other things—but people continue to hear about his invention and e-mail him every day asking where they can buy his alarm clock.If he decides to produce and sell his aroma alarm, maybe he can build a successful business out of it.Y)What was Matty Sallin?Z)How many people worked together on the new alarm clock?AA)How long did it take to cook the bacon?BB)W hy are light bulbs used for cooking bacon instead of a fire? BecauseCC)W hat's special about Sallin's alarm clock?_____________ wakes people up instead of the loud alarm.98. What do you think of the new alarm clock?VII. Writing99.Write at least 60 words about the topic “ I felt _________ at that moment”.(以“在那一刻我感到……”为题写一篇不少于60个词的短文,标点符号不占格)Use the following points as a reference.(以下问题仅供参考)DD)What happened at that moment?EE)How did you feel at that moment? Why?参考答案:31-35: CCDAB 36-40:BCBCB41-45:ADCBC 46-50:DCBCD51-58:FIEABHDC59-66: wooden, mine, Nine, quickest, including, disappeared, probably, success.67: no, or 68: Why does 69: What a70 :Though can't 71: how to 72: so that74-79: CBADBC 80-85: ADCABD86-92: different, easily, with, sometimes, influenced, interesting, less93:He was a student studying art and technology in New York University.94:Three people.95:Ten minutes.96:Because it can turns of f automatically and there's no danger of burning.97:The pleasant smell of bacon98: Any reasonable reason is acceptable.。

青浦一中2013年第一学期9年级10月月考(附答案)

青浦一中2013年第一学期9年级10月月考(附答案)

青浦一中2013年第一学期9年级10月月考(青浦校区提供)Part 2 Vocabulary and Grammar (第二部分词汇和语法)V. Choose the better answer (选择最恰当的答案) (共20分)31. Nowadays more and more people are happy to help______ old.A. aB. theC. anD. /32.Your bedroom is very big, but _____ is rather small.A. sheB. herC. herselfD. hers33. There aren’t any ______ in the fridge. Let’s go and buy some peas, carrots and cabbages.A. meatB.fruitC. vegetablesD. eggs34.—Do you like talking with your friends on the telephone or the mobile phone?--_______. I enjoy using QQ.A. NeitherB. EitherC. BothD. None35. The Trojans had to drag the wooden horse into the city with ropes, ______ they?A. didB. didn’tC. hadD. hadn’t36. You _______ worry about difficulties. We will all help you.A. needn’tB. needn’t toC. don’t needD. not need37. Good service ______ by China during the Expo in 2010.A. providedB. was providedC. has been providedD. has provided38. Mr. and Mrs. Wang are saving money _____ they can pay for their son’seducation in the future.A. so thatB. in order toC. thoughD. because39. Without the help of computers, we can’t do some difficult work_________.A. easilyB.easyC. easierD. much easily40. The boy made the baby ______ by making a face at him.A. to laughB. laughedC. laughingD. laugh41. To prevent the spread of H1N1, one good way is to keep windows _______ so as to keep the indoor air fresh.A. openedB. openC. openingD. to open42. All the students went to the movie _______ Peter. He looked after his mother at home.A. besideB. besidesC. exceptD. except for43. When I was young, my dream was ______ around the world.A. travelB. to travelC. travelsD. traveled44. The professor ______ hometown for many years.A. leftB. has been leftC. has leftD. has been away from45. _____ fun it is to have a picnic on a sunny day!A. WhatB. What aC. HowD. How a46. He felt ______ and hurriedly left the office without saying anything.A. friendlyB. gentlyC. angryD. angrily47. The traffic will be much better if everybody ______ the rules.A. obeysB. will obeyC. obeyedD. is obeying48. --- Why didn’t you come to his birthday party yesterday afternoon?--- I _____ a wonderful football match at that time.A. watchB. am watchingC. was watchingD. will watch49. Winter holidays are coming! Can you tell me_____?A. how you will spend your winter holidaysB. how will you spend your winter holidaysC. how your winter holidaysyou will spendD.how your winter holidays will you spend50. A piece of grass it tied around the cormorant’s neck _____ it can’t swallow the fish.A. soB.becauseC. untilD. so that51. His English improved after he stayed with the British family for 3 months.The underlined part in the sentence means “______”.A.became largerB. became betterC. became easierD. became different52. --- Will you help me with the luggage? It’s too heavy.---________.A. With pleasureB. Never mindC. Not at allD. That’s right.53. --- Tony, I’m terribly sorry. I’ve forgotten to bring your dictionary back to you today.--- _________.A. I’m sorry to hear thatB. You’re welcomeC. Not at allD. That’s all right.VI. Complete the following passage with the words or phrases in the box. Each can only be used once (将下列单词或词组填入空格。

上海青浦区实验中学初三化学初三化学上册期中模拟试题(含答案)

上海青浦区实验中学初三化学初三化学上册期中模拟试题(含答案)

上海青浦区实验中学初三化学上册期中模拟试题(含答案)一、选择题(培优题较难)1.实验小组用如图装置测定空气中的氧气含量,锥形瓶中空气的体积为100mL,注射器中水的体积为25 mL,装置气密性良好。

下列说法不正确的是A.气球的作用是缓冲装置中的压强变化B.瓶底的细沙可以防止红磷燃烧时集气瓶炸裂C.红磷熄灭后应该等装置冷却到室温,再打开弹簧夹观察现象D.打开弹簧夹后,注射器中的水一定全部进入锥形瓶中2.四种物质在一定的条件下充分混合反应,测得反应前后各物质的质量分数如图所示.则有关说法中不正确的()A.丁一定是化合物B.乙可能是这个反应的催化剂C.生成的甲、丙两物质的质量比为8: 1D.参加反应的丁的质量一定等于生成甲和丙的质量之和3.豆腐是人们喜爱的食物,营养丰富,能为人体提供所需的多种氨基酸,其中含量最多的是亮氨酸(C6H13NO2),关于亮氨酸的说法正确的是()A.亮氨酸是氧化物B.亮氨酸中碳元素的质量分数为27.3%C.一个亮氨酸分子由22个原子构成D.亮氨酸中碳、氢、氮、氧四种元素的质量比为6:13:1:24.下列有关实验操作的“先”与“后”的说法中,正确的是()A.制取气体时,先装药品,后检查装置的气密性B.用托盘天平称量10g固体药品时,先放砝码,后放药品C.用滴管取细口瓶内的液体时,先将其伸入液体内,后挤压取液D.加热KMnO4并用排水法收集O2的实验结束时,先熄灭酒精灯,后移出导管5.碳酸乙烯酯(C3H4O3)可用作锂电池电解液,下列有关碳酸乙烯酯的说法正确的是A.碳酸乙烯酯的相对分子质量为(12×3+1×4+16×3)gB.碳酸乙烯酯中C、H、O三种元素的质量比为12:1:16C.碳酸乙烯酯中C、H、O三种原子的个数比为3:4:3D.碳酸乙烯酯中碳元素的质量分数= ×100%6.用如图装置测定空气中氧气的含量,下列说法正确的是( )A.点燃白磷后缓慢将燃烧匙插入瓶中,塞紧瓶塞B.用细铁丝代替白磷进行实验,不会影响实验结果C.不打开瓶塞,聚焦太阳光引燃足量白磷,实验效果更好D.若实验过程中,没有将弹簧夹夹紧,不会影响实验结果7.下列事实不能作为相应观点的证据的是( )A.尘土飞扬,说明分子是运动的B.电解水得到氢气和氧气,说明分子是可分的C.气体被压缩后体积发生了较大变化,说明气体分子间距较大D.将两个干净平整的铅柱紧压在一起会结合起来,说明分子间存在引力8.河水净化的主要步骤如下图所示。

