What happens when a body of cold air approaches a body of warm air
1 answer:
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Answer:
V1 =8.1 m/s
Explanation:
height at highest point (h2) = 4.1 m
height at lowest point (h1) = 0.8 m
acceleration due to gravity (g) = 9.8 m/s^{2}
from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore
mgh1 + = mgh2 +
where
- speed at highest point = V2
- speed at lowest point = V1
- mass of the girl and swing = m
- at the highest point, the speed is minimum (V1 = 0)
- at the lowest point the speed is maximum (V2 is the maximum speed)
- therefore the equation becomes mgh1 +
= mgh2
m(gh1 + ) = m(gh2)
gh1 + = gh2
V1 =
now we can substitute all required values into the equation above.
V1 =
V1 =
V1 =8.1 m/s