A bag of sugar weighs 3.00 lbs on earth. what would it weigh in newtons on the moon, where free-fall acceleration is one-sixth t
1 answer:
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Answer:
The initial velocity is 38.46 m/s.
Explanation:
The horizontal distance travel by the tennis ball = 13 m
The height at which the tennis ball dropped = 56 cm
Now calculate the initial speed of tennis ball.
The vertical velocity is zero.
Below is the calculation. Here, first convert centimetre into kilometre. So, height at which ball dropped is 0.56 km.
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( w₀)²
Kf = ½ w₀²
we look for the relationship between the kinetic energy
=
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision