技术的近义词是什么?一定要是两个字的呀!

答案在最后一页
（同学你是仙游一中的吧）

解题思路：A．氮化硅陶瓷是一种超硬物质，具有耐磨、耐高温、抗腐蚀能力强等优点；
B．光缆是良好的通讯材料，其主要成分是二氧化硅；
C．根据重金属盐可以使蛋白质变性来分析；
D．NO和CO能反应生成氮气和二氧化碳．

A．氮化硅陶瓷是一种超硬物质，具有耐磨、耐高温、抗腐蚀能力强等优点，可用来制造汽车发动机、轴承、汽轮机叶片等，故A正确；
B．制造光缆的主要材料是二氧化硅，故B错误；
C．重金属盐可以是蛋白质变性，所以误食后会中毒，服用大量的蛋清和牛奶后，可以和重金属反应，变为不被吸收的物质，减弱对肠胃的影响，故C正确；
D．采用纳米二氧化钛光触媒技术，能将NO和CO能反应生成氮气和二氧化碳，故D正确．
故选：B．

点评：
本题考点： 硅和二氧化硅；常见的生活环境的污染及治理；氨基酸、蛋白质的结构和性质特点．

考点点评： 本题考查二氧化硅的用途，无机非金属材料，蛋白质的性质等，题目难度不大，注意制造光缆的主要材料是二氧化硅．

B

根据氧化物的定义（含有两种元素，其中一种是氧元素）可判断A正确；C ₂ O ₃ 与CO中C的化合价分别为+3、+2，都没有达到最高价态，都具有还原性，C ₂ O ₃ 与C在燃烧后C的化合价都会变为+4价，而生成CO ₂ ，因此，C、D也是正确的；C ₂ O ₃ 、H ₂ CO ₃ 中C的化合价不同，C ₂ O ₃ 不是碳酸的酸酐，B错误。

解题思路：A、红外光谱图可以确定有机物的分子结构中氢原子的种类以及数目的相对多少；
B、pH计可以测定溶液的pH；
C、质谱仪可以测定有机物的相对分子质量；
D、同位素示踪法用来研究有机反应的机理．

A、红外光谱图可以确定有机物的分子结构中氢原子的种类以及数目的相对多少，不可以确定有机物的结构，故A错误；
B、pH计可以用来准确测定溶液的pH，粗略测定可以用pH试纸，故B正确；
C、质谱仪可以用来测定有机物的相对分子质量的大小，故C正确；
D、有机反应的机理取决于原子的来源和归宿，可以采用同位素示踪法用来研究其过程，故D正确．
故选A．

点评：
本题考点： 有机物结构式的确定；测定溶液pH的方法；探究化学反应机理．

考点点评： 本题属于化学应用方面知识的题目，要求学生注意所学知识的归纳和整理，难度不大．

Virtual reality technology development and application Abstract:virtual reality is a new field of computer technology,has the vital role.Foreign economic developed countries as its defense technology focus on the development of the key technology.The paper introduces the application of virtual reality technology overview of the basic concept,characteristics and composition,this paper expounds the application of virtual reality technology and its development trend.Keywords:virtual reality,The virtual environment,Immersion,Interactive,imagine

（P/A,i,n）＝（1＋i）^0＋（1＋i）^-1＋.＋（1＋i）^-（n-1)＋（1＋i）^-n
（P/A,i,n-1）＝（1＋i）^0＋（1＋i）^-1＋.＋（1＋i）^-（n-1)
即：（P/A,i,n）＝（P/A,i,n-1）＋（1＋i）^-n
＝（P/A,i,n-1）＋（P/F,i,n)

解题思路：（1）已知洗涤时的电压和电流，可利用公式P=UI计算电功率，已知脱水时的电压和电流，可利用公式P=UI计算电功率．
（2）首先判断洗涤时完成一次周期性变化，洗涤电动机工作的时间，已知洗涤时的电功率，再利用公式W=Pt计算电能．
（3）首先判断洗涤时完成一次周期性变化所需要的时间，再判断7min内完成几次周期性变化，再让次数乘以一次消耗的电能即可．

