9702_s06_qp_3

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凌华工业主板常见问题分析处理

1、电源在AT或ATX之间变换时，主板可能无法开机，需BIOS放电或power switch短接2、机箱电源一定要接地，否则机箱上会带110V感应电压，电流在1mA之内3、使用PIII CPU时，需上导热硅脂，否则系统不稳定，易死机4、带网卡的CPU卡，安装winNT Workstation 4.0，在安装网卡前，先做如下处理，否则“事件查看器”会出现“存储器空间不足提示”一、a、进入CMOS,PnP/PCI ConfigurationsPNP OS Installed >[NO]Resources Controlled By > [Manual] 保存重启b、安装完网卡后，再打一次SP6补丁即可c、这时CMOS可更改为其它设置二、或网卡驱动应先于Service Pack来安装.否则有时会不能启动Workstation Service,并且报错:存储空间不足.出现这种错误时,可以重新安装一下Service Pack5、主板不开机，换BIOS或检查电池电量、CMOS放电检查电源PG、5VSB+0.01V6、KB、MS线最长5米，HDD IDE线最长75cm（标准45mm）。

nupro760LV可以支持到20米KB、MS延长线7、CPU卡直接使用LCD时，在系统关机时LCD屏闪动历害，可把屏幕“刷新频率”设高为：75或85Hz8、主板带两个网口时使用WinNT4.0+SP6.0时，网络要设置为两个网段，与其相连的机器也要设置为两个网段。

否则网口只能使用一个。

9、主板安装OS过程死机，除主板、内存及HDD外，也可能是CPU坏或HDD线坏如果是安装win2000出现*.DLL文件无法复制，先考虑内存坏，依次光盘、CDROM、HDD10、win98 SE无法关机或不能重启a、BIOS配置不正确。

在PnP/PCI Configurations选项中，有一项Assign IRQ VGA，如果你装的是Win98SE版本，关不了机，跟这个选项可能有关，要把它打开b、打win98简体关机补丁c、msconfig中“禁用快速关机”win98或win98SE开始不能安装或过程中死机A、如无软驱，可增加一个软驱B、BIOS中软驱设置为无11、在使用ATX电源时，也可当AT电源使用，但效果可能会不好，最好不要使用。

JMB352U

JMB352UHigh-Speed USB & SATA II Combo to Dual SATA II PortMultiplier(1 to 2) BridgeDescriptionThe JMB352U is a bridge controller, which translates the communication protocol between SATA II 3.0G. This chip integrated 60MIPS 8051, and provide eSATA II 3.0G to dual SATA II 3.0G port multiplier applications in a single chip.In this chip, the SATA controller could be configured as host or device. The 3-port SATA II 3.0G controllers further supports the eSATA to dual SATA communication. Integrated these two technologies, the user could perform JMB352U to the following applications.¾ eSATA to dual SATA bridge port multiplier application¾ One-button system backup application for above applications. (With SATA Vender Software Application)FeaturesQuick ReferenceSignal Bit Rate 3.0 Gbps Power Supply 3.3V and 1.2V GPIO Support UP to 9 ESD Protection 2000 V Package 64-pin LQFP RAID Level Support JBOD, Port MultiplierSATA Port 3 • Compliance with Serial ATA International Organization: Serial ATA Revision 2.6• Compliance with Gen1i, Gen1m, Gen2i and Gen2m of Serial ATA International Organization: Serial ATA Revision 2.6• Support SATA 1.5G/3.0G Speed Negotiation • Support SATA II Asynchronous Signal Recovery (Hot Plug) feature• Support SATA Host/Device controller configuration • Support USB HID Interrupt Transfer• Support ATA/ATAPI PACKET command feature set • Support SATA 30MHz or 40MHz external crystalUSB Port1• Integrated 60MIPS 8051 microprocessor with 64k-byte mask ROM• Support one-button system backup operation • Support one-button disk copy• Support external NVRAM for customer VID/PID, Manufacture and Product String.Applications• Mass storage devices • Optical storage • Storage system • Port MultiplierFunctional Block DiagramMCUUSB Device Controller Hi-Speed USB PHY3-port SATA II Controller3-port SATA II PHYDMA ControllerProgram ROM GPIO I/F Voltage RegulatorGPIOSATA_ASATA_CUSBFig. 1 Functional Block Diagram of JMB352UApplicationsNVRAM KeypadSATA AFig. 2 eSATA to dual SATA port multiplier applicationNVRAM LED KeypadSATAFig. 3 Hi-Speed USB to SATA with 3 LUN bridge applicationSATA DeviceNVRAM LED KeypadSATA BFig. 4 Dual eSATA(Port B & Port C) & Hi-Speed USB to SATA applicationProduct Information Name DescriptionJMB352 Serial ATA Bridge ChipDesign KitContact Information 1 JMB352 Data Sheet Department Email 2 JMB352 Design Guide Sales sales@ 3 Application EVBTech. Support fae@。

基于单片机的抽油烟机控制系统设计毕业论文答辩.ppt

系统硬件电路的设计
2、油烟煤气浓度监测电路设计
功能：主要检测煤气泄漏，当有煤气泄漏时，并超过了设定的预警值，经过MQ-2的检测，输出较强的电压信号，再经 ADC0832的转换成数字信号，触发AT89S52的中断程序，启动抽油烟机并伴随报警。
VCC 5678 76 11 I O D D CLK VCC ADC0832 P3.7P3.6 D P3.5 AT89S52 N S ADC0832 AT89S52 CCH0CH1G 5 1 1234 FROMMQ-2 TOADC0832 K 7 0 1R 21 T BB U O MQ-2 12 N MQ-2 AIA VCC
毕业设计中完成的主要任务
1 选题背景 2 系统的总体构思 3 系统中各个元器件的选择 4 硬件电路的设计 5 单片机程序的编写 6 系统软硬件的仿真
选题背景
随着我国第三产业的发展和城市化进程的加快，饮食服务业在城市经济发展中所占的比重越来越大。未经处理的油烟直接排到室外，对建筑物外墙面和空气造成严重污染。油烟污染不仅影响城市公共环境卫生，对人体健康也有很大危害。饮食油烟含有许多有毒有害成分，不同食品在高温下会产生不同的挥发性物质，其中含有许多突变和致癌性物质。现代厨房一般都离不开煤气灶和抽油烟机(通风设备)，一旦使用煤气灶，则一定会产生油烟、水汽与乙烷等影响厨房空气环境的混合气体，为了健康生存的需要，往往用通风设备如抽油烟机来改善充满油烟气的厨房环境。由于目前国内厨房油烟净化的研究还处于起步阶段，人们对油烟危害认识不足以及对其处理方法缺乏重视，我国餐饮企业和家庭厨房对油烟净化的效率都远远达不到要求。因此，分析油烟的污染状况，提出油烟治理新技术，已经刻不容缓。
S1：手动开关；S2：调时间，加1min； S3：调时间，减1min；S4：显示温度

性能级SDD推荐

性能级SDD推荐作者：来源：《微型计算机》2020年第04期普及大容量NVMe SSD 金士顿A2000 M.2NVMe固态硬盘500GB这款NVMe SSD采用慧荣的SM2263EN 4通道PCle 3.0 x4主控，每通道支持4CE片选信号。

同时该主控支持NVMe l.3规范，并配备了512MB缓存。

这款SSD板载了来自美光、FPGA编号为“NW951”的96层堆叠3DNAND TLC闪存，500GB容量产品的质保可写容量为350TB。

性能上，这款SSD配有一个接近75G B容量的超大SLC CACHE缓存空间，在ASSSD BENCHMARK中的总分逼近3000分。

在实际的游戏载入体验中，金士顿KC2000则能在20秒以内快速地启动各类游戏大作，比如《坦克世界》的启动时间只有不到13秒。

值得一提的是，目前金士顿A2000 M.2 NVMe固态硬盘正以豪华套装的形式销售，附送由散热器厂商Tt 专门为金士顿M.2NVMe SSD定制的铝合金散热器，颇具性价比。

兼得容量与性能西数蓝盘SN550 1TB这款SSD采用了来自闪迪的主控，是一款DRAM-Iess无缓存主控，支持HMB技术，可以使用少量电脑内存作缓存。

同时西数为其采用了96层堆叠的3DNAND TLC闪存颗粒，单颗闪存容量就达到了1TB。

质保方面，西部数据为SN550 SSD提供了5年质保加可写容量的质保政策（以先到为准），其1TB容量产品的质保可写容量达到600TB。

在测试中，西数蓝盘SN550 1TB的AS SSD BENCHMARK测试成绩达到了3689分，在PCle 3.O SSD中还是达到了中上水平。

它的连续读取速度为2233M B/s，高队列深度随机4KB 写入性能达到23.6万IOPS，全盘平均写入速度也能达到850.3M B/s。

其表现远远优于掉速后写入速度不到300M B/s的主流TLC SSD，以及实际写入速度不到100M B/s的QLC SSD。

中兴交换机配置

一、系统的启动过程如下。

， Ltd.System Booting..。

.。

CPU: S3C45010 ARM7TDMIBSP version: 1.2/0Creation date: Feb 11 2004， 09：37：01Press any key to stop auto-boot。