实验中学2013学年第一学期九年级月考数学201310

实验中学2013学年第一学期九年级月考数学201310

一、选择题(每小题4分,共48分)1.在Rt ABC△中,ACB∠=Rt∠,1BC=,2AB=,则下列结论正确的是()A.tan3B= B.1tan2A= C.3cos2B= D.3sin2A=2.如图,AB是O⊙的直径,点C、D在O⊙上,110BOC∠=°,AD OC∥,则AOD∠=()A.70°B.60°C.50°D.40°3.二次函数xxy22-=与坐标轴交点个数为()A.3 B.2 C.1 D.04.已知圆锥的侧面展开图的圆心角为120°,则这个圆锥的侧面积是底面积的()A.3倍 B.2倍 C.31D.215.如图,在平行四边形ABCD中,E为CD上一点,连接AE、BD,且AE、BD交于点F,S△DEF:S△ABF=4:25,则DE:EC=()A.2:5 B.2:3 C.3:5 D.3:2(第2题)(第5题)(第6题)6.如图,将边长为2cm的正方形ABCD沿直线l向右翻动(不滑动),当正方形连续翻动8次后,正方形的中心O经过的路线长是()cm.A.8B.8 C.3πD.4π7.直角三角形纸片的两直角边长分别为6,8,现将ABC△如图那样折叠,使点A与点B重合,折痕为DE,则tan CBE∠的值是()A.247B.73C.724D.136 8CEABD第7题慈溪实验中学九年级数学阶段性测试试卷2013学年第一学期8.二次函数2()y a x m n=++的图象如图,则一次函数y mx n=+的图象经过()A.第一、二、三象限B.第一、二、四象限C.第二、三、四象限D.第一、三、四象限(第8题)(第9题)(第10题)9.如图,在△ABC中,以BC为直径的圆分别交边AC、AB于D、E两点,连接BD、DE.若BD平分∠ABC,则下列结论不一定成立的是()A.BD⊥ACB.AC2=2AB·AEC.△ADE是等腰三角形D. BC=2AD.10.如图所示,在⊙O内有折线OABC,其中OA=8,AB=12,∠A=∠B=60°,则BC 的长为()A.19 B.16 C.18 D.2011.小敏在作⊙O的内接正五边形时,先做了如下几个步骤:(1)作⊙O的两条互相垂直的直径,再作OA的垂直平分线交OA于点M,如图1;(2)以M为圆心,BM长为半径作圆弧,交CA于点D,连结BD,如图2.若⊙O的半径为1,则由以上作图得到的关于正五边形边长BD的等式是()A.BD2=OD B.BD2=OD C.BD2=OD D.BD2=OD12.如图,AB是半圆直径,半径OC⊥AB于点O,AD平分∠CAB交弧BC于点D,连结CD、OD,给出以下四个结论:①AC∥OD;②OECE=;③△ODE∽△ADO;④ABCECD⋅=22.其中正确结论的个数是().A.1个B.2个C.3个D.4个第11题第12题ADCOE二、填空题(第13—18题,每题4分,共24分) 13.若2:a = a :8,则a = 。

上海市青浦区中考物理一模(即期末)试题(扫描版) 沪科版

上海市青浦区中考物理一模(即期末)试题(扫描版) 沪科版

(17)不规则物体丁
(19)F(牛)
(20)F(牛)
8
9.(1)220;
(2)并联。
10.(3)大气压; (4)奥斯特;
(5)重力大小。
11.(6)1;
(7)1×10-3。
12.(8)增大;
(9)减小;
(10)水内部的压强越大。
13.(11)小于;
(12)1∶1。
二、28 分
14.(13)0.3;
(14)10;
(15)0.9。
15.(16)9.8×104; (17)9.8×103。
上海市青浦区 2013 年中考物理一模(即期末)试题(扫描版) 沪科 版
1
2
3
4
5
答案和评分参考
13.01
题号
答案及评分参考
6
一、16 分
1.A。 2.B。 3.C。 4.B。 5.C。 6.D。 7.C。 8.D。
说明:第 1—20 格,每格 1 分;第 21—24 格,每格 2 分。
26.(1)断开;
(2)并联;
(3)液体内部压强的规律。
27.(4)不同;
(5)相同;
(6)左。
28.(7)略; 29.(10)B
(8)L1;
(9)干路。
(11) 物理量
序号 1 2 3
电压 (伏)
4.5
电流 (安)
0.2
电阻 (欧)
12.1 12.5 12.5
说明:第 12—15 格,每格 1 分;第 16、17 格,共 1 分;第 18、19、20 格,共 2
16.(18)4.4×106; (19)4.9×107;
(20)2.94×107。
17.(21)A1;

