工频正弦电流的Im=50A,初相φ=30°,它的数学解析式是i=()A。

工频正弦电流的Im=50A,初相φ=30°,它的数学解析式是i=()A。


相关考题:

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【 】。

如果存在如下过程:Private Function FMax(a() As Integer)Dim First As Integer, Last As Integer, i As IntegerFirst=LBound(A)Last=UBound(A)Max=a(First)For i=First To LastIf a(i) Max Then Max=a(i)Next iFMax=MaxEnd Function在窗体上添加一个命令按钮,然后编写如下事件过程:Private Sub Command1_Click()ReDim m(1 To 4) As Integerm(1) =20: m(2) =30: m(3) =50: m(4) =100c=FMax(m)Print cEnd Sub单击命令按钮,其输出结果为 【12】 。

下面是用冒泡法将5个整数按升序排列,请将程序补充完整。Sub GASR()Dim mm=Array(10, 1, 5, 6, 7)For i=0 To 3For j= 【13】If m(i) =m(j) Then【14】m(i) =m(j)m(j) =tEnd IfNext j【15】For i=0 To 4Print m(i)Next iEnd Sub

在窗体上画一个命令按钮,然后编写如下事件过程:Sub BTC(oldCopy() As Integer, newCopy() As Integer)Dim i As IntegerFor i=LBound(oldCopy() ) To UBound(oldCopy() )newCopy(i) =oldCopy(i)Next iEnd SubPrivate Sub Command1_Click()Dim m(10) As Integer, n(10) As IntegerFor i=1 To 10m(i) =i  iNext iCall BTC(m, n)Print n(5)End Sub程序运行后,单击命令按钮输出结果为( )。A.0B.2C.10D.25

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。

某工频正弦交流电电流有效值为IA初相角为300,则该电流解析表达式为()A。 A.i=sin50tB.f=sin(314t+30°)C.i=√2min(314t+30°)D.i=√2sin(5Ot+30°)

正弦电流的最大值Im=50A,频率f=50HZ,初相角Φi=30°,写出瞬时表达式,并要求t=0.02秒电流的相位角和瞬时值。 A.50sin(314t+30°),390°,25AB.50sin(314t+30°),360°,25AC.50sin(314t+30°),390°,250A

有如下程序: Private Sub Commandl_Click( ) Dim i As Integer For i=1 To 2 DS Next i End Sub Sub DS( ) Dim x As Integer,m As String Static y,n X=X + 1 y=y + 1 m=m "*”:n=n"#" Print x,y,m,n End Sub 程序运行后,输出的结果是A.1 1 * #B.1 1 * #C.1 1 * # 1 1 * # 1 2 * #D.1 1 * # 1 1 * ## 1 2 * ##

下面程序的输出结果是( )。 #include"stdio.h" int sub(int m) { static a=4; a+=m; return a; } void main() { int i=2,j=1,k; k=sub(i); k=sub(j); printf("%d",k); }A.6B.7C.8D.9

在窗体上画一个命令按钮,然后编写如下事件过程: Private Sub Commandl_Click() Dim m As Integer, n As Integer, p As Integer m=3: n=5: p=0 Call Y(m, n, p) Print Str(p) End Sub Sub Y(ByVal i As Integer, ByVal j As Integer, k As Integer) k=i+j End Sub 程序运行后,如果单击命令按钮,则在窗体上显示的内容是( )A.4B.6C.8D.10

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 Command1 Click() a = 2 For i=1 To 3 Call funl(i,A)Next i x = i + 2 Print x;a End Sub Private Sub fun1(n,m) m=m + n * 2 n = n + m End Sub 程序运行后,单击命令按钮,则窗体上显示的内容是A.1 5B.4 5C.8 4D.3 4

在窗体上画一个命令按钮,然后编写如下事件过程: Private Sub Command1_Click() Dim in As Integer,n As Integer,p As Integer m=3:n=5:p=0 Call Y(m,n,p) Print Str(p) End Sub Sub Y(ByVal i As Integer,ByVal j As Integer,k As Integer) k=i+j End Sub 程序运行后,如果单击命令按钮,则在窗体上显示的内容是A.4B.6C.8D.10

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

有过程如下:Sub fact (m As Integer, total As Long) Dim i As Integer total = 1 for i = 1 To m total = total*i Next iEnd Sub调用它的事件过程如下:Private Sub Command1_ Click () Dim tot As Long Dim a As Integer a = Val InputBox (“请输入数据”)) Call fact (a, tot) Print totEnd Sub则输入数据5,运行结果为 ______。A.100B.120C.200D.50

单击命令按钮后,下列程序代码的执行结果是______。Public Sub fun (a As Integer, b As Integer, c As Integer)Doa=b+ cn=n+1Loop While n > 3End SubPrivate Sub command1_ click()Dim m As Integer, n As Integer, i As IntegerFor i = 0 To 2: m = n + 1: Next iFor i = 1 To 2: Call fun (m, n, i): Next iFor i = 0 To 3Print m;Next iEnd Sub

在窗体上画一个命令按钮,然后编写如下事件过程: Sub BTC(old Copy( ) As Integer,new Copy( ) As Integer) Dim i AS Integer For i=LBound(oldCopy( ))To UBound(oldCopy( )) newCopy(i)=oldCopy(i) Next i End Sub Private Sub Command1 _ Click( ) Dimm(10) As Integer,n(10)As Integer For i=1 To 10 m(i) =i*i Next i Call BTC(m,n) Print n(5) End Sub 程序运行后,单击命令按钮输出结果为A.0B.2C.10D.25

有如下程序:Private Sub Form_Click()Dim i As Integer, Sum As IntegerSum=0For i=2 To 10If i Mod 2=0 And i Mod 30 ThenSum=Sum+iEnd IfNextPrint SumEnd Sub程序运行后,单击窗体,输出结果为A.12B.30C.24D.18

单击一次命令按钮之后,在对话框中输入12,21,30,3,窗体中的输出结果为______。 Private Sub Command1_ Click() For i = 1 To 4 s = InputBox (“请输入”) Call per(i, s) Next i Print s End Sub Sub per(a, ByValB)b = 1 For i = 1 To 3 a = a * i Next i End SubA.12 21 30 3B.24 42 60 6C.3 30 21 12D.只能输入一个数

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

某正弦电流的初相为30°,在t=T/2时,瞬时值为-0.4A,求这个正弦电流的有效值。

已知工频正弦交流电流在t=0时的瞬时值等于0.5A,计时始该电流初相为30°,求这一正弦交流电流的有效值。

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.

正弦电流的最大值Im=50A,频率f=50HZ,初相角Φi=30°,写出瞬时表达式,并要求t=0.02秒电流的相位角和瞬时值。

普通晶闸管的额定电流的大小是以()来标志的。A、工频正弦全波电流的平均值B、工频正弦半波电流的平均值C、工频正弦全波电流的有效值D、工频正弦半波电流的有效值

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”