Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, 
Initial angular velocity of the platform, 
Part A,
Let 
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :
Here, 
Solving the above equation, we get the value as :
Part B,
The initial rotational kinetic energy is given by :
The final rotational kinetic energy is given by :
Hence, this is the required solution.
Answer:
Weather is the conditions (temperature, wind etc.) at a given time, like on that day. Climate, which is what his data would show, is the conditions over an extended period of time like the 3 months he collected data
