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200 K is what degrees C?
200 K is equal to -73.15 degrees Celsius.
What is the formula of kelvin?
There are several formulas -Kelvin to Fahrenheit: [°F] = [K] × 1.8 − 459.67Fahrenheit to Kelvin: [K] = ([°F] + 459.67) × 5⁄9Celsius to Kelvin: [K] = [°C] + 273.15Kelvin to Celsius: [°C] = [K] − 273.15
Is the increase in 1 kelvin greater than 1 celsius?
No, it is not. it can be seen when looking at the conversion factor of Kelvin to Celsius, let K=kelvin and C=celsius, then the equation is K=273.15+C, the units are changed but the value of change is the same.
What is 10 degrees Celsius in Fahrenheit and Kelvin?
The conversion from degrees Celsius to Kelvin is: degrees C + 273.15 K Thus, 10 C equals 283.15 K (10 C + 273.15 K = 283.15 K) and 30 C equals 303.15 K (30 C + 273.15 K = 303.15 K). So the change in temperature when converted to the Kelvin system is 20 K (303.15 K - 283.15 K = 20 K).
How do you convert degees celsius to kelvins?
Conversion formulas are C = K - 273.15 K = C + 273.15 And it is quite handy to use some conversion tool, like this http://www.ghostcalc.com/temperature-converter.html It can convert K to C anf C to K, it also knows Fahrenheit and other measurment units.
C program for optimal merge pattern?
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }
How many C in a K?
1 °C = 1 K and 1 ℃ = 1.8 °F
How do you make a diamond using for loop in C plus plus?
Here is the code to do it: #include<stdio.h> main() { int n, c, k, space = 1; //Here we ask for the number of rows would be : printf("Enter number of rows\n"); scanf("%d",&n); space = n - 1; //This is the first half of the diamond for ( k = 1 ; k <= n ; k++ ) { for ( c = 1 ; c <= space ; c++ ) printf(" "); space--; for ( c = 1 ; c <= 2*k-1 ; c++) printf("*"); printf("\n"); } space = 1; //Here is the second half of the diamond for ( k = 1 ; k <= n - 1 ; k++ ) { for ( c = 1 ; c <= space; c++) printf(" "); space++; for ( c = 1 ; c <= 2*(n-k)-1 ; c++ ) printf("*"); printf("\n"); } return 0; } Hope that helped :)
Write a c programm for pyramid of given character?
#include<stdio.h> main() { int i,j,k,n; char c; printf("enter the # of rows of graphical output\n"); scanf("%d",&n); printf("enter the character you want to print\n"); scanf("%c",&c); for(i=1;i<=n;i++) { for (k=1;k<=(n-i);k++) { printf(" "); } for(j=0;j<i;j++) { printf("%c",c); printf(" "); } for(k=1;k<=(n-i-1);k++) { printf(" "); } printf("\n"); } getch(); }
What is the integral of 1 over c to the power of 1 minus x?
use this strategy: integral of (b^x) dx = (b^x)/ln(b) + K [K is integration constant, b is not a variable]rewrite (1/c)^(1-x) = ((1/c)^1)*((1/c)^(-x)) = (1/c)*(c^x). (1/c) is a constant, so bring outside the integral, then let b = c in the formula above, and you have (1/c)*(c^x)/ln(c) + K
Which atom has the larger atomic radius K or Li?
K has a larger atomic radius than Li. This is because atomic radius generally increases down a group in the periodic table, so potassium (K) being below lithium (Li) in Group 1 will have a larger atomic radius.
How many K are there in C?
1 degree Celsius equates to 274.15 kelvin. Conversion: [K] = [°C] + 273.15
Calculate the multiplier if the consumption function is 12 plus 0.4Y?
c = 12 + 0.4y Multiplier is represented by k Therefore k = 1/(1-MPC) MPC = b = 0.4 recall C = a + by Hence, k = 1/(1 - 0.4) K = 1.67
How many degrees celsius equals 1 kelvin?
Answer: 1 K = -272 ºC
Code of K'maps i n C plus plus?
