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trocare hp 100 物理安全技术说明书
安全技术说明书Trocare HP100修订：20.12.2015 1.产品和企业标识商品名：Trocare HP100检测产品：己脒定二（羟乙基磺酸）盐(EU号:211-533-5;CAS:659-40-5)用途：抗菌剂供应商名称、地址、电话及传真：广州同隽医药科技有限公司中国广东省广州市萝岗区科学城科丰路31号华南新材料创新园G11栋501号邮编510663电话：+86(20)82350801传真：+86(20)82003600紧急联络信息：+86(20)823508022.危险性标识依据(EC)1272/2008对人和环境有特别危害：警示词：警告危险性说明：H302–食入有害H319–对眼睛有严重刺激H411–与水生生物有害防范说明：P262–不要接触眼睛，皮肤或衣服P280e–穿戴保护眼睛/脸部的防护器具P301+P312–如果吞食：若感不适，立即打电话给毒物咨询中心或送医P305+P351+P338–若接触眼睛：小心翻转眼脸，用水冲洗数分钟。
若方便，摘除隐形眼镜后继续冲洗P273–避免泄露到环境中3.成分/组分信息化学通用名称：己脒定二（羟乙基磺酸）盐(EU号:211-533-5;CAS:659-40-5)4.急救措施如吸入：移至空气新鲜处，就医诊治皮肤接触：清水冲洗。
脱下污染衣物眼睛接触：立即用大量流动清水清洗10分钟以上，出现刺激，咨询眼科医生摄食：立即就医诊治并展示此包装容器或标签5.消防措施适宜的灭火介质：泡沫，干粉，二氧化碳水喷雾消防人员特殊保护设备：穿全防护服戴自给式呼吸器燃烧产物：氮氧化物，硫氧化物6.泄露应急处理个人防范：防尘口罩避免皮肤和眼睛接触环境污染预防：不得排入排水沟/地表水系/地下水系中清理或收集方法：器械清除。
用大量水冲洗掉残留7.操作处置和储存操作处置：确保作业场所通风良好防止静电积累8.接触控制/个人防护系统设计的附加信息：如果操作引起的粉尘，则工厂必须配备吸尘设备确保作业场所通风良好特殊防护：注意：一般粉尘限值为6毫克/立方米（细粉尘浓度）呼吸防护：防尘口罩双手防护：合适的防护手套，例如德国的KCL GmbH,D-36124Eichenzell,邮件:***************,应符合以下规格(依据EN374测试)。
天津市第一百中学2022-2023高二上学期期末网测英语试题
天津市第一百中学2022-2023高二上学期期末网测英语试题1. Mary can’t attend the conference at 3 o’clock this afternoon because she ________ an event at that time.A．will beinterviewingB．would interview C．has interviewed D．will interview 2. Last week they went to the school library to do some reading, only to be told that it ________.A．was decorated B．had been decoratingC．was being decorated D．had decorated3. Two months ago the official was sent to the village to see how the plan ________ in the past two years.A．would be carried out B．had been carriedoutC．was carried out D．has been carriedout4. ________I have to give a speech, I get extremely nervous before I start.A．Whatever B．Whenever C．Whoever D．Whichever 5. Integrity is a kind of quality -- and that’s what ________ takes to do anything well.A．they B．that C．this D．it6. MSF is an independent organization mission is to provide medical care to those affected by conflicts, epidemics or disasters.A．which B．when C．where D．whose7. From my perspective, ________ every morning will surely benefit our health in the long run.A．take exercise B．taking exercise C．taken exercise D．being takenexercise8. Our headteacher sincerely expects his students to________and make more contributions to our country.A．be stricken blind and deaf B．follow in the hero’s footstepsC．Think on their feet D．roar with laughter9. The manager encouraged his employees to have a second try before they ________ so that they would have no regret.A．did the trick B．threw in thetowel C．thought on theirfeetD．sharpened theirskills10. The experiment group , ________ about 40 adults of similar age and education, will be assessed by language skills and test.A．be engaged in B．is coexisted with C．consisting of D．be paid attentionto11. Tom has to keep his pet dog outdoors because it often gets________ when he is doing the housework.A．in a row B．in his way C．in vain D．underconstruction12. The college students decided to ________ their life to the construction of their hometown.A．approve B．cherish C．dedicate D．digest13. The headmaster, as well as the headteachers, believes it’s vital to ________ Internet’s influences on children from those of the family.A．distinguish B．update C．approve D．swallow14. The investigation shows that too much love can make boys less independent, weaker and less able to ________ frustration.A．switch on B．cope with C．take advantage of D．develop anappetite for15. As is reported , mothers in the big cities started their ________work at 6 am and finished at around 11pm.A．virtue B．commitment C．digestion D．routine16. Life is one ride. Sometimes you are the passenger, sometimes you are the driver, and sometimes your passengers or driver can become a ________ and lifelong friend.That is ________ what happened to Latonya and Kevin. While ________ for Uber one night to pay for utility bills, Latonya ________ Kevin to drive him home. The two started talking.During the long drive, Latonya and Kevin ________ one another about what was really going on. As they got to know each other more and more, Latonya said that she ________ of receiving higher education. Latonya had enrolled at Georgia State University in 2010 ________ was forced to drop out due to financial reasons. With tuition being too ________ and her dream destroyed, Latonya thought that her dream remained out of reach, but she would go back to school and her dream would become a ________. When Kevin left, he encouraged her to keep going and asked Latonya to tell him the enrollment process.Some weeks later, Latonya attempted to re-enroll at Georgia State University, but there was a________. She was told that there was a $ 693 balance (结欠) from her ________ studies eight years before that had to be paid before she could ________ her studies.Well , not all heroes wear capes(披肩), because Kevin went straight down to the university and paid off the amount without Latonya’s ________ . Once the balance was paid, Latonya could ________ the university to study. She was shocked at Kevin’s ________ and promised to pay him ________ , but he didn’t ________ payment, saying “ the best payment is graduation”.Latonya ________ ! On the graduation day, Kevin was there. Kevin said that when Latonya made her graduation speech he felt ________ . Latonya talked about how one important Uber ride changed her life. “He ________ me,” Latonya says of her Uber passenger.1.A．dependable B．legal C．life-changing D．life-saving2.A．obviously B．expectedly C．generally D．exactly3.A．waiting B．working C．searching D．leaving4.A．picked up B．called on C．caught up with D．came up with 5.A．looked after B．relied on C．put up with D．opened up to6.A．spoke B．dreamed C．heard D．complained7.A．beyond B．regardless of C．but D．on behalf of8.A．high B．average C．minimum D．different9.A．prediction B．ambition C．disaster D．reality10.A．mistake B．pity C．change D．problem11.A．previous B．historic C．independent D．current12.A．check B．continue C．stop D．choose13.A．support B．evidence C．assistance D．permission14.A．visit B．establish C．enter D．finish15.A．kindness B．integrity C．leadership D．thinking16.A．within B．back C．despite D．without17.A．examine B．return C．borrow D．want18.A．succeeded B．attempted C．agreed D．regretted19.A．bittersweet B．embarrassed C．proud D．frustrated20.A．understood B．persuaded C．encouraged D．praised17. Community Development Projects OverseasThe aim of Community Development projects abroad in Temple Conservation, DIY and more is to improve poorer local communities. You can do them as part of your work experience, gap year or career break. Or just as a volunteering holiday.South AfricaTime :1st, March --31st, May 2018Registration fee: $ 65Maintenance / Handyman / DIY skills, this project would love to have you! The project is truly unique ....you will assist an Animal Rehabilitation Center that desperately needs any help it can get and you will be benefiting the animals at the same time! And, of course, the location is very unusual. And finally, conservation are lovely people to work with!Sri Lanka:Time :1st,May --31st, July 2018Registration fee: FreeBuddhist Temple Repair and Construction Project in KandyRebuild and repair Buddhist temple in and around Kandy. This unique and fascinating project offers an opportunity to learn about Sri Lankan construction using traditional manual methods, without using modern machines.Work includes construction, repairing temples, cleaning in and around temples, interacting with children and villagers and helping the Temples arrange Community Centers for the village youths. You will also assist with the repair of paintings and sculptures inside the temples.Zimbabwe:Time :1st, October-31st, December , 2018Registration fee: $ 120Community Healthcare and Medical Project in Gweru gain basic medical experience in local clinics and play a role in HIV/AIDS education and awareness. You will be able to get involved in not only this clinic but others in the nearby area as well. You do not need to be qualified in any particular area. You will work with children from the ages of 3 to 6 years. It is a very worthwhile, enjoyable placement.1. What can we know about Community Development project?A．They can better poorer local community.B．They bring you a chance to travel abroad.C．They can make you understand foreign countries well.D．They can make you more experienced in your future work.2. Which of the following is needed in South Africa?A．People with experience of caring for animals.B．People who are experienced in painting.C．People with experience in treating AIDS.D．People with rich travelling experience.3. How will the temple repair be done in Sri Lanka?A．It will be directed by experienced expert.B．The repair work can only be done by hand.C．All work will be done through modern machines.D．The construction workers do not have to be experienced.4. What is the project of Zimbabwe aimed at?A．The community and education.B．Food and environments.C．Health and medicine.D．Animals and resources.5. What is the similarity of these three projects?A．They all charge some fees.B．They all need particular experience.C．They all need you to be good at foreign languages.D．