I think that by "Classical physics" is meant low speed things. By low speed, I think is meant speed far below very roughly half the speed of light, so that Relativistic, special or general, effects can be ignored. Or at least it is hoped that they can be ignored.
Fire extinguishers and rockets get propelled by forcing out large amounts of material (gases under very high pressure) through a nozzle, and the RECOIL from that propels something forward. So, if the action is the ejection of material, the reaction (recoil) is the ejector moving along the same line in the other direction. And that's an example of Newton's third law.
Given a propulsion system, the magnitude of the force recoiling on the ejector will change the momentum of the ejector, often written as the equation F=ma where F is the force, m is the mass being accelerated, and a being the acceleration.
Just as something will stay still until it is moved - inertia - so once set in uniform motion in a straight line, the thing will continue in that motion, theoretically for ever or until something alters its momentum. Newton's first law is to the effect of "every body continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force". Which, I think, is where the concept of inertia stems from.
I think that the above mostly tcuches on the 3 laws.Any more help needed, please ask.
Running means your body must convert stored glucose into energy through glycolysis.
The famous Newton’s Third Law states that “For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.”
By using this,
10grams or 0.01kg of bullet with speed 400 m/sec and 5kg gun recoil with speed suppose ‘v’.
0.01×400=5×v
4/5=v
v=0.8m/sec ANSWER.
Answer:
Current in outer circle will be 15.826 A
Explanation:
We have given number of turns in inner coil 
Radius of inner circle 
Current in the inner circle 
Number of turns in outer circle 
Radius of outer circle 
We have to find the current in outer circle so that net magnetic field will zero
For net magnetic field current must be in opposite direction as in inner circle
We know that magnetic field is given due to circular coil is given by

For net magnetic field zero

So 

First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m