吸入含5%CO2的空气后,呼吸运动员有何变化?试述其机制。
吸入含5%CO2的空气后,呼吸运动员有何变化?试述其机制。
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在通用声明中定义a,在窗体中添加一个命令按钮Command1,编写如下程序代码:DimaAsIntegerSubtest()a=a+1:b=b+1:c=c+1PrintSub:;a;,;b;,;cEndSubPrivateSubCommand1_Click()a=2:b=3:c=4CalltestCalltestEndSub程序运行后,单击命令按钮,窗体中将显示( )A.Sub:3,4,5Sub:4,5,6B.Sub:2,3,4Sub:2,3,4C.Sub:3,1,1Sub:4,1,1D.Sub:1,1,1Sub:1,1,1
以下程序的运行结果是sub(int x,int y,int *z){*z=y-x;}main(){ int a,b,c;sub(10,5,a);sub(7,a,b);sub(a,b,c);printf("M,M,M\n",a,b,c);}A.5,2,3B.-5,-12,-7C.-5,-12,-17D.5,-2,-7
II型呼吸衰竭合并代谢性酸中毒()。 A、Pa0/sub2/supsubno60mHl.PaCO/sub2/supsubno50mmlgB、Pa0/sub2/supsubno60mHgPaCO/sub2/supsubno45mmgC、Pa0/sub2/supsubno60mmHgPaCO/sub2/supsubno50mmHD、Pa0/sub2/supsubno60mHg、PaCO/sub2/supsubno45mmHgE、Pa0/sub2/supsubno60mHg、PaC0/sub2supsubno45mmHg
有以下程序: #includestdio.h #define SUB(a)(a)-(a) main( ) {int a=2,b=3,c=5,d; d=SUB(a+b)*C; printf("%d\n",d); } 程序运行后的输出结果是( )。A.0B.-l2C.-20D.10
有如下程序: Private Sub Command1_Click() Dim a As Single Dim b As Single a=5:b=4 Call Sub1(a,b) End Sub Sub Sub1(x As Single,y As Single) t=X X=t\Y Y=t Mod y End Sub 在调用运行上述程序后,a和b的值分别为A.0 0B. 1 1C.2 2D.1 2
A、PH7.38,PaOSUB2/SUB50mmHg,PaCOSUB2/SUB40mmHgB、PH7.30,PaOSUB2/SUB50mmHg,PaCOSUB2/SUB80mmHgC、PH7.40,PaOSUB2/SUB60mmHg,PaCOSUB2/SUB65mmHgD、PH7.35,PaOSUB2/SUB80mmHg,PaCOSUB2/SUB20mmHgE、PH7.25,PaOSUB2/SUB70mmHg,PaCOSUB2/SUB20mmHg血气分析结果符合代偿性代谢性酸中毒( )
在窗体上画一个命令按钮,然后编写下列程序 Private Sub Command3_Click( ) Tcl 2 Tcl 3 Tel 4 End Sub Sub Tcl(a As Integer) Static x As Integer x=x + a Print x; End Sub 程序运行后,单击命令按钮,输出结果为A.2 3 4B.2 5 9C.3 5 4D.2 4 3
有如下程序。 Private Sub Commandl_Click() Dim a As Single Dim b As Single a=5:b=4 Call Sub1 ( a,B)End Sub Sub Subl(x As Single, y As Single) t=x x=t\y y = t Mod y End Sub 在调用运行上述程序后,a和b的值分别为A.0 0B.1 1C.2D.1 2
在窗体上画一个命令按钮,然后编写如下程序: Private Sub Comrnand3_Click( ) Cop 2 Cop 3 Cop 4 End Sub Sub Cop (a As Integer) Static x As Integer x=x + a Print x; End Sub 程序运行后,单击命令按钮,输出结果为A.2 3 4B.2 5 9C.3 5 4D.2 4 3
有程序如下: Sub subP(b() As Integer) For i=1 To 4 b(i)=2*i Next i End Sub Private Sub Command1_Click() Dim a(1 To 4) As Integer a(1)=5 a(2)=6 a(3)=7 a(4)=8 subP a() For i=1 To 4 Print a(i) Next i End Sub运行上面程序,单击命令按钮,输出结果为______ 。A. 2 4 6 8B.5 2 2 2C.10 12 14 16D.出错
有如下程序。Private Sub Command1_Click()Dim a As SingleDim b As Singlea=5: b=4Call Sub1(a, b)End SubSub Sub1(x As Single, y As Single)t=xx=t \ yy=t Mod yEnd Sub在调用运行上述程序后,a和b的值分别为A.0 0B.1 1C.2 2D.1 2
