The moon lacks an atmosphere compared to the Earth. Hope this helps!

Assuming vertical acceleration of
, the speed after x seconds of falling is 
Answer:
The question is missing something it doesn't say how fast down its going and doesn't show the figure sorry for wasting an answer
Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R = 
Gauge Pressure inside, 
Blood Pressure outside, 
Now,
Change in pressure, 
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:


And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.