..72、出现上述信息后，等待大约7 秒，用户可以在这段时间内按任意键进入boot 状态,修改启动参数.当系统在规定时间未检测到用户输入时，系统便开始自动加载版本，并提示下列信息：auto—booting。

..boot device ： secEndunit number ： 0processor number ： 0host name ： tigerfile name : vxWorksinet on ethernet (e) ： 10.40.92。

106host inet （h) : 10。

40.92.105flags (f） : 0x80Attaching to TFFS.。

done。

Loading version：/kernel.。

.1459932 + 75292 + 6358852Starting at 0x1656e0...Attaching interface lo0。

.。

done（省略)Welcome ！ZTE Corporation。

二、配置开始工作1．打开超级终端，输入连接的名称，如ZXR10，并选择一个图标。

机房和施工组织模板

江西新昌电厂“上大压小〞新建工程厂级管理信息系统〔MIS〕网络及硬件平台系统集成投标文件投标编号：03-ct-03-2007-156〔第三卷附件1-技术标准书〕投标人：安徽科大恒星电子商务技术二00九年四月目录第一章概述 (1)工程建设需求 (1)设计原那么 (2)实用性与先进性相结合 (2)可扩展性和开放性 (2)可靠性和平安性 (3)可管理性 (3)设计、施工标准 (3)工程建设目标 (6)第二章效劳器系统平台设计 (7)建设需求 (7)数据库效劳器设计 (7)IBM power550方案 (7)效劳器选型 (7)2.2.1.2 IBM Power 550配置 (12)HP 小型机方案 (13)效劳器选型 (13)2.2.2.2 HP Integrity rx6600 配置 (17)系统运行模式设计 (17)并行处理模式 (18)任务分担模式 (18)主机备用模式 (18)其它效劳器设计 (18)效劳器的部署 (18)物理效劳器与逻辑效劳器之间的关系 (19)效劳器选型设计 (20)2.3.3.1 IBM X3850 M2方案 (20)2.3.3.2 HP ProLiant DL580 G5 效劳器方案 (23)效劳器配置 (26)2.3.4.1 IBM X3850 M2配置 (26)2.3.4.2 HP ProLiant DL580 G5配置 (27)SAN存储系统设计 (27)2.4.1 IBM TotalStorage DS4800方案 (28)2.4.1.1 IBM TotalStorage DS4800介绍 (28)2.4.1.2 TotalStorage DS4800配置 (31)2.4.2 HP StorageWorks EVA8100方案 (32)2.4.2.1 HP StorageWorks EV A8100介绍 (32)2.4.2.2 HP StorageWorks EV A8100配置 (35)光纤交换机选型 (36)2.4.3.1 IBM System Storage SAN2005-B16 特性 (36)2.4.3.2 IBM System Storage SAN2005-B16 配置 (38)数据备份系统 (38)数据备份需求 (38)数据备份方案 (39)2.5.2.1 Veritas的优势 (39)2.5.2.2 Veritas配置 (42)磁带库选型配置 (43)虚拟磁带库选型 (44)第三章网络系统平台设计 (49)网络建设需求 (49)网络拓扑结构设计 (49)分层设计 (49)可靠性保证〔冗余设计〕 (49)网络拓扑实现 (50)核心层设计 (51)接入层设计 (51)远程访问层设计 (51)Cisco解决方案 (51)中心交换机选型 (51)3.3.1.1 Cisco® 6500介绍 (52)3.3.1.2 Supervisor Engine 720 (56)3.3.1.3 Cisco Catalyst 6513配置 (58)接入交换机选型 (59)H3C解决方案 (59)中心交换机选型 (59)H3C S7500E介绍 (60)H3C S7510E配置 (63)接入交换机选型 (63)第四章系统平安设计 (65)系统平安需求 (65)系统平安体系结构 (65)物理设备平安 (65)访问平安 (66)应用平安 (66)数据平安 (66)平安策略 (67)传输通道和传输设备平安 (67)应用平台平安 (68)资源访问平安 (68)网络防病毒 (69)多功能防火墙设计 (70)HillstoneSA-5020介绍 (71)HillstoneSA-5020功能规格 (72)内网核心平安防护系统〔主机加固〕设计 (74)S-NUMEN用途 (74)S-NUMEN功能 (75)S-NUMEN配置 (76)内网主机审计及监管系统 (76)系统功能 (77)系统特点 (79)体系架构 (80)配置 (81)防病毒系统设计 (81)网络防病毒配置 (84)第五章综合布线系统设计 (85)综合布线系统需求 (85)设计标准与设计目标 (85)设计标准 (85)设计目标 (86)布线系统设计 (86)光缆敷设 (86)SAN光纤走向 (86)布线系统测试 (87)测试标准与内容 (89)提交文档 (94)第六章工程组织与管理 (95)工程管理与目标 (95)工程组织及人力资源分配 (96)施工组织总体部署 (96)人力资源配置 (98)与相关单位合作 (99)与工程单位合作 (99)与厂商合作 (100)工程管理 (100)管理方法 (100)管理措施 (101)工程风险管理 (104)工程进度管理 (104)工程质量控制与保证 (105)工程阶段性评估 (106)第七章工程实施方案 (107)工程实施环节规划 (107)实施环节规划总表 (107)设计联络 (109)设备订货及到货 (110)实施现场情况调研 (110)制定详细实施方案 (111)培训 (111)设备验收 (111)查验设备 (111)查验方法 (112)查验报告 (117)系统安装调试 (117)系统调试方案制定 (117)小型机调试 (118)中心交换机调试 (118)边缘交换机调试 (118)中心路由器调试 (119)网管软件调试 (119)主机调试 (119)备份系统调试 (120)系统初验收 (120)试运行 (120)技术文档的提交 (121)系统终验 (121)技术支持与售后效劳 (121)实施进度方案 (121)里程碑事件及考核标准 (122)第八章验收方案与文档 (123)验收测试内容 (123)现场验收测试 (124)文档验收 (126)系统初验收 (126)系统终验 (127)测试方案 (127)测试方法 (127)测试工程 (127)测试方法 (128)工程文档 (129)工程文档内容 (129)工程文档提交方案 (132)第九章机房建设工程 (134)工程简述 (134)电子计算机机房组成及使用面积确定 (134)可维护性设备布置 (135)设计原那么 (135)计算机机房平安分类 (135)建设标准 (137)设计依据 (138)设计依据 (138)设计指标 (140)机柜及机房隔断、装饰设计 (141)机柜 (141)图腾机柜概述 (141)图腾机柜选型 (141)机房隔断 (141)装饰设计 (142)装修材料选材 (142)材料表 (143)吊顶 (143)地面 (144)墙、柱面 (145)门、窗 (145)供配电系统〔含UPS、照明等〕 (146)电气概述 (146)供配电系统 (147)设计标准要求 (147)9.5.2.2 机房配电冗余供电系统 (149)配电设备 (149)UPS系统 (149)UPS选型选型 (149)艾默生“UL33〞系列UPS性能参数 (150)美国“艾默生〞系列UPS特点 (151)电池配置 (151)照明系统设计 (152)普通照明系统设计 (153)应急照明、疏导灯具系统设计 (153)配电线路安装技术 (153)空调系统 (154)空调系统设计 (154)空调设备选型 (154)防雷接地 (154)防雷系统概述 (154)对雷电引入的分析 (155)机房防雷设计 (156)防雷验收及保障 (156)接地系统概述 (156)接地系统解决措施 (157)机房的地线系统 (157)局部等电位连接 (157)抗静电保护地 (157)静电防护 (158)KVM设备及机房布线 (158)KVM设计需求 (158)KVM设计方案说明 (159)环境监控、消防报警及其他相关效劳 (159)环境监控 (159)门禁系统 (159)视频监控 (159)动力环境监控 (159)消防报警 (160)9.9.3 控制台桌椅 (161)灾害处理 (161)机房区防水防护措施 (161)机房给水排水技术 (161)防虫、防鼠害 (162)电磁屏蔽 (162)第一章概述江西新昌电厂网络及硬件平台系统集成工程建设主要包括了网络系统、无线网络覆盖、效劳器系统、存储系统、数据备份系统、微软AD域设计及网络部署、管理及系统软件、MIS终端、智能机房、综合布线等系统，按现代先进技术设计，该系统集成完成后,新昌电厂具有统一的生产MIS系统运行平台，能为其信息化建设提供良好的根底效劳。

