湖南省益阳市箴言中学2015届高三第十次模拟考试(5月)化学 Word版含答案

益阳市箴言中学2015届高三第十次模拟考试试题理科数学试题（考试时间120分钟满分150分）第I卷（选择题共50分）一、选择题（本大题共10小题，每小题5分，共50分. 在每小题列出的的四个选项中,选出符合题目要求的一项）1．若集合1{|01},{|lg}xM x x N x yx-=≤≤==，则RM C N =( )A．{0}B．{0,1}C．{|01}x x≤≤D．{|0x x<或1}x>2．已知随机变量X服从正态分布N(3，1)，且(24)P X≤≤=0.6826，则p（X>4）=（）A、0.1588B、0.1587C、0.1586 D0.15853．已知命题p:函数21()sin2f x x=-的最小正周期为π；命题q：若函数)1(+xf为偶函数，则)(xf关于1=x对称.则下列命题是真命题的是（）A．qp∧B．qp∨C．)()(qp⌝∧⌝D．()p q∨⌝4．如下图是某四棱锥的三视图，则该棱锥的体积是( )A．48 B．C．16 D．5. 若执行右上方的程序框图，输出S的值为4，则判断框中应填入的条件是( )A ．?14<k ?14<kB ．?15<k ?15<kC ．?16<k ?16<kD ．?17<k ?17<k 6．在公差不为零的等差数列{}n a 中，23711220a a a -+=，数列{}n b 是等比数列，且77b a =则268log ()b b 的值为 ( )A ．2B ．4C ．8D ．17. 已知132a -=，21211log ,log 33b c ==，则（ ） A ．a b c >> B ．a c b >> C ．c a b >> D ．c b a >>8．已知函数x x x f cos 2)(=，则函数)(x f 的部分图象可以为 （ ）9.已知双曲线22221x y a b -=（0,0a b >>）的渐近线与圆22(2)1x y -+=相切，则双曲线的离心率为（ ）AB ．43 CD ．10.设函数⎩⎨⎧>≤=0,log 0,2)(2x x x x f x ，若对任意给定的),1(+∞∈t ，都存在唯一的R x ∈，满足at t a x f f +=222))((，则正实数a 的最小值是 （ ）A ．2B ．21C ．41D ．81第II 卷 （非选择题 共100分）二 填空题（本大题应答5小题，每小题5分，共25分，请把答案填入答卷中的横线上） （一）选做题（请考生在11、12、13三题中任选两题作答，全做按前两题记分）11．如图，BD 是半圆O 的直径，A 在BD 的延长线上，AC 与半圆相切于点E ，AC BC ⊥，若AD =6AE =，则EC =.12．已知在平面直角坐标系xOy 中圆C的参数方程为：3cos 13sin x y θθ⎧=+⎪⎨=+⎪⎩，（θ为参数），以Ox 为极轴建立极坐标系，直线极坐标方程为：,0)6cos(=+πθρ则圆C 截直线所得弦长为13．若关于x 的不等式|2||2|6x x a -+-<的解集不空，则a 的取值范围是 （二）必做题（14－16题）14．设()0sin cos a x x dx π=+⎰，则二项式6⎛ ⎝的展开式的常数项是_________. 15.已知x ，y 满足⎪⎪⎩⎪⎪⎨⎧≥≤+≥412x y x x y 且y x z +=2的最大值与最小值分别为a 和b ，则b a -的值是16. 对定义在区间D 上的函数和，如果对任意，成立，那么称函数在区间D 上可被替代，D 称为“替代区间”．给出以下命题： ①在区间②“替代区间”③在区间可被替代，则； ④，则存在实数，使得在区间 上被替代；其中是真命题的有 .)(x f )(x g D x ∈)(x f )(x g 1)(2+=x x f ),(+∞-∞x x f =)(x x f ln )(=],1[e b x x g -=)(22≤≤-b e )(sin )(),)(lg()(212D x x x g D x x ax x f ∈=∈+=)0(≠a a )(x f 21D D ⋂)(x g三、解答题 （本大题共6小题，共75分. 解答应写出文字说明、演算步骤或证明过程）17. (本小题满分12分)己知函数2()2sin cos cos sin sin (0)2f x x x x ϕϕϕπ=+-<<在x π=处取最小值．（I ）求ϕ的值。

湖南省益阳市箴言中学化学自主招生试卷一、选择题1．某单质X能从某溶液中置换出单质Y，由此推断下列说法中正确的是 ( )A．X是金属时，Y一定比X活泼B．X可能是铁，Y一定是铜C．X是金属时，Y可能是金属，也可能是非金属D．X一定是排在金属活动顺序表中氢以前的金属2．t℃时，Na2CO3溶解度为Ag，现有饱和Na2CO3溶液(100+A)g，溶质质量分数为a%，向该溶液中投入无水碳酸钠Ag，静置后析出碳酸钠晶体(Na2CO3·10H2O)Bg，加水使晶体全部溶解，所得溶液质量分数为a%，则加入水的质量为( )A．(100+A)g B．100gC．100180286Ag D．(10 -A·a%)g3．如图所示，将液体X加入到集气瓶中与固体Y作用，观察到气球逐渐变大，如表中液体X和固体Y的组合，符合题意的是（）①②③④⑤X稀盐酸水水双氧水水Y铁粉氢氧化钠氯化钠二氧化锰硝酸铵A．①②⑤B．①③④C．①②④D．②③⑤4．在硝酸银、硝酸铜的混合溶液中加入一定量锌粉，反应停止后过滤，滤液仍为蓝色，有关判断正确的是（）A．滤渣中一定有银、没有铜和锌B．滤渣中一定有银和锌，可能有铜C．滤液中一定有硝酸锌、硝酸铜、硝酸银D．滤液中一定有硝酸锌、硝酸铜，可能有硝酸银5．下列鉴别两种不同物质的方法，不正确的是（）序号待鉴别的物质鉴别方法A CO2与O2燃着的木条，观察燃着的情况B酒精与白醋闻气味C CaCO3与NaCl加水，观察是否溶解D NaOH与Na2CO3滴加酚酞溶液，观察溶液颜色的变化A．A B．B C．C D．D6．下列所示的四个图像，能正确反映对应变化关系的是A．向一定量的硝酸铜和硝酸镁的混合溶液中加入铁粉B．向pH=2的盐酸中加水稀释C．向一定量的含有盐酸的氯化铜溶液中滴加氢氧化钠溶液D．等质量的镁和铁分别与等质量、等浓度足量的稀硫酸反应7．下列叙述中不符合实验事实的是A．稀硫酸中滴加石蕊试液，溶液变红B．在K2CO3、K2SO4、AgNO3三种溶液中滴入BaCl2溶液，都有白色沉淀生成C．将CO2气体通入CaCl2溶液中有白色沉淀D．将铁丝浸入硫酸铜溶液中，铁丝表面会覆盖一层红色物质8．下列图像不能正确反映对应变化关系的是A．向等质量的氧化钙、氢氧化钙中分别加入等质量分数的稀盐酸至过量B．向一定质量氯化亚铁和氯化铝的混合溶液中加入镁粉至过量C．向盐酸和氯化钙的混合溶液中逐滴加入纯碱溶液至过量D．向等质量的镁、铝中分别加入等质量分数的稀硫酸至过童9．某实验小组将Ba(OH)2溶液逐滴滴入硫酸溶液中，溶质的质量与加入的Ba(OH)2溶液的质量关系如图所示。

益阳市箴言中学2014—2015学年高一10月月考 化 学 时量：60分钟 满分100分 相对原子质量：H 1 C 12 O 16 Na 23 N 14 S 32 Fe 56 Cl 35.5 第I卷（选择题 共60分） 选择题部分共15小题，每小题给出四个选项中只有一个选项正确，每小题4分，共60分。

