Tài liệu Chapter 2 Protocols and Architecture docx
... access point or SAP (OSI) Address Concepts William Stallings Data and Computer Communications Chapter 2 Protocols and Architecture Addressing Scope ❚ Global nonambiguity ❙ Global address ... changes in other layers The OSI Environment OSI as Framework for Standardization Layer Specific Standards Elements of Standardization ❚ Protocol specification ❙ Operates betwee...
Ngày tải lên: 16/02/2014, 08:20
... sodium chloride contains 22 .99 g of sodium and 35.44 g of chlorine 541.1 g 22 .99 g 5.443 Na of mass Cl of mass == Chemistry, Julia Burdge, 2 nd e., McGraw Hill. 12 Law of Multiple Proportions • When ... chloride contains 39.3 g of sodium and 60.7 g of chlorine a 20 0.0 g sample of sodium chloride contains 78.6 g of sodium and 121 .4 g of chlorine 54.1 g 78.6 g 21 .41 Na o...
Ngày tải lên: 28/11/2013, 01:11
... overview of the section topics and demonstrations. 2 Module 2: Designing and Modeling # ## # Introduction to Analysis and Design ! Importance of Analysis and Design ! The MSF Application ... Review 57 Module 2: Designing and Modeling 38 Module 2: Designing and Modeling # ## # Creating a Physical Design ! Mapping to the Real World ! Componentization...
Ngày tải lên: 11/12/2013, 14:15
Tài liệu Chapter-2-Before you install docx
... PS /2 mice, and USB mice are becoming more popular. concepts.mm,v v4 .21 (20 03/04/ 02 06:37: 12) 42 Chapter 2: Before you install 10 April 20 03, 06:13:07 The Complete FreeBSD (concepts.mm), page 42 The ... on the market for over20 years, and it has changed a lot in that time. In particular: concepts.mm,v v4 .21 (20 03/04/ 02 06:37: 12) 25 44 Chapter 2: Before you install...
Ngày tải lên: 24/01/2014, 14:20
Tài liệu Plant physiology - Chapter 2 Energy and Enzymes docx
... of ∆GRT= C C 2 1 23 . log EE RT nF h oxidant [reductant] = ′ + 0 23 . log [] 2Fe O H Fe H O 2+ 2 +3+ ++⇔ + 1 2 2 22 1 2 2 22 OHEHO 2 +± ++⇔ Fe Fe e 2+ 3+ ± 22 2⇔+ CHAPTER 2 8 * The standard hydrogen ... groups Hydrophobic (nonpolar) R groups :NH CH HC :N H CH 2 CH CH 2 CH CH 2 CH 2 CH 2 CH 2 CH 2 C NH 3 NH NH 2 H 2 N CH 2 CH CH 2 CH CH 2 Glutam...
Ngày tải lên: 20/02/2014, 01:20
Tài liệu Chapter 2 Internet Protocols ppt
... 127 2. B 类地址: 前 2 个 octets 代表网络号,剩下的 2 个代表主机位. 范围是 10xxxxxx,即 128 到191 3.C 类地址: 前 3 个octets 代表网络号,剩下的1 个代表主机位 . 范围是 110xxxxx,即 1 92 到 22 3 4.D 类地址:多播地址,范围是 22 4 到 23 9 5.E 类地址:保留,实验用,范围是 24 0 到 25 5 ... 地址用于广播.所谓广播地址指同时 向网上所有的主机发送报文,也就是说,不管物理网络特性如何,Internet 网支持广播...
Ngày tải lên: 09/12/2013, 17:15
Tài liệu CHAPTER 1: Greeting and Getting Acquainted docx
... ______________________________ Lớp _________________________ © Lê Phạm Thuý-Kim Chapter 1 1 CHAPTER 1: Greeting and Getting Acquainted TẬP VIẾT N A. Họ tên người Việt You are helping ... ______________________________________________________________________ 2 Workbook Manual to accompany Let’s Speak Vietnamese C. Họ là người nước nào? 1. First, complete the...
Ngày tải lên: 10/12/2013, 04:15
Tài liệu Chapter 2 - Communicating over the Network CCNA Exploration 4.0 docx
... www.bkacad.com Protocol Suites and Industry Stardards • Many of the protocols that comprise a protocol suite reference other widely utilized protocols or industry standards. • A standard is a process ... endorsed by the networking industry and ratified by a standards organization, such as IEEE or IETF. • The use of standards in developing and implementing protocols ensures tha...
Ngày tải lên: 12/12/2013, 14:15
Tài liệu Chapter 6: Force and Motion II docx
... this point the net force and thus the acceleration become zero and the cat moves with constant speed v t known the the terminal speed 2 1 2 t D C Av mg ρ = = (6-4) 2 t mg v C A ρ = C Uniform ... Chapter 6 Force and Motion II In this chapter we will cover the following topics: Describe the frictional force between two objects. Differentiate between static and kinet...
Ngày tải lên: 13/12/2013, 05:15
Tài liệu Chapter 2: Motion Along a Straight Line docx
... t 1 , x 1 ) with point ( t 2 , x 2 ). In the plot below t 1 =1 s, and t 2 = 4 s. The corresponding positions are: x 1 = - 4 m and x 2 = 2 m 2 1 2 1 2 ( 4) 6 m 2 m/s 4 1 3 s avg x x v t ... at time t 1 to a new position x 2 at time t 2 by determining the average velocity between t 1 and t 2 . Here x 2 and x 1 are the positions x(t 2 ) and x(t 1 )...
Ngày tải lên: 13/12/2013, 05:15