IBM Storwize V7000 硬盘微码对照表

Drive Microcode Package for IBM Storwize V7000Ask the IBM Support Agent ToolDownloadable filesAbstractThis document contains details of the content of theIBM2072_DRIVE_20170901 Drive Microcode package and supporting documentation.Download DescriptionContents of Drive Microcode PackageNote:vendor_id IBM-207x is used on drives for IBM Storwize V7000 Gen1 (models 2076-112, 2076-124, 2076-312, 2076-324, 2076-212, 224)vendor_id IBM-D050 is used on drives for IBM Storwize V7000 Gen2 (models 2076-524, 2076-12F, 2076-24F)Level85Y5862v7.4.0.1 v6.3.0.04K BS** It is strongly recommended that customers with these drive model numbers (HUC106030CSS60, HUC106045CSS60, HUC106060CSS60) in their configuration, upgrade to this latest level of drive firmware.* If upgrading from 291E, please first upgrade to 2920 and then 2936. Please note: Ralston Peak 300 GB product_id=HUSML4030ASS60 is immune from this problem.2920 can be found on Fix Central in thispackage: StorageDisk-2076-DriveMicrocode-20121210Important Information:Drive Firmware upgrades are not supported on SAN Volume Controller or Storwize V7000 systems running V6.1 or V6.2. Please refer to the following Flash for more information.Drive Firmware Upgrades May Result in Temporary Loss of Host Access to Volumes on SAN Volume Controller and Storwize V7000Please refer to the svctask applydrivesoftware CLI command help in the IBM Storwize V7000 Information Center for installation instructions regarding this Drive Microcode Package.To check that your drive firmware is up to date download and run the Software Upgrade Test UtilityKeep Informed of UpdatesKeep up to date with the latest V7000 information by subscribing toreceive support notifications.Legal AgreementTHE FOLLOWING DOWNLOADS CONTAIN UPDATES AND FIXES TO CODE THAT WAS ORIGINALLY PROVIDED WITH THE IBM STORWIZEV7000 SOLUTION, INCLUDING THE IBM STORWIZE V7000 SOFTWARE. THE UPDATED CODE IS SUBJECT TO THE TERMS AND CONDITIONS OF THE LICENSE AGREEMENTS APPLICABLE TO THE CODE THAT IT UPDATES INCLUDING, AS APPROPRIATE, THE IBM AGREEMENT FOR MACHINE CODE AND IBM INTERNATIONAL PROGRAM LICENSE AGREEMENT. BY DOWNLOADING THE FOLLOWING FILES, YOU ARE AGREEING TO TREAT THE UPDATED CODE IN ACCORDANCE WITH THE APPLICABLE LICENSE AGREEMENTS.Download packageDESCRIPTION DOCUMENTATION LABEL DownloadOptionsPlatform IBMStorwize V7000Version IndependentEnglishByte Size 96217827Date 02 Sep 2017DriveMicrocodePackageHTTP。

NBU信息系统数据集中备份解决方案

NBU信息系统数据集中备份解决方案1.1 概述ABC客户采用了Symantec VERITAS NetBackup作为信息系统数据集中备份解决方案。

整个方案涉及整个系统的软硬件系统，包括数量众多的Unix 和Windows服务器、磁带库、磁盘阵列、光纤网络和各种数据库系统。

本文档讲述了NetBackup软件在各种服务器的补丁要求、安装、配置和管理，还包括如何配置NetBackup管理下的磁带库机械手和磁带机。

同时还设计了备份系统所需的各种表格，方便用户填写和管理。

服务器补丁安装、磁带库安装配置等不在本文档讨论范围。

1.2 名词解释从技术层面划分，NetBackup采用三层结构进行数据的备份和管理，包括NetBackup Master Server主备份服务器、NetBackup Server备份服务器（包含SAN Media Server）、以及NetBackup Client备份客户端和NetBackup Agent备份代理模块。

其他的还包括各种选件模块。

集中备份方案主要包括以下几个基本模块：●备份主服务器模块：NetBackup Master Server●备份服务器：NetBackup Server●SAN介质服务器模块：NetBackup SAN Media Server●备份客户端：NetBackup Client●数据库联机备份代理：NetBackup Database Agent●手提电脑和桌面机备份模块：Desktop and Laptop Option●Windows智能灾难恢复模块：Intelligent Disaster Recovery Option●磁带库驱动：NetBackup Tape Drive Support●SAN磁带机共享模块：NetBackup Share Storage Option●磁带容灾模块：NetBackup Vault Option●NDMP模块：NDMP OptionNetBackup Master Server是备份系统主控服务器。