1．如果你家里的食用花生油混有水份，你将采用下列何种方法分离( )B．蒸馏C．分液D．萃取 2．下列实验操作中，不合理的是( ) ① ② ③ A．洗涤沉淀时(如图①)，向漏斗中加适量水，搅拌并滤干 B．用CCl4提取碘水中的碘，选③ C．蒸馏时蒸馏烧瓶中液体的体积不能超过容积的2/3，液体也不能蒸干 D．粗盐提纯，选①和② ( )( )( ) B．CH4C．CO2 D．SO2 6．下列实验操作中，主要不是从安全因素考虑的是( ) A．点燃氢气前一定要检验氢气的纯度 B．未使用完的白磷要随时收集起来，并与空气隔绝 C．用氢气还原氧化铜时，要先通一会儿氢气，再加热氧化铜 D．酒精灯不用时，必须盖上灯帽 7．鉴别SO42－时所选用的试剂及顺序最合理的是( )用NA表示阿伏加德罗常数的值，下列叙述中正确的是( ) A．1 mol·L－1 NaCl溶液中含有NA个Na＋ B．25 ℃，101 kPa64 g SO2中含有的原子数为3NA C．在常温常压下.4 L Cl2含有的分子数为 NA D．标准状况下11.2 L CCl4含有的分子数为0.5NA Na2CO3俗名纯碱，下面对纯碱采用不同分类法进行的分类中正确的是( ) A．Na2CO3是碱 B．Na2CO3是酸式盐 C．Na2CO3是难溶性盐 D．Na2CO3是钠盐 Na2SO4溶液，正确的方法是( )Na2SO4 溶于100ml水中 ② 将32.2g Na2SO4·10H2O溶于少量水中，再用水稀释至100 ml ③ 将20 ml 5.0 mol/L Na2SO4溶液用水稀释至100 ml A．①②B．②③C． ③D．①②③ 11．医学上对血液中毒最常用的净化手段是血液透析。

益阳市箴言中学2015—2016学年高二10月月考 化学试题 时量：90分钟 满分：100分 一、选择题（本题包括17个小题，每小题3分，共51分，每小题只有一个答案符合题意） 1．在下列反应中，反应物的总能量低于生成物的总能量的是 A．2H2＋O22H2O B．CaO＋CO2=CaCO3 aCO3CaO＋CO2D．C2H5OH＋3O2 2CO2＋3H2O 2．在 xA(气)＋yB(气) zC(气)＋wD(气) 的可逆反应中，经1分钟后A减少a mol/L，B减少a/3 mol/， C增加2a/3 mol/L， D增加a mol/L， 则x、y、z、w的比例关系是 A．3 : 1 : 2 : 3 B．1 : 3 : 3 : 1 C．3 : 2 : 1 : 3 D．1 : 2 : 3 : 2 3．反应A(g)+3B(g) 2C(g)+2D(g)，在不同情况下测得反应速率，其中反应速率最快的是 A．υ(D)=0.4 mol / ?L·s? B．υ(C)=0.5 mol / ?L·s? C．υ(B)=0.6 mol / ?L·s? D．υ(A)=0.15 mol / ?L·s? 4．下列条件的改变，一定能加快化学反应速率的是 A．增大压强 B．升高温度C．增大反应物的量 D．减小生成物的浓度 2NH3的正逆反应速率可用各反应物或生成物浓度的变化来表示。

下列关系中能说明反应已达到平衡状态的是 A．υ正(N2)=υ逆(NH3) B．3υ正(N2)=υ正(H2)C．υ正(N2)=3υ逆(H2) D．2υ正(H2)=3υ逆(NH3) 6．能增加反应物分子中活化分子的百分数的是 A．使用催化剂 B．增大固体表面积 C．增大压强 D．增加浓度 7．某化学反应其△H=—122 kJ/mol，?S=231 J/(mol·K)，则此反应在下列哪种情况下可自发进行 A．在任何温度下都不能自发进行 B．在任何温度下都能自发进行 C．仅在高温下自发进行 D．仅在低温下自发进 8. 对于3Fe(s)＋4H2O(g) Fe3O4(s)＋4H2(g)，反应的·L－1,则化学平衡常数为 A．1/54 B．1/6.75 C．1/27 D．1 9．在密闭容器中，2(g)+3H2(g) 2NH3（g） ΔH＜0，达到甲平衡。