The length of the working time of the three is the same.18. Sabera Hossain’s passion for helping others is endless.Hossain, a senior at East Meadow High School in New York, has taken part in humanitarian (慈善的) tasks bo th at home and abroad, from fighting poverty as president of her school’s World Hunger Action Club to teaching English to children in Bangladesh.While in the South Asian country last summer, she spent six weeks volunteering at a school she called “a shack on the banks of a lake”. She provided a group of about 30 students, aged 4 — 8, with homemade educational materials, including cards and game booklets, to help them learn numbers and words.But Hossain’s humane efforts don’t end there. She has also volunt eered at senior-living communities and Nassau University Medical Center in New York. She’s even worked with her local Volunteer Ambulance Corps, where she was trained to help give emergency treatment, alongside nursing and medical technicians.“I feel like everyone needs to go out and see what the world has to offer,” Hossain said. “That’s why I’ve done so much; I wanted to explore what I like and don’t like.”You won’t be surprised to know she’s also a star student. Hossain is an AP Scholar with Distinctio n and earned a score of 1,550 on the SAT.She is also a member of the math team and Pre-Medicine Club. Sabera also earned a membership in the National Academy of Finance, which came after she passed four standardized exams and held a 120-hour internship in the business field.“Sabera is someone who is always looking for a challenge, eager to travel the world, experience new cultures, meet new people and make a difference in the lives of others,” said her teacher, Joanna Silberman. “I can think of no one more mature and ready to begin the next journey of her life.”This year, Hossain will attend college, but she has not decided on a major. She is most looking forward to widening her horizons and finding new things that she likes. Big colleges offer big benefits to students.1. What can we learn about Hossain according to paragraph 2 ?A．Hossain has participated in charity tasks only at home.B．Hossain has never joined in charity tasks abroadC．Hossain has devoted herself to the different charity tasks full of passion for helping othersD．Hossain has seldom succeeded in fighting poverty2. What was Hossain’s intention when she was in the South Asian country?A．To spend six weeks at a school.B．To help students there to learn numbers and words.C．To provide students with homemade educational materials.D．To teach students how to make cards and game booklets.3. Where has Hossain also volunteered apart from the schools?A．Hossain has also volunteered at Nasssan University Medical Center.B．Hossain has volunteered in Pre-Medicine Club.C．Hossain has ever worked in the National Academy of Finance.D．Hossain has also volunteered at senior-living communities and in her Local VolunteerAmbulance.4. What is Hossain’s ideal college like?A．It can broaden her viewpoint and provide new things which will interest her.B．It can offer a chance to experience the new cultureC．It can make a great difference in the lives of others.D．It can provide a platform to meet new people.5. What is the main idea of the passage?A．Hossain’s background.B．Hossain’s encouragement.C．Worldwide volunteering D．Volunteering’s popularity19. Does the lemon, the famous fruit of Limone, contain a secret ingredient that fights heart disease? Limone a small lemon-growing town on the shores of Lake Garda in northern Italy, holds a mystery which has made it popular with tourists in search of health. About 30 years ago scientists discovered that many people from the town had a unique protein in their blood. Thanks to it, the town’s p eople remained unaffected by heart disease even if they smoked, drank alcohol or ate large quantities of animal fat.The discovery was wonderful for the people who lived in Limone; nowadays a million tourists visit their town between March and November each year. That makes 4,000 tourists per day - four times the actual population of the town. It is popularly believed that the protein must come from the lemons, which the town has grown for centuries. Until the discovery of the protein, the fruit was the only real source of income for the townspeople.Lemons, which were brought to Europe from the Middle East in the 12th century, have many well-documented power. Over the years they have been used to treat all sorts of illnesses. At its peak in the 19th century , the lemon industry in Limone produced as many as 15 million of the fruit annually. The crop was so successful because of the situation of Lake Garda: although it is relatively far north, it is protected from the winner cold by the mountains which surround it.Nowadays, sadly, competition from southern lemon producers has meant that Limone concentrates more on its tourist industry than on lemon growing. The few remaining growers still produce the best quality lemon, however. Signor Ezio Ceruti, a lemon pr oducer, says, “ To grow these lemons you need to love the trees and learn from the old people who still remember how it was oncedone .The trees respond by being healthy and producing beautiful fruit .” Although scientists do not yet know for certain whether this fruit contains the magic ingredient that protect the people of Limone against heart disease, a local lemon producer’s recommendation for health is simple: each day slice a whole lemon into a cup, fill with boiling water, cover, leave overnight, then strain and drink.1. What do scientists believe about the unique protein?A．It protects people against heart disease.B．It can treat all sorts of illness.C．It comes from the lemon.D．It keeps people away from smoking and drinking.2. After the discovery of the unique protein , ________ .A．people in Limone ate a whole lemon each day.B．the tourist industry developed rapidly in LimoneC．more people in Limone started to grow lemons.D．Lemon became the main source of income for Limone.3. What can we learn about Limone from the text?A．It has a population of about 1,000.B．It is located at the top of a high mountain.C．It produced more than 15 million lemons every year.D．Its lemon-growing history dates back to the 19th century4. Ceruti’s words suggest ________ .A．we’re within easy reach of healthB．a lemon a day keeps the doctor away.C．Lemon trees in Limone bear beautiful fruit .D．one must work heart and soul to plant superior5. What can be the best title for the text?A．Magic fruit of LimoneB．History of lemon growingC．Lemon producers in ItalyD．Tourist industry in Limone20. According to the National Audubon Society, about 60 percent of all North American bird species have experienced northward shift during migration over the past four decades. This is symbolic of habitat loses from human development, as well as climate change. Without taking action to protect migratory bird’s habitats, many of these species will eventually die out, which in turn can destroy the ecosystems that depend on the bird species. While these numbers are based in North American, bird extinction is a worldwide problem.World Migratory Bird Day was established in 2006 as a holiday to help educate the public and raise awareness about the growing problems that are affecting the habitats of bird species around the world. It has since become a holiday recognized by the United Nations, and it helps to organize events in numerous countries. World Migratory Bird Day was established at the height of the age in which humans were starting to understand the influence of climate change.Since World Migratory Bird Day is held over the second weekend of May, the dates can differ. In 2014, the holiday started on Saturday, May 10th. In 2015, the starting date was May 9th. Celebrating the holiday over an entire weekend gives families and bird enthusiasts alike a chance to get outdoors and learn more about migratory birds.Education is at the center of World Migratory Bird Day celebration. The number of programs and festivals continue to grow each year, the content differing by area. World Migratory Bird Day provides people with the opportunity to learn about their favourite birds, and there are often bird watching tours planned for that weekend.While World Migratory Bird Day can be a fun way to learn about birds and their habitats, the weekend holiday is also focused on some serious notes surrounding conservation. At events and festivals, the public can learn exactly how their native birds’ habitats are being destroyed, and m ore importantly, people can learn how to stop the destruction.1. What does the first paragraph mainly talk about?A．The destruction of ecosystems.B．The seriousness of bird extinction.C．The reason to protect migratory birds.D．Causes of migratory bird s’ habitat losses.2. Which of the following does the author agree with?A．Climate change is the main cause of bird extinction.B．Migratory birds’ habitat losses may endanger our ecosystem.C．Bird extinction is a matter of course of historical development.D．There is no doubt that bird migration can speed up global warming.3. What does the author think of the establishment of World Migratory Bird Day?A．Worthy but impractical.B．Timely and educational.C．Simple but beneficial.D．Meaningless and wasteful.4. What do you know about World Migratory Bird Day?A．It is suitable for people of certain ages.B．The dates of celebrations are changeable.C．It is held by bird enthusiasts and experts.D．It aims to provide chances for families to unite.5. What can be inferred from the last two paragraph?A．People can watch the whole process of bird migration.B．People have the chance to have a direct connection with birds.C．Learning to recognize birds is the key purpose of celebration.D．The holiday lets people appreciate and encourage protection of birds.21. 阅读短文，按照题目要求用英语回答问题。