9702_s04_qp_1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSGeneral Certificate of EducationAdvanced Subsidiary Level and Advanced LevelPHYSICS 9702/01Paper 1 Multiple ChoiceMay/June 20041 hourAdditional Materials: Multiple Choice Answer SheetSoft clean eraserSoft pencil (type B or HB is recommended)READ THESE INSTRUCTIONS FIRSTWrite in soft pencil.Do not use staples, paper clips, highlighters, glue or correction fluid.Write your name, Centre number and candidate number on the Answer Sheet in the spacesprovided unless this has been done for you.There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C, and D.Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully.Each correct answer will score one mark. A mark will not be deducted for a wrong answer.Any rough working should be done in this booklet.Dataspeed of light in free space, c=3.00 x 108m s–1=4π x 10–7H m–1 permeability of free space, µ=8.85 x 10–12F m–1 permittivity of free space, εelementary charge, e=1.60 x 10–19Cthe Planck constant, h=6.63 x 10–34J s unified atomic mass constant, u=1.66 x 10–27kgrest mass of electron, m e=9.11 x 10–31kg rest mass of proton, m p=1.67 x 10–27kg molar gas constant, R=8.31J K–1mol–1the Avogadro constant, N A=6.02 x 1023mol–1 the Boltzmann constant, k=1.38 x 10–23J K–1 gravitational constant, G=6.67 x 10–11N m2kg–2 acceleration of free fall, g=9.81m s–2© UCLES 2004 9702/01/M/J/04© UCLES 20049702/01/M/J/04[Turn overFormulaeuniformly accelerated motion, s =ut +2at 21v 2 =u 2 + 2aswork done on/by a gas, W =p ∆Vgravitational potential, r Gm −=φsimple harmonic motion, x a 2ω−= velocity of particle in s.h.m., t v v ω cos 0=ω ±=v √)(220x x −resistors in series, R =R 1 + R 2 + . . . resistors in parallel, 1/R =1/R 1 + 1/R 2 + . . . electric potential, rQ V 0π4ε=capacitors in series, 1/C =1/C 1 + 1/C 2 + . . . capacitors in parallel, C =C 1 + C 2 + . . .energy of charged capacitor, W =QV 21 alternating current/voltage, x =x 0 sin ω t hydrostatic pressure, p =ρghpressure of an ideal gas, ><=231c VNm p radioactive decay, x =x 0 exp(–λt ) decay constant,210.693t =λcritical density matter of the Universe, Gπ08320H =ρ equation of continuity, Av =constantBernoulli equation (simplified) 2222112121v p v p ρρ++= Stokes' law, F =Arη vReynolds' number, ηρvr R =edrag force in turbulent flow,F =Br 2ρv 24© UCLES 2004 9702/01/M/J/041Which pair contains one vector and one scalar quantity? A displacement : acceleration B force: kinetic energyC momentum : velocityD power : speed2Which of the following could be measured in the same units as force? A energy / distance B energy x distance C energy / timeD momentum x distance3 The notation µs is used as an abbreviation for a certain unit of time.What is the name and value of this unit?name value A microsecond 10–6 s B microsecond 10 –3 s C millisecond 10 –6 s D millisecond 10 –3 s4A 2.35 mAB 2.7 mAC 3.4 mAD 3.7 mA55The following trace is seen on the screen of a cathode-ray oscilloscope.The setting of the time base is then changed from 10ms cm–1 to 20ms cm–1and the Y-sensitivity is unaltered.Which trace is now seen on the screen?A BC D6 In a simple electrical circuit, the current in a resistor is measured as (2.50 ± 0.05) mA. Theresistor is marked as having a value of 4.7Ω ± 2%.If these values were used to calculate the power dissipated in the resistor, what would be the percentage uncertainty in the value obtained?A2% B4% C6% D8%© UCLES 2004 9702/01/M/J/04[Turn over6© UCLES 20049702/01/M/J/04m s –1 and, when it passes the next one, its speed is 20 m s –1.What is the car’s acceleration? A 0.67 m s –2B 1.5 m s –2C 2.5 m s –2D 6.0 m s –28 A tennis ball is released from rest at the top of a tall building.Which graph best represents the variation with time t of the acceleration a of the ball as it falls, assuming that the effects of air resistance are appreciable?9 A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.1.25W hat was the speed at take-off? A 5 m s –1 B 10 m s –1 C 15 m s –1 D 20 m s –110 A ball falls vertically and bounces on the ground.The following statements are about the forces acting while the ball is in contact with the ground.Which statement is correct?A The force that the ball exerts on the ground is always equal to the weight of the ball.B The force that the ball exerts on the ground is always equal in magnitude and opposite indirection to the force the ground exerts on the ball.C The force that the ball exerts on the ground is always less than the weight of the ball.D The weight of the ball is always equal in magnitude and opposite in direction to the force thatthe ground exerts on the ball.11The diagram shows a situation just before a head-on collision. A lorry of mass 20000kg is travelling at 20.0m s–1 towards a car of mass 900kg travelling at 30.0m s–1 towards the lorry.mass of lorry 20000kg mass of car 900kg20.0m s_1What is the magnitude of the total momentum?A 373kN sB 427kN sC 3600kN sD 4410kN s12 An object, immersed in a liquid in a tank, experiences an upthrust.What is the physical reason for this upthrust?A The density of the body differs from that of the liquid.B The density of the liquid increases with depth.C The pressure in the liquid increases with depth.D The value of g in the liquid increases with depth.© UCLES 2004 9702/01/M/J/04[Turn over© UCLES 20049702/01/M/J/04N is 3.0 m long and is supported on a pivot situated 1.0 m from oneW is hung from that end, the beam is in equilibrium, as shown in thediagram.1.0 m3.0 mWhat is the value of W ? A 25 NB 50 NC 75 ND 100 N14 The diagram shows a sign of weight 20 N suspended from a pole, attached to a wall. The pole iskept in equilibrium by a wire attached at point X of the pole.The force exerted by the pole at point X is F , and the tension in the wire is 40 N.Which diagram represents the three forces acting at point X?F20 N40 NF40 N20 N20 N40 NF40 NF20 NA BCD© UCLES 20049702/01/M/J/04[Turn over15 What is the expression used to define power?A inputenergy output energyB energy x time takenC force x velocityDtakentime donework16 A ball is thrown vertically upwards.Neglecting air resistance, which statement is correct?A The kinetic energy of the ball is greatest at the greatest height attained.B By the principle of conservation of energy, the total energy of the ball is constant throughoutits motion. C By the principle of conservation of momentum, the momentum of the ball is constantthroughout its motion. D The potential energy of the ball increases uniformly with time during the ascent.17 Car X is travelling at half the speed of car Y. Car X has twice the mass of car Y.Which statement is correct?A Car X has half the kinetic energy of car Y.B Car X has one quarter of the kinetic energy of car Y.C Car X has twice the kinetic energy of car Y.D The two cars have the same kinetic energy.18 A barrel of mass 50 kg is loaded onto the back of a lorry 1.6 m high by pushing it up a smoothplank 3.4 m long.barrel mass = 50What is the minimum work done? A 80 J B 170 J C 780 J D 1700 J© UCLES 2004 9702/01/M/J/0419 Comparing the properties of solids, liquids and gases, which option is correct?property solids liquids gases A ordering of molecules high not so highrandom B spacing of molecules close far far C translation of molecules no no yes Dvibration of moleculesnoyesyes20 Particles of dust, suspended in water, are viewed through a microscope. The particles can beseen to move irregularly.This movement is due toA convection currents in the water.B evaporation of the water near the dust particles.C gravitational forces acting on the particles of dust.D water molecules hitting the dust particles in a random way.21 Two solid substances P and Q have atoms of mass M P and M Q respectively. They have N P andN Q atoms per unit volume.It is found by experiment that the density of P is greater than that of Q.Which of the following deductions from this experiment must be correct?A M P > M QB N P > N QC M P N P > M Q N QD P P N M > QQ N M22The graph shown was plotted in an experiment on a metal wire.YThe shaded area represents the total strain energy stored in stretching the wire.How should the axes be labelled?Y XA force extensionB mass extensionC strain energyD stress strain23 Nylon breaks when the stress within it reaches 1 x 109Pa.Which range includes the heaviest load that could be lifted by a nylon thread of diameter 1mm?A2N to 20NB 20N to 200NC 200N to 2000ND 2000N to 20000N24Which observation indicates that sound waves are longitudinal?A Sound can be reflected from a solid surface.B Sound cannot be polarised.C Sound is diffracted around corners.D Sound is refracted as it passes from hot air to cold air.© UCLES 2004 9702/01/M/J/04[Turn over25The diagram shows a transverse wave on a rope. The wave is travelling from left to right.At the instant shown, the points P and Q on the rope have zero displacement and maximum displacement respectively.direction of waveWhich of the following describes the direction of motion, if any, of the points P and Q at this instant?point P point QA downwards stationaryB stationary downwardsC stationary upwardsD upwards stationary26 A plane wave of amplitude A is incident on a surface of area S placed so that it is perpendicularto the direction of travel of the wave. The energy per unit time reaching the surface is E.1S.The amplitude of the wave is increased to 2A and the area of the surface is reduced to2 How much energy per unit time reaches this smaller surface?1EA4E B2E C E D227What is the approximate range of frequencies of infra-red radiation?A 1 x 103 Hz to 1 x 109 HzB 1 x 109 Hz to 1 x 1011 HzC 1 x 1011 Hz to 1 x 1014 HzD 1 x 1014 Hz to 1 x 1017 Hz28The lines of a diffraction grating have a spacing of 1.6 x 10–6m. A beam of light is incident normally on the grating. The first order maximum makes an angle of 20o with the undeviated beam.What is the wavelength of the incident light?A 210nmB 270nmC 420nmD 550nm© UCLES 2004 9702/01/M/J/04© UCLES 2004 9702/01/M/J/04[Turn over29 The diagram shows an electron in a uniform electric field.In which direction will the field accelerate the electron?electric field30 The diagram shows a thundercloud whose base is 500 m above the ground.The potential difference between the base of the cloud and the ground is 200 MV. A raindrop with a charge of 4.0 x 10–12 C is in the region between the cloud and the ground.What is the electrical force on the raindrop? A 1.6 x 10–6 NB 8.0 x 10–4 NC 1.6 x 10–3 ND 0.40 N31 Two wires made of the same material and of the same length are connected in parallel to thesame voltage supply. Wire P has a diameter of 2 mm. Wire Q has a diameter of 1 mm.What is the ratio Q in current P in current ?A41 B21 C2 D 432 What is an equivalent unit to 1 volt?A 1 J A –1B 1 JC –1C 1 W C –1D 1 W s –133The terminal voltage of a battery is observed to fall when the battery supplies a current to an external resistor.What quantities are needed to calculate the fall in voltage?A the battery's e.m.f. and its internal resistanceB the battery's e.m.f. and the currentC the current and the battery's internal resistanceD the current and the external resistance34The potential difference between point X and point Y is 20V. The time taken for charge carriers to move from X to Y is 15s, and, in this time, the energy of the charge carriers changes by 12J.W hat is the current between X and Y?A 0.040AB 0.11AC 9.0AD 25A35The diagram shows a battery, a fixed resistor, an ammeter and a variable resistor connected in series.A voltmeter is connected across the fixed resistor.The value of the variable resistor is reduced.Which correctly describes the changes in the readings of the ammeter and of the voltmeter?ammeter voltmeterA decrease decreaseB decrease increaseC increase decreaseD increase increase© UCLES 2004 9702/01/M/J/04© UCLES 2004 9702/01/M/J/04[Turn over36 Kirchhoff’s two laws for electric circuits can be derived by using conservation laws.On which conservation laws do Kirchhoff’s laws depend?Kirchhoff’s first lawKirchhoff’s second lawAcharge current B charge energy C current mass Denergy current37 The diagram shows a parallel combination of three resistors. The total resistance of thecombination is 3 Ω.What is the resistance of resistor X? A 2 Ω B 3 ΩC 6 ΩD 12 Ω38 A nucleus of the nuclide Pu 24194 decays by emission of a β-particle followed by the emission of anα-particle.Which of the nuclides shown is formed?A Np 23993BPa 23991CNp 23793DU 2379239 A thin gold foil is bombarded with α-particles as shown.The results of this experiment provide information about theA binding energy of a gold nucleus.B energy levels of electrons in gold atoms.C size of a gold nucleus.D structure of a gold nucleus.40 Isotopes of a given element all have the sameA charge/mass ratio.B neutron number.C nucleon number.D proton number.University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.© UCLES 2004 9702/01/M/J/04。

ThinkServer TS140 用户手册 V1.2

ThinkServer User Guide
T品前请先阅读本手册
目录
声明 ...................................................................................................................................................................................................................................3 商标和版权 ........................................................................................................................................................................................................................4 Safety Information（安全信息）....................................................................................................................................................................................5 第一章产品简介 .......................................................................................................................................................................................................... 10 1.1 产品概述 ................................................................................................................................................................................................................... 10 1.2 产品特色 ................................................................................................................................................................................................................... 10

==序列号==

YM2S-PVC6-RL2T-UT89-SW8T
4EFF-YZNG-6AS4-3QJQ-44YG
QFYY-CEQT-J4G8-3UWQ-48UY
N3XF-MGW6-GLEE-8S2A-YASU
AZNE-P748-W8C4-SSWS-4E8S
87Z9-U64J-Y84U-QA6W-82NJ
Web Server: d42x8-7mwxd-m4b76-mkyp7-cw9fd
CPP Keys - not RTM: cky24-q8qrh-x3kmr-c6bcy-t847y
(.net server rc) : hk4gf-rfkkt-c9h7x-3d748-tyf8m hdrf9-p7xgm-v23yt-wdmr8-ctbpb
暗黑王座 RAC2-RAD2-RAC2-RAC2-3387
暗黑武士道 Throne of Darkness S/N:RAC2-RAD2-RAC2-RAC2-3387
傲世苍龙：赵云传~O............Dragon.ChaoYun-中文版 S/N: LY8882L51PNE
J24DX-G4YRT-2K77H-V3JB3-QTH3Y
CQYRF-W3782-6WD43-JFGGX-9DBQY
D9GQ6-42QFW-Pห้องสมุดไป่ตู้YHJ-BW62B-GF8YB
HDJTW-FGHX9-C666D-G3HDY-G6T7M
B24F8-JPTJW-36YG4-KJ8MV-KT9VM
Microsoft 2003 Storage Server Information
There is not a seperate 2003 Storage Server Edition

