Chapter 4 Questions – Panko

Chapter 7 Questions – Wide Area Networks (WANs)

1. WANs often use the transmission lines of telephone companies.

a. True

b.False

Reference: p.208

2. Which of the following is NOT one of the basic technologies or techniques

normally used to construct a WAN?

a. Ordinary telephone lines and telephone modems

b. A mesh of leased lines

c. Use of a PSDN

d. A VPN

e. A mesh of routers and codecs

Reference: p. 208

3. Most WANs operate at speeds lower than LAN speeds.

a. True

b.False

Reference: p. 209

4. Since WANs cover a larger area than LANs, WANs typically operate at higher

speeds than LANs.

a. True

b.False

Reference: p. 209

5. If you had 3 Megabytes of data that you wanted to send to a far distant location

once a day, the most economical way to link the two sites in a WAN would be to use:

a. frame relay.

b. a VPN.

c. ordinary telephone lines and telephone modems.

d. a low-speed leased lin

e.

Reference: p. 209

6. The V.92 standard standardizes ____.

a. Modem on Hold

b. The possibility of transmitting (sending) faster than 33.6 kbps

c. Both of the above.

d. Neither a. nor b.

Reference: p. 211

7. Asymmetric modem standards should be used when:

a. the modem is capable of performing compression.

b. the modem is capable of performing error detection and correction.

c. the downstream transmission speed should be higher than the upstream

speed.

d. All of the abov

e.

Reference: p. 210

8. The V.34 standard:

a. requires that upstream and downstream transmission speeds are equal, but

may both change based upon line conditions.

b. requires that upstream and downstream transmission speeds are both fixed

at 33.6 kbps.

c. allows downstream transmission speeds to exceed upstream speeds when

line conditions permit.

d. begins transmitting and receiving at 33.6 kbps and increases the speed

based upon line conditions.

e.begins transmitting and receiving at 56 kbps and increases the speed based

upon line conditions.

Reference: p. 210

9. Both parties have V.90 modems. How fast can they send files to each other?

a. 33.6 kbps

b.56 kbps

Reference: p. 211

10. One disadvantage of using a V.90 modem to communicate is that:

a. the service is only available in larger cities.

b. for 56 kbps receiving, the party at the other end must have a digital

connection and special equipment to communicate with you.

c. most ISPs do not support the V.90 standard for communications.

d. you cannot communicate with V.34 standard modems.

Reference: p. 210

11. A communication line that requires you to order it in advance of using it, brings

one pair of UTP to the customer premises, and is available for use at any time is a:

a. digital subscriber line.

b. T1 Leased line.

c. virtual private network.

d. cable modem lin

e.

Reference: p. 214-216

12. Which is an important difference between a trunk-based leased line and a DSL?

a. A trunk line-based leased line requires two-pair copper media; DSL

requires fiber optic media.

b. A trunk line-based leased line requires two-pair copper or fiber optic

media; DSL can operate on single pair of UTP.

c. A trunk based leased line requires fiber optic media; DSL can operate on

either single-pair copper, two-pair copper or fiber optic media.

d. A trunk line-based leased line can operate on copper or fiber media; DSL

requires two-pair copper media.

Reference: p. 213

13. Which of the following can use voice-grade copper?

a. Trunk line-based leased line

b. DSL

c. Both of the above.

d. Neither a. nor b.

Reference: p. 213

14. DSLAMs are used with _____.

a. digital subscriber lines

b.virtual private networks

c.cable modem lines

d. trunk line-based leased lines

Reference: p. 215

15. ADSL requires a splitter at the customer end of the circuit to divide the voice and

data signals.

a. True

b.False

Reference: p. 215

16. Which of the following is used most heavily by businesses and so usually receives

good speed guarantees?

a. ADSL

b. HDSL

c.Both of the above.

Reference: p. 216

17. SHDSL uses _____ UTP.

a. voice-grade

b. data-grade

c. None of the above. SHDSL uses optical fiber.

d.None of the abov

e. SHDSL uses satellite connections.

Reference: p. 213

18. The major functional difference between ADSL and HDSL is that:

a. HDSL data is sent at half the speed of ADSL.

b. HDSL sends data at the same speed in both directions.

c. ADSL uses two-pair copper media, HDSL used only fiber optic media.

d. ADSL is faster than HDSL but does not provide error correction.

e.HDSL is a lower-cost service geared towards home users.

Reference: p. 215

19. A T1 line uses _____ UTP.

a. voice-grade

b. data-grade

c. Either a. or b.

d. None of the abov

e. T1 lines use optical fiber.

e. None of the above. T1 lines use satellite connections.