（1）设洗衣机洗涤时的电功率是P₁，脱水时的电功率为P₂，
P₁=UI₁=220V×0.5A=110W，
P₂=UI₂=220V×0.8A=176W．
答：该洗衣机洗涤时的电功率是110W，脱水时的电功率是176W．
（2）洗衣机洗涤时完成一次周期性变化时，洗涤电动机正转30s、反转30s，
洗涤电动机工作的时间为60s，
W₁=P₁t=110W×60s=6600J．
答：该洗衣机洗涤时完成一次周期性变化所消耗电能为6600J．
（3）t_总=7min=420s，
∵洗涤时完成一次周期性变化所需要的时间为：t₁=30s+5s+30s+5s=70s．
∴该洗衣机洗涤420s完成6次周期性变化，
W=nW₁=6600J×6=39600J．
答：该洗衣机洗涤7min所耗电能为39600J．

点评：
本题考点： 电功率的计算；电功的计算．

考点点评： 本题考查了利用公式P=UI计算电功率，以及利用公式W=Pt计算电能，与生活相连，使学生觉得学了物理有用，且加强了学生的节能意识，注意计算时的单位变换要正确．

Modern human civilization and progress of contemporary science and technology in particular,the rapid development of electronic computers and electronic information network technology to human social life has brought great convenience.It will not only enable us to achieve the global scope of the instant messaging,but also the global e-commerce possible,but due to limitations of network technology to cyber crime can flourish and the relevant laws and the building of the use of network lag indulgence of crime,busy busy The social life of many hidden behind the crisis.From the virtual reality community to extend the network society.Along with the rapid development of network technology to propagate a variety of network crime,no doubt become a host of social reality many of the most important risk factors.Therefore,we should do a good job in preparation for the prevention,strengthen the building of networks to strengthen moral education in the legal system improve legislation and strengthen their self-protection awareness of all mankind together to meet this challenge.
Key words:Internet crime prevention network against cyber-crime

解题思路：先依据实际需要的时间=原计划需要的时间-提前时间，求出实际需要的时间，再依据工作效率=工作总量÷工作时间，分别求出原计划和实际每天生产皮卡车台数，最后用实际每天生产台数减原计划每天生产台数即可解答．

2000÷（25-5）-2000÷25
=2000÷20-2000÷25
=100-80
=20（台）
答：该厂实际平均每天比原计划多生产20台家用皮卡车．

点评：
本题考点： 有关计划与实际比较的三步应用题．

考点点评： 依据等量关系式：工作效率=工作总量÷工作时间，分别求出原计划和实际每天生产皮卡车台数，是解答本题的关键．

两电容并联,可以认为是上下两个导线各相接,也可以认为是上下两个极板分别直接拼接.两个一样的电容的两个极板各自拼起来,面积增大为原来的两倍.电容量与极板面积成正比,所以两个一样的电容并联,容量增加为原先的两倍,耐压不变.同样两个一样的电容串联,容量增加为原先的一半,但耐压增加为原先的两倍.
因此您只要准备4只500V4uF电容,先把两个两个并联起来得到两只500V8uF电容,再串联起来,就得到1000V4uF电容.