益阳市箴言中学2015届高三第十次模拟考试试题英语试题（时量：120分钟满分：150分）PART ONE: LISTENING COMPREHENSION（30 points）Section A (22.5 points)Directions: In this section, you will hear 6 conversations between 2 speakers. For each conversation, there are several questions and each question is followed by 3 choices. Listen to the conversations carefully and then answer the questions by marking the corresponding letter（A, B or C）on the question booklet. You will hear each conversation TWICE.Conversation 11. What is the probable relationship between the two speakers?A. Teacher and student.B. Salesgirl and customer.C. Doctor and patient.2. What does the man think will help the woman?A. Some medicine.B. Breathing slowly.C. Doing some tests. Conversation 23. Why does the man request the girl to stop watching TV?A. Because she should finish her homework first.B. Because she should protect her eyes.C. Because they will go out for a tour soon..4. What can we know about the man?A. He is a heavy smoker.B. He likes reading books in bed.C. He always breaks his promise.Conversation 35. What does the man like to do this evening?A. Hold a party.B. Watch TV.C. Go to the cinema.6. When will the company party start?A. At 5:30.B. At 7:30.C. At 8:00.Conversation 47. What are the two speakers mainly talking about?A. People working in shops.B. Goods in various qualities.C. Shopping in different places.8. What is the man?A. A salesman in a small shop.B. A manager of a supermarket.C. A staff of a department store.9. What might the woman think of supermarket staff?A. They’re unfriendly.B. They’re very nice.C. They’re well paid. Conversation 510. Where did the man see the ad for the bike sale?A. On TV.B. In a department store.C. In the newspaper.11. Which bike is the latest model?A. The Curzon.B. The Anderson.C. The Instant.12. What does the man decide to do in the end?A. Buy the cheapest one.B. Go to the shop to have a look first.C. Buy the one recommended by the woman.Conversation 613. What is the purpose of the woman’s visit?A. Business.B. Pleasure.C. Business and pleasure14. What are in the woman’s luggage?A. Clothes, a computer and books.B. A CD player, clothes and books.C. Some gifts, books and a CD player.15. What do we know about the woman?A. Her parents are on the same trip.B. She enjoys travelling to different countries.C. She was born in that country.Section B (7.5 points)Directions: In this section, you will hear mini-talk. Listen carefully and then fill in the numbered blanks with the information you’ve got. Fill in each blank with NO MORE THAN 3 WORDS .YouPAER TWO: LANGUAGE KNOWLEDGE （45 points）Section A（15 points）Directions: Beneath each of the following sentences there are 4 choices marked A, B, C and D. Choose one answer that best completes the sentence.1.Experts suggest that parents should let their child do tasks that he is physically and mentally capable of doing by himself, ______ will help to build a self-confident kid.A. thatB. whoC. whichD. what2.I can’t imagine what air we would be breathing in if we __________ anything to stop air pollution.A. hadn’t doneB. didn’t doC. haven’t doneD. don’t do3.We all can feel something unusual about Kate, but she just _________ let us know what it is.A. shouldn’tB. mustn’tC. needn’tD. won’t4.When effectively _______, the feedback we share with students or employees can develop their awareness of their own learning.A. managingB. being managedC. managedD. to manage5.So far, Mike has spent about $28,000 on the house and __________ it to cost about $38,000 when it is finished and furnished.A. expected B .expects C. has expected D. will expect6.Although Beijing didn’t establish its first Antarctic research base until 1985, Chinese efforts _________ its influence across the continent are now outpacing other nation’s plans.A. expandingB. to be expandedC. expandedD. to expand7.It’s only after marriage and having children ________ a person understands others’ feelings.A. whatB. whenC. thatD. Who8.--- Why didn’t you ring me up when Sue arrived?---sorry, I ______ my essay that was to be handed in by noon.A. had writtenB. wroteC. have been writingD. was writing9.The agreements, _______ from culture and education to special training programs, will play a key role in the China-Latin America partnership.A. rangedB. having rangedC. rangingD. to range10. Almost every gym club in Shanghai ________ yoga classes. And there are more than 100 special clubs such as this one.A. has set upB. is setting upC. set upD. will set up11. In your life, sometimes you might end up back_________ you started.A. whichB. whenC. whereD. who12. When we entered, we found lying on the ground _______ along with her parents.A. was a girlB. were a girlC. a girl wasD. a girl were13. Wherever you prefer to work after graduation, _______ close contacts with your family members.A. keepingB. keepC. to keepD. kept14. Peace is necessary to all. After all, it is the United States and China, as the two largest economies in the world, that ________ most from a peaceful and stable Asia-Pacific.A. are benefitedB. will benefitC. will be benefitedD. had benefited15.One main reason comes from the Chinese tradition _______ the mother of the young woman usually requires her future son-in-law to prepare a house for their marriage, which brings financial pressure to the young man and his family.A. whatB. thatC. whichD. WhenSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A,B,C and D. Fill in each blank with a word or phrase that best fits the context.When I was in high school, I didn’t know anything about engin eering. At that time, when the car I drove needed repairing, I was afraid to take it to the 36 . Because honestly, the mechanic could have 37 me an electric can opener and said, ―This is part of your car and it’s38 -----pay me to fix it.‖ Then I wouldn’t have known any better.At the end of my junior year of high school, I heard about a summer program 39 to interest girls in engineering. The six-week program was free, and students were 40 college credits and a dorm room at the University of Maryland. I applied to the program, not because I wanted to be an engineer, but because I was looking forward to 41 and wanted to stay away from my parents for six weeks.I was accepted to the program and I earned six engineering credits. The next year I entered the university as an engineering student. Five years later I had a degree and three good job offers.I can’t help shaking42 I hear about studies that show women are 43 when it comes to math. They imply that I am a little stupid. I’m not, but I44 know that if I hadn’t met with that summer program, I wouldn’t be an engineer.When I was growing up, I was told, as many students were, to do what I was best at. But I didn’t know what that was. Most people think that when you are45 something, it comes easily to you. But this is what I discovered: just because a subject is 46 to learn, it does not mean you are not good at it. You just have to grit your teeth and work hard to get good at it. Once you do, there’s good47 that you will enjoy it more than anything else.36. A. engineer B. salesman C. factory D. mechanic37. A. shown B. given C. brought D. taken38. A. old B. broken C. fixed D. dirtied39. A. designed B. performed C. controlled D. described40. A. sent B. afforded C. offered D. awarded41. A. independence B. success C. happiness D. beauty42. A. until B. when C. since D. before43. A. at a disadvantage B. at a loss C. in a trap D. in a shock44. A. never B. do C. hardly D. nearly45. A. good at B. bad at C. interested in D. devoted to46. A. easy B. funny C. difficult D. interesting47. A. will B. feeling C. desire D. chanceSection C (12 points)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Male fans love to play football. Football is also a bonding experience for fathers and sons. When male football fans are younger, they probably spend hours 48. ________ watching football games with their fathers. Male football fans also probably spend a lot of time playing football 49.________ their friends.Female fans do love the game, too, 50. ________ most started watching football because a cute player caught their eyes. This is 51. ________ I started watching football. I was 52. ________ interested in football until I saw the movie Goal. While watching the movie, I spotted Kuno Becker and I quickly became interested in 53. ________ game. Other female fans probably became interested after watching a game with a boyfriend 54. ________ their husband. Whatever the reason why people start watching football 55. ________, it’s very addictive and hard to stop once you start.PART THREE: READING COMPREHENSION（30 points）Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are 4 choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.AThe Jungle BookThe Jungle Books were published in 1894 and 1895. Lost in the jungles of India as a child and adopted into a family of wolves, Mowgli is brought up on a diet of Jungle Law, loyalty, and fresh meat from the kill. Regular adventures with his friends and enemies in the jungles improve this child’s strength and cleverness and stir every reader’s imagination.Price: $ 7.79ISBN-13: 978-1613820742Average Customer Review: ★★★★A Stolen LifeThe first work of its kind—Jaycee Dugard’s personal life story, her own story of being kidnapped in 1991. When Jaycee was eleven years old, she was kidnapped from a school bus stop. She was missing for more than eighteen years, and gave birth to two daughters during her imprisonment.Price: $ 19.79ISBN-13: 978-1442344983Average Customer Review: ★★★★★While We’re Far ApartFive-time Christy Award winner Lynn Aus tin is called as ―one of the style’s best historical fiction novelists‖. Set in Brooklyn, New York, during World War I, while a motherless girl longs for her daddy and a young lady hopes for a second chance at love, this tale explores the uncertainty that stays in people in Europe.Price: $ 6.00ISBN-10: 0764204971Average Customer Review: ★★★★The Fashion PoliceThe Fashion Police was runner-up in the Chapter One Promotions Novel Competition 2010 and nominated (提名) Best Novel with Romantic Elements 2010 by The Romance Reviews. It is a strange comedy-mystery, combining murder with romance and chick-lit.For starters, Amber accidently shoots Chief Inspector Janice Skipper and gets thrown off the police force. The only one who knows the truth about the incident is Amber, but no one will believe her. After accepting a job as an insurance investigator from her ex-fiancé, Brad Beckett, it turns out that Brad thinks they’ve still got unfinished business and the job description includes sexual favors that come with a price.Price: $ 12.73ISBN-13: 978-1451555653Average Customer Review: ★★★56. If the book concerning a person who had children in prison appeals to you, you can choose thebook .A. A Stolen LifeB. The Jungle BookC. The Fashion PoliceD. While We’re Far Apart57. The ISBN of the book that gains least popularity among the research is .A. ISBN–13: 978–1613820742B. ISBN–13: 978–1442344983C. ISBN–10: 0764204971D. ISBN–13: 978–145155565358. If you want to buy one book about human’s living with animals and two about historical fiction, you have to pay .A. 19.79 dollarsB. 20.52 dollarsC. 27.58 dollarsD. 