上海青浦区实验中学九年级上册压轴题数学模拟试卷含详细答案

上海青浦区实验中学九年级上册压轴题数学模拟试卷含详细答案

上海青浦区实验中学九年级上册压轴题数学模拟试卷含详细答案一、压轴题1.如图,在平面直角坐标系xOy 中,已知直线AB 经过点A (﹣2,0),与y 轴的正半轴交于点B ,且OA =2OB .(1)求直线AB 的函数表达式;(2)点C 在直线AB 上,且BC =AB ,点E 是y 轴上的动点,直线EC 交x 轴于点D ,设点E 的坐标为(0,m )(m >2),求点D 的坐标(用含m 的代数式表示);(3)在(2)的条件下,若CE :CD =1:2,点F 是直线AB 上的动点,在直线AC 上方的平面内是否存在一点G ,使以C ,G ,F ,E 为顶点的四边形是菱形?若存在,请求出点G 的坐标;若不存在,请说明理由.2.已知函数1221,(21)1y x m y m x =+-=++均为一次函数,m 为常数.(1)如图1,将直线AO 绕点()1,0A -逆时针旋转45°得到直线l ,直线l 交y 轴于点B .若直线l 恰好是1221,(21)1y x m y m x =+-=++中某个函数的图象,请直接写出点B 坐标以及m 可能的值;(2)若存在实数b ,使得||(1)10m b b ---=成立,求函数1221,(21)1y x m y m x =+-=++图象间的距离;(3)当1m 时,函数121y x m =+-图象分别交x 轴,y 轴于C ,E 两点,(21)1y m x =++图象交x 轴于D 点,将函数11y y y =的图象最低点F 向上平移5621m +个单位后刚好落在一次函数121y x m =+-图象上,设12y y y =的图象,线段OD ,线段OE 围成的图形面积为S ,试利用初中知识,探究S 的一个近似取值范围.(要求:说出一种得到S 的更精确的近似值的探究办法,写出探究过程,得出探究结果,结果的取值范围两端的数值差不超过0.01.)3.如图,抛物线214y x bx c =-++经过点()6,0C ,顶点为B ,对称轴2x =与x 轴相交于点A ,D 为线段BC 的中点.(1)求抛物线的解析式;(2)P 为线段BC 上任意一点,M 为x 轴上一动点,连接MP ,以点M 为中心,将MPC 逆时针旋转90︒,记点P 的对应点为E ,点C 的对应点为F .当直线EF 与抛物线214y x bx c =-++只有一个交点时,求点M 的坐标. (3)MPC 在(2)的旋转变换下,若2PC =(如图).①求证:EA ED =.②当点E 在(1)所求的抛物线上时,求线段CM 的长.4.在平面直角坐标系中,抛物线y =ax 2+bx ﹣3过点A (﹣3,0),B (1,0),与y 轴交于点C ,顶点为点D .(1)求抛物线的解析式;(2)点P 为直线CD 上的一个动点,连接BC ;①如图1,是否存在点P ,使∠PBC =∠BCO ?若存在,求出所有满足条件的点P 的坐标;若不存在,请说明理由;②如图2,点P 在x 轴上方,连接PA 交抛物线于点N ,∠PAB =∠BCO ,点M 在第三象限抛物线上,连接MN ,当∠ANM =45°时,请直接写出点M 的坐标.5.已知点P(2,﹣3)在抛物线L :y =ax 2﹣2ax+a+k (a ,k 均为常数,且a≠0)上,L 交y 轴于点C ,连接CP .(1)用a 表示k ,并求L 的对称轴及L 与y 轴的交点坐标;(2)当L 经过(3,3)时,求此时L 的表达式及其顶点坐标;(3)横、纵坐标都是整数的点叫做整点.如图,当a <0时,若L 在点C ,P 之间的部分与线段CP 所围成的区域内(不含边界)恰有4个整点,求a 的取值范围;(4)点M(x 1,y 1),N(x 2,y 2)是L 上的两点,若t≤x 1≤t+1,当x 2≥3时,均有y 1≥y 2,直接写出t 的取值范围.6.如图1,抛物线24y ax bx =+-与x 轴交于(3,0)A -、(4,0)B 两点,与y 轴交于点C ,作直线BC .点D 是线段BC 上的一个动点(不与B ,C 重合),过点D 作DE x ⊥轴于点E .设点D 的横坐标为(04)m m <<.(1)求抛物线的表达式及点C 的坐标;(2)线段DE 的长用含m 的式子表示为 ;(3)以DE 为边作矩形DEFC ,使点F 在x 轴负半轴上、点G 在第三象限的抛物线上. ①如图2,当矩形DEFC 成为正方形时,求m 的值;②如图3,当点O 恰好是线段EF 的中点时,连接FD ,FC .试探究坐标平面内是否存在一点P ,使以P ,C ,F 为顶点的三角形与FCD ∆全等?若存在,直接写出点P 的坐标;若不存在,说明理由.7.已知抛物线y =ax 2+bx+c(a >0),顶点D 在y 轴上,与x 轴的一个交点的横坐标为6.(1)求a 、c 满足的关系式;(2)若直线y =kx-2a 与抛物线交于A 、B 两点(点A 在点B 左侧),以AB 为直径的圆恒过点D .①求抛物线的解析式;②设直线y =kx-2a 与y 轴交于点M 、直线l 1:y =px+q 过点B ,且与抛物线只有一个公共点,过点D 作x 轴的平行线l 2,l 1与l 2交于点N .分别记BDM 、NDM 的面积为S 1,S 2,求12S S . 8.如图1,抛物线221y x x =-+-的顶点A 在x 轴上,交y 轴于B ,将该抛物线向上平移,平移后的抛物线与x 轴交于,C D ,顶点为()1,4E .(1)求点B 的坐标和平移后抛物线的解析式;(2)点M 在原抛物线上,平移后的对应点为N ,若OM ON =,求点M 的坐标;(3)如图2,直线CB 与平移后的抛物线交于F .在抛物线的对称轴上是否存在点P ,使得以,,C F P 为顶点的三角形是直角三角形?若存在,直接写出点P 的坐标;若不存在,请说明理由.9.如图,在平面直角坐标系中,已知抛物线2y x bx c =++与直线AB 相交于A ,B 两点,其中()3,4A --,()0,1B -.(1)求该抛物线的函数表达式;(2)点P 为直线AB 下方抛物线上的任意一点,连接PA ,PB ,求PAB △面积的最大值; (3)将该抛物线向右平移2个单位长度得到抛物线()211110y a x b x c a =++≠,平移后的抛物线与原抛物线相交于点C ,点D 为原抛物线对称轴上的一点,在平面直角坐标系中是否存在点E ,使以点B ,C ,D ,E 为顶点的四边形为菱形,若存在,请直接写出点E 的坐标;若不存在,请说明理由.10.如图①,在ABC 中,AB AC =,BAC α∠=,点D 、E 分别在边AB 、AC 上,AD AE =,连接BE ,点M 、P 、N 分别为DE 、BE 、BC 的中点.(1)观察猜想:图①中,线段PM 与PN 的数量关系是_____________,用含α的代数式表示MPN ∠的度数是________________________;(2)探究证明:把ADE 绕点A 顺时针方向旋转到图②的位置,连接MN ,BD ,CE ,当120α=︒时,判断PMN 的形状,并说明理由;(3)拓展延伸:把ADE 绕点A 在平面内任意旋转,若90α=︒,3AD =,7AB =,请直接写出线段MN 的最大值和最小值.11.新定义:在平面直角坐标系中,过一点分别作坐标轴的垂线,若与坐标轴围成的长方形的周长与面积相等,则这个点叫做“和谐点”.例如,如图①,过点P 分别作x 轴、y 轴的垂线,与坐标轴围成长方形OAPB 的周长与面积相等,则点P 是“和谐点”.(1)点M (1,2)_____“和谐点”(填“是”或“不是”);若点P (a ,3)是第一象限内的一个“和谐点”,3x a y =⎧⎨=⎩是关于x ,y 的二元一次方程y x b =-+的解,求a ,b 的值. (2)如图②,点E 是线段PB 上一点,连接OE 并延长交AP 的延长线于点Q ,若点P (2,3),2OBE EPQ S S ∆∆-=,求点Q 的坐标;(3)如图③,连接OP ,将线段OP 向右平移3个单位长度,再向下平移1个单位长度,得到线段11O P .若M 是直线11O P 上的一动点,连接PM 、OM ,请画出图形并写出OMP ∠与1MPP ∠,1MOO ∠的数量关系.12.如图所示,在Rt ABC ∆中,90B ∠=︒,43BC =,30C ∠=︒,点D 从点C 出发沿CA 方向以每秒2个单位长度的速度向点A 匀速运动,同时点E 从点A 出发沿AB 方向以每秒1个单位长度的速度向点B 匀速运动,当其中一点到达终点时,另一个点也随之停止运动.