I think the question should be the below:
<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:
<span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>
<span>x (max) = a(max) /ω² </span>
<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
Answer:
Tha ball- earth/floor system.
Explanation:
The force acting on the ball is the force of gravity when ignoring air resistance. At the moment the player releases the ball, until it reaches the top of its bounce, the small system for which the momentum is conserved is the ball- floor system. The balls exerts and equal and opposite force on the floor. <u>Here the ball hits the floor, because in any collision the momentum is conserved. Moment of the ball -floor system is conserved</u>. Mutual gravitation bring the ball and floor together in one system. As the ball moves downwards, the earth moves upwards, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound and separate.
It is the last number minus 3
A).
It would decrease because the speed of sound and temperature are proportional.