18.73 dollars59. What makes the book The Fashion Police distinguish itself from others?A. It wins the award Christy Award five timesB. It is a great book on violence and family educationC. It is named Best Novel with Romantic Elements 2010D. It has many elements borrowed from stories of true policemen.60. In which part of a newspaper would the information of the books be most likely to appear?A. ScienceB. OpinionC. LifestyleD. AdvertisementBAre morning people born or made? In my case it was definitely made. In my early 20s, I hardly went to bed before midnight, and I would always get up late the next morning.But after a while I couldn’t ignore the high relationship between success and rising early. On those rare occasions where I did get up early, I noticed that my productivity (效率) was always higher. So I set out to become a habitual early riser. But whenever my alarm went off, my first thought was always to stop that noise and go back to sleep. Eventually some sleep research showed that my strategy was wrong.The most common wrong strategy is this: sup posing you’re going to get up earlier, you’d bettergo to bed earlier. It sounds very reasonable, but will usually fail.There are two main schools (流派) of thought on sleep patterns. One is that you should go to bed and get up at the same time every day. The second school says you should go to bed when you’re tired and get up when you naturally wake up. However, I have found both are wrong if you care about productivity. If you sleep at fixed hours, you’ll sometimes go to bed when you aren’t sleepy enou gh. You’re wasting time lying in bed awake.My solution is to combine both methods. I go to bed when I’m sleepy and get up with an alarm clock at a fixed time. So I always get up at the same time (in my case 5 a.m.), but I go to bed at different times every night.However, going to bed only when I’m sleepy, and getting up at a fixed time every morning are my ways. If you want to become an early riser, you can try your own.61. According to the passage, the underlined phrase refers to ________.A. people who stay up until the next morningB. people who get up early in the morningC. people who feel sleepy in the morningD. people whose productivity is the lowest in the morning62. Why did the author want to become a habitual early riser?A. Because he / she found that the productivity was higher.B. Because he / she wanted to do morning exercise.C. Because he / she wanted to test which school is better.D. Because he / she wanted to have more sleep time.63. The author experienced all the following EXCEPT ________.A. going to bed after midnightB. asking scholars for advice on sleeping habitsC. getting up early occasionallyD. pressing off the alarm to go on sleeping64. What’s the author’s sleep pattern?A. Going to bed early and getting up early.B. Going to bed late and getting up late.C. Going to bed when sleepy and getting up at a fixed early time.D. Going to bed early and getting up late.65. The passage is mainly about ________.A. main schools of thought on sleep patternsB. how to have a good sleepC. wrong strategies for getting up earlyD. how to become an early riserCHere’s one number to keep in mind during your next cell phone conversation: 50. A new experiment shows that spending 50 minutes with an active phone pressed up to the ear increases activity in the brain. This brain activity probably doesn't make you smarter. When cell phones are on, they emit (发出) energy in the form of radiation that could be harmful, especially after years of cell phone usage. Scientists don't know yet whether cell phones are bad for the brain. Studies like this one are attempting to find it out.The 47 participants in the experiment may have looked a little strange. Each one had two Samsung cell phones attached to his or her head — one on each ear. The phone on the left ear was off. The phone on the right ear played a message for 50 minutes, but the participants couldn't hear itbecause the sound was off.With this set-up, the scientists could be sure they were studying brain activity from the phone itself, and not brain activity due to listening and talking during a conversation. After 50 minutes with two phones strapped to their heads, the participants were given PET scans.The PET scan showed that the left side (the side with the phone turned off) of each participant's brain hadn't changed during the experiment. The right side of the brain, however, had used more glucose, which is a type of sugar that provides fuel to brain cells. These right-side brain cells were using almost as much glucose as the brain uses when a person is talking. This suggests that the brain cells there were active ― even without the person hearing anything. That activity, the scientists say, was probably caused by radiation from the phone.Henry Lai, who works at the University of Washington in Seattle, is uncomfortable with the data related to cell phones. Holding a cell phone to your ear during a conversation is ―not really safe,‖ Lai told Science News. L ai is a bioengineer at the University of Washington in Seattle. He wrote an article about the new study for a journal, but he did not work on the study. Bioengineers bring together ideas from engineering and biology.For those who don't want to wait to find out for sure whether cell phones are bad for the brain, there are ways to talk more safely. You can have short and sweet conversations, use a speakerphone or keep the phone away from your head.66. Which of the following statement is true?A. Scientists are sure that cell phones are bad for the brain.B. In the experiment, the left side of the brain used more glucose.C. Radiation from the phone probably causes the change in the brain.D. Henri Lai wrote a lot of articles about this new study.67. Why weren’t the participants allowed to have a conversation on the phone during the experiment?A. Because the scientists want to be sure of the accuracy of the experiment.B. Because they really looked strange and no one wanted to talk to others.C. Because they were given PET scans and they lost the ability to talk.D. Because that would be too noisy and bad for the experiment.68. What is glucose?A. A type of sugar that provides vitamin to brain cells.B. Something that the right side of the brain used.C. A type of sugar that gives energy to brain cells.D. Something that makes a human excited.69. According to the last two paragraphs, which is the safest way to use a cell phone?A. Holding the cell phone close to your head.B. Using a cell phone more than three hours a day.C. Taking the most powerful cell phone.D. Keeping the cell phone at a distance.70. Where is this article probably taken from?A. Literature magazine.B. Science News.C. Story books.D. Art Journal.PART FOUR: WRITING（45 points）SECTION A (10 marks)Directions: Read the following passage. Complete the diagram / Fill in the numbered blanks by using the information for the passage.Write NO MORE THAN 3 WORDS for each answer.Lead poisoning (铅中毒) occurs when lead is present in high levels in the blood of a person. It is one of the most common medical problems found in children. And it’s important to recognize the presence of lead in children’s bodies because lead is very h armful.Children have the tendency to go around and put things in their mouths like paint peelings (油漆层剥皮). This is one of the main causes of lead poisoning. It has also been found that children born in poor families have higher chances of getting exposed to lead.Some of the common symptoms of lead poisoning in children are headaches, stomach pains, hair loss, muscle weakness and weight loss, etc.Lead poisoning can have severe effects and can lead to heart diseases, kidney diseases and neurological (神经系统的) problems. It can also cause mental dullness and a low IQ level. There have been cases where lead poisoning proved to be terminal for children.It’s difficult to easily recognize whether a child is affected by lead poisoning, as the symptoms of this disease can occur due to some other medical problems. Therefore, children under the age of six should regularly be checked to know the level of lead in their blood through blood tests. It’s possible to completely cure a child with lead poisoning through treatment. Usually the oral medicine chelator (螯合剂)is given to the children. Sometimes, children are given an injection(注射)for reducing the lead level. Children who have very high level of lead in their blood would need treatment for a long period of time and would also be put on a special diet.Parents have to take certain precautions to lower the chance of lead poisoning in children. Cleaning the house regularly and removing shoes at the door after coming in from outside will prevent the lead particles from entering the house. Children should be put on a healthy diet, containing high proportions (比例)of calcium and iron, such as eggs, milk, fruits, potatoes, etc, which helps to keep the lead levels in control in the children’s bodies.In short, it’s a must to be cautious and careful and protect your children from the horrible consequences of lead poisoning.Lead Poisoning in ChildrenⅠ. 71.* Children’s going around and 72. in their mouths.* High chances of blood lead.Ⅱ. Symptoms*Headaches, stomach pains, hair loss, muscle weakness, and weight loss, etc.Ⅲ. 73.* Leading to heart diseases, kidney diseases and neurological problems.* 74. in mental dullness and a low IQ level.Ⅳ. Treatment* Making children take the chelator or 75. an injection to reduce the lead level.* Putting children with a very high lead level on 76. .Ⅴ. Suggestions / Preventions* Cleaning the house regularly* 77. shoes at the door after coming in from the outside.* Letting children eat healthy food rich in 78. .Ⅵ. 79.* Being cautious and careful to 80. from lead poisoning.Section B (10 points)Section BRead the following passage. Answer the following questions according to the information given in the passage.Abby and Ally are my dear daughters. Last week, we had an outing and we were really tired later that day. After enjoying good food, we headed back. Abby pushed the button as we awaited the elevator and suddenly Ally let out a sigh of disappointment.“Oh wait, we can’t use the elevator,” Ally said, pointing to a sign near the elevator. Too tired to challenge her, both Abby and I followed her without looking around the corner to another set of elevators. “Oh, my god,” Ally said, visibly upset. “we can’t use the elevators either！They’re also reserved for the firefighters.” Abby looked at me. I looked at Abby. We both looked at the sign and then we stared at Ally to see if she was serious. She was. .“Ally, the sign does n’t say the elevators are reserved for firefighters,” Abby explained. “It says not to use the elevators in case of fire！With that, Ally took a longer look at the sign and actually read it instead of skimming it this time.What’s the lesson here? You have to spend time reading signs, maps or recipes instead of just assuming you know what they say, or you might go through life climbing many flights of stairs when you could be taking the elevator.It also got me thinking about how often we do this with reading. I wonder how many people just assume they know what a book says instead of actually spending time reading it, thus making wrong assumptions. How often have I neglected a book simply because I think I “know” what it says when I really just need to slow down, open the book, and focus on the reading?81. Why did Abby believe what Ally said about the first elevator? (No more than12 words)_______________________________________________________82. What did the sign really say? (No more than 9 words)________________________________________________________83. How did Ally read the sign at first? (No more than5 words)_________________________________________________________84. How should we read a book according to the author? (No more than 8 words)___________________________________________________________Section C (25 points)Directions: Write an English composition according to the instructions given below .Section C (25 marks)Directions: Write an English composition according to the instructions given below in Chinese.假设你是李华，在一个礼品网站为你的姐姐订购了一份生日礼物。

