Answer:
newtons 3rd law of motion
Explanation:
Answer:
529.2 N
Explanation:
As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.
The force that is acting on the student will be due to gravitational force, that is equal to his weight.
F=mg: 54kg x 9.8m/s^2 =529.2 N
So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.
This is because the student is sitting in a constant state and all the weight is exerted on the stool.
Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N. However, since the question asked normal force, it would be weight of student in general terms.
Hope it helps!
Answer:
class sum (
public static void sumofvalue (int m, int n, int p)
{
System.out.println(m);
System.out.println(n);
System.out.println(p);
int SumValue=m+n+p;
System.out.println("Average="+Sumvalue/3);
}
)
Public class XYZ
(
public static void main(String [] args)
{
sum ob=new sum();
int X=3;
int X=4;
int X=5;
ob.sumofvalue(X,Y,Z);
int X=7;
int X=8;
int X=10;
ob.sumofvalue(X,Y,Z);
}
)
Explanation:
The above program is made in Java, in which first we have printed value in a separate line. After that, the average value of those three values has been printed according to the question.
The processing of the program is given below in detail
* The first one class named 'sum' has been created which contains the function to print individual value and the average of those three values.
* In seconds main class named 'XYZ', the object of that the above class had been created which call the method of the above class to perform functions.
* In the main class values are assigned to variables X, Y, Z.
Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
