尖峰电流Ipk是指持续时间1-2S的短时()。

尖峰电流Ipk是指持续时间1-2S的短时()。


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有一过程如下:Sub Sub1(m As Integer,total As Long)Dim i As Integertotal=1For i=1 To mtotal=total*iNextEnd Sub调用它的事件过程如下:Private Sub Command1_Click()Dim tot As Long,a As Integera=Val(InputBox("请输入数据"))Call Sub1(a,tot)Print totEnd Sub则输入数据5,运行结果为【 】。

以下能够正确计算n!的程序是( )。A.Private Sub Command1_Click()n=5: x=1Dox=x * II=I + 1Loop While I nPrint xEnd SubB.Private Sub Command1_Click()n=5: x=1: I=1Dox=x * II=I + 1Loop While I nPrint xEnd SubC.Private Sub Command1_Click()n=5: x=1: I=1Dox=x * II=I + 1Loop While I =nPrint xEnd SubD.Private Sub Command1_Click()n=5: x=1: I=1Dox=x * II=I + 1Loop While I nPrint xEnd Sub

程序执行结果s的值是【 】.Private Sub Command l-Click ()I =0DoS= I +SI = I + lLoop Until I >=4End Sub

阅读程序: Sub p( b () As Integer)For i =1To 4 b(i) = 2(iNext 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 call p (a) For i=1 To 4 Print a(i) Next iEnd Sub运行上面的程序,单击命令按钮,输出结果为。

下面程序的输出结果是【 】。Private Sub Form_Click()i=0Do Until 0i=i+1if i>10 then Exit DoLoopPrint iEnd Sub

以下能够正确计算n!的程序是A.Private Sub Command1 Click() n=5:x=1 Do x=x * I I=I + 1 Loop While I < n Print x End SubB.Private Sub Command1_Click() n=5:X=1:I=1 Do X=X*I I=I + 1 Loop While I <n Print x End SubC.Private Sub Command1_Click() n=5:X=1:I=1 Do X=X * I I=I + 1 Loop While I<=n Print X End SubD.Private Sub Command1_Click() n=5:X=1:I=1 Do x=x * I I=I + 1 Loop While I>n Print X End Sub

以下程序的运行结果是( ) #define MAX 10 int a[MAX],i; main() { printf("\n");sub1();sub3(A) ,sub2(),sub3(A) ; } sub2() { int a[MAX],i,max; max=5; for(i=0;i<max;i++)a[i]=i; } sub1() {for(i=0;i<MAX;i++)a[i]=i+i; } sub3(int a[]) { int i; for(i=0;i<MAX,i++)printf("%d",a[i]); printf("\n"); }A.0 2 4 6 8 10 12 14 16 18 0 1 2 3 4B.0 1 2 3 4 0 2 4 6 8 10 12 14 16 18C.0 1 2 3 4 5 6 7 8 9 0 1 2 3 4D.0 2 4 6 8 10 12 14 16 18 0 2 4 6 8 10 12 14 16 18

设有如下程序: Private Sub search(a()As Variant,ByVal key As Variant,index%) Dim I% For I = Lbound(a)To Ubound(A)If key=a(I)Then index=I Exit Sub End If Next I index=-1 End Sub Private Sub Form_Load() Show Dim b()As Variant Dim n As Integer b=Array(21,64,92,15,72,38,45,72) Call search(b, 45, n) Print n End Sub 程序运行后,输出的结果是A.2B.6C.10D.12

下列程序的输出结果是【 】。Private Sub Command l Click()Dim a(1 TO 20)Dim iFor i=1 TO 20a(i)=iNext iFor Each i In a()a(i)=20Next iPrint a(2) End Sub

请阅读程序: Sub subP(b( )As Integer) For i=1 To4 b(i)=2*i Next i End Sub Private Sub Commandl 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 运行上面的程序,单击命令按钮,则输出结果是( )。A.2 4 6 8B.5 6 7 8C.10 12 14 16D.出错

下面程序的输出结果是【 】。Private Sub Form_Click()i=0Do Until 0i=i+1if i10 then Exit DoLoopPrint iEnd Sub

在窗体上画一个名称为CoilTlilandl的命令按钮,然后编写如下事件过程: Private Sub command1 Click() Dim m As Integer, i As Integer, x(3)As Integer For i=0 To 3:x(i)=i:Next i For i = 1 To 2: Call sub1(x,i):Next i For i = 0 To 3: Print x(i);: Next i End Sub Private Sub sub1(a()As Integer,k As Integer) Dim i As Integer Do a(k)=a(k)+a(k+1) j = j + 1 Loop While j < 2 End Sub 程序运行后,单击命令按钮,则窗体上显示的内容是A.0 3 7 5B.0 1 2 3C.3 2 4 5D.0 5 8 3