箴言中学高三第十次模拟考试理科综合试题可能用到的相对原子质量：H:1 C:12 N:14 O:16 S: 32 Na: 23 Si:28 Cl: 35.5 K:39 Ag:108 Fe:56第I卷一、选择题：本题包括13小题，每小题6分，共78分。

在每小题给出的四个选项中，只有一个选项符合题目要求。

1．20世纪80年代，Cech发现四膜虫的某种RNA具有催化活性。

20世纪90年代，Cuenoud发现某种单链DNA分子也具有催化活性。

下列相关说法中，不正确的是A．有催化活性的RNA和DNA分子在化学反应前后的质量不会发生变化B．组成RNA和单链DNA的单体共有8种，而碱基共有5种C．在该单链DNA分子中每个脱氧核糖上都连接有两个磷酸和一个碱基D．某些蛋白质也具有催化活性，这些蛋白质与题干中的核酸都是酶2.如图表示某植物叶肉细胞内光合作用的生理过程，下列相关叙述正确的是A．叶黄素缺失突变体与正常植株相比，若给予红光照射，则光吸收差异显著B.过程Ⅱ为暗反应，在类囊体薄膜上产生的2和3只能用于暗反应还原C3C．若降低周围环境中的CO2浓度，则释放O2的速率不变D．若该细胞净光合速率大于0，推测该植物体能积累有机物从而使植株干重增加3.下列关于变异及育种的说法正确的是A．染色单体之间的片段交换均属于基因重组B．单倍体植株的体细胞中可能存在同源染色体C．基因突变和染色体变异均可用光学显微镜观察到D．人工选择获得的优良品种都能适应不同的自然环境4.神经损伤后产生的信号会激活脑室下区的神经干细胞，这些细胞可以向脑内病灶迁移和分化，从而实现组织修复。

下列有关神经细胞的说法，不正确的是A．实现组织修复的过程是基因选择性表达的过程B．组成神经细胞膜的磷脂分子和大多数蛋白质分子是可以运动的C．损伤信号使神经干细胞膜内K+／Na+的值增大D．神经细胞释放神经递质的方式与胰岛素的分泌方式相同5.生物学是以实验为基础的一门学科，下列有关实验的叙述，不正确的是A．探究酵母菌细胞呼吸方式实验中，在酸陛条件下，重铬酸钾与酒精反应呈现灰绿色B．提取绿叶中的色素时，向研钵中加入碳酸钙的目的是防止研磨中色素被破坏C．在低温诱导植物染色体数目的变化实验中，利用卡诺氏液固定细胞形态后，需用体积分数为95％的酒精冲洗两次D．在土壤中小动物类群丰富度的研究实验中，记名计算法需按预先确定的多度等级来估计个体数目6.下图是外源性致敏原引起哮喘的示意图。

湖南省益阳市箴言中学化学自主招生试卷一、选择题1．有一镁的合金2.4g，在合金中加入100 g一定溶质质量分数的稀盐酸后，金属与稀盐酸恰好完全反应，产生氢气的质量为m，下列说法错误的是（）A．若是镁铝合金，则 m＞0.2 gB．若是镁铁合金，m=0.16g，则合金中铁的质量分数是 50%C．若是镁铜合金，则稀盐酸溶质的质量分数小于7.3%D．若是镁锌合金，则 m＜0.2g2．在AlCl3溶液中逐滴加入NaOH溶液至过量，发生如下反应：3NaOH+AlCl3=Al(OH)3↓+3Na Cl, Al(OH)3+NaOH=NaAlO2+2H2O。

已知NaAlO2易溶于水，则下列图像不正确的是( )A．B．C．D．3．有一包固体粉末，可能含碳、铝、铜、氧化铝、氧化铜中的一种或几种。

为探究该固体粉末的组成，某化学兴趣小组进行了如下图所示实验。

下列结论正确的个数是①固体B中的物质为碳②蓝色溶液乙为硝酸铜溶液③原固体样品中一定含有的物质是碳、铝、铜④蓝色溶液丙中一定含有的溶质是硝酸铝、硝酸铜、硝酸A．1个B．2个C．3个D．4个4．用数形结合的方法表示某些化学知识直观、简明、易记．下列用数轴表示正确的是（）A．不同物质的着火点：B．硫及其化合物与化合价的关系：C．50g19.6%的稀硫酸与足量的金属反应产生氢气的质量：D．物质形成溶液的pH：5．下列图像不能正确反映其对应变化关系的是A B C DA．用等质量、等浓度的过氧化氢溶液在有无催化剂条件下制氧气B．一定质量的红磷在密闭容器中燃烧C．向等质量、等浓度的稀硫酸中分别逐渐加入锌粉和铁粉D．向一定质量的氯化铜和稀盐酸的混合溶液中逐滴加入氢氧化钠溶液6．甲烷和水反应可以制备水煤气（混合气体），其反应的微观示意图如图所示，根据微观示意图得出的结论中，正确的是（）A．反应前后碳元素化合价没有发生变化B．反应中甲和丙的质量之比为4：7C．水煤气的成分是一氧化碳和氧气D．反应中含氢元素的化合物有三种7．小亮同学在实验室中制取CO2气体后，对废液进行后续探究，他向一定质量的含CaCl2和HCl的溶液中逐滴加入溶质质量分数为10%的Na2CO3溶液．实验过程中加入Na2CO3溶液的质量与生产沉淀或者气体如图1所示；加入Na2CO3溶液的质量与溶液的pH变化关系如图2所示，下列说法正确的是：( )A．图1中b→c段表示生成气体的过程B．图1中b点的值为106C．图1中0→a段反应过程中溶液的pH变化情况可用图2中d→h段曲线表示 D．图1中c点时，溶液中的溶质有两种8．许多物质在溶液中都以离子形式存在。