QTZ100
QTZ100(TC6013)塔吊使用说明书QTZ100（TC6013）塔式起重机（以下简称起重机）是济南建工机械有限公司根据国家标准，开发研制的新型建筑用起重机。
该机为水平臂架，小车变幅，上回转自升式多用途起重机。
其标准臂长为50米，加长臂长可达55米、60米，最大起重量为8吨，额定起重力矩为960千牛.米，最大起重力矩为1000千牛.米。
该机主要特点如下：爬升架主要由套架、平台、液压顶升装置及标准节引进装置等组成。
套架是套在塔身标准节外部，上端用螺栓与下支座相连，高6.93米，截面2.3米×2.3米，是由型钢和钢板组焊成的框架结构。
为了便于顶升安装的安全需要，特设有工作平台。
爬升架内侧沿塔身主弦杆安装有16个可调节的滚轮，支撑在塔身主弦杆的外侧。
在爬升架的横梁上，焊上两块耳板与液压系统油缸铰接承受油缸的顶升载荷。
爬升架下部有两个杠杆原理操纵的摆动爬爪，在液压油缸回收活塞以及引进标准节的过程中作为爬升架承托上部结构重量之用。
起重臂上、下弦杆都是用两个角钢拼焊成的方管，整个臂架为三角形截面，高1.2米，宽1.4米，总长为63.20米，共分为十节，节与节之间用销轴连接，拆装方便。
为了提高起重性能，减轻吊臂重量，吊臂采用双吊点、变截面空间桁架结构。
在起重臂第一节放置小车牵引机构和悬挂吊蓝，便于安装和维修。
臂架根部第一节与回转塔身用销轴连接。
平衡臂是由槽钢（Q235-C［32b］和角钢拼焊而成，全长13.425米，宽1.494米，上有扶栏和走道，为了便于运输，分两节制作，两节间用销轴连接，起升机构和平衡重均安装在平衡臂尾部，根据不同的臂长，配备不同的平衡重。
为了保证平衡臂水平，在它尾部一节有两个吊点，用销轴通过平衡臂拉杆与塔顶连接，平衡臂前节根部用销轴与回转塔身相连。
本机构采用YZRW250M2-6，45kW电机，通过柱销联轴器带动变速箱（此变速箱采用电磁离合器换档，可达三种传动比）再驱动卷筒，使卷筒获得三种绳速。
X100说明书
感谢您购买本公司记录仪，本机采用高性能低照度“cmos wxga hd ”高清感光芯片，有独特的超便携式设计，使它可以应用在各个领域，为您带来方便，安全，丰富多彩的日常生活。
主用于车载摄像，是行车中安全事故取证的最佳帮手。
在使用本产品之前，请详细阅读此用户手册，并请保管好此手册。
我们希望本产品，能满足您的需求并长期服务于您！本使用手册在编写和印刷中，可能存在一定文字错误，请谅解。
使用手册印刷过程中，机器因外观和软件的修复过程导致了同该使用手册的部分操作的非一致性，本公司保持最终解释权。
安装记录仪1.关闭引擎，将锁匙从点火器上取出。
2.将tf卡插入记录仪卡槽中。
3.将记录仪用支架垫黏贴安装在汽车的挡风玻璃上或驾驶仪表台上。
4.用电源线将记录仪的usb终端和汽车点烟器连接起来。
5.调整摄像位置，以便获取最佳的拍摄范围。
6.发动引擎，检查机器是否已安装正确。
【注意】（1）请在光线充足、全安的地方安装记录仪。
（2）当机器安装正确，其系统状态led会变成蓝色及显示屏会显示。
使用简介一．使用自动录像功能1.安装完毕后，启动汽车引擎，记录仪将会自动开启并开启自动录像功能。
2.关闭汽车引擎，在6秒后，记录仪保存档并将会关闭。
【注意】记录仪摄像镜头正/反相切换方法，长按【mic键】并保持3秒钟进行切换反相摄像。
长按【录像键】并保持3秒钟进行切换正相摄像，切换正或反相后，关机会自动保存其操作设置。
二．使用手动录像功能1.长按记录仪上的【开启/关闭键】并保持3秒钟。
2.当记录仪启动后，自动开启录像功能。
3.记录仪摄像镜头正/反相切换方法，长按【mic键】并保持3秒钟进行切换反相摄像。
长按【录像键】并保持3秒钟进行切换正相摄像。
（切换正或反相后，关机会自动保存设置）。
4.当再次点擎【录像键】，记录仪将会停止录像功能。
三．使用拍照功能1.长按记录仪上的【开启/关闭键】并保持3秒钟。
2.当记录仪启动后，在自动开启录像功能的状态下。
天津市第一百中学2023-2024学年高一上学期10月过程性诊断(1)英语试卷(不含音频)
天津市第一百中学2023-2024学年第一学期过程性诊断（1）高一英语本试卷满分150分，考试用时120分钟。
第一部分听力(共两节，满分20分)第一节:听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你将有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
(共5小题:每小题1分，满分5分)1. When will the goods arrive?A. On February 13th.B. On February 16th .C. On February 19th .2. How does the woman fight a cold?A. By having a rest.B. By eating a certain diet.C. By taking some medicine.3. What does the man plan to buy for his brother?A. A mobile phone.B. A book.C. A shaver.4. What does the man like best in the zoo?A. The Cat House.B. The Lizard Lounge.C. The Monkey House5. What does the Man mean by saying “it is all Greek to me. I can't follow him at all"?A. The physics problem is difficult.B. He can explain everything clearly.C. Professor Smith can't speak Greek.第二节:听下面几段材料。
每段材料后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
粘度指数的计算公式
Viscosity index 方法A——适用于粘度指数小于100或等于100的石油产品 U 40℃ Y 100℃若小于等于70mm2/s,则根据表1选出合适的L和H；若大于70mm2/s,利用公式L=0.835Y H=0.1684Y2+11.85Y-97，计算L和H VI=【（L-U）/（L-H）】*100 方法B——适用于粘度指数大于或等于100的石油产品 Y 100℃若小于等于70mm2/s,则根据表1选出合适的H，若大于70mm2/s时，则按H=0.1684Y2+11 VI=〈【（antilogN）-1】/0.00715〉+100 N=（logH-logU）/logY VI小于100的计算公式
U Y U Y
73.3 无 22.83
L H
无
119.94 69.48
VI
92.42965
VI大于100的计算公式
L 5.05 H
N 28.97 VI
0.147084 156.3753
Y2+14.67Y-216
11.85Y-97计算H
部编数学八年级上册专题21轴对称之将军饮马基础篇(解析版)含答案
专题21 轴对称之将军饮马基础篇1．如图，30AOB Ð=°，M ，N 分别是边,OA OB 上的定点，P ，Q 分别是边,OB OA 上的动点，记,OPM OQN a b Ð=Ð=，当MP PQ QN ++的值最小时，关于a ，b 的数量关系正确的是（ ）A ．60b a -=°B ．210b a +=°C ．230b a -=°D ．2240b a +=°【答案】B【解析】【分析】如图，作M 关于OB 的对称点M′，N 关于OA 的对称点N′，连接M′N′交OA 于Q ，交OB 于P ，则MP+PQ+QN 最小易知∠OPM=∠OPM′=∠NPQ ，∠OQP=∠AQN′=∠AQN ，KD ∠OQN=180°-30°-∠ONQ ，∠OPM=∠NPQ=30°+∠OQP ，∠OQP=∠AQN=30°+∠ONQ ，由此即可解决问题．【详解】如图，作M 关于OB 的对称点M ¢，N 关于OA 的对称点N ¢，连接M N ¢¢交OA 于Q ，交OB 于P ，则此时MP PQ QN ++的值最小．易知¢Ð=Ð=ÐOPM OPM NPQ ，¢Ð=Ð=ÐOQP AQN AQN ．∵18030Ð=°-°-ÐOQN ONQ ，30Ð=Ð=°+ÐOPM NPQ OQP 30Ð=Ð=°+ÐOQP AQN ONQ ，∴303018030210+=°+°+Ð+°-°-Ð=°ONQ ONQ a b ．故选：B.【点睛】本题考查轴对称-最短问题、三角形的内角和定理．三角形的外角的性质等知识，解题的关键是灵活运用所学知识解决问题，属于中考常考题型．2．如图，△ABC 是等腰三角形，底边BC 的长为4，面积是18，腰AC 的垂直平分线EF 分别交AC ，AB 于点E ，F ．若点D 为BC 边的中点，点M 为线段EF 上一动点，则△CDM 周长的最小值是（ ）A ．11B ．13C ．9D ．8【答案】A【解析】【分析】连接AD ，由于△ABC 是等腰三角形，点D 是BC 边的中点，故AD ⊥BC ，再根据三角形的面积公式求出AD 的长，再再根据EF 是线段AC 的垂直平分线可知，点C 关于直线EF 的对称点为点A ，故AD 的长为CM +MD 的最小值，由此即可得出结论．【详解】解：连接AD ，∵△ABC 是等腰三角形，点D 是BC 边的中点，∴AD ⊥BC ，∴1141822ABC S BC AD AD =×=´´=V ，解得AD =9，∵EF 是线段AC 的垂直平分线，∴点C 关于直线EF 的对称点为点A ，∴CM =AM ，∴CD +CM +DM =CD +AM +DM ，∵AM +DM ≥AD ，∴AD 的长为CM +MD 的最小值，∴△CDM 的周长最短=（CM +MD ）+CD =AD +12BC =9+12×4=9+2=11．故选：A ．【点睛】本题考查的是轴对称-最短路线问题，熟知等腰三角形三线合一的性质是解答此题的关键．3．如图，25AOB Ð=°，点M ，N 分别是边OA ，OB 上的定点，点P ，Q 分别是边OB ，OA 上的动点，记MPQ a Ð=，PQN b Ð=，当MP PQ QN ++的值最小时，b a -的大小=__________（度）．【答案】50【解析】【分析】作M 关于OB 的对称点M ¢，N 关于OA 的对称点N ¢，连接M N ¢¢，交OB 于点P ，交OA 于点Q ，连接MP ，QN ，可知此时MP PQ QN ++最小，此时OPM OPM NPQ OQP AQN AQN ¢¢Ð=Ð=ÐÐ=Ð=Ð，，再根据三角形外角的性质和平角的定义即可得出结论．【详解】作M 关于OB 的对称点M ¢，N 关于OA 的对称点N ¢，连接M N ¢¢，交OB 于点P ，交OA 于点Q ，连接MP ，QN ，如图所示．根据两点之间，线段最短，可知此时MP PQ QN ++最小，即MP PQ QN M N ¢¢++=，∴OPM OPM NPQ OQP AQN AQN ¢¢Ð=Ð=ÐÐ=Ð=Ð，，∵MPQ PQN a b Ð=Ð=，，∴11(180)(180)22QPN OQP a b Ð=°-Ð=°-，，∵QPN AOB OQP Ð=Ð+Ð，25AOB Ð=°，∴11(180)25(180)22a b °-=°+°- ，∴50b a -=° ．故答案为：50．【点睛】本题考查轴对称-最短问题、三角形内角和，三角形外角的性质等知识，灵活运用所学知识解决问题是解题的关键，综合性较强．4．如图，点P 是AOB Ð内任意一点，3cm OP =，点M 和点N 分别是射线OA 和射线OB 上的动点，30AOB Ð=°，则PMN V 周长的最小值是______．【答案】3【解析】【分析】根据“将军饮马”模型将最短路径问题转化为所学知识“两点之间线段最短”可找到PMN V 周长的最小的位置，作出图示，充分利用对称性以及30AOB Ð=°，对线段长度进行等量转化即可．【详解】解：如图所示，过点P 分别作P 点关于OB 、OA 边的对称点P ¢、P ¢¢，连接PP ¢¢、PP ¢、P P ¢¢¢、OP ¢、OP ¢¢，其中P P ¢¢¢分别交OB 、OA 于点N 、M ，根据“两点之间线段最短”可知，此时点M 、N 的位置是使得PMN V 周长的最小的位置．由对称性可知：,PN P N PM P M ¢¢¢==，,P OB POB POA P OA¢¢¢Ð=ÐÐ=Ð 3OP OP OP ¢¢¢===，30POA POB AOB Ð+Ð=Ð=°Q 30P OA P OB ¢¢¢\Ð+Ð=°+=60POA POB P OA P OB P OP ¢¢¢¢¢¢\Ð+ÐÐ+ÐÐ=°P OP ¢¢¢\△为等边三角形=3P P OP OP ¢¢¢¢¢¢\==\PMN V 的周长=PN PM MN ++=P N P M MN P P ¢¢¢¢¢¢++==3故答案为：3【点睛】本题是典型的的最短路径问题，考查了最短路径中的“将军饮马”模型，能够熟练利用其原理“两点之间线段最短”作出最短路径示意图是解决本题的关键．5．如图，ABC V 是等边三角形，AD 是BC 边上的高，E 是AC 的中点，P 是AD 上的一个动点，当PCE V 的周长最小时，ACP Ð的度数为______．【答案】30°##30度【解析】【分析】连接BP，由等边三角形的性质可知AD为BC的垂直平分线，即得出BP=CP，由此可知要使△PCE 的周长最小，即P点为BE与AD的交点时．最后根据等边三角形三线合一的性质，即得出CP平分ACBÐ，从而可求出1==302ACP ACBÐа．【详解】如图连接BP．∵ABCV为等边三角形，∴AD为BC的垂直平分线，∴BP=CP，∵△PCE的周长=PE+CP+CE= PE+BP+CE，∴当PE+BP最小时，△PCE的周长最小，∵PE+BP最小时为BE的长，即此时BE与AD的交点为P，如图．又∵点E为中点，AD为高，ABCV为等边三角形，∴P点即为等边ABCV角平分线的交点，∴CP平分ACBÐ，∴1==302ACP ACBÐа．故答案为：30°【点睛】本题考查等边三角形的性质，线段垂直平分线的判定和性质，两点之间线段最短等知识．理解要使△PCE的周长最小，即P点为BE与AD的交点是解题关键．6．如图，在四边形ABCD中，∠BCD＝50°，∠B＝∠D＝90°，在BC、CD上分别取一点M、N，使△AMN的周长最小，则∠MAN＝_____°．【答案】80【解析】【分析】作点A关于BC、CD的对称点A1、A2，根据轴对称确定最短路线问题，连接A1、A2分别交BC、DC于点M、N，利用三角形的内角和定理列式求出∠A1+∠A2，再根据轴对称的性质和角的和差关系即可得∠MAN．【详解】如图，作点A关于BC、CD的对称点A1、A2，连接A1、A2分别交BC、DC于点M、N，连接AM、AN，则此时△AMN的周长最小，∵∠BCD＝50°，∠B＝∠D＝90°，∴∠BAD＝360°﹣90°﹣90°﹣50°＝130°，∴∠A1+∠A2＝180°﹣130°＝50°，∵点A关于BC、CD的对称点为A1、A2，∴NA＝NA2，MA＝MA1，∴∠A2＝∠NAD，∠A1＝∠MAB，∴∠NAD+∠MAB＝∠A1+∠A2＝50°，∴∠MAN＝∠BAD﹣（∠NAD+∠MAB）＝130°﹣50°＝80°，故答案为：80．【点睛】本题考查了轴对称的最短路径问题，利用轴对称将三角形周长问题转化为两点间线段最短问题是解决本题的关键．7．如图，在锐角△ABC中，∠BAC = 40°，∠BAC的平分线交BC于点D，M，N分别是AD和AB 上的动点，当BM +MN有最小值时，ABMÐ=_____________°．【答案】50【解析】【分析】在AC上截取AE=AN，可证△AME≌△AMN，当BM +MN有最小值时，则BE是点B到直线AC的距离即BE⊥AC，代入度数即可求∠ABM的值；【详解】如图，在AC上截取AE=AN，连接BE，∵∠BAC 的平分线交BC 于点D ，∴∠EAM =∠NAM ，∵AM =AM ，∴△AME ≌△AMN ，∴ME =MN ，∴BM +MN =BM +ME ≥BE ．∵BM +MN 有最小值．当BE 是点B 到直线AC 的距离时，BE ⊥AC ，∴∠ABM =90°-∠BAC =90°-40°=50°；故答案为：50．【点睛】本题考查的是轴对称－最短路线问题，通过最短路线求出角度；解答此类问题时要从已知条件结合图形认真思考，通过角平分线性质，垂线段最短，确定线段和的最短路线，代入即可求出度数．8．如图，直线1l ，2l 交于点O ，点P 关于1l ，2l 的对称点分别为1P ，2P ．若4OP =，127PP =，则12POP △的周长是______．【答案】15【解析】【分析】根据对称的性质可知，OP 1=OP =OP 2=3，再根据P 1P 2=7即可求出△P 1OP 2的周长．【详解】∵P 关于l 1、l 2的对称点分别为P 1、P 2，∴OP 1=OP =OP 2=4，∵P 1P 2=7，∴△P 1OP 2的周长=OP 1+OP 2+P 1P 2=4+4+7=15．故答案为15【点睛】本题考查的是轴对称的性质，熟知轴对称的性质是解答此题的关键．9．如图，等腰三角形ABC 的面积是18，底边BC 长为4，腰AC 的垂直平分线EF 分别交AC ，AB 于点E ，F ．若D 为BC 的中点，G 为线段EF 上一动点，则CDG V 周长的最小值为___________．【答案】11【解析】【分析】连接AD ，由于ABC D 是等腰三角形，点D 是BC 边的中点，故AD BC ^，再根据三角形的面积公式求出AD 的长，再再根据EF 是线段AC 的垂直平分线可知，点C 关于直线EF 的对称点为点A ，故AD 的长为CM MD +的最小值，由此即可得出结论．【详解】解：连接AD ，△ABC 是等腰三角形，点D 是BC 边的中点，AD BC \^，∴S △ABC =1141822BC AD AD ×=´´= ，解得9AD =，EF 是线段AC 的垂直平分线，\点C 关于直线EF 的对称点为点A ，CM AM\=，CD CM DM CD AM DM\++=++，AM+DM≥AD，AD\的长为CM MD+的最小值，CDM\D的周长最短11()94921122CM MD CD AD BC=++=+=+´=+=．故答案为11．【点睛】本题考查的是轴对称-最短路线问题，熟知等腰三角形三线合一的性质是解答此题的关键．三、解答题10．问题：如图①，要在一条笔直的路边l上建一个燃气站，向l同侧的A、B两个城镇分别铺设管道输送燃气．试确定燃气站的位置，使铺设管道的路线最短．（1）如图②，作出点A关于l的对称点A'，线段A'B与直线l的交点C的位置即为所求，即在点C 处建燃气站，所得路线ACB是最短的．为了证明点C的位置即为所求，不妨在直线l上另外任取一点C'，连接AC'、BC'，证明AC＋CB＜AC'＋C'B．请完成这个证明．（2）如图③，点P为∠MON内的一个定点，在OM上有一点A，ON上有一点B．请你作出点A 和点B的位置，使得△PAB的周长最小．（保留作图痕迹，不写作法）在上述条件下，若∠MON＝40°，则∠APB＝°．【答案】（1）证明见解析；（2）作图见解析，100【解析】【分析】（1）如图②，连接A C ¢¢，由轴对称的性质可得,,AC A C AC A C ¢¢¢¢== 再证明：,A B AC BC ¢=+ 再利用三角形的三边关系可得结论；（2）分别作点P 关于,OM ON 的对称点,,P P ¢¢¢ 连接P P ¢¢¢交OM 于,A 交ON 于,B 则PAB △的周长最短，再由轴对称的性质可得：,,OPB OP B OPA OP A ¢¢¢V V V V ≌≌ 证明,APB OP B OP A ¢¢¢Ð=Ð+Ð 80,P OP ¢¢¢Ð=° 再求解50,OP P OP P ¢¢¢¢¢¢Ð=Ð=° 从而可得答案．