Question

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一个香水喷洒器包含一个伸到香水里的管子。当空气被吹过管子顶部时，香水上升并被喷洒出去。（见图). 假设该管子总长为6.0cm,其中5.0cm在液面以上（香水密度为900km/m^3). 空气的密度为1.3kg/m^3.当流动的空气打倒何速度时可以使香水被吸至管子顶部？忽略因管子细小而产生的阻力。

以下为此题原题：
A perfume dispenser consists of a tube that idps into the liquid. Air blown acrossthe top of the tube cuases the perfume to rise in the tube and be dispersed out the nozzle (see fugure). THe device is called an aspirator. The tube is 6.0cm long with its top 5.0cm above the level of the liquid perfume (of density 900 kg/m^3)/ The density of air is 1.30kg/m^3. What is the air speed needed to raise the perfumeto the top of the tube? Neglect capillary action.

Answer 1

(1)香水恰好吸至顶部，此过程可以考虑为只有势能的变化：E1=ρ水×g×h
(2)空气以损失其动能使香水吸至顶部，恰好提升香水，因求最小速度，则可理解为动能完全转化：E2=0.5×ρ空×V^2
(3)系统能量守恒：E1=E2
即：ρ水×g×h=0.5×ρ空×V^2
V^2=2×ρ水×g×h÷ρ空
=2×1000kg/m^3×10N/kg×0.005m÷1.3kg/m^3
=76.9m^2/s^2
故：V=8.77m/s

Answer 2

SB！！！