2015年湖南省益阳市箴言中学高考物理十模试卷（5月份）学校:___________姓名：___________班级：___________考号：___________一、单选题(本大题共5小题，共30.0分)1.亚里士多德在其著作《物理学》中说：一切物体都具有某种“自然本性”，物体由其“自然本性”决定的运动称之为“自然运动”，而物体受到推、拉、提、举等作用后的非“自然运动”称之为“受迫运动”．伽利略、笛卡尔、牛顿等人批判的继承了亚里士多德的这些说法，建立了新物理学；新物理学认为一切物体都具有的“自然本性”是“惯性”．下列关于“惯性”和“运动”的说法中不符合新物理学的是（）A.一切物体的“自然运动”都是速度不变的运动--静止或者匀速直线运动B.作用在物体上的力，是使物体做“受迫运动”即变速运动的原因C.可绕竖直轴转动的水平圆桌转得太快时，放在桌面上的盘子会向桌子边缘滑去，这是由于“盘子受到的向外的力”超过了“桌面给盘子的摩擦力”导致的D.竖直向上抛出的物体，受到了重力，却没有立即反向运动，而是继续向上运动一段距离后才反向运动，是由于物体具有惯性【答案】C【解析】解：A、一切物体的“自然运动”都是速度不变的运动--静止或者匀速直线运动，故A正确；B、作用在物体上的力，是使物体做“受迫运动”即变速运动的原因，故B正确；C、可绕竖直轴转动的水平圆桌转得太快时，放在桌面上的盘子会向桌子边缘滑去，这是由于“盘子需要的向心力”超过了“桌面给盘子的摩擦力”导致的，故C错误；D、竖直向上抛出的物体，受到了重力，却没有立即反向运动，而是继续向上运动一段距离后才反向运动，是由于物体具有惯性，故D正确；本题选不正确的，故选：C．惯性是物体的固有属性，一切物体都惯性，与物体的运动状态无关．力不是维持物体运动的原因，力是改变物体运动状态的原因．惯性是物体的固有属性，一切物体都有惯性，惯性的大小取决于物体的质量．关键在于平时的积累，对课本提到的各个定律，各种现象，要知道做出这个贡献的科学家．2.汽车甲和乙质量相等，以相等速率沿同一水平弯道做匀速圆周运动，甲车在乙车的外侧．两车沿半径方向受到的摩擦力分别为f甲和f乙．以下说法正确的是（）A.f甲小于f乙B.f甲大于f乙C.f甲等于f乙D.f甲和f乙大小均与汽车速率无关【答案】A【解析】解：汽车在水平弯道做圆周运动，靠静摩擦力提供向心力，根据牛顿第二定律得，f=m，因为两车的速率相等，质量相等，甲的轨道半径大，则甲的摩擦力小．故A正确，B、C、D错误．故选：A．汽车在水平弯道上做匀速圆周运动，靠静摩擦力提供向心力，根据轨道半径的大小，通过牛顿第二定律比较摩擦力的大小．解决本题的关键知道圆周运动向心力的来源，结合牛顿第二定律进行求解．3.17世纪，英国天文学家哈雷跟踪过一颗彗星，他算出这颗彗星轨道的半长轴约等于地球公转半径的18倍，并预言这颗彗星将每隔一定的时间飞临地球，后来哈雷的预言得到证实，该彗星被命名为哈雷彗星．哈雷彗星围绕太阳公转的轨道是一个非常扁的椭圆，如图所示．从公元前240年起，哈雷彗星每次回归，中国均有记录．它最近一次回归的时间是1986年．从公元前240年至今，我国关于哈雷彗星回归记录的次数，最合理的是（）A.24次 B.30次 C.124次 D.319次【答案】B【解析】解：设彗星的周期为T1，地球的公转周期为T2，由开普勒第三定律得：=，可知哈雷彗星的周期大约为76年，．所以最合理的次数是30次．故B正确，A、C、D错误．故选：B．因为地球和彗星的中心天体相等，根据开普勒第三定律（常数），通过半径关系求出周期比，从而得出哈雷彗星的周期，求出哈雷彗星回归记录的次数．解决本题的关键掌握开普勒第三定律（常数），通过该定律得出彗星与地球的公转周期之比．4.一足够长的倾角为θ的斜面固定在水平面上，在斜面顶端放置一长木板，木板与斜面之间的动摩擦因数为μ，木板上固定一力传感器，连接传感器和光滑小球间是一平行于斜面的轻杆，如图所示，现由静止释放木板，木板沿斜面下滑，稳定时传感器的示数为F1，当木板固定时，传感器的示数为F2．则下列说法正确的是（）A.稳定后传感器的示数一定为零B.tanθ=C.cotθ=D.cotθ=【答案】C【解析】解：当木板沿斜面下滑时，对整体分析，加速度a=gsinθ-μgcosθ，隔离对小球分析，mgsinθ-F1=ma，解得F1=μmgcosθ，当木板固定时，对小球分析，根据共点力平衡有：F2=mgsinθ，则，解得．故C正确，A、B、D错误．故选：C．当木板沿斜面下滑时，对整体分析，求出加速度，隔离对小球分析，求出传感器示数的表达式，当木板固定时，对小球分析，根据共点力平衡求出传感器示数的表达式，从而分析判断．本题考查了共点力平衡和牛顿第二定律的基本运用，掌握整体法和隔离法的灵活运用，知道木板沿斜面下滑时，小球和木板具有相同的加速度．5.如图所示，理想变压器原、副线圈匝数之比为n1：n2=22：1，原线圈接220V的正弦交变电压，副线圈连接理想交流电压表V、交流电流表A、理想二极管D和电容器C．则下述结论错误的是（）A.电压表的示数为10 VB.稳定后电流表的读数为零C.电容器不断地充电和放电，电量不断变化D.稳定后电容器两极板间电势差始终为10V【答案】C【解析】解：A、根据电压与匝数成反比可知，副线圈的电压为10V，所以A正确．BC、在电路没有稳定之前，由于二极管的作用，只有正向的电流可以通过，在电路稳定之后，由于电容器的隔直流的作用，就没有电流通过了，所以电容器不会反复的充电和放电，所以B正确，C错误．D、电容器两极板间电势差为副线圈的最大的电压，即为V，所以D正确．本题选错误的，故选：C．由电压与匝数成反比可以求得副线圈的电压的大小，二极管的作用是指允许正向的电流通过，电容器的作用是通交流隔直流．本题需要掌握变压器的电压之比和匝数比之间的关系，同时对于二极管和电容器的作用要了解．二、多选题(本大题共3小题，共18.0分)6.如图，一个人站在商场内自动扶梯的水平踏板上，随扶梯向上加速运动，下列说法正确的是（）A.踏板对人做的功等于人的机械能的增加量B.踏板对人的支持力做的功等于人的机械能的增加量C.克服人的重力做的功等于人的重力势能的增加量D.重力和踏板对人的支持力做的总功等于人的动能的增加量【答案】AC【解析】解：A、人随扶梯一起斜向上加速运动的过程中，人相对于扶梯有向后运动趋势，人要受到扶梯的静摩擦力．则人受到重力、踏板对人的支持力和静摩擦力三个力作用，根据功能关系得知，除了重力以外的力对人做功等于人的机械能的增加，即踏板对人做的功等于人的机械能增加量．故A正确，B错误；C、根据功能关系可知，重力势能的增加量等于克服重力做的功，故C正确；D、根据动能定理得知，人所受合力做的功等于人的动能的增加量．故D错误．故选：AC人随扶梯一起斜向上加速运动的过程中，受到重力、踏板对人的支持力和静摩擦力三个力作用，根据竖直方向的加速度方向分析人对踏板的压力大小与人所受到的重力大小关系．踏板对人做的功等于人的机械能增加量．人所受合力做的功等于人的动能的增加量．对于受力分析，一般按重力、弹力和摩擦力的顺序进行分析．对于功能关系，要理解并掌握几对常见的对应关系：合力做功等于动能的变化，除重力以外的力做功等于机械能的变化，等等7.两点电荷q1、q2固定在x轴上，在+x轴上每一点的电势φ随x变化的关系如图所示，其中x=x0处的电势为零，x=x1处的电势最低．下列说法正确的是（）A.x=x0处的电场强度为零 B.x=x1处的电场强度为零C.q1带正电、q2带负电D.q1的电荷量比q2的大【答案】BD【解析】解：A、电势φ随x变化的关系图线上每点切线的斜率大小等于电场强度，即E=，x=x0处的电场强度不为零，x=x1处的电场强度为零，故A错误，B正确．C、因为x1点的场强为0，所以两点的电荷在x1点产生的场强大小相等，方向相反，两电荷为异种电荷，根据沿电场线方向电势逐渐降低，q1带负电、q2带正电．故C错误D、因为x1点的场强为0，所以两点的电荷在x1点产生的场强大小相等，方向相反，根据E=k，知距离大的电量大，所以q1的电荷量大于q2的电荷量．