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Before we move on, I just wanted
to make sure that you

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understood that last point that
I made at the end of that

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last video.

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We said that the pressure
inputting into this, that we

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could view this cup with a hole
in it as essentially a

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pipe, where the opening on the
top of the cup is the input to

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the pipe, and this little
mini-hole is the output to the

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pipe, and we said that
this is a vacuum.

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Let's say this is vacuum
all around.

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I know when I drew it last time,
I closed it, but we have

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a vacuum everywhere.

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Since there's a vacuum
everywhere, the pressure at

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this point P1 is
equal to zero.

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The point I wanted to make is
because we have a hole here,

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the pressure at that point at
P2 is also equal to zero.

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You can almost view it as maybe
the atmospheric pressure

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at that point, but since
we're in a vacuum,

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that pressure is zero.

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That might have been a little
confusing to you, because you

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said, well, wait, I thought at
depth, if I had a point at

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that same height, that I would
actually have a pressure at

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that point of rho gh.

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That's true.

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That's completely true.

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You do have an innate pressure
in the liquid at that point of

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rho gh, and actually, that's
what's causing the

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liquid to come out.

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But that's actually taken care
of in the potential energy

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part of the equation.

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Let me rewrite Bernoulli's
equation.

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The input pressure plus rho g h1
plus rho V1 squared over 2

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is equal to the output pressure
plus rho g h2 plus

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PV2 squared over 2.

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I think you understand that this
term is pretty close to

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zero if the rate at which the
surface moves is very slow if

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this surface area is much
bigger than this hole.

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It's like if you poked a hole in
Hoover Dam, that whole lake

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is going to move down very, very
slowly, like 1 trillionth

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of the speed at which the
water's coming out at the

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other end, so you could
ignore this term.

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We also defined that the hole
was at zero, so the height of

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h2 is zero.

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It simplified down to the input
pressure, the pressure

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at the top of the pipe, or at
the left side of the pipe,

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plus rho gh1.

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This isn't potential energy,
but this was kind of the

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potential energy term when we
derived Bernoulli's equation,

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and that equals the output
pressure, or the pressure at

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the output of the hole, at the
right side of the hole, plus

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the kinetic energy PV2.

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It's the kinetic energy term,
because it doesn't actually

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doesn't add up completely to
kinetic energy, because we

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manipulated it.

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I just wanted to really
make the point that

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is definitely zero.

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I think that is clear
to you, because we

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have a vacuum up here.

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The pressure at that point is
zero, so we can ignore that.

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The question is what is
the pressure here?

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This pressure is zero, because
we have a vacuum here.

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If I were to say that the
pressure over here at this

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hole is equal to pgh, then I
would have the situation where

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pgh is equal to pgh plus
PV squared over 2.

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What does that mean?

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When I say that that pressure
at the output of the pipe is

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pgh, that means that
I'm applying some

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pressure into that hole.

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Essentially, that pressure I'm
applying into the hole is

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exactly just enough offset
to offset the

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pressure at this depth.

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Because of that, none of
the water will move.

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You could imagine that if this
is the hole, let's say that's

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the opening of the hole, and I
have some water particles, or

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some fluid particles, let's say
that these are the atoms,

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we're saying innately at any
point that there is a pressure

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at this point that's equal to
rho gh, but this is P2.

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How much pressure
am I exerting on

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this end of the hole?

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If I exert rho gh at this end,
then these molecules that were

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just about to exit the hole
aren't going to exit, because

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they're going to get the same
pressure from every direction.

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What we said in the last video,
and I really want to--

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because this is a subtle point--
is that the outside

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pressure, being on the outside
part of the hole, is zero, and

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because of that, we end up--
this term is zero, and we

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essentially end up with that
the change in the potential

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energy all becomes kinetic
energy, which is something

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we're familiar with from
just our kinematics

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and our energy equations.

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With that out of the way, let
me do another problem.

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Actually, I will do that next
problem in the next video,

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just so we have a clean
cut between videos.

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See you soon.

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