【详解】证明：（1）如图②，连接A C ¢¢，∵点A ，点A ¢关于l 对称，点C 在l 上，∴CA CA ¢=，∴AC BC A C BC A B ¢¢+=+=，同理可得：AC C B A C BC ¢¢¢¢¢+=+，∵A B ¢＜A C C B ¢¢¢+，∴AC ＋BC ＜AC C B ¢¢+；（2）如图所示，点A 、B 即为所求，由轴对称的性质可得：,,OPB OP B OPA OP A ¢¢¢V V V V ≌≌,,,,PO P O PO P O OPB OP B OPA OP A ¢¢¢¢¢¢\==Ð=ÐÐ=Ð,,POB P OB POA P OA ¢¢¢Ð=ÐÐ=Ð,APB OPB OPA OP B OP A ¢¢¢\Ð=Ð+Ð=Ð+Ð40,POB POA P OB P OA ¢¢¢Ð+Ð=Ð+Ð=°404080,P OP ¢¢¢\Ð=°+°=°,OP OP ¢¢¢=Q()11808050,2OP P OP P ¢¢¢¢¢¢\Ð=Ð=°-°=° 100,APB OP P OP P ¢¢¢¢¢¢\Ð=Ð+Ð=°故答案为：100°．【点睛】本题考查的是轴对称的作图，利用轴对称的性质求解线段和或周长的最小值，同时考查线段的垂直平分线的性质，等腰三角形的性质，掌握以上知识是解题的关键．11．如图，在平面直角坐标系中，已知点(2,5)A ，(2,1)B ，(6,1)C ．(1)画出ABC V 关于y 轴对称的111A B C △；(2)在x 轴上找一点P ，使PB PC +的值最小（保留作图痕迹），并写出点P 的坐标．【答案】(1)见解析；(2)见解析，P 的坐标为(4,0)．【解析】【分析】（1）根据轴对称的性质结合坐标系，分别确定点A 、B 、C 关于y 轴的对称点A 1、B 1、C 1，即可作出111A B C △；（2）作出点B 关于x 轴的对称点B 2，连接B 2C ，交x 轴于P ，点P 即为所求做的点．(1)解：解：（1）如图所示，111A B C △即为ABC V 关于y 轴对称的三角形．(2)解：如图所示，点P 即为所求做的点，点P 的坐标为(4,0)．【点睛】本题考查了平面直角坐标系中的轴对称图形，将军饮马问题，熟知轴对称的性质是解题关键，注意坐标系中两个点关于x 轴对称，则横坐标不变，纵坐标互为相反数，两个点关于y 轴对称，则横坐标互为相反数，纵坐标不变．12．如图，在锐角∠AOB的内部有一点P，试在∠AOB的两边上各取一点M，N，使得△PMN的周长最小．（保留作图痕迹）【答案】见详解【解析】【分析】作点P关于直线OA的对称点E，点P关于直线OB的对称点F，连接EF交OA于M，交OB于N，连接PM，N，△PMN即为所求求作三角形．【详解】解：如图，作点P关于直线OA的对称点E，点P关于直线OB的对称点F，连接EF交OA于M，交OB于N，连接PM，PN，△PMN即为所求作三角形．理由：由轴对称的性质得MP＝ME，NP＝NF，∴△PMN的周长＝PM+MN+PN＝EM+MN+NF＝EF，根据两点之间线段最短，可知此时△PP1P2的周长最短．【点睛】本题考查轴对称﹣最短问题、两点之间线段最短等知识，解题的关键是学会利用对称解决最短问题，属于中考常考题型．13．如图，在Rt△ABC中，∠ACB＝90°，∠ABC＝30°，AC＝2，以BC为边向左作等边△BCE，点D 为AB 中点，连接CD ，点P 、Q 分别为CE 、CD 上的动点．（1）求证：△ADC 为等边三角形；（2）求PD +PQ +QE 的最小值．【答案】（1）证明见解析；（2）4．【解析】【分析】（1）先根据直角三角形的性质可得60,BAC AD CD Ð=°=，再根据等边三角形的判定即可得证；（2）连接,PA QB ，先根据等边三角形的性质可得12ACE ACD Ð=Ð，再根据等腰三角形的三线合一可得CE 垂直平分AD ，然后根据线段垂直平分线的性质可得PA PD =，同样的方法可得QB QE =，从而可得PD PQ QE PA PQ QB ++=++，最后根据两点之间线段最短即可得出答案．【详解】证明：（1）Q 在Rt ABC V 中，90,30,2ACB ABC AC Ð=°Ð=°=，60,24BAC AB AC Ð\=°==，Q 点D 是Rt ABC V 斜边AB 的中点，2AD AC \==，ADC \V 是等边三角形；（2）如图，连接,PA QB ，BCE QV 和ADC V 都是等边三角形，60BCE \Ð=°，60ACD Ð=°，1302ACE ACB BCE ACD \Ð=Ð-Ð=°=Ð，CE \垂直平分AD ，PA PD \=，同理可得：CD 垂直平分BE ，QB QE \=，PD PQ QE PA PQ QB \++=++，由两点之间线段最短可知，当点,,,A P Q B 共线时，PA PQ QB ++取得最小值AB ，故PD PQ QE ++的最小值为4．【点睛】本题考查了等边三角形的判定与性质、含30°角的直角三角形的性质等知识点，熟练掌握等边三角形的性质是解题关键．14．如图，在正方形网格中，每个小正方形的边长都是1，每个小正方形的顶点叫做格点．网格中有一个格点ABC V （即三角形的顶点都在格点上）．(1)在图中作出ABC V 关于y 轴对称的111A B C △，并写出点1C 的坐标．(2)在y 轴上求作一点P ，使得PA PC +最短（保留作图痕迹，不需写出作图过程）．(3)求ABC V 的面积．【答案】(1)画图见解析；()11,4C (2)画图见解析(3)6【解析】【分析】（1）利用网格，根据轴对称的性质画出点A 、B 、C 关于y 轴的对称点A 1、B 1，C 1，再连接A 1B 1，A 1C 1，B 1C 1即可；（2）连接A 1C 交y 轴于点P ，即可；（3）利用网格，用矩形面积减去三个直角三角形面积求解即可．(1)解：如图所示，111A B C △就是所要求画的．()11,4C ．(2)解：如图所示，点P 就是所要求作的点．(3)解：111353322156222ABC S =´-´´-´´-´´=△．【点睛】本题考查利用轴对称性质作轴对称图形，利用轴对称求最短路径问题，熟练掌握轴对称的性质是解题的关键．15．如图所示的方格纸中，每个小方格的边长都是1，点A （-4，1）、B （-3，3）、C （-1，2）．(1)请作出△ABC向右平移5个单位长度，下移4个单位长度后的△A₁B₁C₁；(2)作△ABC关于y轴对称的△A₂B₂C₂；(3)在x轴上求作点N，使△NBC的周长最小（保留作图痕迹）．【答案】(1)答案见详解；(2)答案见详解；(3)答案见详解；【解析】【分析】（1）分别作出点A，B，C向右平移5个单位长度，下移4个单位长度后的对应点A₁，B₁，C₁再顺次连接A₁B₁C1；(2)分别作出点A，B，C关于y轴的对称点，再首尾顺次连接可得；(3)作点B关于x轴的对称点B3，再连接B3C交y轴于点N，顺次连接点NB，NC，即可；(1)如图所示：分别作出点A，B，C向右平移5个单位长度，下移4个单位长度后的对应点A₁，B₁，C₁再顺次连接A₁B₁C1；(2)如图所示：分别作出点A，B，C关于y轴的对称点A2，B2，C2，再首尾顺次连接可得；(3)作点B关于x轴的对称点B3，再连接B3C交y轴于点N，顺次连接点NB，NC，△NBC的周长最小；【点睛】本题主要考查作图-轴对称变换，图形的平移，解题的关键是熟练掌握轴对称变换的定义和性质及最短路线问题．。
Abandon100%preventable
Parenting Strategies•Keys to working successfully with children with FASDs:–Structure–Consistency–Variety–Brevity–Persistence and repetition•Families might need education, counseling, and/or parenting classes. For parent with FASD’s must be en vivo, modeling, real life approach to parenting skills.•Birth families might need intervention and encouragement to pursue treatment for their addiction.Disability Services•Individuals with an FASDs might qualify for:–Supported employment/job coach–Transportation–Assisted living–Respite care–Social Security disability benefits–Supplemental Security Income (SSI)Legal System •Adolescents and adults with FASDs can experience issues in the legal system both as victims and perpetrators of a crime.–Individuals with FASDs are at risk for victimization, poor judgment, not understanding cause andeffect.–This is due to the nature of cognitive deficitsassociated with FASDs.•Need mentor or advocate to navigate the legal system.Informed Approaches•Don’t ascribe lack of follow through to a motivational issue•Identifies buddies/natural supports to get them to appointments•Informed mentoring/life coaching•Change reward‐based point/level systems (find ways individual can earn rewards to reinforce expectations, tangible, visual, antecedent strategies) •Re‐assess concepts of “dependency” and “enabling”3 A’s for Care Planning Adapt techniques, strategies, and services.Adopt what works. Abandon what doesn’t work!All FASDs are 100% preventableScreen ALL Women of Childbearing Age•Screening•Brief Intervention,•Dynamic Case Mgt for Women of Childbearing AgeScreening for alcohol use in women of childbearing ageWhat constitutes a standard drink?12 ounces ofbeer5 ounces of wine1.5 ounces of hard alcoholScreening Instrument I t tshouldthe“Brief interventionsare Not designedto treat personspwith alcoholdependence.”(Babor & Higgins-Biddle, 2001)•PictureFeedback and Responsibility。
富了么 FX100+ 吉他效果器中文说明书
/CLEAR功能； 08 PLAY：切换回主界面； 09 SAVE：保存当前音色； 10 EXP：踏板模块； 11 RHYTHM：鼓机模块； 12 TAP：TAPTEMPO打点设速； 13 SYSTEM：系统设置，可调节INPUTLEVEL(输入音量)、SCREEN(屏幕亮度)、CABSIMTHRU
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9V 300mA
01 AUXIN：1/8'立体声音频输入接口，可通过外置音频输入进行同步练习； 02 INPUT：1/4'单声道输入接口，连接至吉他或其他效果器的输出口； 03 LEFT/MONO：1/4'单声道（左）输出接口，连接至音箱、声卡、调音台等音频设备或其他效果器
www.fl深圳市富了么电子科技有限公司
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音色编辑
FX100共有9个效果模块，FX、DS、AMP、CAB、NS、EQ、MOD、DELAY、REVERB，共包括151个效果类型， 支持9个效果模块同时使用，每个效果模块支持选择使用一种类型。
安全注意事项： 1.请仔细阅读本说明书 2.请保存好本说明书 3.注意所有警告事项 4.遵循说明书指示 5.请勿将设备靠近水源或湿度大的环境 6.用干毛巾擦拭清洁 7.请确保使用时远离收音机和电视机或者其他有磁场产生的设备，以免造成干扰 8.请勿暴力使用开关及控制元件 9.请勿让纸屑、金属制品及其他物体落入机内 10.请勿摔落以及让本机受到冲击和过度按压
100HA系列伺服驱动器使用说明书
——驱动器端子排 R、S、T 动力电源为单相交流 220V（接 R、S、T 任意两端）或 三相交流 220V（接 R、S、T），推荐使用单相或三相隔离变压器。PE 为接地 端子。
——r、t 为驱动器控制电路电源，接单相交流 220V。 ——驱动器不能直接接交流 380V，否则会造成驱动器损坏。 ——驱动器端子排 U、V、W、PE 端子必须与电机的 U、V、W、PE 一一对应连接。 ——电机后部装有精密编码器，安装时严禁敲击电机。 ——电机后部的精密编码器已调整零点，用户不允许自行拆卸反馈元件。 ——为避免水、油等流入电机内，安装电机时尽可能请将电缆插头朝下。 ——断电后，驱动器内部仍有残余高压，因此 5 分钟内不得接触强电端子；不允
6.2 第 2 层···································································································45 6.2.1 监视方式 ·····················································································45 6.2.2 参数设置 ·····················································································47 6.2.3 参数管理 ·····················································································47 6.2.4 速度试运行 ·················································································50 6.2.5 JOG 运行 ·····················································································50
pt-100
pt100设计原理：pt100是铂热电阻，它的阻值会随着温度的变化而改变。
PT后的100即表示它在0℃时阻值为100欧姆，在100℃时它的阻值约为138.5欧姆。
它的工业原理：当PT100在0摄氏度的时候他的阻值为100欧姆，它的阻值会随着温度上升而成匀速增长的。
应用范围：医疗、电机、工业、温度计算、阻值计算等高精温度设备，应用范围非常之广泛。
PT100分度表-50度80.31欧姆-40度84.27欧姆-30度88.22欧姆-20度92.16欧姆-10度96.09欧姆0度100.00欧姆10度103.90欧姆20度107.79欧姆30度111.67欧姆40度115.54欧姆50度119.40欧姆60度123.24欧姆70度127.08欧姆80度130.90欧姆90度134.71欧姆100度138.51欧姆110度142.29欧姆120度146.07欧姆130度149.83欧姆140度153.58欧姆150度157.33欧姆160度161.05欧姆170度164.77欧姆180度168.48欧姆190度172.17欧姆200度175.86欧姆组成的部分常见的pt1oo感温元件有陶瓷元件，玻璃元件，云母元件，它们是由铂丝分别绕在陶瓷骨架，玻璃骨架，云母骨架上再经过复杂的工艺加工而成薄膜铂电阻薄膜铂电阻：用真空沉积的薄膜技术把铂溅射在陶瓷基片上，膜厚在2微米以内，用玻璃烧结料把Ni（或Pd）引线固定，经激光调阻制成薄膜元件常用温度传感器可以分成三大类：热敏电阻、热电阻、热电偶。
热电阻顾名思义，它的电阻的阻值是随着温度变化而变化的，比如，用线性比较好的铂丝、铜丝作的电阻。
工业用热电阻一般采用Pt100,Pt10,Pt1000、Cu50,Cu100,铂热电阻的测温的范围一般为零下200-800摄氏度，铜热电阻为零下40到140摄氏度。
如用铂丝做成的热电阻，其分度号称Pt100。
就是说它的阻值在0度时为100欧姆，负200度时为18.52欧姆，200度时为175.86欧姆，800度时为375.70欧姆。
阳光电源调eward逆变器SG225HX用户手册说明书
4.4.2 起吊搬运 ........................................................................................... 22 4.5 安装挂板..................................................................................................... 23
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1.1 拆包检查....................................................................................................... 1 1.2 安装安全....................................................................................................... 2 1.3 电气连接安全 ................................................................................................ 2 1.4 运行安全....................................................................................................... 3 1.5 维护安全....................................................................................................... 4 1.6 报废安全....................................................................................................... 4
六年级上册数学百分数一1百分数的意义和读写法人教新课标(10张)标准课件
比像较上分 面数这、样百的分数数，的如异14同%、65.
既百可分以 数表示一个具数体是的另数一、个数的百分之几。
百既分可数 以也表叫示做一百个分具率体或的百数分、比。
既 百可分以数表 也示 叫一 做个 百具 分体 率的 或数百、 分比。 百分分数数表示一个数是另一个数的百分之几。 百所分以数 百表分示数一的个后数面是不另能一写个单数位的名百称分。之几。
百像分上数 面表这示样一的个数数，是如另14一%、个6数5.的百分之几。
百既分可数 以也表叫示做一百个分具率体或的百数分、比。
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百分数也叫做百分率或百分比。
百分数也叫做百分率或百分比。
120% 读作： 百分数通常不写成分数形式，而在原来的分子后面加上百分号“%”来表示，
百分之一百二十
百分数的意义和读写法
比较分数、百分数的异同
百分数
意义
表示两个数之间的 倍数关系。
分数
既可以表示一个具体的数、 又可以表示两个数之间的 倍数关系。
同学，下节再见
所( 以)百分数的后面不能写单位名称。
5判%断、：……一叫个做苹百果分的数重。量是 千克。
百分分数数通常不写成分数形式，而在原来的分子后面加上百分号“%”来表示，读作“百分之……”。
像百上分面 数这表样示的一数个，数如是另14一%、个6数5.的百分之几。
百分数也叫做百分率或百分比。
像既上可面 以这表样示的一数个，具如体的14数%、、65.
