在两级齿轮传动中,若传动比的分配方案是i1=i2,则其遵循的原则是()

在两级齿轮传动中,若传动比的分配方案是i1=i2,则其遵循的原则是()


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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

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。

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

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

在直齿圆柱齿轮传动的弯曲强度计算中,小齿轮齿形系数Y与()有关A、齿数Z1B、齿宽系数υaC、模数mD、传动比i

设计齿轮传动时,若保持传动比i与齿数和z1=z2=z3不变,而增大模数m,则齿轮的弯曲强度(),接触强度()。

计算题:在一组定轴齿轮传动系统中,齿轮1带动齿轮2、齿轮2带动齿轮3完成传动,齿轮1为主动轮、齿轮3为从动轮,其齿数分别为Z1=30、Z2=60、Z3=90,计算系统的传动比i13?

一对齿轮传动,若齿轮传动传动比i12=4.5,小齿轮转速n1=990r/min,则从动轮转速n2为多少?

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.

一对齿轮传动,若两齿轮齿数z1=25,z2=50,则齿轮传动的传动比i12为多少?

传动比是主动齿轮的转数n1与从动齿轮的转数n2之比,以i表示,i=n1/n2,i>1时是加速。

机床常用传动副中,齿轮传动的传动比为()。A、i=k/zB、i=z1/z2C、i=z2/z1D、i=d1/d2

在多级齿轮传动中,为了减小传动误差,传动比的分配应遵守前小后大原则。

在直齿圆柱齿轮传动的弯曲强度计算中,小齿轮齿形系数YF1与()有关。A、齿数Z1B、齿宽系数υaC、模数mD、传动比i

在两级齿轮传动中,若传动比的分配方案是i1=i2,则其遵循的原则是()A、重量最轻B、等效转动惯量最小C、输出轴转角误差最小D、加速度响应最快

设计齿轮传动时,若保持传动比i和齿数和Z=Z1+Z2不变,而增大模数m,则齿轮的弯曲强度(),接触强度()。

根据最小等效转动惯量原则,小功率齿轮传动的传动比分配为()。A、i1=i2=…=inB、i2≈i12/2C、前小后大D、前大后小

填空题设计齿轮传动时,若保持传动比i与齿数和z1=z2=z3不变,而增大模数m,则齿轮的弯曲强度(),接触强度()。

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.

填空题在两级齿轮传动中,若传动比的分配方案是i1=i2,则其遵循的原则是()

单选题在两级齿轮传动中,若传动比的分配方案是i1=i2,则其遵循的原则是()A重量最轻B等效转动惯量最小C输出轴转角误差最小D加速度响应最快

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”

多选题在过程控制中,若TL=-3,Tu=3,u=1,σ=1,则(  )。ACsubP/sub=1 BCsubP/sub=2/3 CCsubPk/sub=1 DCsubPk/sub=2/3