故D正确故选：BD解答本题应抓住：因为电势φ随x变化的关系图线上每点切线的斜率为（△x→0），表示电场强度E，所以可知x1点的场强为0．根据P点的场强为0，由点电荷场强公式分析两电荷的电量大小．根据各点场强为0，知两电荷是同种电荷，根据电势的变化，判断电荷的电性．根据电场力做功判断电势能的变化解决本题的关键掌握电势φ随x变化的关系图线上每点切线的斜率表示电场强度E．以x1点场强为0作为突破口，展开分析8.如图所示，电阻不计的平行导轨竖直固定，上端接有电阻R，高度为h的匀强磁场与导轨平面垂直．一导体棒从磁场上方的A位置释放，用x表示导体棒进入磁场后的位移，i表示导体棒中的感应电流大小，υ表示导体棒的速度大小，E K表示导体棒的动能，a表示导体棒的加速度大小，导体棒与导轨垂直并接触良好．以下图象可能正确的是（）A. B.C. D.【答案】AC【解析】解：A、导体棒进入磁场后，受到向上的安培力和重力，如果安培力大于重力，导体棒将做减速运动，安培力F=减小，当安培力减小到等于重力时，做匀速运动；电流I=，由于速度先减小后不变，故电流也是先减小，后不变，故A正确；B、导体棒进入磁场后，受到向上的安培力和重力，如果安培力大于重力，导体棒将做减速运动，根据动能定理，有：-F A x+mgx=E k-E K0故E k=-F A x+mgx+E K0由于减速运动时安培力是变力，故对应的E k-x图不是直线，故B错误；C、导体棒进入磁场后，受到向上的安培力和重力，如果安培力等于重力，导体棒的加速度为零；离开磁场后只受重力，加速度为g；故C正确；D、导体棒离开磁场后只受重力，加速度为g，即x＞h部分的v-t图象的斜率不可能为零，故D错误；故选：AC．导体棒进入磁场后，受到向上的安培力和重力，可能加速、匀速、减速运动，根据牛顿第二定律和动能定理列式分析各个图象即可．本题关键明确导体棒的受力情况和运动情况，要分情况讨论；注意推导出表达式进行分析，不难．五、多选题(本大题共1小题，共6.0分)13.若以V表示在标准状态下水蒸气的摩尔体积，ρ表示在标准状态下水蒸气的密度，M表示水的摩尔质量，M0表示一个水分子的质量，V0表示一个水分子的体积，N A表示阿伏加德罗常数，则下列关系式中正确的是（）A.V=B.V0=C.M0=D.ρ=E.N A=【答案】ACE【解析】解：A、体积等于摩尔质量除以密度．则A正确B、因气体间有大的空隙，则B错误C、阿伏加德罗常数N A个原子的质量之和等于摩尔质量，则C正确D、因N A v0并不等于摩尔体积，则D错误E、ρV为摩尔质量，摩尔数等于质量与摩尔质量之比，则E正确故选：ACE密度等于摩尔质量除以摩尔体积，摩尔数等于质量与摩尔质量之比．阿伏加德罗常数N A个原子的质量之和等于摩尔质量．而对水蒸气，由于分子间距的存在，N A v0并不等于摩尔体积本题的解题关键是建立物理模型，抓住阿伏加德罗常数是联系宏观与微观的桥梁，也可以将水分子看成立方体形七、多选题(本大题共1小题，共4.0分)15.已知氘核的比结合能是1.09M e V，氚核的比结合能是2.78M e V，氦核的比结合能是7.03M e V．在某次棱反应中，1个氘核和1个氚核结台生成1个氮核，则下列说法中正确的是（）A.这是一个裂变反应B.核反应方程式为H+H→H e+nC.核反应过程中释放的核能是17.6M e VD.目前核电站都采用上述核反应发电E.该核反应会有质量亏损【答案】BCE【解析】解：A、1个氘核和1个氚核结台生成1个氮核，这是聚变反应．故A错误；B、1个氘核和1个氚核结台生成1个氮核，根据质量数与质子数守恒知同时有一个中子生成，反应方程为H+H→H e+n．故B正确；C、根据质能方程△E=△mc2得一次聚变释放出的能量：△E=E2-E1=7.03×4-（2.78×3+1.29×2）=17.6M e V；故C 正确；D、目前核电站都采用核裂变发电．故D错误；E、该反应放出热量，所以一定有质量亏损．故E正确．故选：BCE1个氘核和1个氚核结台生成1个氮核，这是聚变反应；由质量数守恒和电荷数守恒判定反应方程是否正确；由质能方程判断出释放的核能；目前核电站都采用核裂变发电．本题考查了核反应方程的书写以及质能方程的简单应用，属于简单基础题目，平时练习中对这类问题注意多加训练，不可忽视．三、实验题探究题(本大题共2小题，共15.0分)9.某实验小组用如图甲所示装置测量木板对木块的摩擦力所做的功．实验时，木块在重物牵引下向右运动，重物落地后，木块继续向右做匀减速运动．图乙是重物落地后打点计时器打出的纸带，纸带上的小黑点是计数点，相邻的两计数点之间还有4个点（图中未标出），计数点间的距离如图乙所示．已知打点计时器所用交流电的频率为50H z．（1）根据纸带提供的数据可计算出打点计时器在打下A点、B点时木块的速度v A、v B，其中v A= ______ m/s．（结果保留两位有效数字）（2）要测量在AB段木板对木块的摩擦力所做的功W AB，还应测量的物理量是______ ．（填入物理量前的字母）A．木板的长度l B．木块的质量m1C．木板的质量m2 D．重物的质量m3E．木块运动的时间t F．AB段的距离x AB （3）在AB段木板对木块的摩擦力所做的功的关系式W AB= ______ ．（用v A、v B和第（2）问中测得的物理量的字母表示）【答案】0.72；B；【解析】解：（1）重物落地后，木块由于惯性继续前进，做匀减速直线运动，相邻计数点间的距离逐渐减小，故纸带向右运动，故其右端连着小木块；计数点间的时间间隔t=0.02s×5=0.1s，纸带上某点的瞬时速度等于该点前后相邻两个点间的平均速度，打A点时的速度v A=m/s=0.72m/s，（2）木块在运动过程中，克服摩擦力做的功等于木块动能的减小量，由动能定理得：木块克服摩擦力做的功为：W f=因此实验过程中还需要用天平测出木块的质量m1，故ACDEF错误，B正确．故选：B．（3）在AB段对木块，由动能定理得：W AB=，因此在AB段木板对木块的摩擦力所做的功的关系式W f=故答案为：（1）0.72（2）B；（3）（1）重物落地后，木块由于惯性继续前进，做匀减速直线运动，相邻计数点间的距离逐渐减小；纸带上某点的瞬时速度等于该点前后相邻两个点间的平均速度；（2）克服摩擦力做的功等于动能的减小量，故需要天平测量质量．（3）由动能定理可以求出木板对木块的摩擦力所做的功．本题关键要明确实验的原理和实验的具体操作步骤，然后结合匀变速直线运动的规律和动能定理进行分析判断．10.某物理兴趣小组要精确测量一只电流表G（量程为1m A、内阻约为100Ω）的内阻．实验室中可供选择的器材有：电流表A1：量程为3m A内阻约为200Ω；电流表A2：量程为0.6A，内阻约为0.1Ω；定值电阻R1：阻值为10Ω；定值电阻R2：阻值为60Ω；滑动变阻器R3：最大电阻20Ω，额定电流1.5A；直流电源：电动势1.5V，内阻0.5Ω；开关，导线若干．（1）为了精确测量电流表G的内阻，你认为该小组同学应选择的电流表为______ 、定值电阻为______ ．（填写器材的符号）（2）在方框中画出你设计的实验电路图．（3）按照你设计的电路进行实验，测得电流表A的示数为I1，电流表G的示数为I2，则电流表G的内阻的表达式为r g= ______ ．【答案】A1；R2；【解析】解：（1）待测电流表G量程是1m A，因此可以选电流表A1：量程为3m A，内阻约为200Ω；当通过电流表A1的电流等于其量程3m A时，电路最小电阻约为R===500Ω，则定值电阻最小应为R定=R-R A1-R G=500Ω-200Ω-100Ω=200Ω，为保证电路安全，定值电阻阻值应大一点，定值电阻应选R2：阻值为60Ω；（2）待测电流表与定值电阻并联，然后由电流表测出并联电流，滑动变阻器采用分压接法，电路图如图所示．（3）电流表A的示数为I1，电流表G的示数为I2，通过定值电阻的电流为I=I1-I2，电流表G两端的电压U=IR2，待测电流表内阻r g==；故答案为：（1）A1；R2；（2）电路图如图所示；（3）=．