National Instruments PXIe系列产品数据手册说明书

Board Assembly Part Number(s)Part Number Description159040B-01L or later NI PXIe-7846R159040B-02L or later NI PXIe-7847R159040B-03L or later NI PXIe-7856R159040B-04L or later NI PXIe-7857R159040B-05L or later NI PXIe-7858RManufacturer: National InstrumentsVolatile MemoryUser Accessible/ BatteryType Size System Accessible1Backup? Purpose Method of Clearing2(7846R/7847R/7856R/7857R)FPGA Block RAM 11,700 KB Yes/Yes No Data storage during Cycle PowerVI Execution(7858R)FPGA Block RAM 16,020 KB Yes/Yes No Data storage during Cycle PowerVI Execution (7847R/7857R/7858R)DRAM 512MB Yes/Yes No Onboard Memory Storage Cycle PowerNon-Volatile MemoryUser Accessible/ BatteryType Size System Accessible Backup? Purpose Method of Clearing(7846R/7847R/7856R/7857R)Flash 64 Mb No- Device information No/Yes Product Identification None available to user- Calibration data3No/Yes Calibration Information None available to user- Calibration metadata Yes/Yes User-defined See Clearing Notes- FPGA bitstream Yes/Yes User LV FPGA VI Bitstream See Clearing Notes(7858R)Flash 128 Mb- Device information No/Yes No Product Identification None available to user- Calibration data3No/Yes Calibration Information None available to user- Calibration metadata Yes/Yes User-defined See Clearing Notes- FPGA bitstream Yes/Yes User LV FPGA VI Bitstream See Clearing Notes1 Items are designated No for the following reason(s):a) Hardware changes or a unique software tool from National Instruments are required to modify contents of the memorylisted.b) Hardware-modifying software tools are not distributed to customers for any personal access or customization, also knownas non-normal use.2 The designation None Available to User indicates that the ability to clear this memory is not available to the user undernormal operation. The utilities required to clear the memory are not distributed by National Instruments to customers fornormal use.3 Calibration constants that are stored in device calibration memory include information for the device’s full operating range. Calibration constants do not maintain any unique data for specific configurations at which the device is used unlessotherwise specified.Media StorageUser Accessible/ BatteryType Size System Accessible Backup? Purpose Method of Clearing NONEClearing Notes:Calibration metadata: There are two items in the calibration metadata that need to be cleared:1.To clear the user-defined information, you can use the Calibration Utility to write a known value to the user stringfield.2.To clear the calibration password, you can use the Calibration Utility to change the password to a known value.FPGA bitstream: You can use the NI-RIO Device Setup utility to erase the FPGA bitstream data. For more details, visit /info and enter the infocode fpgaflashclr.Terms and DefinitionsUser Accessible Allows the user to directly write or modify the contents of the memory during normal instrument operation. System Accessible Does not allow the user to access or modify the memory during normal instrument operation. However, system accessible memory may be accessed or modified by background processes. This can be something that is not deliberate by the user and can be a background driver implementation, such as storing application information in RAM to increase speed of use.Cycle Power The process of completely removing power from the device and its components. This process includes a complete shutdown of the PC and/or chassis containing the device; a reboot is not sufficient for the completion of this process. Volatile Memory Requires power to maintain the stored information. When power is removed from this memory, its contents are lost.Non-Volatile Retains its contents when power is removed. This type of memory typically contains calibration or chip configuration information, such as power up states.。

存储及双活测试方案

融通基金存储升级及双活数据中心项目测试方案V1.0深圳市桑威科技有限公司月5年2016．文档信息 1.0 文档版本号：融通基金项目名称：VPLEXMetro项目测试报201文档作者生成日期彭世文档审核者审核日期：文档维护记录描维护日作版本维护1.0创建初2011彭世录目未指定书签。

............................................................................................................................ 错误!1. 环境配置未指定书签。

............................................................................................................................ 错误!2. 测试目的未指定书签。

..................................................................................................... 错误!3. VPLEX测试环境的建立未指定书签。

...................................................................................................... 错误!3.1. SANSwitch物理连线未指定书签。

............................................................................................................. 错误!3.2. SANSwitch配置. 未指定书签。

Lenovo ThinkSystem服务器新模型与选项介绍-2018年6月说明书

Figure 1. Flex System SN550 serverServer optionsOn June 12, we announced the following new options for ThinkSystem servers: Samsung PM983 Entry NVMe PCIe 3.0 x4 SSDs - in capacities 1.92TB and 3.84TB Samsung PM883 Entry SATA 6Gb SSDs - available in capacities 240GB to 3.84TB Intel P4500 Entry NVMe PCIe3.0 x4 Flash AdapterNVIDIA Tesla V100 32GB PCIe Passive GPU - additional 32GB memory GPUFigure 2. ThinkSystem NVIDIA Tesla V100Click here to check for updatesFigure 3. Lenovo B300 Fibre Channel SAN SwitchNew interface options for Lenovo V Series storageOn June 5, Lenovo announced new interface connectivity options for the Lenovo Storage V Series - Lenovo Storage V5030 25 GbE (iWARP) and 25 GbE (RoCE) Adapter Pairs.For more information, refer to the following product guides that were updated with these options:Figure 4. Lenovo Storage V5030Figure 5. Lenovo ThinkSystem DS6200TrademarksLenovo and the Lenovo logo are trademarks or registered trademarks of Lenovo in the United States, other countries, or both. A current list of Lenovo trademarks is available on the Web athttps:///us/en/legal/copytrade/.The following terms are trademarks of Lenovo in the United States, other countries, or both:Lenovo®Flex SystemThinkSystem®TopSellerThe following terms are trademarks of other companies:Intel® is a trademark of Intel Corporation or its subsidiaries.Other company, product, or service names may be trademarks or service marks of others.。

