在[tex=1.643x1.071]KRN39XNzhNKLNW8UykCR8mvj5a3P/xYKFrnIZ1EqNkw=[/tex]和[tex=2.286x1.0]BXyl3jrscB0ti56TOaCXKELkO5N4uYrIUdKLdAtNSK4=[/tex]下,空气的密度为[tex=5.143x1.429]Sd5X1+pVVEvPuuo/P6k6GGsmauvpIdo9Phz0yOHDKAvzkgO/GodDLkKWbZideWq8[/tex] 空气的比定压热容 [tex=15.0x1.5]Wc7vGQbn2EjkwWiNF72N0rYkpMwlQ33VkpYhbbEfifSQWF4WGLiP1mimo/Z+/RrNp99CqAZ3HPKJcVh5e26B9PuwsVGYsH57R8n4ZFn0zaz45Oy5swpwnVJSqTsm4EHWWMGwGcaD4I/VOOiw1QG0mg==[/tex]今有[tex=2.286x1.214]mLh77wdZ8zZ7aZGrJbZKjq/Ke9CW5G7/yRGGGkBm7xc=[/tex]的空气,试计算:[tex=1.357x1.357]TWUgLpDrEXIKICMuiEQPjw==[/tex]若维持体积不变,将空气由 [tex=1.643x1.071]KRN39XNzhNKLNW8UykCR8mvj5a3P/xYKFrnIZ1EqNkw=[/tex]加热至 [tex=2.143x1.071]5EnNnQAH4EmVEukBezotBFR0Fi5MFGaqWMC31mH7Qvg=[/tex] 所需的热量.[tex=1.214x1.357]vzdGmXlbw83hTiK2SebvEA==[/tex] 若维持压强不变,将空气由 [tex=1.643x1.071]KRN39XNzhNKLNW8UykCR8mvj5a3P/xYKFrnIZ1EqNkw=[/tex]加热至[tex=2.143x1.071]5EnNnQAH4EmVEukBezotBFR0Fi5MFGaqWMC31mH7Qvg=[/tex]所需的热量.[tex=1.214x1.357]zs4t7aUJaV7q8Vd+b4EZVA==[/tex]若容器有裂缝，外界压强为[tex=2.286x1.0]6B0O4QOMWRagm33Q5DG9gw==[/tex],使空气由[tex=1.643x1.071]KRN39XNzhNKLNW8UykCR8mvj5a3P/xYKFrnIZ1EqNkw=[/tex] 缓慢地加热至[tex=2.143x1.071]5EnNnQAH4EmVEukBezotBFR0Fi5MFGaqWMC31mH7Qvg=[/tex]所需的热量. 

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。

共有四个站进行CDMA通信，现收到这样的码片序列（-3 +1 -1 -1 +1 +1 +1 -3），则下列站点发送了数据1的是（ ） 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）

共有四个站进行码分多址CDMA通信。四个站的码片序列为：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) , 现在收到这样的码片序列：（-1 -1 +1 +1 +3 +3 -1 -1 ）

A、B、C三个站点采用CDMA技术进行通信，A、B要向C发送数据，设A的码片序列为+1，-1，-1，+1，+1，+1，+1，-1。则B可以选用的码片序列为: -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