（1）根据待测电流表的量程选择电流表，由欧姆定律求出电路最大电流时电路的最小电阻，根据该电阻选择定值电阻．（2）没有电压表，可以把待测电流表与定值电阻并联，然后由电流表测出并联电流，然后由并联电路特点及欧姆定律求出待测电流表内阻；为了多次测量，可以使用滑动变阻器的分压接法；据此设计实验电路．（3）由并联电路特点及欧姆定律求出电流表G的内阻．本题考查了实验器材的选取、实验电路的设计、求电阻等问题；选择实验器材时，首先要保证电路安全，在保证安全的情况下，为使读数准确，电表量程及电阻阻值应选小的．实验电路的设计是本题的难点，没有电压表，利用并联电路特点求出待测电流表两端电压是常用的方法．四、计算题(本大题共2小题，共32.0分)11.某汽车训练场地有如图设计，在平直的道路上，依次有编号为A、B、C、D、E的五根标志杆，相邻杆之间的距离△L=12.0m．一次训练中，学员驾驶汽车以57.6km/h的速度匀速向标志杆驶来，教练与学员坐在同排观察并记录时间．当教练经过O点时向学员发出指令：“立即刹车”，同时用秒表开始计时．忽略反应时间，刹车后汽车做匀减速直线运动，停在D标杆附近．教练记录自己经过C杆时秒表的读数为t C=6.0s，已知L OA=36m，教练距车头的距离△s=1.5m．求：（1）刹车后汽车做匀减速运动的加速度大小a；（2）汽车停止运动时，车头离标志杆D的距离△x．【答案】解：（1）汽车从O到标志杆B的过程中：L OA+△L=v0△t+v0（t B-△t）-（t B-△t）2汽车从O到标志杆C的过程中：L OA+2△L=v0△t+v0（t C-△t）-（t C-△t）2联立方程组得：△t=0.5s；a=2m/s2（2）汽车从开始到停下运动的距离：可得x=72m因此汽车停止运动时车头前端面在CD之间离DL OA+3△L-△s-x=44+36-1.5-72=6.5m．答：（1）车开始刹车后做匀减速直线运动的加速度大小a为2m/s2；（2）汽车停止运动时车头前端面离D的距离为6.5m．【解析】（1）学员甲在反应时间△t内，汽车做仍匀速运动，刹车后做匀减速运动．汽车从O 到标志杆B的过程中和汽车从O到标志杆C的过程中分别列位移方程，联立求解速度和加速度．（2）先求出汽车从开始到停下运动的距离，在根据位移关系求汽车停止运动时车头前端面离D的距离．此题要理解反应时间内汽车继续做匀速运动，还要养成画运动过程示意图，找位移之间的关系．此题有一定的难度12.如图所示的xoy坐标系中，x轴上方，y轴与MN之间区域内有沿x轴正向的匀强电场，场强的大小E1=1.5×105N/C；x轴上方，MN右侧足够大的区域内有垂直于纸面向里的匀强磁场，磁感应强度大小B=0.2T．在原点O处有一粒子源，沿纸面向电场中各方向均匀地射出速率均为v0=1.0×106m/s的某种带正电粒子，粒子质量m=6.4×10-27kg，电荷量q=3.2×10-19C，粒子可以无阻碍地通过边界MN进入磁场．已知ON=0.2m．不计粒子的重力，图中MN与y轴平行．求：（1）粒子进入磁场时的速度大小；（2）求在电场中运动时间最长的粒子射出后第一次到达坐标轴时的坐标；（3）若在MN右侧磁场空间内加一在xoy平面内的匀强电场E2，某一粒子从MN上的P点进入复合场中运动，先后经过了A（0.5m，y A）、C（0.3m，y c）两点，如图所示，粒子在A点的动能等于粒子在O点动能的7倍，粒子在C点的动能等于粒子在O点动能的5倍，求所加电场强度E2的大小和方向．【答案】解：（1）由动能定理得：代入数据解得：v=2×106m/s（2）粒子在磁场中，由解得：r=0.2m在电场中运动时间最长的粒子沿+y轴出发作类平抛运动，后从Q点进入磁场，进入磁场的方向与NM的夹角为θ，由类平抛运动的规律得：，则θ=60°粒子在磁场中作匀速圆周运动从T点回到电场，由对称规律可得将在H点第一次与y 轴相切，轨迹如图．对于O到Q的类平抛运动，有：NQ=v0tON=at2由牛顿第二定律得q E1=ma联立解得：NQ=m弦长：QT=2rsinθ=m所以：y H=2NQ+QT解得：y H=m（3）粒子从P点进入磁场时的动能为：MN右侧磁场空间内加一在xoy平面内的匀强电场后，设A点的电势为U A，C点的电势为U C，取P点零电势点，则由动能定理得：q（U P-U A）=E k A-E k P q（U P-U C）=E k C-E k P解得：，在AP连线上取一点D，设，则由匀强电场特性可知U P-U A=3（U P-U D）由几何知识可得：x A-x P=3（x D-x P）解得：，x D=0.3m=x C，即x坐标相同的两点为等势点，A点电势低于P点的电势，所加电场沿x轴正方向则有：U P-U C=E2（x C-x P）联立以上各式并代入数据解得：N/C答：（1）粒子进入磁场时的速度大小是2×106m/s；（2）在电场中运动时间最长的粒子射出后第一次到达坐标轴时的坐标是m；（3）所加电场强度E2的大小为1.0×105N/C，方向沿x轴正方向．【解析】（1）粒子先在电场中加速运动，由动能定理求解加速后的速度，即粒子进入磁场时的速度大小；（2）在电场中运动时间最长的粒子沿+y轴出发作类平抛运动，后从Q点进入磁场，进入磁场的方向与NM的夹角为θ，由类平抛运动的规律求出θ．由牛顿第二定律和分位移公式、几何关系结合求解．（3）设A点的电势为U A，C点的电势为U C，取P点零电势点，由动能定理求得U A、U C与动能的关系，由电场的特性和几何关系求解．本题考查带电粒子在电场中和磁场中的运动，理清粒子的运动规律是解决本题的关键，处理粒子在磁场中运动问题，要会确定粒子做圆周运动的圆心、半径和圆心角．六、计算题(本大题共1小题，共9.0分)14.如图1所示，左端封闭、内径相同的U形细玻璃管竖直放置，左管中封闭有长为L=20cm的空气柱，两管水银面相平，水银柱足够长．已知大气压强为p0=75cm H g．（1）若将装置翻转180°，使U形细玻璃管竖直倒置（水银未溢出），如图2所示．当管中水银静止时，求左管中空气柱的长度；（2）若将图1中的阀门S打开，缓慢流出部分水银，然后关闭阀门S，右管水银面下降了H=35cm，求左管水银面下降的高度．【答案】解：（1）设左管中空气柱的长度增加h，由玻意耳定律：p0L=（p0-2h）（L+h）代入数据解得：h=0或h=17.5cm所以，左管中空气柱的长度为20cm或37.5cm（2）设左管水银面下降的高度为x，左、右管水银面的高度差为y，由几何关系：x+y=H由玻意耳定律：p0L=（p0-y）（L+x）联立两式解得：x2+60x-700=0解方程得：x=10cm x=-70cm（舍去）故左管水银面下降的高度为10cm答：（1）左管中空气柱的长度为20cm或37.5cm（2）左管水银面下降的高度为10cm【解析】（1）根据玻意耳定律求的即可（2）气体发生等温变化，由玻意耳定律求出气体的压强，然后再求出水银面下降的高度本题考查了求水银面下降的高度，根据题意求出气体的状态参量，应用玻意耳定律即可正确解题，解题时要注意几何关系的应用八、计算题(本大题共1小题，共10.0分)16.如图所示光滑水平直轨道上有三个滑块A、B、C质量分别为m A=m C=2m和m B=m，A、B用细绳相连，中间有一压缩的弹簧（弹簧与滑块不栓接），开始时A、B以共同速度V0向右运动，C静止，某时刻细绳突然断开，A、B被弹开，然后B又与C发生碰撞并粘在一起，最终三者的速度恰好相同．求：（1）B与C碰撞前B的速度（2）弹簧释放的弹性势能多大．【答案】解：（1）设三者最后的共同速度为v共，由动量守恒得：（m A+m B）v0=m A v A+m B v Bm B v B=（m B+m C）v共三者动量守恒得：（2m+m）v0=（2m+m+2m）v共所以得共（2）弹簧释放的弹性势能．答：（1）B与C碰撞前B的速度为．（2）弹簧释放的弹性势能为．【解析】（1）A、B组成的系统，在细绳断开的过程中动量守恒，B与C碰撞过程中动量守恒，抓住三者最后速度相同，根据动量守恒定律求出B与C碰撞前B的速度．（2）根据能量守恒定律求出弹簧的弹性势能．。