设点D 、E 运动的时间是t 秒(0)t >,过点D 作DF BC ⊥于点F ,连接DE 、EF .(1)求证:AE DF =;(2)四边形AEFD 能够成为菱形吗?若能,求出t 的值;若不能,请说明理由; (3)当t =________时,DEF ∆为直角三角形.13.如图,抛物线y =mx 2﹣4mx+2m+1与x 轴交于A (x 1,0),B (x 2,0)两点,与y 轴交于点C ,且x 2﹣x 1=2.(1)求抛物线的解析式;(2)E是抛物线上一点,∠EAB=2∠OCA,求点E的坐标;(3)设抛物线的顶点为D,动点P从点B出发,沿抛物线向上运动,连接PD,过点P做PQ⊥PD,交抛物线的对称轴于点Q,以QD为对角线作矩形PQMD,当点P运动至点(5,t)时,求线段DM扫过的图形面积.14.如图1,抛物线M1:y=﹣x2+4x交x正半轴于点A,将抛物线M1先向右平移3个单位,再向上平移3个单位得到抛物线M2,M1与M2交于点B,直线OB交M2于点C.(1)求抛物线M2的解析式;(2)点P是抛物线M1上AB间的一点,作PQ⊥x轴交抛物线M2于点Q,连接CP,CQ.设点P的横坐标为m,当m为何值时,使△CPQ的面积最大,并求出最大值;(3)如图2,将直线OB向下平移,交抛物线M1于点E,F,交抛物线M2于点G,H,则EGHF的值是否为定值,证明你的结论.15.如图,在平面直角坐标系xOy中,直线y=12x+2与x轴交于点A,与y轴交于点C.抛物线y=ax2+bx+c的对称轴是x=32且经过A、C两点,与x轴的另一交点为点B.(1)求抛物线解析式.(2)若点P为直线AC上方的抛物线上的一点,连接PA,PC.求△PAC的面积的最大值,并求出此时点P的坐标.(3)抛物线上是否存在点M,过点M作MN垂直x轴于点N,使得以点A、M、N为顶点的三角形与△ABC相似?若存在,求出点M的坐标;若不存在,请说明理由.16.在平面直角坐标系中,抛物线2y ax bx c =++经过点A 、B 、C ,已知A (-1,0),B (3,0),C (0,-3).(1)求此抛物线的函数表达式;(2)若P 为线段BC 上一点,过点P 作y 轴的平行线,交抛物线于点D ,当△BCD 面积最大时,求点P 的坐标;(3)若M (m ,0)是x 轴上一个动点,请求出CM+12MB 的最小值以及此时点M 的坐标.17.在平面直角坐标系xoy 中,点A (-4,-2),将点A 向右平移6个单位长度,得到点B .(1)若抛物线y =-x 2+bx +c 经过点A ,B ,求此时抛物线的表达式;(2)在(1)的条件下的抛物线顶点为C ,点D 是直线BC 上一动点(不与B ,C 重合),是否存在点D ,使△ABC 和以点A ,B ,D 构成的三角形相似?若存在,请求出此时D 的坐标;若不存在,请说明理由;(3)若抛物线y =-x 2+bx +c 的顶点在直线y =x +2上移动,当抛物线与线段AB 有且只有一个公共点时,求抛物线顶点横坐标t 的取值范围.18.如图①,在矩形ABCD 中,3AB =cm ,AD AB >,点E 从点A 出发,沿射线AC 以a (cm/s)的速度匀速移动.连接DE ,过点E 作EF DE ⊥,EF 与射线BC 相交于点F ,作矩形DEFG ,连接CG .设点E 移动的时间为t (s),CDE ∆的面积为S (cm 2), S 与t 的函数关系如图②所示.(1) a = ;(2)求矩形DEFG 面积的最小值;(3)当CDG ∆为等腰三角形时,求t 的值.19.如图,在平面直角坐标系中,四边形ABCD 的顶点A 、B 在函数(0)m y x x =>的图象上,顶点C 、D 在函数(0)n y x x=>的图象上,其中0m n <<,对角线//BD y 轴,且BD AC ⊥于点P .已知点B 的横坐标为4.(1)当4m =,20n =时,①点B 的坐标为________,点D 的坐标为________,BD 的长为________.②若点P 的纵坐标为2,求四边形ABCD 的面积.③若点P 是BD 的中点,请说明四边形ABCD 是菱形.(2)当四边形ABCD 为正方形时,直接写出m 、n 之间的数量关系.20.如图,⊙O 经过菱形ABCD 的三个顶点A 、C 、D ,且与AB 相切于点A .(1)求证:BC为⊙O的切线;(2)求∠B的度数.(3)若⊙O半径是4,点E是弧AC上的一个动点,过点E作EM⊥OA于点M,作EN⊥OC 于点N,连接MN,问:在点E从点A运动到点C的过程中,MN的大小是否发生变化?如果不变化,请求出MN的值;如果变化,请说明理由.【参考答案】***试卷处理标记,请不要删除一、压轴题1.(1)y=12x+1;(2)2m,0m2D⎛⎫⎪-⎝⎭;(3)(2,4)或(﹣2,2)或523(,)36【解析】【分析】(1)利用待定系数法即可解决问题;(2)求出点C坐标,利用待定系数法求出直线DE的解析式即可解决问题;(3)求出点E坐标,分两种情形分别讨论求解即可;【详解】(1)∵A(﹣2,0),OA=2OB,∴OA=2,OB=1,∴B(0,1),设直线AB的解析式为y=kx+b,则有201k bb-+=⎧⎨=⎩解得1 k2 b1⎧=⎪⎨⎪=⎩∴直线AB的解析式为y=12x+1.(2)∵BC=AB,A(﹣2,0),B(0,1),∴C(2,2),设直线DE 的解析式为y =k ′x +b ′,则有22k b b m '''+=⎧⎨=⎩解得2m k 2b m -⎧'='=⎪⎨⎪⎩∴直线DE 的解析式为22m y x m -=+ 令y =0,得到2m m 2x =- ∴2m ,0m 2D ⎛⎫ ⎪-⎝⎭(3)如图1中,作CF ⊥OD 于F .∵CE :CD =1:2,CF ∥OE ,∴23CF CD OE DE == ∵CF =2,∴OE =3.∴m =3.∴E (0,3),D (6,0),①当EC 为菱形ECFG 的边时,F (4,3),G (2,4)或F ′(0,1),G ′(﹣2,2). ②当EC 为菱形EF″CG ″的对角线时,F ″G ″垂直平分线段EC ,易知直线DE 的解析式为132y x =-+,直线G ″F ″的解析式为122y x =+由122112y x y x ⎧=+⎪⎪⎨⎪=+⎪⎩,解得1376x y ⎧=⎪⎪⎨⎪=⎪⎩∴F ″17,36⎛⎫ ⎪⎝⎭, 设G ″(a ,b ),则有17023236,2222a b ++++== ∴523,36a b == ∴G ″523,36⎛⎫ ⎪⎝⎭【点睛】本题考查一次函数综合题、平行线分线段成比例定理、菱形的判定和性质、待定系数法等知识,解题的关键是灵活运用所学知识解决问题,学会用分类讨论的思想思考问题,属于中考压轴题.2.(1)(0,1);1或0 (2(3)348131200010S << 【解析】【分析】(1)由题意,可得点B 坐标,进而求得直线l 的解析式,再分情况讨论即可解的m 值; (2)由非负性解得m 和b 的值,进而得到两个函数解析式,设1y 与x 轴、y 轴交于T ,P ,2y 分别与x 轴、y 轴交于G ,H ,连接GP ,TH ,证得四边形GPTH 是正方形,求出GP 即为距离;(3)先根据解析式,用m 表示出点C 、E 、D 的坐标以及y 关于x 的表达式为()221221421y y y m x m x m =⋅+++-=,得知y 是关于x 的二次函数且开口向上、最低点为其顶点()222212,2121m m F m m ⎛⎫- ⎪-- ⎪++⎝⎭,根据坐标平移规则,得到关于m 的方程,解出m 值,即可得知点D 、E 的坐标且抛物线过D 、E 点,观察图象,即可得出S 的大体范围,如:ODE S S <,较小的可为平行于DE 且与抛物线相切时围成的图形面积.