创新微课
所百以分百 数分数的后面不能写单位名称。
既14可%表以示表已示经一格个式具化体的的部数分、占所要格式化的总量的 。
像14上%表面示这已样经的格数式，化如的14部%分、占65所. 要格式化的总量的 。
MullerBeltex可调PT100轴承温度传感器说明书
Solid partners for powder and bulk handling componentsAdjustable PT100 Bearing Temperature Sensor with grease nippleHot bearing WARNING device, for use in hazardous areasInstallation guideCertificationUSC ®PT100V3CSKF PT100K3DNBearing Temperature Sensor With Grease nippleThe PT100 bearing temperature sensor with grease nipple is specially designed for accurate measuring of universal bearing housings.The PT100 sensor is very easy to mount in thebearing housing without need for modifications. Use this installation method to avoid losing the guarantee on the bearing.The advantage of measuring with a PT100 is that it sends a continuous output signal to the operating system.PT100V3C Taper thread G1/8 SKF Taper thread M6The bearing temperature sensor with grease nipple is specially designed for accurate measuring of universal bearing housings. The bearing temperature sensor emits a standard analogue output signal. The signal can be implemented very simply in an operating system. If the analogue signal is interrupted, the reason for this must be investigated (failsafe). Early detection of high temperature can prevent the bearing from becoming overheated, thereby reducing the risk of possible dust explosions.Prior to fitting the Rub Block it is of great importance to ensure that:- Fitting is implemented by qualified personnel only.- Proper and safe work conditions are provided.- Bearings have a correct design and are in good condition.- The spot where the PT100 sensor is to be mounted is clean and free of dust.- A good earth wire is available to ensure precise reading.- If the supply leads are terminated in a hazardous area, the termination arrangement must comply with the Zone/Category/EPL of the hazardous area that it is to beinstalled.PT100V3C SKF PT100K3DN1. Rub-Block RB100DN or RB200DN belt misalignment monitoring location (both sides)2. Bearing temperature sensor monitoringPT100 location (both sides)3. Speed monitoring4. Junction box small or largeBefore the Bearingtemperature sensor can be mounted, it is important that the following steps will be taken:First of all, the electrical signal for the PT100 sensors has to be operational. If this is not the case, you must put back the existing grease nipple till the electrical signal is operational. For fieldcabling we recommend using a shielded core cable in order to avoid any electrical interference.STEP 1:1. Ensure the installation and surroundings are quite dust-free and clean.2. Remove the existing grease nipple from the bearing house.3. Clean the threaded hole with airflow.STEP 2:1. Measure the depth (A) from the top of bearing house till the inner bearing ring2. Before mounting the PT100 brass body please make sure you will use the gasket and the PTFE tape on the threadfrom the brass body to ensure that the IP 67 class will prevented.3. Screw the sensor brass body in to the existing threaded hole in the bearing housing.STEP 3:1. Insert the temperature probe with the olive nut into the brass body til the end of the bearing ring.STEP 4: 1. Tighten the olive nut using an appropriate spanner. The maximum recommended torque for the olive nut is 2Nm (18 lb-in) this is not done properly, no exact measurement can be taken and the risk that the sensor will break is very high. 2. lubricate the bearing by using the grease nipple on the PT100 sensor body. Use the right grease.STSTEP 4STEP 2STEP 3STEP 1The equipment used in areas with an explosion hazard only contains intrinsically safe circuits. A circuit is intrinsically safe if, both in normal operation and in the event of a fault, a short-circuit in the circuit does not produce a spark capable of causing ignition, and the current flowing through the equipment does not heat any part of the surface above the level specified for the temperature class (see also EN 60079-11).In order for a circuit to be designated as intrinsically safe, every individual device included in the circuit must be designed to be intrinsically safe. Furthermore, it is necessary to test that the complete circuit of the (individually) intrinsically safe devices forms, as a whole, an intrinsically safe circuit.The fact that a circuit is assembled from devices that are (individually) intrinsically safe, does not guarantee that the circuit as a whole is intrinsically safe.For a RTD temperature probe the measuring current (or in case of malfunction the residual current) flows through the sensor element. The result is self-heating of the element and ultimately temperature increase on the surface area of the protection fitting. It is essential to ensure that the limit of the defined temperature class is not exceeded.Adjustible bearing temperature sensor, type PT100V3C installed on a bearing house.Please read this fitting installation instruction carefully before beginning the installation.For the correct compression and force that will put on the compression fitting of the PT100 adapter. Please follow the stepsand instructions very carefully.Step 1: First start with the three components above. Step 2: Put the PT100 probe sensor in de brass bodyStep 3: Tighten the bolt onto the brass body hand-tight Step 4:Now use a 14mm spanner wrench.Brass bodyOlive NutPT100 ProbePlease hold the lower part from the PT100V3C brass body tide with the adjustable spanner. Step 5: Turn the spanner wrench (size 14) clockwise 1½ PLEASE DO NOT OVER-TIGHTEN.When this is done correctly as above described the PT100 sensor is tighten and secured by the compression ring (olive) into the brass adapter.Before mounting the PT100 brass adapter please make sure you will use PTFE tape on the thread from the adapter to ensure that the IP 67 class will prevented.If you loosen or disconnect a fitting, remove the old tape and re-wrap it with a fresh piece.Extra information before mouting the Bearing temperature sensor:Please hold the lower part from the SKF PT100K3DN brass body tide with an spanner (size 14).Step 5: Turn the spanner wrench (size 8) clockwise 1½ PLEASE DO NOT OVER-TIGHTEN.Compression ring SKF PT100K3DNCompression ring PT100V3CMeasuringThe PT100 produces a standard analogue output signal. It is easy to implement this signal in a PLC system, for example a SIEMENS PLC S7. If it is not possible to implement an analogue signal, an analogue card is required or a HEAD transmitter with an analogue output signal of 4-20mA can be used. Software for the visualisation can be programmed by a qualified software engineer.Transmitter• PT100 connected directly to an RTD I/O card in the I/O room via an shielded cable in order to avoid any electrical interference or a remote I/O RTD input card.• Local transmitter• IndicatorThe transmitter can be either 4..20mA or bus technology (e.g. Profibus PA, Foundation Fieldbus, Modbus, ect.)Required functional specification for the transmitter, 4..20mA version:• Output 4-20mA Hart• Calibration Range 4mA (LSL) = 100°C (200°F), 20mA (USL) = 0°C (32°F)• Fail – Low signal (<4mA) on internal failure• Electrical approvals – Hazardous approval according to local zoningBus Technology:• Broken wire detection• Fail detection on internal failure• Electrical approvals – Hazardous approval according to local zoningTransmitter Failsafe Configuration:Power loss failure configuration: The transmitter range shall have a high value at 4mA. This assures broken wire Alarming in case of a broken wire between the transmitter and the I/O card and/or transmitter power loss.The analog input card shall detect a fault on the analog input, and go to “error”.Internal instrument failure configuration: The failure mode alarm shall be set to low or below 4mA either via Hart handheld program-mer or internal hardware switch.In case of a broken cable in the connection between the PT100 and the transmitter, the transmitter will read the maximum resistance, giving max temperature value on the analog output. The PLC will generate alarm.Alarm configuration PLCBest maximum delay for the sum of the timer and the filter is 5 seconds.The following alarm points are recommended in a ATEX environment:• Absolute temperature PRE-ALARM : 60 °C• Absolute temperature STOP-ALARM : 80 °CThe electrical conductivity of metal (i.e. platinum) is based on the mobility of free electrons. With increasing temperaturesthe atoms in the