9702_s05_er

9702 Physics June 2005 CONTENTS FOREWORD (1)PHYSICS (2)GCE Advanced Level and GCE Advanced Subsidiary Level (2)Paper 9702/01 Multiple Choice (2)Paper 9702/02 Structured Questions (3)Paper 9702/03 Practical 1 (6)Paper 9702/04 Core (7)Paper 9702/05 Practical 2 (10)Paper 9702/06 Options (12)FOREWORDThis booklet contains reports written by Examiners on the work of candidates in certain papers. Its contents are primarily for the information of the subject teachers concerned.PHYSICSGCE Advanced Level and GCE Advanced Subsidiary LevelPaper 9702/01Multiple ChoiceQuestion Number KeyQuestionNumberKey1 C 21 D2 C 22 D3 B 23 A4 C 24 B5 A 25 B6 D 26 B7 B 27 C8 D 28 B9 A 29 B10 A 30 D11 A 31 C12 C 32 C13 A 33 D14 D 34 D15 B 35 B16 C 36 D17 B 37 B18 A 38 C19 C 39 B20 A 40 CGeneral commentsThe paper provided good syllabus coverage and proved to have good discrimination and a relatively high overall facility. The mean mark of 24.7 with a standard deviation of 6.6 showed that while some candidates found many of the questions difficult there were enough straightforward questions for the majority to find plenty to do. This year there were some very able candidates, with around 5% able to score 90% or more. This is an outstanding achievement, especially bearing in mind the pressure of time in answering forty questions in only one hour. At the bottom end of the ability range only 8.4% scored less than 40%. This again shows that the vast majority of candidates were well prepared for the examination and had reasonable knowledge of the topics which were being tested. There is a tendency with a multiple choice examination for candidates to work on numerical questions just with their calculator. They do need to see that this is a dangerous habit as many of the distractors are obtained simply by manipulating the data given in a plausible way. Candidates do need to work carefully and use units as a check if they are to avoid pitfalls. It is difficult to ascertain how the timing of the paper seemed to the candidates, but there was no direct evidence of candidates being unable to complete the paper in the allotted time.Comments on specific questionsThe questions which generally did not cause candidates any problems were Questions1, 2, 6*, 12, 17, 19*, 23, 30, 31, 35*, 38* and 40* where the marks were correspondingly high. The starred questions have correct answers from more than 90% of candidates. This is particularly good for questions such as 38 and 40 as it indicates that candidates are familiar with standard terminology. It does however, mean that the discrimination is low for these questions.The following questions were answered poorly.Question 4, where candidates showed that they were often unable to work with percentage uncertainties. A 2% uncertainty in the time and a ¼% uncertainty in the distance gives a total uncertainty of 2¼% in the speed which is therefore 16.0 ± 0. 4.Question 7, where candidates do not seem to appreciate that a body under conditions of free fall has a constant acceleration of g, and even when stationary at the top of its path its acceleration is still g because its velocity is still changing at the same rate.Question 13, where particular care is required to get the moment of the 20 N force and to subtract the moments provides by the 5 N and 10 N forces. Many candidates ignored the 10 N force altogether.Question 33, where C was the most popular answer. This implies that many candidates ignored the fact that if the length is doubled and the volume remains the same then the area of cross-section must be reduced to half its former value, giving a new resistance of 4R.Question 39 was one of a very few questions where one of the distractors (A) had as many selecting it as the correct answer, but this may be because by this stage candidates were in a hurry and resorted to guessing.Paper 9702/02Structured QuestionsGeneral commentsThere were sections of questions that were accessible to all candidates. On the other hand, some parts provided a challenge to the more-able candidates. There were some excellent scripts and several Centres where large numbers of candidates achieved a uniformly high standard.Question 7 highlighted once again two areas of general weakness. First, the drawing of unjustified inferences from quoted formulae by considering two variables in isolation. Second, a lack of appreciation of significant figures both as regards premature ‘rounding’ at intermediate stages in calculations and also the number of significant figures that can reasonably be quoted in a final answer. Both aspects have been mentioned previously in reports.There was no evidence that candidates were short of time. In scripts where the last two questions had not been completed, then almost invariably there were gaps in answers to earlier questions.Comments on specific questionsQuestion 1The speed of sound produced the most reasonable estimates and the density of air the least. For all four quantities there was a very wide range of inadequate estimates, many of which showed that the candidates had little or no appreciation for the physical magnitudes of the various quantities.(a) Estimates ranged from 1.5 m s–1 to 3 × 1022 m s–1.(b) Estimates ranged from 1.7 × 10–31 kg m–3 to 8 × 1023 kg m–3.(c) Estimates ranged from 0.5 mg to 800 kg.(d) Estimates ranged from 10–6 cm3 to 1012 cm3.(a) It seemed that few candidates had actually observed the Brownian motion of smoke particles.Although most candidates did make a reference to ‘random’ or ‘haphazard’ motion, they frequently then went on to state that this motion was due to collisions between smoke particles or between smoke particles and the walls of the container. Only rarely did a candidate mention that specks of light are observed, not the smoke particles themselves.(b) Most answers did include a statement that the motion would still be random, but slower. Often, noexplanation was given. Where explanation was provided, this was usually confined to a statement that greater mass would lead to smaller velocity changes. Very few candidates made any reference to greater surface area and that, because of the random distribution of velocities of air molecules, the effects of collisions of the smoke particle with air molecules would tend to average out. This greater rate of collision would lead to a smaller, rather than a larger, randomness of collision and hence motion of the smoke particle.Question 3(a)(i) Most candidates were able to calculate the change in energy. E rrors were mostly due toinappropriate units for mass and/or vertical displacement.(ii) In most answers, the change in gravitational potential energy was associated correctly with thekinetic energy of the block and bullet. A few candidates obtained the correct answer by unjustified use of equations representing uniform motion in a straight line.(b) Candidates were instructed to use the principle of conservation of momentum and consequently,most answers were correct.(c)(i) Very few candidates failed to calculate a value for the kinetic energy, based on their answers to (b).(ii) The majority of answers included a correct deduction that there would be a loss of kinetic energyand thus the collision is inelastic. However, some asserted that the evidence for a loss of kinetic energy was that some had been converted into potential energy of the block and bullet.Answers : (a)(i) 0.51 J; (b) 390 ms –1; (c)(i) 152 J.Question 4 (a) With few exceptions, the glass was said to be brittle.(b)(i) This calculation presented very few problems for candidates. The most common error wasarithmetical.(ii) E xpressions quoted for the Young modulus were usually correct. Again, most errors werearithmetical, particularly amongst those candidates who did not make a separate calculation of the strain.(iii) The majority did calculate an appropriate area of the graph provided or did use an equivalentcorrect formula. However, a significant minority either used the expression21 × stress × strain ,without realising that this is energy per unit volume, or merely calculated the maximum strain. (c) In general, this was poorly answered. Despite the explicit information given in the question, many answers were based on an assumption that either the mass, the density or the kinetic energy would be different. Frequently, wording was imprecise. The hard ball was said to ‘have more force’ or the soft ball ‘to use up its force in squashing’. Nevertheless, there were some good answers, based on either the times of collision affecting the rate of change of momentum and thus the force on the glass or the conversion of kinetic energy to strain energy in the balls affecting the strain energy within the glass.Answers : (b)(i) 7.6 x 107 Pa, (ii) 6.1 x 1010 Pa, (iii) 9.0 x 10–3 J.(a) Candidates should be encouraged to be as precise as possible with definitions and explanations.Most answers included a statement, with varying degrees of clarity, that diffraction is the bending or spreading of waves through a narrow gap or at an edge. However, wording was frequently ambiguous so that the explanation could apply to refraction. Statements such as ‘change of direction when meeting an obstacle’ are not acceptable.(b)(i) This simple calculation caused many problems. Not only were there many errors involvingpowers-of-ten in otherwise correct calculations but also, many candidates thought that they must use the grating formula given in the question.(ii)In most scripts where d had been calculated correctly, then the maximum value of n was also correct. However, candidates who had made errors in d by as much as factors of up to 1012 usually also calculated equally ridiculous answers for n, without comment.(iii) Very few candidates recognised that the correct use of the formula relies on the fact that light incident on the grating has no path difference. Most candidates repeated the question by stating that the light is not normal to the grating.(c) Many candidates gave at least one relevant difference, based on either angles of diffraction orintensity. A surprisingly large number of answers ignored the fact that the question specified the two wavelengths involved. Consequently, answers referred to different wavelengths, frequencies or colour. In others, there were vague mentions of angles, without a clear indication as to which angle reference was being made.Answers: (b)(i) 1.33 x 10–6 m, (ii) 2.Question 6(a)(i) Candidates should be encouraged to use a ruler when drawing straight lines. In this case, it wasexpected that, by eye, the lines would be straight and equally spaced. A significant number of diagrams were unacceptable free-hand sketches.(ii) Almost all candidates were able to derive the given result.(b)(i) A surprisingly large proportion of candidates drew arrows that were either normal to the electricfield or normal to the axis of the particle.(ii)Although the majority of calculations were correct, there were significant numbers of answers with incorrect physics. Calculations of the Coulombic forces between two charged particles were not uncommon. Others introduced either a sine or a cosine term into the calculation.(iii) Most candidates did multiply the force calculated in (ii) by a distance. However, very frequently this was not the perpendicular distance between the forces. In most of these scripts, this was not a matter of confusion between use of sine or cosine terms.(iv) In the vast majority of scripts, it was realised that the forces would cause rotation. However, most answers gave the impression that the rotation would be continuous, with relatively few stating that the particle would align itself along the direction of the field.Answers: (b)(ii) 2.4 x 10–12 N, (iii) 3.4(4) x 10–15 Nm.Question 7(a) Many candidates gave unsatisfactory answers in terms of ‘the hindrance’ or ‘the opposition to thecurrent’. Of those who did attempt a definition in terms of a ratio, many were imprecise, either defining the unit rather than the property or, frequently, using the unqualified term ‘voltage’ when what was intended is the potential difference across the resistor.(b)(i) The vast majority of answers were correct for the data point.(ii)With few exceptions, the calculation was completed successfully.(iii) This part of the calculation was more challenging and there were many well-expressed correct solutions. However, there were significant numbers of answers where work was laid out poorly. In such cases, many candidates failed to appreciate whether they were dealing with the e.m.f. of the battery, the p.d. across the resistor or the p.d. across the internal resistor. A small but significant number of candidates treated the external and internal resistors as if they were in parallel.(c)This part was poorly answered with very few showing clarity of thought. Many candidates made nomention of internal resistance, simply quoting a formula for power dissipation, and concluding that the lower value of the p.d. across R would give a smaller power, ignoring the fact that both the value of the current and the resistance would change. Others stated that the larger p.d. across R would give greater power dissipation in R thus the power dissipation in the inernal resistor must be smaller. A minority of candidates argued, quite correctly, that a larger p.d. across R would give a smaller p.d. across the internal resistance and since internal resistance is constant and P = V2/r, then the power dissipation in the internal resistor would be lower.Answers: (b)(i) 1.13 W, 1.50 V, (ii) 1.99 Ω, (iii) 1.99 Ω.Question 8(a) Few candidates failed to plot correctly the position for the isotope of protactinium.