【详解】解:(1)由题意可得点B 坐标为(0,1),设直线l 的表达式为y=kx+1,将点A (-1,0)代入得:k=1,所以直线l 的表达式为:y=x+1,若直线l 恰好是121y x m =+-的图象,则2m-1=1,解得:m=1,若直线l 恰好是2(21)1y m x =++的图象,则2m+1=1,解得:m=0,综上,()0,1B ,1m =或者0m =(2)如图,()110m b b ---=()110m b b ∴+--=0m ≥,10b -≥0m ∴=,10b -=0m ∴=11y x ∴=-,21y x =+设1y 与x 轴、y 轴交于T ,P ,2y 分别与x 轴、y 轴交于G ,H ,连接GP ,TH1OG OH OP OT ====,PH GT ⊥∴四边形GPTH 是正方形//GH PT ∴,90HGP ∠=︒,即HG GP ⊥2HP =2GP ∴=(3)121y x m =+-,()2211y m x =++121y x m =+-分别交x 轴,y 轴于C ,E 两点()12,0C m ∴-,()0,21E m -()2211y m x =++图象交x 轴于D 点1,021D m -∴+⎛⎫ ⎪⎝⎭()()()22122121121421y y y x m m x m x m x m =⋅=+-++=+++-⎡⎤⎣⎦1m >210m ∴+>∴二次函数()2221421y m x m x m =+++-开口向上,它的图象最低点在顶点∴顶点()222212,2121m m F m m ⎛⎫- ⎪-- ⎪++⎝⎭抛物线顶点F 向上平移5621m +,刚好在一次函数121y x m =+-图象上 ()()2222156*********m m m m m m -∴-+=-+-+++且1m2m ∴=2125163(3)(51)y y y x x x x =⋅=+=∴+++,∴13y x =+,251y x =+∴由13y x =+,251y x =+得到1,05D ⎛⎫- ⎪⎝⎭,()0,3E , 由25163y x x =++得到与x 轴,y 轴交点是()3,0-,1,05⎛⎫- ⎪⎝⎭,()0,3, ∴抛物线经过1,05D ⎛⎫- ⎪⎝⎭,()0,3E 两点 12y y y ∴=⋅的图象,线段OD ,线段OE 围成的图形是封闭图形,则S 即为该封闭图形的面积探究办法:利用规则图形面积来估算不规则图形的面积.探究过程:①观察大于S 的情况.很容易发现ODE S S <1,05D ⎛⎫- ⎪⎝⎭,()0,3E 11332510ODE S =⨯⨯=,310S ∴< (若有S 小于其他值情况,只要合理,参照赋分.)②观察小于S 的情况.选取小于S 的几个特殊值来估计更精确的S 的近似值,取值会因人而不同,下面推荐一种方法,选取以下三种特殊位置:位置一:如图当直线MN 与DE 平行且与抛物线有唯一交点时,设直线MN 与x ,y 轴分别交于M ,N1,05D ⎛⎫- ⎪⎝⎭,()0,3E ∴直线:153DE y x =+设直线1:15MN y x b =+25163y x x =++21530x x b ∴++-=()1430b ∴∆=-⨯-=,15920b = ∴直线59:1520MN y x =+∴点59,0300M ⎛⎫- ⎪⎝⎭15959348122030012000OMN S =⨯⨯=∴,348112000S ∴> 位置二:如图当直线DR 与抛物线有唯一交点时,直线DR 与y 轴交于点R设直线2:DR y kx b =+,1,05D ⎛⎫- ⎪⎝⎭∴直线1:5DR y kx k =+ 25163y x x =++()21516305x k x k +-∴+-= ()211645305k k ⎛⎫∴∆=--⨯⨯-= ⎪⎝⎭,14k = ∴直线14:145DR y x =+∴点140,5R ⎛⎫ ⎪⎝⎭1141725525ODR S ∴=⨯⨯=,725S ∴> 位置三:如图当直线EQ 与抛物线有唯一交点时,直线EQ 与x 轴交于点Q设直线:3EQ y tx =+25163y x x =++()25160x t x +∴-=()2160t ∴∆=-=,16t = ∴直线:163EQ y x =+∴点3,016Q ⎛⎫- ⎪⎝⎭ 139321632OEQ S =⨯⨯=∴,932S ∴> 348197120003225>> 我们发现:在曲线DE 两端位置时的三角形的面积远离S 的值,由此估计在曲线DE 靠近中间部分时取值越接近S 的值探究的结论:按上述方法可得一个取值范围348131200010S << (备注:不同的探究方法会有不同的结论,因而会有不同的答案.只要来龙去脉清晰、合理,即可参照赋分,但若直接写出一个范围或者范围两端数值的差不在0.01之间不得分.)【点睛】本题是一道综合性很强的代数与几何相结合的压轴题,知识面广,涉及有旋转的性质、坐标平移规则、非负数的性质、一次函数的图象与性质、二次函数的图象与性质、一元二次方程、不规则图形面积的估计等知识,解答的关键是认真审题,找出相关信息,利用待定系数法、数形结合法等解题方法确定解题思路,利用相关信息进行推理、探究、发现和计算.3.(1)2134y x x =-++;(2)(32,0);(3)①见解析;②CM =231或CM =123+【解析】【分析】(1)根据点C 在抛物线上和已知对称轴的条件可求出解析式;(2)根据抛物线的解析式求出点B 及已知点C 的坐标,证明△ABC 是等腰直角三角形,根据旋转的性质推出直线EF 与x 轴的夹角为45°,因此设直线EF 的解析式为y=x+b ,设点M 的坐标为(m ,0),推出点F (m ,6-m ),直线EF 与抛物线2134y x x =-++只有一个交点,联立两个解析式,得到关于x 的一元二次方程,根据根的判别式为0得到关于m 的方程,解方程得点M 的坐标.注意有两种情况,均需讨论.(3)①过点P 作PG ⊥x 轴于点G ,过点E 作EH ⊥x 轴于点H ,设点M 的坐标为(m ,0),由2PC =及旋转的性质,证明△EHM ≌△MGP ,得到点E 的坐标为(m-1,5-m ),再根据两点距离公式证明EA ED =,注意分两种情况,均需讨论;②把E (m-1,5-m )代入抛物线解析式,解出m 的值,进而求出CM 的长.【详解】(1)∵点()6,0C 在抛物线上,∴103664b c =-⨯++, 得到6=9b c +,又∵对称轴2x =,∴2122()4b b x a =-=-=⨯-, 解得1b =,∴3c =,∴二次函数的解析式为2134y x x =-++; (2)当点M 在点C 的左侧时,如下图:∵抛物线的解析式为2134y x x =-++,对称轴为2x =,()6,0C∴点A (2,0),顶点B (2,4),∴AB=AC=4,∴△ABC 是等腰直角三角形,∴∠1=45°;∵将MPC 逆时针旋转90︒得到△MEF ,∴FM=CM ,∠2=∠1=45°,设点M 的坐标为(m ,0),∴点F (m ,6-m ),又∵∠2=45°,∴直线EF 与x 轴的夹角为45°,∴设直线EF 的解析式为y=x+b ,把点F (m ,6-m )代入得:6-m=m+b ,解得:b=6-2m ,直线EF 的解析式为y=x+6-2m ,∵直线EF 与抛物线2134y x x =-++只有一个交点, ∴262134y x m y x x =+-⎧⎪⎨=-++⎪⎩, 整理得:213204x m +-=, ∴Δ=b 2-4ac=0,解得m=32, 点M 的坐标为(32,0). 当点M 在点C 的右侧时,如下图:由图可知,直线EF 与x 轴的夹角仍是45°,因此直线EF 与抛物线2134y x x =-++不可能只有一个交点.综上,点M 的坐标为(32,0). (3)①当点M 在点C 的左侧时,如下图,过点P 作PG ⊥x 轴于点G ,过点E 作EH ⊥x 轴于点H ,∵2PC 2)知∠BCA=45°,∴PG=GC=1,∴点G (5,0),设点M 的坐标为(m ,0),∵将MPC 逆时针旋转90︒得到△MEF ,∴EM=PM ,∵∠HEM+∠EMH=∠GMP+∠EMH =90°,∴∠HEM=∠GMP ,在△EHM 和△MGP 中,EHM MGP HEM GMP EM MP ∠=∠⎧⎪∠=∠⎨⎪=⎩, ∴△EHM ≌△MGP (AAS ),∴EH=MG=5-m ,HM=PG=1,∴点H (m-1,0),∴点E 的坐标为(m-1,5-m );∴22(12)(50)m m --+--221634m m -+又∵D 为线段BC 的中点,B (2,4),C (6,0),∴点D (4,2),∴22(14)(52)m m --+--221634m m -+∴EA= ED .当点M 在点C 的右侧时,如下图:同理,点E 的坐标仍为(m-1,5-m ),因此EA= ED .②当点E 在(1)所求的抛物线2134y x x =-++上时, 把E (m-1,5-m )代入,整理得:m 2-10m+13=0,解得:m=523+m=523-,∴CM =231或CM =123+.