This is an example of an elastic collision. The two objects collide and return to their original shapes and move separately. In such a collision, kinetic energy is conserved. I think we can agree that this represents Newton's third law by demonstrating conservation of momentum.
Answer:
The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Explanation:
Given;
weight of the man, W = 700 N
Weight of the woman, W = 440 N
momentum is given by;

Kinetic energy of the man;

Momentum of the man is calculated as;

The kinetic energy of the woman is given by;

The momentum of the woman is given;

Since, momentum of the man = momentum of the woman


mass of the mas = 700 / 9.8 = 71.429
mass of the woman is = 440 / 9.8 = 44.898

Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Answer:
An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes.
Explanation:
The transit method requires watching the light output of a star over long periods of time. A transit occurs when the planet crosses in front of its star from earths point of view. Since there is a small object (the planet) now blocking some of the star, it appears to dim a little bit for a while until the planet passes. If we are in a position where that occurs regularly (most paths of planets do not happen to be on the line of sight between earth and their star) we can deduce the period of orbit. From the amount of dimming and the period you can estimate the mass
Explanation:
It is given that,
Mass of the rim of wheel, m₁ = 7 kg
Mass of one spoke, m₂ = 1.2 kg
Diameter of the wagon, d = 0.5 m
Radius of the wagon, r = 0.25 m
Let I is the the moment of inertia of the wagon wheel for rotation about its axis.
We know that the moment of inertia of the ring is given by :


The moment of inertia of the rod about one end is given by :

l = r


For 6 spokes, 
So, the net moment of inertia of the wagon is :


So, the moment of inertia of the wagon wheel for rotation about its axis is
. Hence, this is the required solution.