以下能够正确计算1+2+3+…+10的程序是A.Private Sub Command1_Click() Sum=0 ForI=1 To 10 Sum=Sum+I Next I Print Sum End SubB.Private Sub Command1_Click() Sum=0,I=1 Do While I<=10 Sum=Sum+I I=I+1 Print Sum End SubC.Private Sub Command1_Click() Sum=0: I=1 Do Sum=Sum+I I=I+1 Loop While I<10 Print Sum End SubD.Private Sub Command1_Click() Sum=0: I=1 Do Sum=Sum+I I=I+1 Loop Until I<10 Print Sum End Sub

编写如下事件过程: Private sub sub1 (ByVal x1 As String, y1 As String) Dim xt As String Dim i As Integer i = Len(x1) Do While i>= 1 xt = xt + Mid(x1, i, 1) i=i-1 Loop y1 = xt End Sub Private Sub Form. Click() Dim s1 As String, s2 As String s1= "teacher" sub1 s1, s2 Print s2 End Sub 程序运行后,单击窗体,则窗体上显示的内容是A.rehcaetB.tahreeeC.themeeD.eerthea

编写如下事件过程: Option Base 1 Private Sub Form Click() Dim x1()As Integer Dim i As Integer Dim s As Integer ReDim x1(3) For i = 1 To UBound(x1) x1(i)=i + 1 Next i Call sub1(x1) For i = 1 To UBound(x1) s = s + x1(i) Next i Print s End Sub Private Sub sub1(n()As Integer) Dim i As Integer ReDim Preserve n(5) For i = 3 To 5 n(i)=n(i-1)*2 Next i End Sub 程序运行后,单击窗体,则窗体上显示的内容是A.6B.12C.24D.47

以下能够正确计算n!的程序是______。A.Private Sub Commeadl_Click() n=5: x=1 Do x=x*i i=i+1 Loop While i<n Print x End SubB.Private Sub Command1_Click() n=5: x=1: i=1 Do x=x*i i=i+1 Loop While i<n Print x End SubC.Private Sub Command1_Click() n=5: x=1: i=1 Do x=x*i i=i+1 Loop Whilei<=n Print x End SubD.Private Sub Commsndl_Click() n=5: x=1: i=1 Do x=x*i i=i+1 Loop While i>n Print x End Sub

以下能够正确计算n!的程序是( )。A.Private Sub Commandl_C1ick()B.hiVate Sub Commandl_C1ick() n=5:x=1 n=5:x=1:i=1 DO DO X=x*1 X=X*1 i=i+1 i=i+1 Loop while i<n Loop While<n Print x Ptinte x End Sub End SubC.Private Sub Commandl_Click ()D.Pdvate Sub Commandl C1ick() n=5:X=1:i=1 n=5=:x=1:i=1 DO DO X=X*1 X=X*1 i=i+1 i=i+1 Loop While i>n Print x Print x End Sub End Sub

尖峰电流Ipk是指持续时间()的短时最大负荷电流。A10min以内B10~30minC10s以内D1~3s

下列对尖峰电流的说法错误的是()。A、是持续时间1-2s的短时最大负荷电流B、用来来选择熔断器和低压断路器C、用于相间短路灵敏度的校验D、用来整定继电保护装置及检验电动机自启动条件等

尖峰电流是指持续时间为()的短时最大负荷电流。A、1~2sB、2~5sC、5~210sD、10~15s

尖峰电流是持续时间为1—2s的短时()负荷电流。

class Super {  public int i = 0;  public Super(String text) {  i = 1; }  }  public class Sub extends Super {  public Sub(String text) {  i = 2;  }   public static void main(String args[]) {  Sub sub = new Sub(“Hello”);  System.out.println(sub.i);  }  }  What is the result?()  A、 0B、 1C、 2D、 Compilation fails.

尖峰电流Ipk是指持续时间()的短时最大负荷电流。A、10min以内B、10~30minC、10s以内D、1~3s

对于尖峰电流,下列叙述中错误的是()。A、尖峰电流是指持续1—2S的短时最大负荷电流B、尖峰电流主要用来选择熔断器C、单台用电设备的尖峰电流就是其起动电流D、多台用电设备的尖峰电流就是其起动电流

尖峰电流Ipk是考虑设备工作持续时间为5s的短时最大负荷电流。

单选题class Super {  public int i = 0;  public Super(String text) {  i = 1; }  }  public class Sub extends Super {  public Sub(String text) {  i = 2;  }   public static void main(String args[]) {  Sub sub = new Sub(“Hello”);  System.out.println(sub.i);  }  }  What is the result?()A 0B 1C 2D Compilation fails.

单选题class super (   public int I = 0;   public super (string text) (   I = 1   )   )     public class sub extends super (   public sub (string text) (   i= 2   )   public static void main (straing args) (  sub sub = new sub (“Hello”);   system.out. PrintIn(sub.i);  )   )   What is the result?()A Compilation will fail.B Compilation will succeed and the program will print “0”C Compilation will succeed and the program will print “1”D Compilation will succeed and the program will print “2”