【点睛】本题是二次函数综合题,熟练掌握二次函数的图象和性质、旋转的性质、分类讨论的思想是解题的关键.4.(1)y =x 2+2x ﹣3;(2)①存在,点P 的坐标为(1,﹣2)或(﹣5,﹣8);②点M (﹣43,﹣359) 【解析】【分析】(1)y =ax 2+bx ﹣3=a (x +3)(x ﹣1),即可求解;(2)①分点P (P ′)在点C 的右侧、点P 在点C 的左侧两种情况,分别求解即可; ②证明△AGR ≌△RHM (AAS ),则点M (m +n ,n ﹣m ﹣3),利用点M 在抛物线上和AR =NR ,列出等式即可求解.【详解】解:(1)y =ax 2+bx ﹣3=a (x +3)(x ﹣1),解得:a =1,故抛物线的表达式为:y =x 2+2x ﹣3①;(2)由抛物线的表达式知,点C 、D 的坐标分别为(0,﹣3)、(﹣1,﹣4), 由点C 、D 的坐标知,直线CD 的表达式为:y =x ﹣3;tan ∠BCO =13,则cos ∠BCO 10 ①当点P (P ′)在点C 的右侧时,∵∠P′AB=∠BCO,故P′B∥y轴,则点P′(1,﹣2);当点P在点C的左侧时,设直线PB交y轴于点H,过点H作HN⊥BC于点N,∵∠PBC=∠BCO,∴△BCH为等腰三角形,则BC=2CH•cos∠BCO=2×CH1022 3110 +解得:CH=53,则OH=3﹣CH=43,故点H(0,﹣43),由点B、H的坐标得,直线BH的表达式为:y=43x﹣43②,联立①②并解得:58 xy=-⎧⎨=-⎩,故点P的坐标为(1,﹣2)或(﹣5,﹣8);②∵∠PAB=∠BCO,而tan∠BCO=13,故设直线AP的表达式为:y=13x s+,将点A的坐标代入上式并解得:s=1,故直线AP的表达式为:y=13x+1,联立①③并解得:43139xy⎧=⎪⎪⎨⎪=⎪⎩,故点N(43,139);设△AMN的外接圆为圆R,当∠ANM =45°时,则∠ARM =90°,设圆心R 的坐标为(m ,n ),∵∠GRA +∠MRH =90°,∠MRH +∠RMH =90°,∴∠RMH =∠GAR ,∵AR =MR ,∠AGR =∠RHM =90°,∴△AGR ≌△RHM (AAS ),∴AG =m +3=RH ,RG =﹣n =MH ,∴点M (m +n ,n ﹣m ﹣3),将点M 的坐标代入抛物线表达式得:n ﹣m ﹣3=(m +n )2+2(m +n )﹣3③, 由题意得:AR =NR ,即(m +3)2=(m ﹣43)2+(139)2④, 联立③④并解得:29109m n ⎧=-⎪⎪⎨⎪=-⎪⎩, 故点M (﹣43,﹣359). 【点睛】本题考查的是二次函数综合运用,涉及到一次函数的性质、三角形全等、圆的基本知识等,其中(2)①,要注意分类求解,避免遗漏.5.(1)k=-3-a ;对称轴x =1;y 轴交点(0,-3);(2)2y=2x -4x-3,顶点坐标(1,-5);(3)-5≤a <-4;(4)-1≤t ≤2.【解析】【分析】(1)将点P(2,-3)代入抛物线上,求得k 用a 表示的关系式;抛物线L 的对称轴为直线2a x==12a--,并求得抛物线与y 轴交点; (2)将点(3,3)代入抛物线的解析式,且k=-3-a ,解得a=2,k=-5,即可求得抛物线解析式与顶点坐标;(3)抛物线L 顶点坐标(1,-a-3),点C ,P 之间的部分与线段CP 所围成的区域内(不含边界)恰有4个整点,这四个整点都在x=1这条直线上,且y 的取值分别为-2、-1、0、1,可得1<-a-3≤2,即可求得a 的取值范围;(4)分类讨论取a >0与a <0的情况进行讨论,找出1x 的取值范围,即可求出t 的取值范围.【详解】解:(1)∵将点P(2,-3)代入抛物线L :2y=ax -2ax+a+k ,∴-3=4a 4a a+k=a+k -+∴k=-3-a ;抛物线L 的对称轴为直线-2a x=-=12a,即x =1; 将x=0代入抛物线可得:y=a+k=a+(-3-a)=-3,故与y 轴交点坐标为(0,-3); (2)∵L 经过点(3,3),将该点代入解析式中,∴9a-6a+a+k=3,且由(1)可得k=-3-a ,∴4a+k=3a-3=3,解得a=2,k=-5,∴L 的表达式为2y=2x -4x-3;将其表示为顶点式:2y=2(x-1)-5,∴顶点坐标为(1,-5);(3)解析式L 的顶点坐标(1,-a-3),∵在点C ,P 之间的部分与线段CP 所围成的区域内(不含边界)恰有4个整点,这四个整点都在x=1这条直线上,且y 的取值分别为-2、-1、0、1,∴1<-a-3≤2,∴-5≤a <-4;(4)①当a <0时,∵2x 3≥,为保证12y y ≥,且抛物线L 的对称轴为x=1, ∴就要保证1x 的取值范围要在[-1,3]上,即t ≥-1且t+1≤3,解得-1≤t ≤2;②当a >0时,抛物线开口向上,t ≥3或t+1≤-1,解得:t ≥3或t ≤-2,但会有不符合题意的点存在,故舍去,综上所述:-1≤t ≤2.【点睛】本题考查二次函数的图象及性质;熟练掌握二次函数的图象及性质,数形结合解题是关键.6.(1)211433=--y x x , (0,4)C -;(2)4m -;(3)①m 的值为54;②存在;点P 的坐标为(4,2)--或1422(,)55--或42(,)55. 【解析】【分析】(1)将(3,0)A -、(4,0)B 代入24y ax bx =+-,得到关于a 、b 的二元一次方程组,解方程组即可求出a 、b 的值,进而可得到抛物线的表达式和点C 的坐标;(2)设直线BC 的解析式为y kx b =+即可求出解析式的表达式,令x=m ,即可得到线段DE 的长用含m 的式子表示为4m -;(3)①由点D 的横坐标为m ,且04m <<,可得OE m =,再根据四边形DEFG 是正方形求出点G 的坐标,代入函数解析式即可求出m 的值;② 利用①中的方法求出点D 的坐标、CF 、CD 的值,再分不同情况讨论,利用两点间距离公式和全等三角形对应边相等列方程组求解即可.【详解】(1)将(3,0)A -、(4,0)B 代入24y ax bx =+-中,得934016440a b a b --=⎧⎨+-=⎩, 解,得1313a b ⎧=⎪⎪⎨⎪=-⎪⎩, ∴抛物线的表达式为211433=--y x x . 将0x =代入,得4y =-,∴点(0,4)C -.(2)设直线BC 的解析式为y kx b =+,将点(4,0)B 、(0,4)C -代入可得,404k b b +=⎧⎨=-⎩, 解得14k b =⎧⎨=-⎩, ∵直线BC 的表达式为4y x =-,当x=m 时,4y m =-,即线段DE 的长用含m 的式子表示为4m -.故答案为:4m -;(3)①∵点D 的横坐标为m ,且04m <<,∴OE m =,∵四边形DEFG 是正方形,∴4DE EF FG m ===-,∴442OF EF OE m m m =-=--=-,∵点G 在第三象限,∴点G 的坐标为(24,4)m m --,∵点G 在抛物线211433=--y x x 上, ∴211(24)(24)4433m m m ----=-, 解14m =(不符合题意,舍去),254m =, ∴当矩形DEFG 成为正方形时,m 的值为54. ②存在;理由如下:由①可知FG=DE=4-m ,∵点O 是线段EF 的中点,∴点G 的坐标为(-m ,m -4),∵点G 在抛物线211433=--y x x 上, ∴211(24)(24)4433m m m ----=-, 解10m =(不符合题意,舍去),22m =,∴点D 的坐标为(2,-2),∴222425CF =+=,22(20)(24)22CD =-+-+=,如图,设点的坐标为(x ,y ),分以下三种情况:I 、当位于点P 时,可得PF=CD ,PC=CF ,∴22(2)25PF x y =++=22(4)22PC x y =++=解得1142x y =-⎧⎨=-⎩,224525x y ⎧=-⎪⎪⎨⎪=-⎪⎩(不合题意,舍去), ∴点P 的坐标为(4,2)--;II 、当位于点P '时,方法同I 可得点P 的坐标为1422(,)55--; III 、当位于点P ''时,方法同I 可得点P 的坐标为42(,)55;综上,点P 的坐标为(4,2)--或1422(,)55--或42(,)55. 