2,3,-5
If you dcrease the temperature you will decrease the pressure proportionately. So, T1 over T2 will equal P1 over P2. We can derive the formula P1 x T2 = P2 x T1. Substitue the values and we get 50.0 mm x 200K = P2 x 540K 10,000 mmK = P2 x 540K 10,000mmK / 540K = P2 P2 = 18.52 mm of Mercury in a constant volume
60kpa
P(total) = P1 + P2 + P3
Gay-Lussac's Law states that the pressure of a sample of gas at constant volume, is directly proportional to its temperature in Kelvin. The P's represent pressure, while the T's represent temperature in Kelvin. P1 / T1 = constant After the change in pressure and temperature, P2 / T2 = constant Combine the two equations: P1 / T1 = P2 / T2 When any three of the four quantities in the equation are known, the fourth can be calculated. For example, we've known P1, T1 and P2, the T2 can be: T2 = P2 x T1 / P1
p1-p2 = Q8nL/ Pi R^4
A pointer can point to address of another pointer. consider the exampleint x=456, *p1, **p2;p1 = &x;p2 = &p1;Copyright Einstein College of EngineeringDepartment of Civil EngineeringTOPprintf("%d", *p1); will display value of x 456.printf("%d", *p2); will also display value of x 456. This is because p2 point p1, and p1 points x.Therefore p2 reads the value of x through pointer p1. Since one pointer is points towards anotherpointer it is called chain pointer. Chain pointer must be declared with ** as in **p2
You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.
Price elasticity demand formula end point formula epd= [q2-q1/q1]/[p2-p1/p1] midpoint formula epd= [q2-q1/(q2+q1)/2] / [p2-p1/(p2+p1)/2]
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
The composition of Buffer P2 is:200 mM NaOH1% SDS (w/v)Buffer P2 is the lysis buffer
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
A P2 costs $9.00 per user/month.
p2+10d+7
p2 + 3p = p (p + 3)
#include<stdio.h> #include<conio.h> #include<string.h> void main() { char str1[]="ravi kant yadav"; char str2[20],*p1,*p2; clrscr(); p1=str1+strlen(str1)-1; //Make p1 point to end of str1. p2=str2; while(p1>=str1) *p2++=*p1--; *p2='\0'; printf("Original string=%s. Reversed String=%s",str1,str2); getch(); }
p2 - 25 = (p - 5)(p + 5)
The only possible geometry of a diatomic molecule such as P2 is linear.