Reference: p. 213

20. A major disadvantage associated with cable modem service is that:

a. cable modem service is more error prone than DSL.

b. cable modem speed cannot approach DSL speed.

c. subscribers too far from the nearest switch cannot get cable modem

service.

d. cable modem speed is shared.

e. All of the above.

Reference: p. 216

21. A GEO user’s terminal uses a(n) _______ antenna.

a. dish

b. omnidirectional

c. Either of the above.

d.Neither a. nor b.

Reference: p. 217

22. Which type of satellite does not require handoffs between satellites to sustain

communications between fixed locations?

a. FEO

b. GEO

c. LEO

d. MEO

e.ALL types of satellite communications REQUIRE that handoffs occur.

Reference: p. 217

23. How many leased lines are required to connect every pair of sites with a leased

line mesh if you have a total of four sites?

a. 1

b. 3

c. 4

d. 6

e. 8

Reference: p. 219

24. How many leased lines are required to connect every pair of sites with a PSDN if

you have a total of four sites?

a. 1

b. 3

c. 4

d. 6

e. 8

Reference: p. 219

25. One area of savings that is normally associated with the decision to use a PSDN is:

a. a reduction of staff needed to manage network.

b. the elimination of leased lines.

c. a reduction of the number of POPs.

d. All of the abov

e.

Reference: p. 219

26. Which is NOT a type of PSDN service?

a. ATM

b. Frame Relay

c. ISDN

d. T1

e. X.25

Reference: p. 221

27. Which is the most widely used PSDN service?

a. ATM

b. Frame Relay

c. ISDN

d. X.25

Reference: p. 221

28. The fastest nation-wide PSDN services use _____ technology.

a. ISDN

b. X.25

c. Frame Relay

d. ATM

e. T3

Reference: p. 222

29. Which of the following PSDNs is best used for backup links?

a. ATM

b. Frame Relay

c. Ethernet

d. ISDN

e. X.25

Reference: p. 224

30. As inter-city WAN traffic increases, which of the following is likely to grow most

in use?

a. ATM

b. Frame Relay

c. Ethernet

d. ISDN

e. X.25

Reference: p. 223

31. Ethernet is most likely to be a serious _____ competitor.

a. MAN

b. Inter-city WAN

c.MAN and inter-city WAN

Reference: p. 224

32. The maximum transmission and reception speed for Internet access with ISDN is

_____.

a. 56 kbps

b. 64 kbps

c. 128 kbps

d. 144 kbps

e. 1.544 Mbps

Reference: p. 224

33. Which network layer(s) must be considered when connecting an organization’s

internal LAN to a PSDN service?

a. Data link

b. Physical

c. Internet

d. Physical, data link

e. Physical, data link, internet

Reference: p. 225

34. The most expensive element in Frame Relay pricing usually is _____.

a. leased lines

b. the port speeds at the POPs

c. the collective cost of PVCs

d. installation

e. the FRAD

Reference: p. 225

35. PVCs are used to:

a. enclose network media on their way to POPs.

b. define potential communications paths in a PSDN.

c. define predetermined communications paths in a PSDN.

https://www.sodocs.net/doc/595175120.html,pute the potential volume of traffic on a PSDN.

Reference: p. 227

36. In a Frame Relay network, the available bit rate is computed by subtracting the

committed bit rate from the maximum bit rate.

a. True

b.False

Reference: p. 228

37. Frame Relay customers with lots of bursty data should include a(n) ________

clause in their service agreement.

a. ABR

b. Bandwidth sharing

c. CIR

d. POP

e.PVC

Reference: p. 228

38. Frame Relay SLAs are used to_____.

a. offer services to a new geographic area

b. define maximum latency that will occur

c. guarantee that Frame Relay customers commit to using services for at

least a specified time (usually one year)

d. protect vendor equipment installed at the customer’s site

e. prevent viruses from being transmitted over the network inadvertently

Reference: p. 229

39. Compared to PSDNs, VPNs offer _____.

a. lower cost

b. lower latency

c. Both of the above.

d. Neither a. nor b.

Reference: p. 230

40. RASs call _____ servers to decide whether to accept connections.

a. DHCP

b. Directory

c. IPsec

d. PPTP

e. RADIUS

Reference: p. 232

41. For remote access VPN service, the _____ security standard is the most popular.

a. PPTP

b. IPsec

c. Tunnel mode

d. Frame Relay

e. Firewalls

Reference: p. 232

42. For site-to-site VPNs, the _____ security standard is the most popular.

a. PPTP

b. IPsec

c. Frame Relay

d. STS Control Protocol

e. PPP Control Protocol

Reference: p. 234

43. Which of the following is included in all recent versions of Microsoft Windows?

a. PPTP

b. IPsec

c.Both of the above.

