朗伯-比耳定律的数字表达式为lgI0/I=Kbc。(I0为入射光强度、I为透过光强度)
朗伯-比耳定律的数字表达式为lgI0/I=Kbc。(I0为入射光强度、I为透过光强度)
相关考题:
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 3=0 ThenSum=Sum+iEndIfNextPrint SumEnd Sub程序运行后,单击窗体,输出结果为A.12B.30C.24D.18
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
有如下程序: Private Sub Form_Click( ) Dim i As Integer, Sum As Integer Sum = 0 For i = 2 To 10 If i Mod 2=0 And i Mod 3<>0 Then Sum = Sum + i End If Next Print Sum End 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,窗体中的输出结果为______。 Private Sub Command1_ Click() Dim a As Integer, b As Integer For i = 1 To 6 a=i*i+i Next i Call writein(a,B)Print a, b End Sub Sub writein(a, ByValB)b = 1 For i = 1 To 6 a=b* 4 b = b + 1 Next i End SubA.24 6B.24 0C.12 6D.12 0
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
9.程序执行结果s的值是 [9] 。Private Sub Commandl_Click()i = 0Doi = i + ls = i + sLoop Until i = 4Print sEnd Sub
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(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”