(b) The values of A and Z were given and consequently, almost all candidates who attempted this taskdid plot the position correctly. It was realised that the daughter product of plutonium would have the same value for A. However, opinion was divided as to whether Z should be 93 or 95.Paper 9702/03Practical 1General commentsThe overall standard of the work produced by the candidates was generally good, although as in previous years the performance variation was mainly by Centre (i.e. some Centres continue to prepare their candidates very well for this examination). It was pleasing to see fewer low scores (< 12) than in previous years and there were quite a lot of strong candidates scoring 21+. Most Centres had no difficulty with the apparatus requirements, although there were some cases where Centres had used spring balances calibrated in grams instead of Newtons. There were very few reports of Supervisors giving assistance to candidates. There was no evidence that candidates were short of time and no problems with the rubric. Comments on specific questionsQuestion 1In this question candidates were required to investigate how the force required to maintain the equilibrium of a suspended mass depends on the angle between the line of action of the force and the horizontal.In (a)(ii) many candidates stated the uncertainty in θ to be 0.5°,which was considered to be unrealistic as it was difficult to place the protractor in the correct position when measuring the angle.In (b) the difficulties mentioned by candidates included thick string/difficulty of alignment of the Newton meter/holding the protractor steady whilst measuring θ/keeping AB horizontal, all of which were credited. Some answers were vague (e.g. ‘it was difficult to read the scale on the newton meter’ or ‘parallax error’ with no clarification). Oscillation of the string or the mass was not accepted.Most candidates were able to set up the apparatus correctly and use it to obtain six sets of readings for F and θ. A number of candidates misread the scale on the protractor and obtained values of θ which were acute instead of obtuse. A few candidates calculated 1/sin θ using θ in radians instead of degrees.Virtually all candidates presented the results in tabular form. Raw values of θ were sometimes given to an unreasonable degree of precision (i.e. to one or two decimal places). A number of candidates had impossibly large values for F , presumably because they had used spring balances instead of newton meters. These candidates were not penalised.Candidates were required to plot a graph of F against 1/sin θ. Common errors made by the weaker candidates included poor choice of scales (i.e. where the plotted points occupied less than half the graph grid in both the x and y directions) or where the scales were awkward (e.g. one large square on the grid corresponding to three units). Points were usually plotted correctly, although it was sometimes difficult to see where the points had been plotted. It is expected that small crosses will be used. When plotting errors occurred it was usually because awkward scales had been chosen. It is expected that candidates will plot six points since six observations have been made. Candidates who did not plot all their observations were penalised. Most of the better candidates were able to determine a value for the gradient of the line correctly. When the mark was not awarded it was usually because the triangle that had been used was too small or an error had been made in the read-offs (particularly when awkward scales had been used). The y -intercept was often read incorrectly from the graph because a ‘false origin’ had been used (i.e. the value was read from a line were x ≠ 0). The more able candidates substituted values from a point on the line, together with the gradient value, into y = mx + c .Two marks were available for the ‘quality of results’. This was judged on the scatter of points about the line of best fit. Candidates who had done the experiment carefully were able to score here if the scatter of points about a line of best fit was small. Candidates lost marks if they used a narrow range of angles (a spread of < 10° was common).The analysis section continues to differentiate well between candidates. The weaker candidates often did not attempt this section. The better candidates were able to equate mg with the gradient of the graph and kwith the y -intercept. A significant number of candidates equated θsin mgto the gradient. Candidates wereinstructed to use their answers from (e) to determine values for m and k . Many of the weaker candidates did not do this and attempted to substitute two sets of values into the given equation and solve the resulting equations simultaneously. Work of this kind was not credited.It was expected that m and k would be given to a sensible number of significant figures (i.e. two or three significant figures) and that units would also be given. A large number of candidates failed to recognise that the unit of k is the Newton, and gave the value of k to one significant figure only.It was pleasing to see many of the more able candidates scoring full marks in the analysis section.Paper 9702/04CoreGeneral commentsAll questions were accessible to better candidates. Weaker candidates tended to score the majority of their marks on the first four questions.It was pleasing to note that the number of scripts where work is laid out well and adequate explanation is given is increasing. There were some outstanding scripts from a number of Centres.Candidates appeared to have sufficient time to complete their answers. However, there were some instances where it appeared as if the candidate did not realise that a question was printed on the back page of the script, possibly because the penultimate question ended half-way down page 15 or because the subject material of the questions was not in syllabus order. Candidates should be encouraged to read carefully the question paper. On the cover page it stated that there are 16 printed pages in the question paper and on page 15, the instruction ‘[Turn over ’ was given.Comments on specific questionsQuestion 1(a)(i) Most candidates successfully calculated the magnitude of the angular velocity. A minority didconfuse ‘speed’ with ‘angular velocity’.(ii) A correct formula was usually given. There were very few arithmetical errors. Some did substitute w for v in the expression F = mv2/r.(b)(i)It was expected that candidates would mention gravitational force. Some did merely state ‘the Sun’but others gave more than required by stating that the gravitational force between the Sun and the Earth provides the centripetal force.(ii)Most candidates did quote a correct expression for the gravitational force and, generally, the arithmetic was correct. A minority of candidates used the expression GM = r3ω2. This alternative approach could be awarded full credit.Answers: (a)(i) 1.96 x 10–7 rads–1, (ii) 3.46 x 1022 N; (b)(ii) 1.95 x 1030 kg.Question 2(a) Most candidates mentioned either the universal gas equation or individual gas laws. However, itwas disappointing to note that very few stated that, for an ideal gas, the law(s) must be obeyed at all values of pressure, volume and temperature.(b)(i) A significant number stated that <c2> represented the root-mean-square speed. Very few referredto the square of the mean speed.(ii)There were some very pleasing derivations, with clear explanation at all stages. On the other hand, the work of some candidates lacked all meaning. It was common to find that density was defined as M/V, with M being used in the expression for the mean kinetic energy of an atom. The term N was then either ignored, cancelled or stated to be unity.(c)(i)Generally, this calculation was completed successfully. The most common error was a failure tosquare the value for the speed.(ii)It was pleasing to note that more-able candidates made reference to a distribution of speeds. A significant number of candidates thought that atoms with a lower speed would escape. There wasa number of ingenious explanations, including the possibility of isotopes.Answer: (c)(i) 1.9 x 104 K.Question 3(a) Definitions tended to lack precision. It was common to find that any reference to unit mass wasomitted. Furthermore, weaker candidates tended to fail to state what is meant by fusion or thought that energy is required to convert liquid to solid.(b)(i)Although the majority of answers were correct, a significant minority considered only the energyrequired to warm the ice or to melt it.(ii)Most candidates did attempt the calculation, using their answer in (i). However, very few took into account the energy required to heat the melted ice from 0°C to the final temperature.Answers: (b)(i) 8700 J, (ii) 16°C.Question 4(a) In general, definitions were adequate. The most common failing was either to omit to give the relative directions of the displacement and the acceleration or to express them poorly.(b)There were some very clearly expressed derivations. Weaker candidates tended to give some relevant expressions but were unable to link them. Candidates should always be encouraged to give as much relevant information as they can, since credit is often given for such expressions.(c)(i)The majority of answers did include an acceptable value for either the period or the frequency of the oscillations. However, there were many answers where it appeared that the expression for the area was unknown. Besides confusing diameter and radius, it was not uncommon to find the expression A = 4πr 2. Candidates should be advised to use data given in the question paper. The use of g = 10 ms –2 is not acceptable where data for a calculation is given to two or more significant figures.(ii)It was pleasing to note that very few candidates referred to displacement, rather than amplitude or peak height.Answer : (c)(i) 0.0384 kg. Question 5(a)Answers were disappointing. Very few made reference to potential gradient or x V ∆∆/. Most who attempted this part of the question stated E = V/x . This is, of course, incorrect since V is defined as potential and x as the distance from the centre of the sphere.(b)Surprisingly few candidates knew that the electric field within the conductor must be zero. Most candidates drew a sketch similar to that in Fig. 5.2 or merely a curve starting at x = O.Question 6(a)(i)Generally satisfactory but it was evident that a minority of candidates had no real appreciation of concepts involving electromagnetic induction.(ii)Most answers were based, quite correctly, on the equation P = VI . However, many failed to state that the output power would need to be constant.(b)(i) Generally correct.(ii)Most sketches did show an appropriate sinusoidal wave with the correct frequency. However, veryfew indicated the correct phase. Most indicated no phase difference with a few showing a π rad change between Fig. 6.3 and Fig. 6.4.(iii) It was common to find that this section was not attempted. Of those who did give an answer, themajority failed to state a unit for the angle.Answer : (b)(iii) π21or 90°.Question 7(a) Surprisingly, many graphs were drawn without a scale on the y -axis. Consequently, only the general shape of the line could be given.(b)(i)The initial number was calculated correctly in the majority of scripts. However, a significant number of candidates appeared to have little idea as to how to proceed with this basic calculation.(ii)A significant number of candidates thought that they had to use an equation involving exponential decay. Of those who did use the correct equation, it was pleasing to note that most did determine the decay constant in s –1.(c)Many candidates did not appear to know how to proceed with this problem. Of those who did, the majority took the ratio N/N o to be 1/9, rather than 1/10.Answers : (b)(i) 1.5 x 1016, (ii) 1.11 x 1012 Bq; (c) 8.63 hours.Question 8 (a) Definitions were disappointingly poor. Many failed to make it clear that a ratio is involved.Consequently, in such statements it appeared that capacitance is an electric charge.(b)(i) Most answers included, in some way, that there would be charge separation. However, very rarelywas any explanation given as to how this charge separation leads to the storage of electrical energy.(ii) Correct answers were in a minority. Many quoted 221CV as the energy stored in a chargedcapacitor but then assumed ()2122V V − is equal to ()212V V −. Others used the expression QV 21but assumed that Q would remain constant when V changed.Answer : (b)(ii) 1.4 J.Paper 9702/05 Practical 2General commentsThe general standard of the work done by the candidates was similar to last year, with quite a wide spread of marks. Question 1 was relatively straightforward, although some of the weaker candidates found the analysis section challenging and gave very brief answers to Question 2. As in previous years, there was a significant range in performance. It would be helpful to candidates generally if attention could be drawn to the published mark schemes.There were no reported difficulties from Centres in obtaining the necessary equipment for Question 1. Very little help was given to candidates from Supervisors in setting up the apparatus in Question 1. Supervisors are reminded that under no circumstances should help be given with the recording of results, graphical work or analysis.A small number of weaker candidates appeared to be short of time. Answers to Question 2 from these candidates was often very brief, or finished in mid-sentence.There were no common misinterpretations of the rubric.Comments on specific questionsQuestion 1In this question candidates were required to investigate how the period of oscillation of a loaded steel blade varies with the length of the blade. (a) Virtually all candidates were able to set up the equipment without help from the Supervisor. (b) In this part many candidates did not repeat the readings of raw times. (c) Virtually all candidates were able to record six measurements of d and t . A few candidates did notdivide the raw times by the number of oscillations and calculated log t instead of log T . Some of the raw times were too small (i.e. less than 10 seconds) because candidates did not allow the blade to perform a sufficient number of oscillations. Values of d were usually given to the nearest millimetre, although weaker candidates recorded their values of d to the nearest centimetre. It is expected that d will be recorded to the nearest millimetre as a millimetre scale had been used to make the measurement.。