Reference: p. 231

44. Which of the following has security weaknesses?

a. PPTP

b.IPsec

Reference: p. 233

45. PPTP works at the Layer __; IPsec works at Layer ___.

a. 1 and 2 ; 1, 2, and 3

b. 2; 2

c. 2; 3

d. 3; 2

e. 3; 3

Reference: p. 232, 234

46. IPsec SAs are the same in both directions.

a. True

b.False

Reference: p. 235

2014科学知识与能力训练六年级(下册)练习册参考答案

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一．填一填 1．在放大镜下我们可以看到蟋蟀的“耳朵”在足的内侧。 2.触角是昆虫主要的感觉器官,有识别气味功能,也有平衡、帮助呼吸、识别异性等作用。不同昆虫触角的形状不同,科学研究表明昆虫触角就是的它“鼻子”。 3.草蛉是蚜虫的天敌。蚜虫在植物嫩枝上吸食汁液,每个蚜虫只有针眼般大小,我们用肉眼只能看见它们是密密麻麻的一片,但在１0倍放大镜下我们可以看清它们的肢体。 4.在放大镜下我们观察到蝴蝶的翅膀表面上布满了彩色小鳞片,这些鳞片其实是扁平的细毛。 3. 放大镜下的晶体 探究接力棒 一、填一填 1．将溶液风干或加热使其水分蒸发可使物质重新结晶析出,得到的是这种溶液的晶体。２.像食盐、白糖、碱面、味精的颗粒都是有规则几何外形，称为晶体。常见的晶体有立方体、金字塔形、针形等形状。 二、选一选 1．选出相应的晶体形状：水晶（Ａ)雪花( B )维生素C( C）。 A、立方体 B、六角形Ｃ、针形 2.下面全是晶体的一组是( Ａ)。 A、糖、碱、维生素C B、水晶、雪花、玻璃 C、盐、味精、珍珠 三、判一判 1.自然界中所有的固体都是晶体。(×） ２．碱面的晶体像花瓣,而食盐的晶体像树枝。 (×） 3.许多岩石是由矿物晶体集合而成的。(√） 4.晶体的形状多种多样,没有规则。( ×） 4．怎样放得更大 探究接力棒 一、填一填 1．两个不同放大倍数的凸透镜组合起来,调整它们之间的距离来观察物体，这样物体的图像被放得更大。 2．显微镜的发明,是人类认识世界的一大飞跃，把人类带入一个崭新的微观世界。

语言学-Chapter4-课后练习答案

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语言学C h a p t e r4课后练习答案

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2014小学科学知识与能力训练五年级下册练习册参考答案 第一单元沉和浮１物体在水中是沉还是浮探究接力棒一、１．泡沫块充了气的游泳圈塑料玩具石块砖头鸡蛋２．体积轻重二、× √ × ２沉浮与什么因素有关探究接力棒一、√ √ √ 二、ＢＡ３橡皮泥在水中的沉浮探究接力棒ＣＢＣＢＢ４造一艘小船探究接力棒１．浸入水中的体积增大，更容易浮起来；增加轮船浸入水的体积２．体积大、质量轻、容易上浮５浮力探究接力棒一、浮力乙二、√ √ √ ６下沉的物体会受到水的浮力吗探究接力棒一、√ × √ √ × 二、ＢＣ７马铃薯在液体中的沉浮探究接力棒√ √ × √ ８探索马铃薯沉浮的原

因探究接力棒一、√ × 二、ＡＣ单元练习一、１．浮力２．轻重体积密度３．物体排开的水量二、√ × 三、选择题ＢＢＡＢ四、往水里放盐改变橡皮泥的形状，增大排开水量第二单元热１热起来了探究接力棒一、热热热二、太阳辐射矿物燃烧物体摩擦２给冷水加热探究接力棒一、不变变大不变变小二、不能。冷水受热时体积会变大，太满容易溢出来浇天火焰发生危险３液体的热胀冷缩探究接力棒一、上升增大下降变小热胀冷缩二、１．铺木地板要留伸缩缝；野外的电线夏天时会粘一起；煮饺子时，饺子鼓起来２．防止热胀损坏包装４空气的热胀冷缩探究接力棒一、１．变大变小２．大二、丘丘球放入热水后变球壁变软，同时内部空气受热膨胀就把凹下去的地方鼓起来了。