C

呵呵,这么大的土方量,一般是没有预留的,因为没有那么大的堆土场地,呵呵,所以,应该是一次运出后,回填时,在回运,就按你说的给你计算下.
你的这个应该是地下室工程,那就是有防水的要求,在这也把挖土的工作面说下
砌筑类 150毫米 支模板类 300 防水 800
1、挖土 S1=（85+0.8*2）*（60+0.8*2）=5334.56M2
S2=（85+0.8*2+8*0.5*2）*（60+0.8*2+8*0.5*2）=6584.16M2
V=（S1+S2+S1S2的开平方）*H/3=47587.30M3
2、预留=挖土-外ks=1.14 ks'=1.05运=47587.3-21000=26587.3M3
3、外运土=21000M3
4、这个问题涉及到土的松散系数,一般的土是1.3~~1.4之间,只是不知道你的土是几类土,就按一二类考虑,那就是1.3的系数,则V松=21000*1.3=27300M3
3.5立方米的汽车外运要运27300/3.5=7800车
呵呵,这个ks=1.14 是土的自然松散系数 ks'=1.05 这个是松散土的压实系数,呵呵,这2个问题很容易让人迷糊,你看下面的计算
1、外运土=21000M3 那虚体积就是21000*1.14=23940M3
2、预留土的虚体积就应该先计算出挖土后的松散体积,在减去回填压实的虚体积,则计算是预留土的虚体积=26587.3M3*1.05=27916.7M3
这部分超出的计算就是外运部分的虚体积=26587.3*1.14-27916.7=2392.8M3
需要外运的总虚土体积就是=23940+2392.8=V
需要的车数=V/3.5

我觉得吧,这个问题还是主要是思路的问题,而不是列一大堆计算式.思路对了这个问题还是挺简单的.
从0级升级到9级需要的平均次数,也就是升级次数的数学期望,肯定是等于从0级升级到1级的平均次数+1级到2级升级的平均次数+...+8级到9级升级的平均次数.这个应该比较容易想,因为从n升到n+1和n+1升级到n+2不相干,并且都是必要的步骤,所以肯定是这样.
Ok,下面只要计算从n级升级到n+1级的所需次数的数学期望就好啦,最后把他们加起来就是结果.
从第n级升级到n+1级成功的概率是p,那么你看这个升级次数的分布：
1次升级成功概率：p
2次升级成功概率：(1-p)p
3次升级成功概率：(1-p)^2*p
...
第k次升级成功概率：(1-p)^(k-1)*p
这个分布的期望你想算也行,不想算也有公式,就是E(n) = 1/p,
那这个问题不就很简单了,把p = 1,0.9,0.8.,0.2带入1/p然后把它们加起来就OK啦：
E(n)升级到9级就是：1/1+1/0/9+1/0.8+...+1/0.2 = 19.2897,你要想写成那种分数的也成,我就用计算器算了,算出小数的这种.
所以计算期望这种问题,重点还是要写出分布列,或者算出分布函数才行.

