用方解石分析 [tex=0.857x1.0]N7iCrOsS+NNEUUlnsYCi1g==[/tex] 射线谱,已知方解石的晶面间距为[tex=6.643x1.357]cBBdsAE8/7X11xeskKxzqYS5sZmLF4DHRj73AT9lSMN98bVgLMuGA7fqlhf8BwHt[/tex],今在[tex=2.643x1.143]58E7z8vLn63ID4hlSuwda+e3RmGNSMSXp9wojdoqyYY=[/tex]和 [tex=2.643x1.143]YpFF1/GrrxFzgLoypHVQ+Pa2fXVioCXnBzmcqflr6SY=[/tex]的掠射方向上观察到两条主最大谱线,求这两条谱线的波长。

用方解石分析[tex=0.857x1.0]N7iCrOsS+NNEUUlnsYCi1g==[/tex] 射线谱, 已知方解石的晶面间距为[tex=6.714x1.357]bMKgKuno+U7kU0ApYElAaY8AUpkPzKwTjtzCVPKGgtI=[/tex], 今在 [tex=2.643x1.143]cVwoPPTwlF6/izxUn/f2363WRAIdaL2R+okzmazwfSA=[/tex]和 [tex=2.643x1.143]YpFF1/GrrxFzgLoypHVQ+Pa2fXVioCXnBzmcqflr6SY=[/tex] 的掠射 方向上观察到两条主最大谱线，求这两条谱线的波长.

用晶格常数等于 [tex=6.714x1.357]bMKgKuno+U7kU0ApYElAaY8AUpkPzKwTjtzCVPKGgtI=[/tex] 的方解石来分析 X 射线的光谱，发现入射光与晶面的夹角 [tex=0.5x1.0]qm+hGi0qngLh1B7HsENMPg==[/tex]为[tex=2.643x1.143]cVwoPPTwlF6/izxUn/f2363WRAIdaL2R+okzmazwfSA=[/tex] 和 [tex=2.643x1.143]YpFF1/GrrxFzgLoypHVQ+Pa2fXVioCXnBzmcqflr6SY=[/tex] 时，各有一条主极大的谱线。求这两谱线的波长。

用晶格常数等于[tex=6.643x1.357]bMKgKuno+U7kU0ApYElAaZcCzYrmkGDeX3+TAaydo+mhCZ9Shk4xPDSnUa+u0G0u[/tex]的方解石来分析[tex=0.786x1.0]yFLhNWXdy+71qunyuRVv1A==[/tex]射线的光谱,发现人射光与晶面的夹角[tex=0.5x1.0]qm+hGi0qngLh1B7HsENMPg==[/tex]为[tex=2.643x1.143]cVwoPPTwlF6/izxUn/f2363WRAIdaL2R+okzmazwfSA=[/tex]和[tex=2.643x1.143]YpFF1/GrrxFzgLoypHVQ+Pa2fXVioCXnBzmcqflr6SY=[/tex]时,各有一条主极大的谱线。求这两谱线的波长。

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4

共有4个站进行码分多址通信。4个站的码片序列为：a：（-1 -1 -1 +1 +1 -1 +1 +1） b：（-1 -1 +1 -1 +1 +1 +1 -1） c：（-1 +1 -1 +1 +1 +1 -1 -1） d：（-1 +1 -1 -1 -1 -1 +1 -1） 现收到这样的码片序列：（-1 +1 -3 +1 -1 -3 +1 +1），则（ ）发送1。