#include<iostream> #include<string.h> //#include<graphics.h> //#include<dos.h> #include<math.h> using namespace std; int main() { // clrscr(); int c=1,c1=4,c2=1,c3=1,r[2000][10],c4=1,c5=1,p[2000][10],q[2000][10],o[2000][10],t[2000][10],s[2000][10],w[2000][10],i,l=1,j,k,m,u[2000][10],a[2000][10],v,e=0,g=1,f=1,h=0,b[2000][10],z[2000][10],y[6000][5],x[2000][10]; cout<<"Enter 0 for not selecting and 1 for selecting\n"; for(i=0;i<=15;i++) { cout<<"Enter the value for "<<i<<":"; cin>>v; if(v==1) { a[c][5]=i; m=i; while(c1!=0) { a[c][c1]=m%2; m=m/2; c1--; } c++; c1=4; } } for(i=1;i<c;i++) { for(j=1;j<c;j++) { for(k=1;k<=4;k++) { if(a[i][k]+a[j][k]==1) e++; } if(e==1) { for(k=1;k<=4;k++) { if(a[i][k]+a[j][k]!=1) b[f][g]=a[i][k]; else b[f][g]=3; g++; } b[f][5]=a[i][5]; b[f][6]=a[j][5]; g=1; f++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) x[c2][k]=a[i][k]; c2++; } h=0; } c=0; for(i=1;i<f;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(b[i][k]==b[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=6;k++) u[h+1][k]=b[i][k]; h++; } e=0; } for(i=1;i<=h;i++) { if(u[i][7]!=4) { for(j=1;j<=h;j++) { if(u[j][7]!=4) { if((u[i][6]==u[j][5])(u[i][6]==u[j][5])(u[i][5]==u[j][5])(u[i][5]==u[j][6])) { for(k=1;k<=h;k++) { if(u[j][6]==u[k][5]u[j][6]==u[k][6]u[j][5]==u[k][5]u[j][5]==u[k][6]) { for(e=1;e<=h;e++) { if(i!=j&&i!=k&&i!=e&&j!=k&&j!=e&&k!=e) { if(u[k][6]!=u[e][5]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][5]) u[j][7]=4; } } } } } } } } } cout<<endl; f=h+1; e=0;h=0;g=1;l=1;c=1; for(i=1;i<f;i++) { for(j=1;j<f;j++) { for(k=1;k<=4;k++) { if(u[i][k]+u[j][k]==1) e++; if((u[i][k]+u[j][k]==3)(u[i][k]+u[j][k]==4)) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(u[i][k]+u[j][k]!=1&&u[i][k]+u[j][k]!=6) z[l][g]=u[i][k]; else z[l][g]=3; g++; } g=1; l++; h++; } e=0; } if(h==0) { for(k=1;k<=7;k++) w[c3][k]=u[i][k]; c3++; } h=0; } c=0; for(i=1;i<l;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(z[i][k]==z[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) t[h+1][k]=z[i][k]; h++; } e=0; } l=h+1; e=0;h=0;g=1;m=1;c=1;c1=4; for(i=1;i<l;i++) { for(j=i+1;j<l;j++) { for(k=1;k<=4;k++) { if(t[i][k]+t[j][k]==1) e++; if(t[i][k]+t[j][k]==3t[i][k]+t[j][k]==4) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(t[i][k]+t[j][k]!=1&&t[i][k]+t[j][k]!=6) y[m][g]=t[i][k]; else y[m][g]=3; g++; } g=1; m++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) r[c4][k]=t[i][k]; c4++; } h=0; } c=0;h=0; for(i=1;i<m;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(y[i][k]==y[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) s[h+1][k]=y[i][k]; h++; } e=0; } f=h+1; e=0;h=0;g=1;l=1;c=1; for(i=1;i<f;i++) { for(j=1;j<f;j++) { for(k=1;k<=4;k++) { if(s[i][k]+s[j][k]==1) e++; if((s[i][k]+s[j][k]==3)(s[i][k]+s[j][k]==4)) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(s[i][k]+s[j][k]!