This paper spark plasma sintering (SPS) of AlN / BN composite ceramics,as the BN in different adjustment sintering additives Y2O3 additives.Significantly improve the efficiency of the sintering and improve flexural strength materials,due to the introduction of effective mitigation BN ceramic thermal conductivity of the composite performance degradation and realizing the reunification of the material properties.On the initial grain size and amount of sintering aids AlN ceramics phase composition,microstructure and properties of.Focused on the impact of AlN / BN composite ceramic microstructure and properties of factors.Through sintering process and the amount of sintering aids change adjustment AlN / BN microstructure of composite ceramics improve the thermal performance to improve machinability in the same time,maintaining a relatively high thermal conductivity and mechanical properties.
Preparation of single-phase SPS AlN ceramics,the sintering temperature in 1800 ℃,soaking time 10 min,the relative density of samples up to more than 98.5 percent,about 11.5 Gpa hardness,bending strength of more than 300 Mpa.AlN initial particle size and amount of sintering aids,density and mechanical properties of almost no impact,but on the material microstructure and thermal conductivity have a significant impact on performance.Initial diameter of 1.3 μ m of AlN powder add 3% Y2O3 wet samples,thermal conductivity to 148 W / m ∙ k.AlN powder particle size and the initial Y2O3 impact of the addition of AlN crystal growth,the distribution of grain boundary phase,as well as the contact between the AlN grains,thus affecting the thermal conductivity.
AlN / BN composite ceramic sintering temperature in the 1600-1800 ℃,and the holding time of 0-10 min under the conditions can quickly compact,vol BN content in the 5-30% range,the density reached more than 98%.With the hardness of the material content of the BN to increase linearly decreased,thereby improving machinability,BN content of 10 vol%,the composite ceramics has been lower than the hardness of the hardness of cemented carbide cutting tools,which can be mechanical processing.
BN content in 10 vol%,the role of grain refinement toughening of deterioration is far greater than the weak,the strength of composite ceramics than single-phase AlN ceramics has improved greatly,and continue to increase the content to 25 vol%,a declining trend strength,but still higher than the single-phase high-strength ceramics.
BN content reaches more than 5 vol%,the composite ceramic thermal conductivity of a single-phase AlN ceramics significantly decreased in 15 vol% 25vol%-within the scope of the thermal conductivity show the trend of changes in relative stability,rate of decline slows.BN second phase of the initial particle size of the ceramic composite microstructure and thermal conductivity properties of the structure has had a notable impact.BN initial particle size too small,microscopic structure of the reunion phenomenon obviously not conducive to optimizing the thermal conductivity.
Further study AlN / BN composite ceramic thermal conductivity factors and the impact of its optimization.BN under the specific content of the adjustment of sintering aids Y2O3 content on the sintering temperature and soaking time can be controlled composite ceramic microstructure,effectively enhance the composite ceramic thermal conductivity.BN content of 15 vol% Y2O3 wet content is 8%,the sintering temperature 1800 ℃,soaking time 10 min,access to thermal conductivity of 141 W / m ∙ k,excellent machinability of composite ceramics.
Preparation of this paper SPS AlN / BN composite ceramics obtained by the high thermal conductivity have been far more than at home and abroad by hot-pressing of the AlN / BN composite ceramics of the highest level,but also has the same excellent processability and mechanical properties of strength .The SPS can help in achieving the AlN / BN composite ceramics rapid densification on the basis of the composition and microstructure of the regulation.SPS process of sintering temperature than hot-pressing process of sintering temperature (1850-1900 ℃) low 100 200 ℃,the SPS holding time than the hot-pressing the holding time 1-3 h significantly reduced,thus greatly enhancing the efficiency of sintered In this regard AlN / BN Machinable Ceramic Composites,high-performance,high efficiency,preparation and application of great importance
软件翻译!仅供参考!

对,三个是一个意思的,细胞核移植技术,是指将一个动物细胞的细胞核移植至去核的卵母细胞中,产生与供细胞核动物的遗传成份一样的动物的技术.

设原价是x元
则(x-a)×(1-20%)=b
x-a=b/0.8=1.25b
x=a+1.25b
答：原价是a+1.25b元
C是AB中点
所以BC=1/2AB
同理,BD=1/2BC=1/4AB
所以AD=1/2AB+1/4AB=3/4AB
E是AD中点
所以AE=1/2*3/4AB=3/8AB
所以AE等于AB的8分之3

1×1×1＝1
2×2×2＝3＋5
4×4×4＝（13+15+17+19）关于4*4=16对称
5×5×5＝（21+23+25+27+29）关于5*5=25对称

科技是第一生产力

1,控制温度,它对微生物生长期很大作用
2,PH的调控 由于在生长过程中培养液会有PH的明显变化要控制好,看你得发酵种类
3,发酵过程中的溶氧（搅拌速度,发酵液粘度,温度也对溶氧影响）
4 ,CO2和呼吸商（RQ）RQ=OUR（摄氧率）/CER（co2释放率）
5,基质浓度,通气搅拌还有泡沫的控制
实际操作的看你得发酵要求还有你得菌种类型

功亏一篑 越俎代庖 蔚然成风 淋漓尽致

从材料来看,前者是指重实践经验的第一次工业革命,后者是重科学的第二次及以后的工业革命.选项中只有①蒸汽机是第一次工业革命的产物,所以答案选 C.②③④.

科学范围

主要有自然源和人为源两种,但危害较大的是后者.u 自然来源包括：风扬尘土、火山灰、森林火灾、漂浮的海盐、花粉、真菌孢子、细菌.u 人为源包括：固定源和流动源.固定源包括各种燃料燃烧源,如发电、冶金、石油、化学...