Synology DiskStation DX513 扩展卷和备份说明书

DiskStation DX513Volume Expansion & Backup When running out of hard drive capacity on the Synology DiskStation, Synology DX513 provides an easy way to immediately scale capacity by additional 5 hard drives. The RAID volume on the Synology DiskStation can be expanded directly without having to reformat the existing hard drives, ensuring the Synology DiskStation continues its service during the capacity expansion.Synology DX513 can be used as a dedicated local backup solution for the Synology DiskStation. When created as a separate volume, the Synology DX513 provides a great backup solution to its local hard disks in case of system failure.Reliable Plug-n-use DesignThe Synology DX513 is connected to the Synology DiskStation using an expansion cable with custom-designed connectors on both ends, ensuring a reliable connection and maximum throughput. The 3.0 Gb/sec connection speed allows the hard drives in the connected Synology DX513 to operate as the internal ones of Synology DiskStation.Green & Easy to ManageDeep Sleep can be manually configured to take effect automatically when the system has been inactive for a period of time. This not only saves energy but also extends the lifespan of the hard disk. The DX513 will also smartly power on/off with the attached DiskStation, so no userintervention is needed to keep this system running smartly and efficiently.●Online Volume Expansion ●Dedicated Local Backup Solution ●Reliable Plug-n-use Design ●Hot-swappable HDD Design ●Auto power on/off withDiskStation Models HighlightsSynology ® DX513 delivers an effortless solution for volume expansion and data backup for selected Synology DiskStation models. The DX513 seamlessly scales up the storage capacity of the DiskStation by additional 5 hard drives on-the-fly whenconnected directly to an expansion cable.Technical Specifications HardwareInternal HDD/SSD 3.5” or 2.5” SATA(II) X 5 (hard drives not included)Expansion PortX 1Hot Swappable HDDYes Volume TypeSynology Hybrid RAID, Basic, JBOD, RAID 0, RAID 1, RAID 10, RAID 5, RAID 5+Spare, RAID 6Size (HxWxD)157mm X 248mm X 233mm Weight3.91KG System Fan80x80mm X 2Noise Level 130dB(A)Power RecoverySync with the connected DiskStation (Default Mode)AC Input Power Voltage100V to 240V Power Frequency50Hz to 60Hz, Single Phase Operating Temperature5°C to 35°C (40°F to 95°F)Storage Temperature-10°C to 70°C (15°F to 155°F)Relative Humidity 5% to 95%RH Power Consumption 245.1W (Access), 2W (Deep Sleep)Supported Models 3DS710+, DS712+, DS1010+, DS1511+, DS1512+, DS1812+ Use as independent volume: DS112, DS112+, DS212+, DS213+, DS412+, DS413Environment and PackagingEnvironmentRoHS Compliant Package ContentDX513 Main Unit, User’s Guide, Assembling Kit, AC Power Cord, Expansion Cable CertificationFCC: Class B, CE: Class B, BSMI: Class B Warranty 3 years*Specifications are subject to change without notice. Please refer to for the latest information.1. Fully loaded with Seagate ® ST31000520AS hard drive(s) in operation; Two G.R.A.S. Type 40AE microphones, each set up at 1 meter away from the DiskStation front and rear; Background noise: 17.1 dB(A); Temperature: 24.2˚C; Humidity: 63%.2. Measured with 5 Western Digital ® 3TB WD30EZRS hard drive. The figures could vary on different environments.3. For info about newest supported models, please visit for the latest information.Connections & ButtonsFans Power PortPower ButtonSYNOLOGY INC.Synology is dedicated to taking full advantage of the latest technologies to bring businesses and home users reliable and affordable ways to centralize data storage, simplify data backup, share and sync files across different platforms, and access data on-the-go. Synology aims to deliver products with forward-thinking features and the best in class customer services.Copyright © 2012, Synology Inc. All rights reserved. Synology, the Synology logo are trademarks or registered trademarks of Synology Inc. Other product and company names mentioned herein may be trademarks of their respective companies. Synology may make changes to specification and product descriptions at anytime, without notice. Printed in Taiwan.DX513-2012-ENU-REV005Headquarters Synology Inc.3F-3, No. 106, Chang An W. Rd., Taipei, Taiwan Tel: +886 2 2552 1814 Fax: +886 2 2552 1824UK Synology UK Ltd.Etheridge Avenue, Brinklow, Milton Keynes, MK10 0BP, UK Tel: +44 1908 587422Germany Synology GmbH Königsallee 92a 40212 Düsseldorf, Germany Tel: +49 (0)211 5403 9658North & South America Synology America Corp.13343 NE Bel-Red Road, Bellevue, WA 98005, USA Tel: +1 425 818 1587Expansion PortLED indicatorsRESETHDD TrayDefault/Manual Switch。

存储HCIP复习题及答案

存储HCIP复习题及答案一、单选题（共38题，每题1分，共38分）1.以下哪个协议不属于NAS协议A、CIFSB、NFSC、FTPD、FC正确答案：D2.华为18000v3系列产品引擎间的控制信息流和业务数据流通过以下哪个部件来传输的A、kvmB、pcie交换机C、svpD、10ge交换机正确答案：B3.在oceanstorv3存储中，NFS不支持如下哪一种验证方式？A、NISB、LADPC、本地认证D、AD域正确答案：D4.关于华为Oceanstor9000系统的组网,以下哪个选项描述是正确的?A、无论是前端业务网络、还是后端存储网络,都推荐使用两台交换机连接所有的节点,实现冗余B、仅需要将存储节点连接到管理网络,其他设备不需要连接到管理网络C、前端业务网络的交换机可进行堆叠,后端存储网络交换机不推荐进行堆叠D、为满足管理需求,仅需要集群的前三个节点连接到管理网络正确答案：A5.下面属于华为存储danger类型的高危命令的时哪个命令？A、rebootsystemB、importconfiguration_dataC、changalarmclearsequencelist=3424D、showalarm正确答案：A6.某客户单位购买了9台OceanStor9000存储节点,规划部署时要求任意三块硬盘一个存储节点故障,不影响数据的完整性,且整体存储空间利用率不低于80%,以下选项中元余配比合适的是:A、6+3B、6+3:1C、8+1D、16+3:1正确答案：D7.同事为客户定制华为双活容灾方案，客户已经有一套OceanStorS5500TV2和一套第三方存储系统，把这两套设备分别放置到两个数据中心实施双活容灾，该方案：A、不可行，主备容灾方案需要至少套OceanStorV3存储。

B、不可行，主备容灾方案需要至少套OceanStor5300V3及以上型号的存储系统才能实现。

C、可行，存储设备满足双活容灾方案要求。

Samsung_DRAM_PartNumberDecoder_2010

U : TSOP2 (Lead-Free)
100TQFP (Lead-Free) only for 128Mb GDDR Z : FBGA (Lead-Free) V : 144FBGA (Lead-Free) only for 128Mb GDDR L : TSOP2 (Lead-Free & Halogen-Free) H : FBGA (Lead-Free & Halogen-Free) F : FBGA (Lead-Free & Halogen-Free) for 64Mb DDR,
Divide Component
Module
Packing Type TAPE & REEL Other ( Tray, Tube, Jar ) Stack MODULE TAPE & REEL MODULE Other Packing
New Marking T 0 ( Number) S P M
17~18. Customer "Customer List Reference"
2/Total
K4XXXXXXXX - XXXXXXX
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
14~15. Speed (Wafer/Chip Biz/BGD: 00) 75 : 7.5ns, PC133 (133MHz CL=3) 60 : 6.0ns (166MHz CL=3) 50 : 5.0ns (200MHz CL=3) 40 : 4.0ns (250MHz CL=3) B0 : DDR266 (133MHz@CL=2.5, tRCD=3, tRP=3) B3 : DDR333 (166MHz@CL=2.5, tRCD=3, tRP=3) CC : DDR400 (200MHz@CL=3, tRCD=3, tRP=3) E6 : DDR2-667 (333MHz@CL=5, tRCD=5, tRP=5) E7 : DDR2-800 (400MHz@CL=5, tRCD=5, tRP=5) F7 : DDR2/3-800 (400MHz@CL=6, tRCD=6, tRP=6) F8 : DDR2/3-1066 (533MHz@CL=7, tRCD=7, tRP=7) H9 : DDR3-1333 (667MHz@CL=9, tRCD=9, tRP=9) K0 : DDR3-1600 (800MHz@CL=11, tRCD=11, tRP=11) 7A : GDDR3-2.6Gbps (0.77ns) 08 : GDDR3-2.4Gbps (0.8ns) 1A : GDDR3-2.0Gbps (1.0ns) 12 : GDDR3-1.6Gbps (1.25ns) 14 : GDDR3-1.4Gbps (1.4ns)