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denis-greek [22]
3 years ago
14

Can a tank of oxygen gas ever be half empty?

Physics
2 answers:
alisha [4.7K]3 years ago
7 0
No. Any amount of gas always expands to completely fill
whatever container it's in.

So, as you take oxygen out of the tank, the pressure of what's
left in the tank certainly becomes less and less, but whatever is
left in the tank always expands and fills the whole tank. 

(And then, eventually, when the pressure inside the tank drops
to equal the pressure outside it (atmospheric pressure), you could
cut off the whole top of the tank and nothing more would come out.)

SCUBA divers regularly talk about having a 'full tank', and 'half a tank',
and maybe a 'quarter tank'.  But that refers to the amount of time that
the tank can still deliver air to them under water.  It's not the amount
of volume inside the tank that's full or empty.  The inside of the tank
is always full. 

Think about it this way:  ==>  There's no way to take half of the air
out of a balloon, leaving it half empty and still inflated. Doesn't happen.
labwork [276]3 years ago
6 0
Yes depending on the rate of oxygen intake. Also, the amount of pressure may limit the amount of air in a tank.
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3 years ago
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
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Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

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3 years ago
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