New answers tagged pressure
1
The pressure inside an empty inner tube is zero. It should be "easy" to fill it to one atmosphere.
I would recommend using very large rubber tubes so that you can collect a large volume of gas at low pressure, which you can then change to high pressure via a pump, if needed or desired.
1
An increase in pressure is not what causes condensation and rain.
Besides, the formula $pV = \text{const}$ applies to an isolated sample of a fixed amount of gas at a fixed temperature. Those conditions don't hold true for air in the atmosphere.
The real reason it rains is quite complicated, but the gist of it is that upward air currents can carry air with ...
2
Generally speaking low and high pressure areas are associated with vertical movement of the air. Air rises in a low pressure area and falls in a high pressure area. In a low pressure area the rising air cools and this is likely to condense water vapour and form clouds, and consequently rain. The opposite is true in a high pressure area, which is why high ...
0
Firstly, as Peter mentioned, your equations implicitly assume that the id fluinside is causing the acceleration. Otherwise you need to specify that the system is being pulled. I'll deal with both situations here.
If the water causes the acceleration
Unfortunately, we can't directly deal with the individual cylinder caps because the cylinder tube exerts a ...
0
See the cylinder as a whole object . The second law of motion gives that the cylinder does not accelerate untill it have a net external force.
So, the external pressures on both sides of the cylinder is not equal.The external agent which accelerates the cylinder applies force ($ma$) on left face towards right so that pressure on left fave exeeds by a ...
1
The other answers already treat decompression sickness, but in addition to the submarine scenario:
How long would one have to hold one's breath? A good diver can apparently maintain something around 0.5 m/s (plus some Buoyancy due to the air in their lungs), so even for a Civil submarine at around 30 m depth, they'd have to hold their breath for a minute. ...
1
The compressive effect on the body itself is not a problem because we are composed of incompressible fluids and solids. All gas filled cavities in a healthy person are connected by tubes to the outside so that the internal pressure can be equalized as external pressure increases.
A deep diver is affected by a variety of other problems that can be corrected ...
7
The pressure itself doesn't really matter as Mark pointed out. Neither does the bends if you stay at the same pressure - it is caused by gas coming out of solution and forming bubbles as you rise (and the pressure drops)
The real challenge at very deep deaths is the chemical effects of gases at high pressure.
Normal air scuba is limited at around 40m ...
3
A diver's body is basically made of water, and water is incompressible to a good approximation over the relevant pressure range. So the physiological effect of the water pressure itself will be negligible: I expect there is no problem with diving at 700m in soft suits in that sense.
However, in order for the diver to inflate her lungs, she must have air ...
6
The pressure inside a can of soda is around 2 to 3 atmospheres (above ambient i.e. 3 - 4 atmospheres total) depending on who you believe. The solubility of CO$_2$ increases with pressure, so when you open the can the pressure immediately drops by 2 - 3 atmospheres, i.e. by a factor of 3 to 4. As a result the solubility of the CO$_2$ decreases and the excess ...
12
The more interesting question is, "What does air feel like when it is moving away from me?" The answer is that there is really no sensation at all. You feel all the air coming into the vehicle, because it has a bulk momentum with respect to your frame of reference. However, air being sucked out of those same windows and doors is being pulled from a large ...
2
The answer to the valve-question,1 according Pilotfriend, seems to be the "floppy walled Eustachian tubes".
During ascent the gas (air) in the middle ear cavity expands and a small amount of pressure builds up against the ear drum causing them to bulge outwards ever so slightly (that ‘fullness’ you feel in your ears just before they ‘Pop’). This pressure ...
3
If the cylinder is stationary, it is according to the hydrostatic pressure equation:
$\vec{\nabla} p = \rho \vec{g}$
You can derive this equation by eliminating all velocities in the Navier-Stokes equations. If gravity is oriented in the negative $y$ direction, the equation becomes:
$\frac{dp}{dy} = -\rho g$
EDIT:
If you integrate this equation you get ...