metal lattice will vibrate more vigorously and thus impede the free flow of electrons towards the positivepole of a voltage source. This impedance causes a resistance in linear proportion to the temperature.The resistance value is in conformance with the European standard: PT100 = 100 + 0.385 055 x T. The positive temperature coefficient of a PT100 is thus 0.385 055 Ohms per Kelvin.PT100V3C and the SKF PT100K3DN technical dataPT100 sensorSensor type : R8-81227320-0115/050.S01 with 1 x 4L connection wires Protection type : IP 67Tolerance class : Class A DIN IEC751Cable type : PFA-PFA-V2A Cn 4 x 0,22 mm²Ambient temperature limit for cable : - 40°C to + 185°CCable Length : 3 meter, (several cable lengths available)Measuring temperature limit : - 40°C to + 280°CMeasuring current : 1mAOutput signal : AnalogueProbe lenght PT100V3C: 100 mm, (several probe lengths available)Standard probe Diameter : Stainless Steel 5 mmProbe lenght SKF PT100K3DN: 60 mm, (several probe lengths available)Standard probe Diameter : Stainless Steel 3 mmElectrical Data PT100Measuring voltage : Ui 30 VMaximum current input : li 101 mAMaximum total output : P 750 mWConnection body PT100V3CHousing sensorbody : Brass or Stainless SteelDiameter screw connection head : G1/8Connection body SKF PT100K3DNHousing sensorbody : Brass or Stainless SteelDiameter screw connection head : M6 x 1Hazardous area classificationATEX Class (Ex-i) : Ex II 1D Ex iaD T85°C / Ex II 1G IIC T6Certificate number : IBExU13ATEX1079XIECEx Class (Ex-i) : Ex II 1D Ex iaD T85°C Da, Ex II 1/2 G Ex ia IIC T6* Ga/GBTAMB -40°C to 185°CCertificate number : IECEx IBE 15.0014XGost R (Ex-i) : Ex II 1D Ex iaD T85°C / Ex II 1G IIC T6Certificate number : POCC PL.AF.H00052Test ProcedureThe RTD temperature tester has been designed to test RTD’s and adjust-able depth bearing temperature sensors in the field. This hand held testunit features an integrated heating block specifically designed to have aRTD probe sensor directly inserted. With integral controls and temperaturedisplay, the unit heats the sensor to the desired trip point, and allowsquick and easy real life testing of the sensor and temperature monitoringsystem.Method of operation: During planned maintenance or periodictesting, the RTD Sensor Tester can be used as a diagnostic tool to verifythe alarm and shutdown sequences of the control unit are functioning asexpected. To test, the heater block should be set above the control unitsalarm operating temperature. Remove the RTD sensor probe from thehousing and insert it into the heater block. As the heater block reachesthe alarm temperature, the RTD sensor will relay this data to the controlunit, allowing you to verify that the alarm and shutdown sequences run asexpected.Inspection of the Bearing PT100 sensorThe PT100 needs to be inspected regularly (recommendation: yearly) andafter a trip. For this purpose, the PT100 probe must be extracted from thebrass body.MeasuringThe PT100 produces a standard analogue output signal. It is easy to im-plement this signal in a PLC system, for example a SIEMENS S7. If it is notpossible to implement an analogue signal, an analogue card is requiredor a HEAD transmitter with an analogue output signal of 4-20mA can beused. Software for the visualisation can be programmed by a qualifiedsoftware engineer.MaintenanceThe valid European and national regulations must be observed for maintenance, servicing and testing. Inparticular,all parts on which explosion protection depends must be checked during maintenance.PT100 coversSunlight does not have an impact on the measurements, the PT100 is sprotected with several insulating layers,different temperatures on two sides of the bearing have no effect. Problems in hot climate have not occurred.Tel +31(0)15 369 54 44 Fax +31(0)15 369 78 44 ***************************Do not damage the sealing surfacesor other seals!Thread adapter to resize the thread from the PT100 sensorThis measuring was done on a bearing from a bucketelevator.Diagram (°C) Time (min.)Ambient temperature of 35°CDuration hot bearing measuring by the PT100V3C adjustable bearing temperature sensor.Duration temperature rise to 80°CPT100: Approximately 40 sec.We recommend to use an trending software module. As soon as a difference in temperature occurs within a certain time, there is a slight hot bearing. It prevents unnecessary wear of the bearing and you will be informed for a threatening hot bearing. • Temperature rise rate STOP-ALARM : 2,5°C /min, max allowed 10°C /min • Temperature rise rate STOP-ALARM : 1°C /5sec, max allowed 5°C /5sec • Temperature rise rate PRE-ALARM : 0,2°C /min, max allowed 10°C /min • Temperature rise rate PRE-ALARM : 0,7°C /5sec, max allowed 5°C /5secAdditional Notes: for temperature line monitoringTel +31(0)15 369 54 44 Fax +31(0)15 369 78 44 *************************** A.In order to maximize efficiency and safety, selecting the right equipment for each operation is vital.The correct installation of this equipment, as well as regular maintenance and inspection, are equallyimportant to proper operation and safety of the product. The correct installation and maintenance of theproducts are the responsibility of the user. B.All installation and wiring must conform to governing local and national electrical codes and otherstandards applicable to specific industries. The installation of the wiring should be undertaken by anexperienced and qualified professional electrician.Failure to correctly wire the product to any machinery may result in the product or machine failing tooperate as intended. C.Periodic inspection by a qualified person will help assure that this product performs properly.It is recommended that a documented inspection is carried out at least annually or more frequently in case that the product has been subjected to intensive use.Customer safety responsibilities1.Read all literature provided with the product. Read all user instructions and safety manuals toensure that the product operation is understood and can be used safely and effectively use this product. 2.Select a qualified and competent installer; Correct installation of the product is important for safety andperformance. It is critical for the safety of your operation and of those who may work with yourequipment that a qualified and competent electrical installer is selected to undertake the installation ofthis product. The product must be installed properly to perform to its designed functions.The installer should be qualified, trained, and competent to perform the installation in accordancewith local and National Electrical Codes, all relevant OSHA Regulations, as well as any of the user’sown standards and preventive maintenance requirements, and other product installation informationsupplied with the product. The installer should be provided with all necessary installation information to assist in the installation.This symbol indicates safety measures that definitely must be taken into account in order to prevent personal injury.This symbol indicates ATEX Certified components with a certain zoning designation.The flat sealing surfaces and seals must not be damaged!Installation and commissioning must be performed by qualified personnel. Read the instructions carefully before starting up. The supplier is not liable for personal injury or prop-It is required to check the latest product information on Muller Beltex’ website, before installing this product. Failure to do so, could result in to product failure or damage.Solid partners for powder and bulk handling componentsTel +31(0)15 369 54 44 Fax +31(0)15 369 78 44 ***************************。
2022-2023学年江苏省南京市第四高一年级上册学期期末复习达标检测数学试题【含答案】
