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零状态响应是电路在零初始 状态下,仅由电路的输入引○+Rc+ 起的响应。 在t0时,v(O)=0,仅由直流 电流源F引起的响应就是 零状态响应。 若电路处零状态,=u(为阶跃函数,所求的响 应就是阶跃响应。若i=u(为单位阶跃函数,所 求的响应就是单位阶跃响应s(t)
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SOLUTIONS TO TEXT PROBLEMS 24.(b) A benzyl group(C., CH2) is ortho to the phenolic hydroxyl group in o-benzylphenol. OH CH2C.Hs (c)Naphthalene is numbered as shown. 3-Nitro-I-naphth has a hydroxyl group at C-1 and a nitro group
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1.试将下述化学反应设计成电池 (1)AgCl(s)=Ag+(aAg+)+Cl-(ac-) (2)AgCl(s)+I-(ar-)=AgI()+Cl-(acr-) (3)H2(pH2)+HgO(s)=Hg()+H2O() (4)Fe2+(aFe2+)+Ag+(+)=Fe3+(aFe3+)+Ag(s)
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本文研究非线性自治大系统$\\frac{{{\\rm{d}}{{\\rm{x}}_{\\rm{i}}}}}{{{\\rm{dt}}}}=\\sum\\limits_{{\\rm{i}}=1}^{\\rm{r}} {{{\\rm{f}}_{{\\rm{ij}}}}({\\rm{xj}})({\\rm{i}}=1, \\ldots,{\\rm{r}})} $这里,xi∈Rni,fij∈C(Rnj,Rni,fij(o)=0,得到保证其零解为全局一致渐近稳定的充分条件(见定理1)
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I. Complete the sentences with the best choice. Write your right letter on the Answer Sheet:(10%) 1. The teacher demanded that each student ( to class before seven o'clock. A. came B. come. comes D. would come 2. Seldom ( any mistakes during my past five years of service here. A. would I make
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pH of a buffer and buffer capacity Example: For a buffer solution consisting of 0. 1 M acetic acid and o1M sodium acetate the ph of the solution is 4.75 HAc= i If an amount of hydrochloric acid equivalent to 10% of the acetate present, is added to the buffer, what is the new pH of the solution?
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Groups. Teams and Effectiveness oGroup: two or more people who interact with each other to accomplish a goal. oTeam: group who work intensively with each other to achieve a specific common goal All teams are groups, BUT, not all groups are teams. o Teams often are difficult to form I Takes time for members to work together. Teams can improve organizational performance Irwin/McGraw-Hill
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#include #define null o FILE fp; ar a[81.c char*pl=\%oc\,*p2=\%d,%d\,*p3=\%of%f,%f”,*p4=\%s\ float flf2.13 f(fp=fopen(\:: I paper. dat\,\rb\))NULL)
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(1)∵内能是态函数,故△Eabd=△Eac 故 Oadb AE adb +Wadb =224+42 =266(J) (2) =AEba +Wba =-AEacb Wha =-224-84=-308(J)放热 (3) ad= d +Wad =(Ed-E)+Wadb =168+42=210(J) Qab=△Ea+O=E-Ea =(E-Ea)-(E-Ea)=Ea-168 =△Eacb-168=224-168=56(J) 8-21mol单原子理想气体i=3
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一、填空题(本大题共6小题,每小题4分,总计24分) 1.i2= 2.w=z+4将平面上|<2变为w平面上的 学号: 3.f()=zRe(z)在何处解析 4. f() =e-4 cos6t L[ f()]= 5.()=2(o)则f(t) 6.f()=u+iv为解析函数,u-v=x3+3x2y-3xy2-y3为解析函数,则u=
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