５金属热胀冷缩吗探究接力棒１．热胀冷缩２．防止铁轨因热胀冷缩而发生弯曲或断裂３．缩短伸长６热是怎样传递的探究接力棒１．高低２．高低７传热比赛探究接力棒一、铁、铝、铜、金、银玻璃、塑料、棉布、纸、土、空气、陶瓷、木头、水二、言诚理可８设计制作一个保温杯探究接力棒１．独自完成，说明原理即可２．是的。因为保温和冷藏的原理相同，只是目的相反。单元练习一１．热热热２．不良导体散失３．鼓起来凹下去４．不高小不变大５．热胀冷缩６．大７．高低８．高低９．热的良导体热的不良导体１０．铁、铝、铜、金、银玻璃、塑料、棉布、纸、土、空气、陶瓷、木头、水１１。不锈钢好好二、１．因热胀冷缩，钢制桥梁温度升高时变长，温度下降时变短，放磙子上可防止气温

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2、当支点不断向筷子尖端移动时，越来越容易夹起菜；当支点不断向筷子顶端移动时，越来越难夹起菜。 杠杆的研究 一、1、省力；费力；既不省力又不费力 2、省力 3、省力；费力 二、1、B 2、AC；B；D 三、1、×2、×、 四、不是所有杠杆的三个点都在一条直线上，例如起钉锤。 螺丝刀里的科学 一、1、轮轴 2、螺丝刀、门把手、汽车方向盘、水龙头。 3、省力；省力 二、1、√2、√3、√4、× 三、我会选择轮轴B提起重物。将重物 挂在轴上，在轮上用力，因为轮轴的轮 越大越省力。 动滑轮和定滑轮 一、定滑轮，改变力的方向；动滑轮，省力。 二、1、（1）电梯（2）吊车

（3）自行车刹车装置（4）晒衣架 2、 起重机 一、1、滑轮组，省力，改变力的方向 2、A是定滑轮，作用是改变力的方向。B是动滑轮，作用是省力。C是滑轮组，作用是既能省力，又能改变力的方向。D是滑轮组，作用是既能省力，又能改变力的方向。 二、B；C 在斜坡上 一、1、斜面，斧头、剪刀、螺丝钉、指甲钳

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chapter4习题

DC-AC变换器（无源逆变电路） 一、学习目的： 通过本章的学习，学者可以了解逆变器的电路结构、分类、特点及主要性能 指标；对三种基本变换方式——方波变换、阶梯波变换、正弦波变换，有一定的 认识；可以理解采用各种变换方式的逆变器的工作原理；了解空间矢量PWM控制 的基本原理。 二、主要内容： 1、基本概念 DC-AC变换器是指能将一定幅值的直流输入电压（或电流）变换成一定幅值、 一定频率的交流输出电压（或电流），并向无源负载（如电机、电炉、或其它用 电器等）供电的电力电子装置，又称为无源逆变电路，常简称作逆变器（Inverter）。 完成直流电压变换的逆变器称为电压型逆变器，而完成直流电流变换的逆变 器则称为电流型逆变器。 2、变换方式的分类 （1）方波变换方式 逆变器的交流输出有两种基本调制方式：脉冲幅值调制（PAM－Pluse Amplitude Modulation）和单脉冲调制（SPM－Single Pluse Modulation）。 所谓脉冲幅值调制（PAM）是指：逆变器的输出频率可由180°方波或120°方波（如图4-3b 所示）的周期来控制，而逆变器输出基波的幅值则由输出方波的幅值即逆变器直流侧电压（或电流）的幅值来控制。显然，采用PAM控制方式时，其方波的导通角恒定（180°方波或120°方波）。 所谓的单脉冲调制（SPM）是指：逆变器的输出频率仍由方波的周期来控制，而逆变器输出基波的幅值则由逆变器输出方波的导通角进行控制，即可使导通角在0°～180°范围调节。显然，采用SPM控制方式时，逆变器输出方波的幅值即逆变器直流侧电压（或电流）的幅值恒定。 （2）阶梯波变换方式 （3）斩控调制方式：是指逆变器输出的调制脉冲幅值固定不变，而逆变器中的功率管 以一定的控制规律进行调制。斩控调制方式主要有以下二类即：①脉冲宽度调制（PWM ）；②脉冲频率调制（PFM） 3、逆变器的分类 （1）按直流侧储能元件的性质，逆变器可分为电压型逆变器（VSI－Voltage Source Inverter）和电流型逆变器（CSI－Current Source Inverter）。 （2）按逆变器输出波形的不同，逆变器可分为方波逆变器、阶梯波逆变器、以及正弦 波逆变器等。 （3）按逆变器功率电路结构形式的不同，逆变器可分为半桥逆变器、全桥逆变器、推 挽式逆变器等。 （4）按逆变器功率电路的功率器件的不同，逆变器可分为半控型逆变器和全控型逆变 器。 （5）按逆变器输出频率的不同，逆变器可分为工频逆变器、中频逆变器以及高频逆变 器。