2022-2023学年高一期末复习达标检测卷一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1. 设全集，集合，集合，则图中阴影部分所表示的集合是N U *={}2,3,4,6,9=A {}4,N B x x x *=>∈（ ）A.B.C.D.{}6,9{}2,3{}2,3,4{}24x x ≤≤【答案】C 【解析】【分析】根据Venn 图表示的集合计算．【详解】因为全集，所以，N U *=U B {|4,N*}{1,2,3,4}x x x =≤∈=所以图中阴影部分表示．(){2,3,4}U A B = 故选：C ．2. 点落在（ ）()sin100,cos100︒︒P A. 第一象限内 B. 第二象限内C. 第三象限内D. 第四象限内【答案】D 【解析】【分析】根据三角函数的诱导公式和符号，得到，即可求得.sin1000,cos1000︒>︒<【详解】因为，sin100sin(9010)cos100,cos100cos(9010)sin100︒=︒+︒=︒>=︒+︒=-︒<︒所以点落在第四象限内.()sin100,cos100︒︒P 故选: D.3. 同时满足：①，②，则的非空集合M 有（）{}1,2,3,4,5M ⊆a M ∈6a M -∈A. 6个 B. 7个C. 15个D. 16个【答案】B 【解析】【分析】根据所给条件确定M 中元素，再根据M 是所给集合的子集，得到所有的M 即可求解.【详解】时，；时，；时，；时，；，1a =65a -=2a =64a -=3a =63a -=4a =62a -=5a =，61a -=∴非空集合M 为，，，，，，，共7个．{}3{}1,5{}2,4{}1,3,5{}2,3,4{}1,2,4,5{}1,2,3,4,5故选：B 4. 已知，，则（ ）2log 3m =3log 7n =42log 56=A. B. C. D. 31mn mn ++321m n m n ++++31mn mn m +++31mn mn m +-+【答案】C 【解析】【分析】由换底公式和对数运算法则进行化简计算.【详解】由换底公式得：，223log 7log 3log 7mn =⋅=71log 2mn=，其中4242424278log 5678log log log ⨯=+=，4277771111711log 421log 61log 2log 311log mnmn m mn n =====+++++++，故424222233383242log log log log lo 67g 1mn m ====+++42313log 5611mn m mn m mn mn m mn +=++=+++++故选：C5. 函数的图象大致是（ ）()13cos313xxf x x -=+A. B.C. D.【答案】A 【解析】【分析】先判断奇偶性，可排除C ，D ，由特殊值，可排除B ，即可得到答案.()f π【详解】因为，所以函数为奇函数，排除()()()1331cos 3cos31331x x x x f x x x f x -----=⋅-=⋅=-++()f x C ，D ；又，排除B ，()13cos3013f ππππ-=>+故选：A.【点睛】函数图象的识辨可从以下方面入手：(1)从函数的定义域，判断图象的左右位置；从函数的值域，判断图象的上下位置．(2)从函数的单调性，判断图象的变化趋势．(3)从函数的奇偶性，判断图象的对称性．(4)从函数的特征点，排除不合要求的图象．利用上述方法排除、筛选选项．6. 若，的终边（均不在y 轴上）关于轴对称，则（ ）αβx A. B. sin sin 0αβ+=cos cos 0αβ+=C.D. 22sin sin 1αβ+=tan tan 0αβ-=【答案】A 【解析】【分析】因为，的终边（均不在轴上）关于轴对称，则，，然后利用诱导公αβy x 2k αβπ+=Z k ∈式对应各个选项逐个判断即可求解．【详解】解：因为，的终边（均不在轴上）关于轴对称，αβy x 则，，2k αβπ+=Z k ∈选项A ：，故A 正确，sin sin sin sin(2)sin sin 0k αβαπααα+=+-=-=选项B ：，故B 错误，cos cos cos cos(2)2cos 0k αβαπαα+=+-=≠选项C ：，故C 错误，22222sin sin sin sin (2)2sin 0k αβαπαα+=+-=≠选项D ：，故D 错误，tan tan tan tan(2)tan tan 2tan 0k αβαπαααα-=--=+=≠故选：A ．7. 已知函数的部分图象如图所示，若存在，满足()()()sin 0,f x x ωϕωϕπ=+><120x x π≤<≤，则（ ）()()1234f x f x ==()12cos x x -=A.C. D. 3434-【答案】C 【解析】【分析】利用图象求得函数的解析式为，由结合正弦函数的()f x ()sin 26f x x π⎛⎫=- ⎪⎝⎭()()1234f x f x ==()f x 对称性得出，且有，将代入结合诱导公式可求得2123x x π=-13sin 264x π⎛⎫-= ⎪⎝⎭2123x x π=-()12cos x x -的值.()12cos x x -【详解】由图象知函数的最小正周期为，，()f x 137622121212T ππππ⎛⎫=⨯-=⨯= ⎪⎝⎭22T πω∴==又，7135121226πππ+=且，555sin 2sin 1663f πππϕϕ⎛⎫⎛⎫⎛⎫=⨯+=+=- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，，πϕπ-<< 257333πππϕ∴<+<所以，，，，5332ππϕ+=6πϕ∴=-()sin 26f x x π⎛⎫∴=- ⎪⎝⎭当时，，0x π≤≤112666x πππ-≤-≤因为存在，满足，120x x π≤<≤()()1234f x f x ==即，则，可得，且，12226622x x πππ-+-=1223x x π+=2123x x π=-13sin 264x π⎛⎫-= ⎪⎝⎭则.()121123cos cos 2cos 2sin 236264x x x x x ππππ⎛⎫⎛⎫⎛⎫-=-=--=-=⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭故选：C.【点睛】方法点睛：根据三角函数的部分图象求函数解析式的方法：()()sin f x A x bωϕ=++（1）求、，；A ()()max min:2f x f x b A -=()()max min2f x f x b +=（2）求出函数的最小正周期，进而得出；T 2T πω=（3）取特殊点代入函数可求得的值.ϕ8. 已知函数的定义域为 ，且函数的图象关于点对称，对于任意的，总有()y f x =R (1)=-y f x (1,0)x 成立，当时，，函数（），对任意(2)(2)f x f x -=+(0,2)x ∈2()21f x x x =-+2()g x mx x =+x ∈R x ∈，存在，使得成立，则满足条件的实数构成的集合为（ ）R t ∈R ()()>f x g t m A.B.1{|}4≤m m 1{|}4<m m C.D.1{|0}4<≤m m 1{|}4>m m 【答案】A 【解析】【分析】由的特性结合函数图象平移变换可得是奇函数，由可得函数(1)=-y f x ()f x (2)(2)f x f x -=+的周期，由此探讨出的值域，再将所求问题转化为不等式在上有解即可.()f x ()f x 21mx x +≤-R【详解】由函数的图象关于点对称知函数的图象关于原点对称，即函数(1)=-y f x (1,0)()y f x =是奇函数，()y f x =由任意的，总有成立，即恒成立，于是得函数的周期是4，x (2)(2)f x f x -=+(4)()f x f x +=()y f x =又当时，，则当时，，而是奇函数，当(0,2)x ∈2()21f x x x =-+(0,2)x ∈0()1f x ≤<()f x 时，，(2,0)x ∈-1()0f x -<≤又，f (-2)=-f (2)，从而得，即时，，(2)(2)f f -=(2)(2)(0)0-===f f f [2,2)x ∈-1()1f x -<<而函数的周期是4，于是得函数在上的值域是，()y f x =()y f x =R (1,1)-因对任意，存在，使得成立，从而得不等式，即在上x ∈R t ∈R ()()>f x g t ()1g x ≤-21mx x +≤-R 有解，当时，取，成立，即得，0m ≤2x =-4221m -≤-<-0m ≤当时，在上有解，必有，解得，则有，0m >210mx x ++≤R 140m ∆=-≥14m ≤104m <≤综上得，14m ≤所以满足条件的实数构成的集合为.m 1{|}4≤m m 故选：A二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，有选错的得0分，部分选对的得2分．9. 关于x 的不等式的解集可以是（ ）230ax x ++>A. B. {|3}x x >RC. D. ∅3|12x x ⎧⎫-<<⎨⎬⎩⎭【答案】BD 【解析】【分析】A 选项的形式看，则不等式不可能是二次不等式，分析出；BC 选项和的符号与判别式有0a =a 关；D 选项利用韦达定理可求出.a【详解】对于A ，若不等式的解集为，不等式不可能是二次不等式， 则，230ax x ++>{|3}x x >0a =此时，解得．显然不符合题意，∴不等式的解集不会是．故A30x +>3x >-230ax x ++>{|3}x x >错误；对于B ，当即时，20,1120,a a >⎧⎨∆=-<⎩112a >不等式的解集是．故B 正确；对于C ，若不等式的解集为，则有230ax x ++>R 230ax x ++>∅事实上，，与矛盾，∴不等式的解集不可以是0,0,a <⎧⎨∆≤⎩21121(12)0a a ∆=-=+->0∆≤230ax x ++>．故C 错误；对于D ，若不等式的解集是，则方程的两∅230ax x ++>3|12x x ⎧⎫-<<⎨⎬⎩⎭230ax x ++=个实数根分别为和，由韦达定理，此时符合题意．故D 正确．1-3231123312a a ⎧-+=-⎪⎪⎨⎪-⋅=⎪⎩20a =-<故选：BD ．10. 已知点是角终边上一点，则（ ）()(),20P m m m ≠αA. B.tan2α=sin α=C.D.()sin 2sin 2παπα-=⎛⎫+ ⎪⎝⎭223sin cos 5αα-=【答案】ACD 【解析】【分析】由三角函数的定义可得，，然后逐一判断即可.tan 2α=sin α=【详解】因为点是角终边上一点，所以，，A 正确，B 错误.P α2tan 2m m α==sin α=，C 正确.()sin sin tan 2cos sin 2παααπαα-===⎛⎫+ ⎪⎝⎭，D 正确.22222222sin cos tan 13sin cos sin cos tan 15αααααααα---===++故选：ACD11. 已知函数的图象如图所示，则（ ）()()sin 0,2f x A x πωϕωϕ⎛⎫=+>< ⎪⎝⎭A. 函数解析式()2sin 23f x x π⎛⎫=+ ⎪⎝⎭B. 将函数的图象向左平移个单位长度可得函数的图象2sin 26y x π⎛⎫=- ⎪⎝⎭4π()f x C. 直线是函数图象的一条对称轴1112x π=-()f x D. 函数在区间上的最大值为2()f x ,02π⎡⎤-⎢⎥⎣⎦【答案】ABC 【解析】【分析】根据图像得到解析式，利用函数的性质进项判断即可.【详解】由题图知：函数的最小正周期，()f x 453612T πππ⎛⎫=⨯-= ⎪⎝⎭则，，所以函数．22πωπ==2A =()()2sin 2f x x ϕ=+将点代入解析式中可得，,212π⎛⎫⎪⎝⎭22sin 6πϕ⎛⎫=+ ⎪⎝⎭则，得，()262k k Z ππϕπ+=+∈()23k k Z πϕπ=+∈因为，所以，2πϕ<3πϕ=因此，故A 正确．()2sin 23f x x π⎛⎫=+ ⎪⎝⎭将函数的图像向左平移个单位长度可得函数的图像，故B 正2sin 26y x π⎛⎫=- ⎪⎝⎭4π()2sin 23f x x π⎛⎫=+ ⎪⎝⎭确.，当时，，故C 正确.()2sin 23f x x π⎛⎫=+ ⎪⎝⎭1112x π=-()2f x =当时，，所以,02x π⎡⎤∈-⎢⎥⎣⎦23x π+∈2,33ππ⎡⎤-⎢⎥⎣⎦()f x ⎡∈-⎣故D 错误．故选：ABC ．12. 设函数，g （x ）＝x 2-（m ＋1）x ＋m 2-2，下列选项正确的有（ ）ln(2),2()1,2x x f x x x ->⎧=⎨+≤⎩A. 当m ＞3时，f [f （x ）]＝m 有5个不相等的实根B. 当m ＝0时，g [g （x ）]＝m 有4个不相等的实根C. 当0＜m＜1时，f [g （x ）]＝m 有6个不相等的实根D. 当m ＝2时，g [f （x ）]＝m 有5个不相等的实根【答案】BCD 【解析】【分析】作出函数的图象，利用函数的图象和函数的图象分析可解得结果.()f x ()f x ()g x 【详解】作出函数的图象：()f x 令，得；()f x t =[()]()f f x f t m ==当时，有两个根：，方程有1个根，方程有2个根，3m >()f x m =31242e t t <->+，1()f x t =2()f x t =所以A 错误；②当时，，，令，0m =2()2g x x x =--[()]0g g x =()g x t =由得()0g t =，1221t t ==-，，由2122t x x ==--12x x ⇒==由所以B正确；223412t x x x x =-=--⇒==③令，，因为，所以有个实根根，()g x t =()f t m =∴01m <<()f t m =3123,,t t t 设，所以123t t t <<12311ln(2)t m t m t m--=+=-=，，，，22()(1)2g x x m x m =-++-221329(24m m m x +--=-+23294m m --≥，221329329144m m m m t m -----=---23254m m --+=因为在上递减，所以，2325m m --+(0,1)23253250m m --+>--+=所以，所以，2132504m m t --+->213254m m t --+>即方程的最小根大于的最小值，()f t m =1t ()g x 所以、、都有2个不等实根，且这6个实根互不相等，1()g x t =2()g x t =3()g x t =所以当0＜m ＜1时，f [g （x ）]＝m 有6个不相等的实根，所以C 正确；④令，则，()f x t =()g t m =当时，方程化为，得；2m =()2g t =230t t -=1230t t ==，当，得；20()t f x ==1213x x =-=，当得符合题意，所以D 正确.13()t f x ==，3442x x =-=，，352e x =+故选：BCD.【点睛】关键点点睛：作出函数的图象，利用数形结合法求解是解题关键.三、填空题：本题共4小题，每小题5分，共20分．13. 函数的最小正周期是______．tan 23y x π⎛⎫=- ⎪⎝⎭【答案】##2π1π2【解析】【分析】根据题意，结合正切函数图像性质，即可求解.【详解】根据题意，结合正切函数图像性质，易知函数的最小正周期.tan 23y x π⎛⎫=- ⎪⎝⎭2T ππω==故答案为：.2π14. 设：，：（）．若是的必要条件，则m 的取值范围是α13x ≤≤β124m x m +≤≤+m ∈R βα______．【答案】1,02⎡⎤-⎢⎥⎣⎦【解析】【分析】记的解集为，的解集为，因为是的必要条件，所以，13x ≤≤A 124m x m +≤≤+B βαA B ⊆讨论，两种情况，利用包含关系得出m 的取值范围.B =∅B ≠∅【详解】记的解集为，的解集为13x ≤≤A 124m x m +≤≤+B 因为是的必要条件，所以βαA B ⊆当时，即，不满足；B =∅3m <-A B ⊆当时，要使得，则，解得B ≠∅A B ⊆12411243m m m m +≤+⎧⎪+≤⎨⎪+≥⎩12m -≤≤故答案为：1,02⎡⎤-⎢⎥⎣⎦15. 数学中处处存在着美，机械学家莱洛沷现的莱洛三角形就给人以对称的美感．莱洛三角形的画法：先画等边三角形ABC ，再分别以点A ，B ，C 为圆心，线段AB 长为半径画圆弧，便得到莱洛三角形．若线段AB 长为2，则莱洛三角形的面积是________．【答案】##2π-2π-+【解析】【分析】由题意，可先求解出正三角形扇形面积，再利用莱洛三角形与扇形之间的关系转化即可求解.【详解】由已知得， 2π3AB BC AC ===则AB =BC =AC =2，故扇形的面积为，2π3由已知可得，莱洛三角形的面积扇形面积的3倍减去三角形面积的2倍，∴所求面积为．22π3222π3⨯-=-故答案为：或．2π-2π-+16. 已知a 为正数，函数在区间和上的最大值分别记为和，若()sin f x x =[0,]a [,2]a a 1M 2M ，则___________，a 的取值范围为___________.122M ≥1M =【答案】 ①. ②. 127,36ππ⎡⎤⎢⎥⎣⎦【解析】得出大致范围，从而求出的值，再根据的范围即可求出的取122M ≥a 1M 2M a 值范围.【详解】由于函数在区间和上的最大值分别记为和，则()sin f x x =[0,]a [,2]a a 1M 2M 122M ≥，否则，与条件矛盾.所以=1.于是得2a π>12MM <1M 2M ≤所以，结合，所以,所以.sin 2a a ≤≤22a a ππ⎧>⎪⎨⎪<⎩27,233a a ππ≥≤2736aππ≤≤故答案为：1；.27,36ππ⎡⎤⎢⎥⎣⎦四、解答题：本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17. 设全集，已知集合，.U =R {}|228xA x =≤≤B x y ⎧==⎨⎩（1）求，.A B ⋃()UB A （2）已知非空集合，且，求实数的取值范围.{}1C x x a =<<C A C = a【答案】（1），；[)1,A B ⋃=+∞()[]1,2UB A ⋂= （2）.(]1,3a ∈【解析】【分析】（1）由指数函数的单调性求集合A ，由根式、分式的性质求集合B ，再应用集合的交、并、补运算求、、即可.U B A B ⋃()U B A （2）由题设知，可得，即可求的取值范围.C A ⊆13a a >⎧⎨≤⎩a 【小问1详解】由题设，可得，，则，[]1,3A =()2,B =+∞(],2U B ∞=- ∴，.[)1,A B ⋃=+∞()[]1,2UB A ⋂= 【小问2详解】∵，即，C A A ⋂=C A ⊆∴，解得.13a a >⎧⎨≤⎩(]1,3a ∈18. （1）已知角的终边经过点，求的值；α3(4,)P -2sin cos αα+（2）已知角终边上一点P 与x 轴的距离与y 轴的距离之比为，求的值．α3:42sin cos αα+【答案】（1） ；（2）答案不唯一，具体见解析 ．22sin cos 5αα+=-【解析】【分析】（1）利用任意角三角函数的定义即可求解；（2）根据题意，设出终边上一点，然后利用任意角的三角函数分象限即可求解.(4,3)(0)P a a a ±±≠【详解】（1）∵角的终边经过点，∴，α()4,3P -5OP ==∴，，∴；3sin 5α-=4cos 5α=22sin cos 5αα+=-（2）∵角终边上一点P 与x 轴的距离与y 轴的距离之比为，α3:4∴，(4,3)(0)P a a a ±±≠当角终边在第一象限时，，，；α4cos 5α=3sin 5α=2sin cos 2αα+=当角终边在第二象限时，，，；α4cos 5α=-3sin 5α=22sin cos 5αα+=当角终边在第三象限时，，，；α4cos 5α=-3sin 5α=-2sin cos 2αα+=-当角终边在第四象限时，，，．α4cos 5α=3sin 5α=-22sin cos 5αα+=-19. 已知函数图象的相邻两条对称轴间的距离为π()2sin(0)3f x x ωω=->π.2（1）求函数的单调递增区间和其图象的对称轴方程;()f x （2）先将函数的图象各点的横坐标向左平移个单位长度，纵坐标不变得到曲线C ，再把C 上()y f x =π12各点的横坐标保持不变，纵坐标变为原来的，得到的图象，若，求x 的取值范围.12()g x 1()2g x ≥【答案】（1）单调递增区间为，对称轴方程为；π5ππ,π(Z)1212k k k ⎡⎤-+∈⎢⎣⎦π5π(Z)212k x k =+∈（2）πππ,π(Z).62k k k ⎡⎤++∈⎢⎥⎣⎦【解析】【分析】(1)由条件可得函数的最小正周期，结合周期公式求，再由正弦函数性质求函数的单()f x ω()f x 调递增区间和对称轴方程；(2)根据函数图象变换结论求函数的解析式，根据直线函数性质解不等式求()g x x 的取值范围.