【点睛】此题是二次函数综合题,主要考查了待定系数法确定解析式,两点间的距离公式,全等三角形的性质,解本题的关键是确定函数关系式.7.(1)6c a =-;(2)①2132y x =-;②2. 【解析】【分析】(1)先根据二次函数的对称性求出抛物线与x 轴的另一个交点的横坐标,然后根据二次函数与一元二次方程的联系、一元二次方程的根与系数的关系即可得;(2)①先根据(1)可得抛物线的解析式和顶点D 的坐标,再设11222),(2)(,,A x k a B k x x a x --,从而可得直线AD 、BD 解析式中的一次项系数,然后根据一元二次方程的根与系数的关系可得12k x x a+=,124x x =-,最后根据圆周角定理可得AD BD ⊥,从而可得1212144x x k a k a x x +⋅=-+,化简可求出a 的值,由此即可得出答案;②先求出点B 、D 的坐标,再根据直线1l 与抛物线只有一个交点可得出2213,2q p x p --==,然后联立直线1l 与2l 求出点N 的坐标,最后利用三角形的面积公式分别求出12,S S ,由此即可得.【详解】(1)抛物线2(0)y ax bx c a =++>,顶点D 在y 轴上,∴抛物线的对称轴为y 轴,即0x =,0b ∴=,抛物线与x∴抛物线与x 轴的另一个交点的横坐标为是关于x 的一元二次方程20(0)ax bx c a ++=>的两根,(c a∴=, 即6c a =-;(2)①由(1)可得:抛物线的解析式为26y ax a =-,顶点D 的坐标为(0,6)D a -,由题意,设点A 、B 的坐标分别为11222),(2)(,,A x k a B k x x a x --,且21x x >,由点A 、D 的坐标得:直线AD 解析式中的一次项系数为11112064x a x x k x a k a -=-++, 由点B 、D 的坐标得:直线BD 解析式中的一次项系数为22222064x a x x k x a k a -=-++, 联立262y ax a y kx a⎧=-⎨=-⎩可得240ax kx a --=, 则1x 与2x 是关于x 的一元二次方程240ax kx a --=的两根, 由根与系数的关系得:1212,4k x x x x a+==-, 以AB 为直径的圆恒过点D , 90ADB ∴∠=︒,即AD BD ⊥, 则1212144x x k a k a x x +⋅=-+, 整理得:2164a =, 解得12a =或102a =-<(不符题意,舍去), 故抛物线的解析式为2132y x =-; ②由①可知,222(0,3),(,31)2D x x B --, 则直线2l 的解析式为3y =-, 联立2132y x y px q⎧=-⎪⎨⎪=+⎩可得22260px x q ---=, 1l 与抛物线只有一个公共点,∴方程22260px x q ---=只有一个实数根2x ,∴其根的判别式244(26)0p q ∆=++=,且2222260x px q ---=, 解得2132q p --=, 将2132q p --=代入2222260x px q ---=得:2x p =, 联立3y y px q =-⎧⎨=+⎩,解得33q x p y --⎧=⎪⎨⎪=-⎩,即点N 的坐标为3(,3)q N p---, 21322p q p DN p p --∴===, 121122S DM x DM p =⋅=⋅,21112224p S DM DN DM DM p =⋅=⋅=⋅, 1212124DM S p M p S D ⋅⋅∴==. 【点睛】本题考查了二次函数与一元二次方程的联系、一元二次方程的根与系数的关系以及根的判别式、二次函数的对称性、圆周角定理等知识点,较难的是题(2)①,利用圆周角定理得出AD BD ⊥,从而利用一次函数的性质建立等式是解题关键.8.(1)B 点坐标(0,-1),平移后的抛物线为2y=-x +2x+3;(2)点M的坐标为或;(3)存在,1P (1,1),2P (1,6)-,3P (12),,4P (1,8)-,详解见解析.【解析】【分析】(1)将x=0代入抛物线公式2y=-x +2x-1求出y 值,即可得到抛物线与y 轴交点B 的坐标,平移后的抛物线的顶点为E(1,4),可根据顶点式求出平移后抛物线的解析式;(2)因为抛物线向上平移4个单位,所以MN=4,又因为OM=ON ,可知点M 的纵坐标为-2,将y=-2代入原抛物线2y=-x +2x-1,即可求出x 值,点M 的坐标就可以表示出来. (3)要使C 、F 、P 为顶点的三角形为直角三角形,可以画一个以C 、F 为直径的圆(直径对应圆周角为直角),交抛物线对称轴x=-1可得点1P 、2P 的坐标解,另外可以使∠PCF=90°或∠CFP=90°,可分别得出点3P 、4P 的坐标解.【详解】解:(1)抛物线2y=-x +2x-1与y 轴相交于点B ,将x=0代入,求得y=-1,∴B 点坐标(0,-1).∵设平移后的抛物线为2y=-(x-h)+k ,顶点为E(1,4),即h=1,k=4,∴2y=-(x-1)+4,即平移后的抛物线为22y=-(x-1)+4=-x +2x+3.(2)如上图所示,∵原坐标顶点A(1,0),平移后抛物线顶点为E(1,4),∴抛物线向上平移了4个单位,即MN //y 轴,MN ⊥x 轴,又∵OM=ON ,MN=4,∴点O 在垂直平分线上,点M 、N 关于x 轴对称,∴M 点的纵坐标为–2,将y=-2代入2y=-x +2x-1,得:222-x +2x-1=-2-(x -2x+1)=-2(x-1)=2x=12± 解得:x=12±,∴点M 的坐标为(1+2-2),或(1-2-2),. (3)存在,且1P (1,1),2P (1,6)-,3P (12),,4P (1,8)-. 如图所示,点P 一共有四种结果,∵C 点为平移后的解析式与x 轴的左交点,将y=0代入2y=-x +2x+3,得x=-13或, ∴C(-1,0),且点B(0,-1),将点B(0,-1)、C(-1,0)代入直线BC 解析式为:y=kx+b , ∴-k+b=0b=-1⎧⎨⎩,解得:k=-1b=-1⎧⎨⎩,即直线BC 解析式:y=-x-1, 根据题意可知,直线BC 与平移后的解析式相交于点F ,∴2y=-x-1y=-x +2x+3⎧⎨⎩,解得:x=-1(舍)或4,y=-5,即F(4,-5), ∵要使C 、F 、P 为顶点的三角形为直角三角形,可以画一个以C 、F 为直径的圆,该圆与抛物线对称轴x=-1交点即为点P (因为圆的直径对应的圆周角为90°,即∠CPF=90°)∴以C 、F 为直径的圆,圆心为线段CF 的中点(32,5-2),直径为线段CF 的长∴圆的方程为:22235x-+y+=22()(),将x=1代入圆的方程,得:y=1或-6, 即1P (1,1),2P (1,6)-, ∵直线CF 解析式:y=-x-1,即斜率k=-1,即直线CF 与x 轴夹角为45°,要使C 、F 、P 为顶点的三角形为直角三角形,则使∠PCF=90°,直线CP 与x 轴夹角也为45°,即直线CP 斜率为1,直线CP 的解析式为:y=x+1,此时该直线与抛物线对称轴x=1的交点为3P (1,2),又∵直线CF 解析式:y=-x-1,即斜率k=-1,即直线CF 与x 轴夹角为45°,要使C 、F 、P 为顶点的三角形为直角三角形,则使∠CFP=90°,直线FP 与x 轴夹角也为45°,即直线FP 斜率为1,直线FP 的解析式为:y=x-9,此时该直线与抛物线对称轴x=1的交点为4P (1,-8).【点睛】本题考查了一元二次函数与坐标轴、直线的交点,一元二次函数的平移及应用,圆的直径所对应的圆周角为直角等知识点,该题有一定的难度,所以一定要结合图形进行分析,这样才不会把解遗漏.9.(1)241y x x =+-;(2)PAB △面积最大值为278;(3)存在,1234(12)(34(34(13)E E E E ------,,,,,,【解析】【分析】(1)将点A 、B 的坐标代入抛物线表达式,即可求解;(2)设AB y kx b =+,求得解析式,过点P 作x 轴得垂线与直线AB 交于点F ,设点()2,41P a a a +-,则(,1)F a a -,1||2PABB A S PF x x ∆=⋅-23327228a ⎛⎫=-++ ⎪⎝⎭,即可求解;。

