Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
D) A gallon of gasoline definitely. Gas is like, if not a chemical. It will have the most.
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Explanation:
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Answer:
the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s
Explanation:
Let's use the equation that relates speed with acceleration:
vf = vi + a * t
where vf stands for final velocity, vi stands for initial velocity, a for acceleration, and t for the time acceleration is applied. Then, in our case we have:
vf = 0 + 6.5 (3)
vf = 19.5 m/s
Answer:
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Explanation:
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