【小问1详解】因为图象的相邻两条对称轴间的距离为，所以的最小正周期为，()f x π.2()f x π所以，，所以，2ππω=2ω=π()2sin(2)3f x x =-由，可得，，πππ2π22π232k x k -≤-≤+π5πππ1212k x k -≤≤+()k ∈Z所以函数的单调递增区间为，()f x π5ππ,π(Z)1212k k k ⎡⎤-+∈⎢⎥⎣⎦由得，()ππ2πZ 32x k k -=+∈π5π(Z)212k x k =+∈所以所求对称轴方程为π5π(Z)212k x k =+∈【小问2详解】将函数的图象向左平移个单位长度得到曲线，()y f x =π12π:2sin(26C y x =-把C上各点的横坐标保持不变，纵坐标变为原来的得到的图象，12π()sin(26g x x =-由得，所以，，1()2g x ≥π1sin(2)62x -≥ππ5π2π22π666k x k +≤-≤+Z k ∈所以，，所以x 的取值范围为ππππ62k x k +≤≤+Z k ∈πππ,π(Z).62k k k ⎡⎤++∈⎢⎥⎣⎦20. 已知函数，．22()log 11f x x ⎛⎫=+ ⎪-⎝⎭1()2x g x +=-（1）求证：为奇函数；()f x （2）若恒成立，求实数的取值范围；()22()xf kg x - k （3）解关于的不等式．a ()(2)22g a g a a --- 【答案】（1）证明见解析 （2）(],7-∞（3）[)1,+∞【解析】【分析】（1）求得的定义域，计算，与比较可得；()f x ()f x -()f x （2）原不等式等价为对恒成立，运用基本不等式可得最小值，进而得()2322121xxk ≤++--0x >到所求范围；（3）原不等式等价为，设，判断其单调性可得()()()22g a a g a a -≤---()()h x g x x =-的不等式，即可求出.a 【小问1详解】函数，2221()log 1log 11x f x x x +⎛⎫=+= ⎪--⎝⎭由解得或，可得定义域，关于原点对称，101x x +>-1x <-1x >()(),11,-∞-⋃+∞因为，()2211()log log 11x x f x f x x x -+-==-=-+-所以是奇函数；()f x 【小问2详解】由或，解得，21x<-21x>0x >所以恒成立，即，()()22()0xf kg x x -> 221log 12122x x x k ++--≥-则，即对恒成立，121221x x x k ++--- ()1212232212121x x x x xk +++=++--- 0x >因为，当且仅当，即时等号成立，()23221322721x x ++-+⨯=- ()222121xx=--1x =所以，即的取值范围为；7k ≤k (],7-∞【小问3详解】不等式即为，()(2)22g a g a a --- ()(2)(2)g a a g a a ---- 设，即，可得在上递减，()()h x g x x =-1()2x h x x +=--()h x R 所以，则，解得，()(2)h a h a - 2a a ≥-1a ≥所以不等式的解集为.[)1,+∞21. 已知函数（且）.41()log 2x a xf x +=0a >1a ≠（1）试判断函数的奇偶性；()f x （2）当时，求函数的值域；2a =()f x （3）已知，若，使得，求实数的取值范()g x x =-[][]124,4,0,4x x ∀∈-∃∈12()()2f x g x -≥a 围．【答案】（1）函数是偶函数 ()f x （2） [1,+)∞（3）(1,2]【解析】【分析】（1）根据偶函数的定义可判断出结果；（2）根据基本不等式以及对数函数的单调性可求出结果；（3）将，使得，转化为，[][]124,4,0,4x x ∀∈-∃∈12()()2f x g x -≥min [()]f x min [()2]g x ≥+利用换元法求出，分类讨论，利用函数的单调性求出的最小值，代入可求出结果.min [()2]g x +a ()f x ()f x 【小问1详解】因为且，所以其定义域为R ，41()log (02x a x f x a +=>1)a ≠又，4114()log log ()22x xa a x x f x f x --++-===所以函数是偶函数；()f x 【小问2详解】当时，，因为,,当且仅当，即时取等，2a =241()log 2x x f x +=20x >4112222x xx x =+≥+21x=0x =所以,241()log 2x x f x +=2log 21≥=所以函数的值域为.()f x [1,)+∞【小问3详解】，，使得，等价于，1[4,4]x ∀∈-2[0,4]x ∃∈12()()2f xg x -≥min [()]f x min [()2]g x ≥+令，，，t =[0,4]x ∈[0,2]t ∈令，则在上的最小值等于在上的最小值，2()22h t t t =-+()2g x +[0,4]()h t [0,2]在上单调递减，在上单调递增，所以在上的最小值为，所以()h t [0,1][1,2]()h t [0,2](1)1h =.min [()]1f x ≥因为为偶函数，所以在上的最小值等于在上的最小值，()f x ()f x [4,4]-()f x [0,4]设，则，41()2x xv x +=()log ()a f x v x =任取，1204x x ≤<≤，1212124141()()22x x x x v x v x ++-=-12121(22)(12x x x x +=--因为，所以，，，，，1204x x ≤<≤1222x x <12220x x -<120x x +>1221x x +>121102x x +->所以，，12121(22)(1)02x x x x +--<12()()v x v x <所以在上为单调递增函数，41()2x xv x +=[0,4]当时，函数在上为单调递减函数，01a <<()log ()a f x v x =[0,4]所以，所以，得（舍）；4min441()(4)log 2a f x f +==257log 16a=257log 116a ≥25716a ≥当，函数在上为单调递增函数，1a >()log ()a f x v x =[0,4]所以，所以，.min ()f x (0)f =log 2a =log 21a ≥12a <≤综上得：实数的取值范围为.a (1,2]22. 设函数（，且）．()x xf x a a -=-x ∈R 0a >1a ≠（1）若，证明是奇函数，并判断单调性（不需要证明）；01a <<()y f x =（2）若，求使不等式恒成立时，实数的取值范围；()10f <()()24f x tx f x ++-<0t （3）若，，且在上的最小值为，求实数的值．()312f =()()222x xg x a a mf x -=+-()g x [)1,+∞2-m 【答案】（1）证明见解析，是减函数；()f x （2）(－3，5)； （3）2﹒【解析】【分析】(1)f (x )定义域为R 关于原点对称，判断f (－x )与f (x )的关系，以此确定奇偶性；f (x )的单调性可以通过单调性的性质进行判断；(2)利用条件，得到在R 上单调递减，从而将转化为()10f ＜()01a f x ＜＜．()()240f x x f x -＋＋＜，进而得，研究二次函数得到结论；()()24f x txf x -＋＜24x txx -＋＞(3)令，得到二次函数h (t )，分类讨论研究得()22xxt f x --＝＝222322()22t mt t m m t ⎛⎫---≥ ⎪⎝⎭＝＋＝＋到，得到结论．2m ＝【小问1详解】证明：的定义域为，关于原点对称，()f x R 且，()()x x f x a a f x --=-=-∴为奇函数，()f x ∵，∴递减，递减，故是减函数；01a <<xy a =xy a -=-()f x 【小问2详解】(且)，()x x f x a a -=-0a >1a ≠∵，∴，()10f <10a a -<又，且，0a >1a ≠∴，01a <<故在上单调递减，()f x R 不等式化为，()()24f x tx f x +<-∴，即恒成立，24x tx x +>-()2140x t x +-+>∴，()21160t ∆=--<解得；35t -<<【小问3详解】∵，∴，即，()312f =132a a -=22320a a --=解得或(舍去)，2a =12a =-∴，()()()()2222222222x x x x x x g x a a mf x m ---=+-=---+令，由(1)可知为增函数，()22x xt f x -==-()22x xf x -=-∵，∴，1x ≥()312t f ≥=令，()()22232222h t t mt t m m t ⎛⎫=-+=-+-≥⎪⎝⎭若，当时，，∴；32m ≥t m =()2min 22h t m =-=-2m =若时，当时，，解得，无解；32m <32t =()min 2h t =-253122m =>综上，．2m =。
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互联网信息内容安全管理教程 - 题库互联网信息内容安全管理教程 - 习题库一、判断题1.中华人民共和国公民的通信自由和通信秘密受法律的保护。
对2.在计算机程序或图像、文字中含有色情、凶杀等内容的信息，其目的是用于违法活动的，可认为是计算机有害数据。
对3.机关、团体、企业、事业单位违反治安管理的，处罚直接责任人员；单位主管人员指使的，同时处罚该主管人员。
对4.用于违法活动的计算机病毒、木马、间谍软件属于计算机有害数据。
对5.利用互联网侵犯他人合法权益，构成民事侵权的，依法承担民事责任。
对6.制定《中华人民共和国计算机信息网络国际联网管理暂行规定》是为了加强对计算机信息网络国际联网的管理，保障国际计算机信息交流的健康发展。
对7.接入网络必须通过互联网络进行国际联网。
对8.国家对国际联网实行统筹规划、统一标准、分级管理、促进发展的原则。
对9.公安部信息管理小组负责协调、解决有关国际联网工作中的重大问题。
错10.单位和个人的计算机信息网络直接进行国际联网时，可以自由选择信道进行国际联网。
错11.任何单位和个人不得自行建立或者使用其他信道进行国际联网。
对12.接入单位拟从事国际联网非经营活动的，不必批准即可接入互联网络进行国际联网。
错13.未取得国际联网经营许可证的，接人单位不得从事国际联网经营业务。
对14.从事国际联网经营活动和非经营活动的接入单位在不符合条件时，其国际联网经营许可证由发证机关吊销。
错15.国家秘密信息在与国际网络联网的计算机信息系统中存储、处理、传递需经批准。
错16.国际出入口信道提供单位、互联单位和接人单位，应当建立相应的网络管理中心，依照法律和国家有关规定加强对本单位及其用户的管理，做好网络信息安全管理工作，确保为用户提供良好、安全的服务。
对17.国际出入口信道，是指国际联网所使用的物理信道。
对18.《中华人民共和国计算机信息网络国际联网管理暂行规定实施办法》中的个人用户一定具有联网帐号。
对19.企业计算机信息网络，是指为企业服务的专用计算机信息网络。
错20.国家对国际联网的建设布局、资源利用进行统筹规划。
对21.企业计算机信息网络，是指为行业服务的专用计算机信息网络。
错22.《中华人民共和国计算机信息网络国际联网管理暂行规定实施办法》规定的企业计算机信息网络，是指企业内部和外部相连接的计算机信息网络。
错23.国际联网实行分级管理，即对互联单位、接入单位、用户、国际出入口信道逐级管理。
错24.在中国已建立的四个互联网络中，两个经营性互联网络不应享受同等的资费政策和技术支撑条件。
错25.在中国已建立的四个互联网络中，两个公益性互联网络所使用信道的资费应当享受优惠政策。
对26.企业计算机信息网络和其他通过专线进行国际联网的计算机信息网络可不受限制地经营国际互联网络业务。
错27.经营性接入单位凭经营许可证向提供电信服务的企业办理所需通信线路手续。
提供电信服务的企业应当在半年内为接入单位提供通信线路和相关服务。
错28.用户向接入单位申请国际联网时，应当提供有效身份证明或者其他证明文件。
对29.计算机系统运行管理部门必须设有安全组织或安全负责人。
对30.国际出入口信道提供单位与互联单位应当签订相应的协议，严格履行各自的责任和义务。
对31.用户有权获得接入单位提供的各项服务；有义务交纳费用。
对32.计算机病毒疫情，是指某种计算机病毒爆发、流行的时间、范围、破坏特点、破坏后果等情况的报告或者预报。
对33.根据《全国人民代表大会常务委员会关于维护互联网安全的决定》的规定，对由计算机病毒所造成的系统瘫痪、数据破坏等重大事故，应及时向公安机关报告。
对34.企业计算机信息网络和其他通过专线进行国际联网的计算机信息网络，只限于内部使用。
对35.进行国际联网的专业计算机信息网络可经营国际互联网络业务。
错36.制定《中华人民共和国计算机信息系统安全保护条例》是为了促进我国计算机的应用和发展，保障社会主义现代化建设的顺利进行。
对37.公安部、国家安全部、国家保密局、国务院是计算机信息系统安全保护工作的主管部门。
错38.除从事国家安全事务的人员外，任何组织和个人都不得利用计算机信息系统从事危害国家利益、集体利益和公民合法利益的活动。
错39.任何单位和个人不得向社会发布虚假的计算机病毒疫情。
对40.对计算机病毒的认定工作，由公安部公共信息网络安全监察部门批准的机构承担。
对41.计算机信息系统打印输出的涉密文件，应当按相应密级的文件进行管理。
对42.监督、检查、指导计算机信息系统安全保护工作是公安机关对计算机信息系统安全保护工作中的监督职权之一。
对43.查处危害计算机信息系统安全的违法犯罪案件不是公安机关对计算机信息系统安全保护工作的监督职权。
错44.公安部在紧急情况下，可以就涉及计算机信息系统安全的特定事项发布专项通令。
对45.违反计算机信息系统安全等级保护制度及计算机信息系统国际联网备案制度，危害计算机信息系统安全的其他行为的，由公安机关处以警告或者停机整顿。
对46.不按规定时间报告计算机信息系统中发生的案件的行为违反了《中华人民共和国计算机信息系统安全保护条例》的规定，由公安机关作出处理。
对47.任何组织或者个人违反《中华人民共和国计算机信息系统安全保护条例》的规定，给国家、集体或者他人财产造成损失的，应当依法承担民事责任。
对48.计算机病毒，是指编制或者在计算机程序中插入的破坏计算机功能或者毁坏数据，影响计算机使用，并能自我复制的一组计算机指令或者程序代码。
对49.计算机信息系统安全专用产品，是指计算机的软、硬件产品。
错50.公安机关公共信息网络安全监察部门应当掌握互联单位、接入单位和用户的备案情况，建立备案档案，进行备案统计，并按照国家有关规定逐级上报。
对51.制定《计算机信息网络国际联网安全保护管理办法》的目的是为了加强对计算机信息网络国际联网的安全保护，维护公共秩序和社会稳定。
对52.和中华人民共和国境内的计算机信息网络相连的所有国际网络的安全保护管理，都适用《计算机信息网络国际联网安全保护管理办法》。
错53.任何单位和个人不得利用国际联网侵害公民的合法权益，不得从事违法犯罪活动。
对54.利用国际互联网公然污辱他人或者捏造事实诽谤他人的应按国家法律进行处罚。
对55.任何单位和个人不得从事危害计算机信息网络安全的活动。
对56.任何单位和个人都可以无条件进入计算机信息网络或者使用计算机信息网络资源。
错57.已经联网的用户可以对计算机信息网络功能进行删除、修改或者增加。
错58.已经联网的用户有对计算机信息网络中存储、处理或者传输的数据和应用程序进行删除、修改或者增加的权利。
错59.除公安机关外，任何单位和个人不得违反法律规定，利用国际联网侵犯用户的通信自由和通信秘密。
错60.使用国际联网的用户的通信自由和通信秘密受法律保护。
对61.任何单位和个人都可以利用国际联网查阅用户的通信秘密。
错62.国际出入口信道提供单位、互联单位的主管部门或者主管单位，应当依照法律和国家有关规定负责国际出入口信道、所属互联网络的安全保护管理工作。
对63.互联单位、接人单位及使用计算机信息网络国际联网的法人，对委托发布信息的单位和个人进行登记后，就可以为其发布其要发布的信息。
错64.互联单位、接人单位及使用计算机信息网络国际联网的法人和其他组织当发现本网络中有危害国家利益的内容的地址、目录时，应当按照国家规定把它删除。
对65.用户在接人单位办理入网手续时，应当填写用户备案表。
对66.使用公用帐号的注册者应当加强对公用帐号的管理。
用户帐号可以转借、转让。
错67.涉及国家事务、经济建设、国防建设、尖端科学技术等重要领域的单位办理备案手续时，可以不出证明就给予特批。
错68.涉及国家事务、经济建设、国防建设、尖端科学技术等重要领域的单位的计算机信息网络与国际联网，应当采取相应的安全保护措施。
对69.县级公安局可以不设负责国际联网的安全保护管理工作的机构。
错70.省、自治区、直辖市公安厅(局)应当有相应机构负责国际联网的安全保护管理工作，但地(市)、县(市)公安局则不必如此。
错71.督促互联单位、接人单位及有关用户建立健全安全保护管理制度是公安机关公共信息网络安全监察部门的职责。
对72.公安机关公共信息网络安全监察部门在组织安全检查时，有关单位应当派人参加。
对73.公安机关公共信息网络安全监察部门对在安全检查中发现的问题，应当提出改进意见，作出详细记录，存档备查。
对74.经营国际联网业务的单位，有违法行为时，公安机关可以向原发证、审批机构提出吊销其经营许可证或者取消其联网资格的建议。
对75.经营国际联网业务的单位，有违法行为时，公安机关可以吊销其经营许可证或者取消其联网资格。
错76.与香港特别行政区和台湾、澳门地区联网的计算机信息网络的安全保护管理，参照《计算机信息网络国际联网安全保护管理办法》执行。
对77.计算机信息网络国际联网，是指中华人民共和国境内的计算机信息网络为实现信息的国际交流，同外国的计算机信息网络相连接。
对二、单选题1.违反国家规定，对计算机信息系统中存储、处理或者传输的数据和应用程序进行删除、修改、增加的操作，后果严重的，构成( B)。
B.破坏计算机信息系统罪2.故意制作、传播计算机病毒等破坏性程序，影响计算机系统正常运行，( )的，应依法处5年以下有期徒刑或者拘役。
AA.后果严重3.在计算机信息系统中，以计算机文字表示的，含有危害国家安全内容的信息，是属于( B)。
B.计算机有害数据4.( )，是指直接进行国际联网的计算机信息网络。
DD.互联网络5.( )，是指通过接人互联网络进行国际联网的计算机信息网络。
CC.接入网络6.新建互联网络，必须报经( )批准。
CC.国务院7.接入单位从事国际联网经营活动的和从事非经营活动的相比，还应具备( )。
DD.为用户提供长期服务的能力8.用户使用的计算机或者计算机信息网络，需要接人接入网络的，应当征得( )的同意，并办理登记手续。
AA.接入单位9.国际出入口信道提供单位、互联单位和接入单位，应当建立相应的( )中心。
BB.网络管理10.中华人民共和国境内的计算机信息网络进行国际联网，应当依照( )办理。
BB.《中华人民共和国计算机信息网络国际联网管理暂行规定实施办法》11.中华人民共和国境内的计算机互联网络、专业计算机信息网络、企业计算机信息网络，以及其他通过专线进行国际联网的计算机信息网络同外国的计算机信息网络相连接，这称为()。
BB.国际联网12.接入网络，是指通过接入互联网络进行国际联网的计算机信息网络；接人网络可以是( )连接的网络。
DD.多级13.《中华人民共和国计算机信息网络国际联网管理暂行规定实施办法》中的个人用户，是指具有( )的个人。