上海市青浦区九年级(上)月考物理试卷(12月份)

上海市青浦区九年级(上)月考物理试卷(12月份)

上海市青浦区九年级(上)月考物理试卷(12月份)一、选择题(16分)1.(3分)首先用实验测出大气压强值的科学家是()A.托里拆利B.安培C.伏特D.阿基米德2.(3分)下列实例中不是利用连通器原理工作的是()A.抽水机B.船闸C.茶壶D.锅炉水位器3.(3分)如图所示为生活中常用的一些物品,使用时主要用来减小压强的是()A.图钉尖很尖锐B.逃生锤的锤头很尖C.切苹果器的刀片做的很薄D.书包带较宽4.(3分)一瓶未打开的500毫升矿泉水,其质量约为()A.0.005千克B.0.05千克C.0.5千克D.5千克5.(3分)下列事例中,运用的科学研究方法相同的是()(1)探究液体内部压强与液体密度的关系(2)用水压比作电压(3)探究串联电路中电阻的规律(4)探究物质质量与体积的关系。

A.(1)与(3)B.(1)与(4)C.(2)与(3)D.(3)与(4)6.(3分)甲、乙、丙三根镍铬合金丝,其横截面积关系S甲>S乙=S丙,长度关系为L甲=L乙<L丙.将它们分别接在同一电路中,则通过它们的电流大小关系是()A.I甲>I乙>I丙B.I甲>I乙=I丙C.I甲<I乙<I丙D.I甲=I乙<I丙7.(3分)如图所示为汽车油量表的设计电路图,其中R是一段弧形电阻,当汽车加油时,要求油量表指针指向右边,油量减少时指针回到左边。

则下列说法合理的是()A.加油时电路电阻变大,电压表显示变大,应将电压表设计为油量表B.加油时电路电阻变小,电流表示数变大,应将电流表设计为油量表C.油耗尽时总电阻变小,电流表示数变大,应将电流表设计为油量表D.油耗尽时总电阻变大,电压表示数变小,应将电压表设计为油量表8.(3分)如图所示,两个盛有等高液体的圆柱形容器A和B,底面积不同(S A>S B),液体对容器底部的压强相等。

现将甲球浸没在A容器的液体中,乙球浸没在B容器的液体中,容器中均无液体溢出,若此时液体对各自容器底部的压力相等,则一定()A.甲球的质量大于乙球的质量B.甲球的重力小于乙球的重力C.甲球的体积大于乙球的体积D.甲球所受浮力小于乙球所受浮力二.填空题(26分)9.(3分)家庭照明电路电压是伏,各用电器之间是的(选填“串联”或“并联”);同时使用的用电器越多,电路中总电阻越。

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青浦区实验中学2013学年九年级第一阶段学业模拟测试
九年级数学 试卷 2013.10
考试时间:100分钟 满分:150分
一、选择题:(每题4分,满分24分)
1、下列图形一定是相似图形的是( ).
(A )两个矩形; (B )两个正方形; (C )两个直角三角形; (D )两个等腰三角形.
2、已知线段a ,b ,c ,求作线段x ,使b
ac x =,下列作法中正确的是( ).
(A ); (B ); (C ); (D ).
3、如图,在△ABC 中,点D 、E 分别在AB 、AC 上,且DE ∥BC ,那么下列结论中错误的是( ).
(A )EC AE DB AD =; (B )AC AE AB AD =; (C )DB AD BC DE =; (D )AC
AE BC DE =.
(第3题图) (第4题图) (第6题图)
4、如图,点F 在平行四边形ABCD 的边AB 的延长线上,连结DF 交BC 于点E .则图中与△BEF 相似的 三角形有( ).
(A )1个; (B )2个; (C )3个; (D )4个.
5、下列各组条件中,一定能推得△ABC 与△DEF 相似的是( ).
(A )∠A =∠E 且∠D =∠F ; (B )∠A =∠B 且∠D =∠F ;
(C )∠A =∠E 且ED EF AC AB =; (D )∠A =∠E 且DE
FD BC AB =. 6、如图,在△ABC 中,点D 、E 分别在边AB 、AC 上,DE ∥BC ,AD :BD =1:2,那么S △DBE :S △CBE 等于 ( ).
(A )1:2; (B )1:3; (C )1:4; (D )1:6.
二、填空题:(本大题共12小题,每小题4分,满分48分)
7、如果3
2=y x ,那么=+y y x _________. 8、已知线段a =9cm 、b =4cm ,那么线段a 、b 的比例中项c =_________cm .
9、在比例尺为1:2000的地图上测得AB 两地间的图上距离为5cm ,则AB 两地间的实际距离为_________m .
10、如果两个相似三角形的面积之比是25:16,那么它们的对应高之比是_________.
11、如图,已知点D 在△ABC 的边AB 上,点E 在AC 边上,AE :EC =2:5,AB =14cm ,当AD 的长等于
_____ ___cm 时,可以证得DE ∥BC .
(第11题图) (第12题图) (第13题图)
12、如图,△ABC 的两条中线AD 、BE 相交于点G ,如果AD =3,那么GD =_____ ___.
13、如图,△ABC 与△DEF 的顶点均在小正方形方格(边长为1个单位长)的顶点处,则△ABC _____
___△DEF (选择“一定相似”或“不一定相似”或“一定不相似”填入空格).
14、如图,D 、E 两点分别在△ABC 的边AB 、AC 上,DE 与BC 不平行,当满足__________条件(写出 一个即可)时,△ADE ∽△ACB .
(第14题图) (第15题图) (第17题图)
15、如图,平行四边形ABCD 中,E 是边BC 上的点,AE 交BD 于点F ,如果32=BC BE ,那么=FD
BF __________. 16、已知点P 是线段AB 的黄金分割点,AB =4厘米,则较长线段AP 的长是__________厘米.
17、如图,已知AB ⊥BD ,ED ⊥BD ,C 是线段BD 的中点,且AC ⊥CE ,ED =1,BD =4,那么AB =_______.
18、在△ABC 和△DEF 中,∠A =∠D ,AB =4,BC =5,AC =6,EF =10,如果△ABC 与△DEF 相似,那么 DE = .
三、解答题:(本大题共7题,满分78分)
19、(本题满分10分)已知:
,432z y x ==22432=+-z y x ,求代数式x +y -z 的值.
20、(本题满分10分)如图,点D 、E 分别在ABC ∆的AB 、AC 边上,∠AED =∠B ,如果AE =2,且 5
4=∆DBCE ADE
S S 四边形.求AB 的长.
(第20题图)
21、(本题满分10分,每小题满分5分)如图,在梯形ABCD 中,AD ∥BC ,点E 在边AD 上,CE 与 BD 相交于点F ,AD =4,BC =BD =6,DE =3.
(1)求:DF 的长; (2)求证:△DFE ∽△DAB .
(第21题图)
22、(本题满分10分,每小题满分5分)如图,直线l 1、l 2、l 3分别交直线l 4于点A 、B 、C ,交直线l 5于 点D 、E 、F ,且l 1∥l 2∥l 3.已知EF :DF =5:8,AC =24.
(1)求AB 的长; (2)当AD =4,BE =1时,求CF 的长.
(第22题图)
23、(本题满分12分,每小题满分6分)如图,△ABC 是一张锐角三角形的硬纸片.AD 是边BC 上的高,BC =40cm ,AD =30cm .从这张硬纸片剪下一个长HG 是宽HE 的2倍的矩形EFGH .使它的一边EF 在BC 上,顶点G ,H 分别在AC ,AB 上,AD 与HG 的交点为M .
(1)求证:BC
HG AD AM ; (2)求这个矩形EFGH 的周长.
(第23题图)
24、(本题满分12分,每小题满分6分)已知:如图,在△ABC 中,∠ADE =∠B ,∠BAC =∠DAE .
(1)求证:AC
AE AB AD ; (2)当∠BAC =90°时,求证:EC ⊥BC .
(第24题图)
25、(本题满分14分,第(1)小题满分4分,第(2)小题满分4分,第(3)小题满分6分)
如图,在梯形ABCD 中,AB ∥CD ,CD =6,BC =4,∠ABD =∠C ,P 是CD 上的一个动点(P 不与点C 点D 重合),且满足条件∠BPE =∠C ,交BD 于点E .
(1)求证:△BCP ∽△PDE ;(2)如果CP =x ,BE =y ,求y 与x 之间的函数关系式;
(3)P 点在运动过程中,△BPE 能否成为等腰三角形,若能,求x 的值,若不能,请说明理由.
(第25题图)
参考答案
一、选择题
1、B
2、D
3、C
4、B
5、C
6、D。

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