①克隆可以有效地繁殖具有“高附加值的牲畜”；②克隆可以用来挽救珍稀动物；③克隆对于人类疾病的防治、寿命的延长具有重要意义.
举例子,具体真切的说明克隆是什么.作诠释,解释说明克隆是什么,使文章更通俗易懂.

实际用油量=36-12=24.
设重复利用率为x,x=60%*（1+1.6%）的12次方.
需用量=24/(1-x).

Pick to
Auto repair is a study of the malfunction car by technical means grate failure reason,and take some measures to make its trouble-shooting and restore to achieve a certain performance and safety standards.Now along with the consumer's consumption level and the change of consumption idea,can provide consumers with superior and satisfied service 4s store,maintenance enterprises can fast establish own brand and the prestige,occupies more market share.This reflects the importance of customer service,need customer service attitude sincere courtesy,process is simple and direct,charge fair and reasonable,in order to make the customer satisfied.For 4s shops and maintenance of enterprise's development in the future lay a solid foundation.

技术好 技术（精湛） 技术（娴熟 ） 技术（ 高超 ）
天气好 天气（ 宜人 ） 天气（ 美好 ） 天气（ 晴朗 ）
态度好 态度 （ 积极 ） 态度（ 良好 ） 态度（ 端正 ）
读书好 读书（ 上进 ） 读书（ 勤奋 ） 读书（ 认真 ）技术好 技术（精湛） 技术（超群 ） 技术（ 尖端 ）
天气好 天气（ 晴朗 ） 天气（明朗 ） 天气（凉爽 ）
态度好 态度 （认真 ） 态度（诚恳 ） 态度（正确 ）
读书好 读书（ 刻苦 ） 读书（ 进步 ） 读书（ 勤奋 ）

6496

成分都是蛋白质；RNA聚合酶就是合成RNA的酶,PCR技术中的热稳定聚合酶是在高温下任然有很高活性的合成DNA的酶

解题思路：第一次降价后的价格为原价-m，第二次降价后的价格为第一次降价后的价格×（1-降低的百分数），把相关数值代入即可．

∵第一次降价后的价格为x-m，
∴第二次降价后的价格为（x-m）（1-20%），
∴根据第二次降价后的价格为n元可列方程为（x-m）（1-20%）=n，
∴x=[5/4]n+m．
故答案为：[5/4]n+m．

点评：
本题考点： 一元一次方程的应用．

考点点评： 考查列一元一次方程；得到第二次降价后的价格的等量关系是解决本题的关键．

SPWM调制方式有单极性调制方式和双极性调制方式

He is the master of a roller coaster ride, and his technology is better than me.

简单地说,就是在生物体外将编码某特定蛋白的DNA序列重组至表达载体或者经转染转移到生物体内,使该编码基因以核外或稳定整合到染色体上的形式借用宿主细胞的转录和翻译元件,表达外源基因的一种方式.

1.功亏一篑
2.越俎代庖
3.蔚然成风
4.淋漓尽致

1.胚胎干细胞(为方便,下简称ES) 定向分化研究的经典途径是:扩增ES后,撤去LIF,继续悬浮或悬滴培养2--5天,形成拟胚体(E B ),然后用不同的措施定向诱导分化为目的细胞.之所以经EB途径,目的是模拟体内胚胎发育过程,EB中...

