Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so
(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'
Solve for v₂' :
24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'
(1.0 kg) v₂' = 9 kg•m/s
v₂' = (9 kg•m/s) / (1.0 kg)
v₂' = + 9 m/s
which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.
Explanation:
They will have not much control over their speed or rotational energy. they will carry a lot of gravitation potential energy which will get converted to kinetic energy as they fall through the atmosphere. They will reach their terminal velocity, the fastest they can travel with earth's gravity before they pull their parachute, when they use a parachute to extend their surface area, increasing wind resistance. this allows them to land safely.
Explanation:
Given that,
The vertical motion of mass A is defined by the relation as :
At t = 1 s
x = 115.33 mm
(a) We know that,
Velocity, 
At t = 1 s
v = 18.94 mm/s
We know that,
Acceleration, 
At t = 1 s
(b) For maximum velocity, 
t = 45 seconds
For maximum acceleration, 
t = 61.8 seconds
Hence, this is the required solution.
PITCH
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Answer:
e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.
Explanation:
A free-fall is a fall just under force of gravity. The rocket;s upward motion is result of engine push - even if it was shut down - and rocket free of engine push effect when it reaches it's maximum height after shutting down of engine. Then rockets stops at it's maximum height for a moment and rtuens back as free fall with only force of gravitation pulling it back to ground with acceleration 'g'.
