A 2.65-kg block is tied to a string that is wrapped around the rim of a pulley of radius 6.72 cm. The block is released from rest. Assuming the pulley is a uniform disk with a mass of 0.540 kg, find the speed of the block after it has fallen through a height of 0.780 m.
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Ответ:
Δh_air=714m
Explanation:
Given data
Solution
ΔP=P₁-P₂
=(ΔhHg)×pHg×g
=(Δh_air)× p_air ×g
Then
Δh_air=(pHg+ΔhHg)÷p_air
Δh_air=714m