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一、填空题 1．像开瓶器那样能绕着一个固定支点将物体撬起的简单机械也是_杠杆 。 2．杆秤是 杠杆 类工具，提绳是 支点 ，秤盘是 阻力点 ，秤砣是 用力点 。 3.杠杆有 省力 杠杆，费力 杠杆， 不省力不费力 杠杆。 4.“四两”拨千斤，称为 省力 杠杆。 5.杠杆省力与否与杠杆的三个点的位置有关，当用力点到支点的距离大于阻点到支点的距离时是__省力__；当用力点到支点的距离小于阻点到支点的距离时是_费力_。 二、选择题 1.杠杆上，用力点到支点1米，支点到阻力点0.5米，那么它（ B ）。 A 、费力 B 、省力 C 、既不费力，也不省力 2.以下工具属于省力杠杆的是（ A 、C ），属于费力杠杆的是（ B ），属于既不费力，也不省力的杠杆是（ D ）。 A 、钳子 B 、镊子 C 、钉锤 D 、翘翘板 三、判断题（正确的打 “√”，错误的的打“×” ） 1.使用杠杆时，不论支点在什么位置都能省力。（ × ） 2.凡是杠杆都是用来省力的。（ × ） 四：简答题 1．说一说，为什么有些杠杆类工具要设计成费力的呢？ 答：如：镊子 属于费力杠杆，主要用来取轻小物体，使用镊子是利用了方便。 2．提绳的位置对称的最大称重有什么影响？ 答：设计提绳的位置越靠近所称的重物，所称量的重物越多。 4．多观察一些工具，是不是所有杠杆的三个重要点都一定在一条直线上？ 答：不是，如使用钉锤起钉子时，三个重要点不在一条直线上。

小学科学教科版四年级《知识与能力训练》上册答案

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不同点是：食盐溶解后，水的颜色无变化，但有味道的变化；而高锰酸钾溶解后水的颜色有变化。 探究接力棒 1．√2．×3．√4．√5．×6．√ 3．液体之间的溶解现象 探究起跑线 1．（其他项由学生根据实验现象进行记录） 2．（其他项由学生根据实验现象进行记录） 探究接力棒 1．√2．×3．√ 4．不同物质在水中的溶解能力 探究起跑线 1．由学生根据实验现象进行记录

2． 探究接力棒 1．×2．√3．√4．√ 5．溶解的快与慢探究起跑线 1．略 2. 3．略 探究接力棒 一、搅拌加热把物质弄碎 二、略

6．一杯水能溶解多少食盐 探究起跑线 1．20（或50 ）用尺子量出一平勺（或称出1克1克的）完全溶解2．略 探究接力棒 1．√2．×3．×4．√ 7．分离盐和水的方法 探究起跑线 1．方法：再加水 结果：略 2．方法：蒸发水分（加热、风吹、晒等） 结果：略

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第四章组合逻辑电路 1. 解： (a)(b)是相同的电路，均为同或电路。 2. 解：分析结果表明图(a)、(b)是相同的电路，均为同或电路。同或电路的功能：输入相同输出为“1”；输入相异输出为“0”。因此，输出为“0”(低电平)时，输入状态为AB＝01或10 3. 由真值表可看出，该电路是一位二进制数的全加电路，A为被加数，B为加数，C为低位向本位的进位，F1为本位向高位的进位，F2为本位的和位。 4. 解：函数关系如下： AB S F+ ⊕ = + + A BS S S A B B 将具体的S值代入，求得F 3 1 2 值，填入表中。

A A F B A B A B A A F B A B A A F A A F AB AB F B B A AB F AB B A B A B A AB F B A A A B F B A B A B A F B A AB AB B A B A F B B A B A B A B A B A B A F AB BA A A B A A B A F F B A B A F B A B A F A A F S S S S =⊕==+==+⊕===+⊕===⊕===⊕===+⊕===+=+⊕===⊕==+==⊕==Θ=+=+⊕===+++=+⊕===+=⊕===⊕==+=+⊕==+=+⊕===⊕==01111 111011010110001011101010011000001110110)(010101001 010011100101000110000 0123

5. （1）用异或门实现，电路图如图(a)所示。 (2) 用与或门实现，电路图如图(b)所示。 6. 解因为一天24小时，所以需要5个变量。P变量表示上午或下午，P＝0为上午，P=1为下午；ABCD表示时间数值。真值表如表所示。 利用卡诺图化简如图(a)所示。 化简后的函数表达式为

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