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If you're talking about Temperature, then 1C is warmer. 1K is only one Kelvin above absolute 0, while 1C is one degree Celsius above the freezing point of water. One Kelvin and One degree Celsius are equal in value.
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200 K is what degrees C?
200 K is -73.15 deg C.
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 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
What is 15.25 degrees K expressed in C?
Zero K = - 273.15 C15.25 K = - 273.15 + 15.25 = - 257.9 C
What is -90.0 f to -5.0 c in Kelvin?
-85
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
What is larger km or cm?
The prefix "k" (kilo) means 1000. The prefix "c" (centi) means 1/100.
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
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 (); }
Which atom has the larger atomic radius K or Li?
K (potassium) has the larger atomic radius.
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(); }