在窗体上画一个名称为Command1的命令按钮,然后编写如下事件过程: Private Sub sub1(ByVal x As Integer, ByVal y As Integer, ByVal z As Integer) z=x * x+ y * y End Sub Private Sub Command1_Click() Dim a As Integer a = 8 Call sub1(1, 2,A)Print a End Sub 程序运行后,单击命令按钮,则窗体上显示的内容是A.8B.2C.5D.11
若有如下程序:sub(int a,int b,int *z){*z=a+b;return;}main(){int a=1,b=2,c=3,d,e;sub(a,b, return;} main() {int a=1,b=2,c=3,d,e; sub(a,b,D) ; sub(c,d,e); printf("%d",e); } 则程序运行后的输出结果是A.3B.4C.5D.6
在窗体中添加一个命令按钮,编写如下程序: Private Sub Sub1(p,m,n) p=p+1:m=m+1:n=n+1 Print "sub1:";p;m;n End Sub Private Sub Command1_Click() a1=1:b=2:c1=3 Call Sub1(a,b1+3,c1) Print"Main:";a1;b1;c1 End Sub 程序运行后,输出结果为A.Sub: 2 6 4 Main: 2 6 4B.Sub: 2 6 4 Main: 2 6 4C.Sub: 2 6 4 Main: 1 2 3D.Sub: 2 6 4 Main: 2 2 3
人体内O2、CO2进出细胞膜是通过A.单纯扩散B.易化扩散C. 人体内O<sub>2</sub>、CO<sub>2</sub>进出细胞膜是通过A.单纯扩散B.易化扩散C.主动转运D.人胞作用E.出胞作用
执行下面的一段C程序后,输出结果变量应为______。sub (int x, int y, int *z) { *z=y-x; } main() { int a, b, c; sub (10, 5, sub(7, a, sub(a, b, printf ("%d, %d, %d\n", a, b, c); }A. 5, 2, 3 B. -5, -12, -7 C. -5, -12, -17 D. 5, -2, -7
单选题变直径圆管流,细断面直径d1,粗断面直径d2=2d1,粗细断面雷诺数的关系是( )。ApResub1/sub=0.5Resub2/sub/pBpResub1/sub=Resub2/sub/pCpResub1/sub=1.5Resub2/sub/pDpResub1/sub=2Resub2/sub /p
单选题干熄焦循环气体的主要成分为N2,其他成分还有H2、CH4、CO等气体,其中属于不可燃组分是( )。ApNsub2/sub /pBpHsub2/sub/pCpCHsub4/sub/pDCO
单选题煤矿井下的有害气体主要是由()、CO2、H2S、NO2、H2、NH3气体组成。ACOBpCHsub4/sub、SOsub2/sub/pCpSOsub2/sub、CO/pDpCO、CHsub4/sub、SOsub2/sub/p
单选题酶促反应速度(v)达到最大反应速度(Vmax)的80%时,底物浓度[S]为Ap1Ksubm/sub/pBp2Ksubm/sub/pCp3Ksubm/sub/pDp4Ksubm/sub/pEp5Ksubm/sub/p
单选题A pIsub2/sub=Isub3/sub/pB pIsub2/sub=4Isub3/sub/pC pIsub2/sub=2Isub3/sub/pD pIsub3/sub=4Isub2/sub/p
单选题下列测量结果的表示中,错误的是()。ApIsubS/sub=10.0413A,Usubrel/sub=5×10sup-5/sup,k=2 /pBpIsubS/sub=10.0413(1±5×10sup-5/sup)A,k=2 /pCpIsubS/sub=(10.0413±5×10sup-5/sup)A,k=2 /pDpIsubS/sub=10.0413A,Usub95rel/sub=5×10sup-5/sup,Vsubeff/sub=9 /p
多选题1. class Super { 2. private int a; 3. protected Super(int a) { this.a = a; } 4. } ..... 11. class Sub extends Super { 12. public Sub(int a) { super(a); } 13. public Sub() { this.a= 5; } 14. } Which two, independently, will allow Sub to compile?()AChange line 2 to: public int a;BChange line 2 to: protected int a;CChange line 13 to: public Sub() { this(5); }DChange line 13 to: public Sub() { super(5); }EChange line 13 to: public Sub() { super(a); }
多选题下列表示中____的表示形式是正确的。ApUsub95/sub= 1%,vsubeff/sub =9 /pBpUsubr/sub= 1%,k=2 /pCpusubC/sub=0. 5% /pDpusubC/sub=±0 5%.k=1 /p