With the continuous development of digital technology,sampling theory has been widely used in communications,control and signal processing applications.In order to ensure undistorted signal from the sample to restore the original continuous signal,then the request was sampled signal energy should be concentrated in a frequency range,while the sampling frequency has a minimum limit,which is discussed in this article Nyquist Special sampling theorem.In the sampling process,the key is this part of the ADC,where aliasing has become the main target of our study.We have the basic theory of the sampling theorem,in-depth study of the pseudo-frequency generation theory and the process to address this problem we choose to leave their frequency filter to be filtered.In this paper,using the second-order low-pass filter 7 in series aliasing filters and DFS-V-based aliasing filter principle and features:second-order low-pass filter in this study,we not only on a single frequency 7 in series of low-pass filter to study effects of different cut-off frequency of 7 Chuanlian into the low-pass filter were compared to study and then apply the MATLAB simulation of the transfer function amplitude frequency characteristic to calculate the degree steep ,by comparing the verification conclusions; in the DFS-V model to study aliasing filter,we study the characteristics of its composition and its simple simulation.Two circuits with over-sampling technology will steepness to 70 decibels or more,that more than corner frequency,the amplitude and frequency characteristics of the circuit of the steepness of the curve decline rate of increase,the frequency can interfere with considerable attenuation.

1、旧设备的年平均成本=3600/(P/A,15%,4)+4000-1000/(F/A,15%,4)=3600/2.855+4000-1000/4.9934=1260.95+4000-200.26=5060.69
2、新设备的年平均成本=14000/(P/A,15%,8)+4000-3000/(F/A,15%,8)=14000/4.4873+4000-3000/13.727=3119.92+4000-218.55=6901.37
由上可以,旧设备的年平均运行成本较低,所有不宜进行更新.

精心的

（X-900）*(1-20%)=4900 所以X=6925 你想问的是原价是多少吧 原价是6925

最大优点是可以无限制地大量产生

In order to solve the problem of the current oil fracturing fluid industry with backward technology and equipment, liquid mixing speed, precision, stability is not good, the base fluid powder is unevenly distributed, the development of China's oil fracturing fluid with rapid and efficient mixing truck, this design is the design and Research on the distribution of liquid car retarder. Gear box design with liquid vehicle adopts a gear transmission. Its main advantage is: the instantaneous transmission ratio constant, stable and reliable work, drive; application of power and wide speed range; high transmission efficiency; reliable work, long service life, this design uses two cylindrical gear reducer. The first design of transmission scheme, selection of the gear reducer as the driving device, and then calculate the design of speed reducer (including the choice of motor, design of gear, shaft structure design, selection and calculation of rolling bearing, selection and checking, checking the coupling flat key connection, select the gear and bearing lubrication the nine part). Planar design of gear reducer using CAXA software, draw gear reducer 2D part drawings and assembly drawings.

船夫的驾驶技术特别好,行船的速度极快.遇到极窄的地方,他总能左拐右拐的拐过去.
我希望?

A

HB+1/2 π R²

设李星投中x个
9×12+x=10(x-1.8)
108+x=10x-18
10x-x=108+18
9x=126
x=14
答：李星投中14个
解析：先设未知数x 然后根据等式列方程 左边9×12+x表示的是10人投中的总数
右边x-1.8 因为题目中说李星比10人平均数多1.8个
换句话说 就是李星投的减去1.8得到的就是10人平均数
再×10 得到的就是10投中的总数
这样两边相同 解出的结果就是李星投中的数量

因为DNA聚合酶不能直接在一条DNA单链上复制,一定要先让一小段寡核苷酸与模板DNA互补配对,然后聚合酶才能在引物的一端继续复制下去

原癌基因（proto-oncogene）是细胞内与细胞增殖相关的基因,【是维持机体正常生命活动所必须的】,在进化上高等保守.当原癌基因的结构或调控区发生变异,基因产物增多或活性增强时,使细胞过度增殖,从而形成肿瘤.
：）

技术等级（级）中位数4.5 众数1 2 3 4 5 6 7 8都是
工人人数（人）中位数30 众数22 38 75 87 64 20 12 7 都是

培育杂交植物的重点是得到杂交融合的杂合原生质体.杂交原生质体的得到需要细胞融合技术.细胞融合技术,将两个细胞融合成一个,从而得到杂合原生质体.如果是植物细胞,需要去细胞壁.得到两个原生质体,也不是我们想想的很快融合成一个.与可能细胞之间存在排除.这和细胞膜上识别蛋白有关.我的感觉.