=1&&s[i][k]+s[j][k]!=6) q[l][g]=s[i][k]; else q[l][g]=3; g++; } g=1; l++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) p[c5][k]=s[i][k]; c5++; } h=0; } c=0; for(i=1;i<l;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(q[i][k]==q[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) o[h+1][k]=q[i][k]; h++; } e=0; } for(i=1;i<=h;i++) { cout<<1; } cout<<endl; for(i=1;i<c2;i++) { if(x[i][1]==0) cout<<"A'"; if(x[i][1]==1) cout<<"A"; if(x[i][2]==0) cout<<"B'"; if(x[i][2]==1) cout<<"B"; if(x[i][3]==0) cout<<"C'"; if(x[i][3]==1) cout<<"C"; if(x[i][4]==0) cout<<"D'"; if(x[i][4]==1) cout<<"D"; cout<<"+"; } c=0; for(i=1;i<c3;i++) { if(w[i][7]!=4) { if(w[i][1]==0) cout<<"A'"; if(w[i][1]==1) cout<<"A"; if(w[i][2]==0) cout<<"B'"; if(w[i][2]==1) cout<<"B"; if(w[i][3]==0) cout<<"C'"; if(w[i][3]==1) cout<<"C"; if(w[i][4]==0) cout<<"D'"; if(w[i][4]==1) cout<<"D"; cout<<"+"; } } cout<<endl; c=0; for(i=1;i<c4;i++) { for(j=1;j<=h;j++) { if(((o[j][1]-r[i][1])==2(o[j][1]-r[i][1])==3(o[j][1]-r[i][1])==0)&&((o[j][2]-r[i][2])==2(o[j][2]-r[i][2])==3(o[j][2]-r[i][2])==0)&&((o[j][3]-r[i][3])==2(o[j][3]-r[i][3])==0(o[j][3]-r[i][3])==3)&&((o[j][4]-r[i][4])==2(o[j][4]-r[i][4])==0(o[j][4]-r[i][4])==3)) c++; } for(j=1;j<c5;j++) { if(((p[j][1]-r[i][1])==2(p[j][1]-r[i][1])==3(p[j][1]-r[i][1])==0)&&((p[j][2]-r[i][2])==2(p[j][2]-r[i][2])==3(p[j][2]-r[i][2])==0)&&((p[j][3]-r[i][3])==2(p[j][3]-r[i][3])==0(p[j][3]-r[i][3])==3)&&((p[j][4]-r[i][4])==2(p[j][4]-r[i][4])==0(p[j][4]-r[i][4])==3)) c++; } if(c==0) { if(r[i][1]==0) cout<<"A'"; if(r[i][1]==1) cout<<"A"; if(r[i][2]==0) cout<<"B'"; if(r[i][2]==1) cout<<"B"; if(r[i][3]==0) cout<<"C'"; if(r[i][3]==1) cout<<"C"; if(r[i][4]==0) cout<<"D'"; if(r[i][4]==1) cout<<"D"; cout<<"+"; } c=0; } cout<<endl; for(i=1;i<c5;i++) { if(p[i][1]==0) cout<<"A'"; if(p[i][1]==1) cout<<"A"; if(p[i][2]==0) cout<<"B'"; if(p[i][2]==1) cout<<"B"; if(p[i][3]==0) cout<<"C'"; if(p[i][3]==1) cout<<"C"; if(p[i][4]==0) cout<<"D'"; if(p[i][4]==1) cout<<"D"; cout<<"+"; } // getch (); }
Write a c program to print all combination of a 4 digits number?
#include<stdio.h> #include<conio.h> void main() { int i,j,k,l,m; clrscr(); for(i=0;i<4;i++) { for(j=i+1;j<=4;j++) { printf("%d",j); } for(k=1;k<=i;k++) { printf("%d",k); } printf("\n"); for(k=i;k>=1;k--) { printf("%d",k); } for(j=4;j>=i+1;j--) { printf("%d",j); } getch(); }
Gaussian elimination in c?
#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