4
There are several ways we can approach this, but I'll argue that the integral of the PV curve is a more general form of the force times distance concept of work:
$$ W = F \Delta x $$
This applies for pretty much any action over a distance. If you compress a spring, lift a box, drive a car, the above equation applies to formalize the work done. To ...
0
Yes, provided you wait an infinite length of time :)
First, forget the tube. It has no effect on the problem, except maybe to minimize splashing.
Assuming the rate of flow out of the top bucket is proportional to the pressure at the hole, you can write a 1-term linear differential equation for the amount of fluid in that bucket, for which the solution is ...
2
Yes, at least if you ignore the little droplets that leave the bottom of the upper bucket a little damp. There will also be some water in the tube, but it won't be much higher than the level of the water in the lower bucket. (There might be some capillary action that raises it a little, but probably not much.)
I suspect the reason you ask this is because ...
0
In this case, $P_2 > P_1$. Therefore, of you look at the side tube, through which gas is supposed to flow, there is a pressure gradient that will oppose the flow of the gas.
But there is one more flaw in the experiment. The liquid itself will rise through the side tube to a height that matches the level in the main vessel. In that case, there will be ...
2
Ordinarily, no. In fact, the liquid level would usually rise in the tube until it is at the same height as in the main tank, so your gas would be very far from the end of the tube. To see why, imagine the surface of the liquid directly above the the end of the tube. The pressure is P1. That pressure gets transmitted through the liquid down to the end of ...
0
Negative pressure is the same thing as positive pressure, simply acting in the opposite direction.
$$\vec{P}=\frac{1}{A}\vec{F}$$
Force is a vector, area is a scalar, $F=PA$ suggests that pressure is a vector. Negative pressures occur when the force is in the negative direction in some arbitrarily determined co-ordinate system (e.g. if decreasing the radius ...
0
The air in the bag will be compressed when a weight is placed on the bag. The change of air pressure in the bag will change according to the bulk modulus of the fluid (air in this case) in the bag.
The equation for bulk modulus is as follows:
$$B=-\frac{\Delta p}{\Delta V/V}$$
In this case, $B$ is the bulk modulus of air, $\Delta p$ is the change of the air ...
0
Well, it sounds like you're assuming that the lid was on, but suddenly got released. So the initial speed was zero. The interesting quantity is the initial acceleration.
Pressure equals force per unit area ($P = F/A$), so force equals pressure times area: $F = P \times A$. And force equals mass times acceleration, $F = m \times a$. We have $m \times a = ...
1
New answer
What you've done here is just dimensional analysis. But you've gone a little too far. In particular, just because two things have the same dimensions doesn't mean that they are equal. If you want to expand $F/A$ a little more, you can choose your favorite from
\begin{equation}
F = \frac{d p}{dt} = \frac{d}{dt}(m\, v) = m\, \frac{d}{dt} v = m\, ...
2
Your question in poorly defined because the concept of sound doesn't extend very nicely to non-atmospheric settings. Are gravity waves sounds? Are the pressure / shock waves in nebula? I don't think there is a unambiguously correct interpretation of sound for your question.
Regarding lethal sound here on Earth, the answer depends on what you consider ...
2
Well, we've classified a whole range of scales for the human hearing (which includes pure tone too). For lethal, we don't use how loud it should be, but instead - we say "how intense it should be" so that it can affect our ears. A quote from Wiki...
Loudness, a subjective measure, is often confused with objective measures of sound strength such as sound ...
1
Sound as we know it is a disturbance of our atmosphere, transmitted as a wave to our ears - and yes, it can absolutely be lethal - shockwaves can hurt people very badly, as anyone who's been to the scene of a large explosion can attest. We typically measure "loudness" on a log scale of the pressure of the sound wave - I admit I'm unsure of how much pressure ...
3
Short answer: No.
As you pump more and more air into the container, the pressure rises and rises. At some point, the molecules are so close to each other that instead of a gas, you get a liquid. If you continue even more, eventually you'll get a solid.
In this solid, atoms/molecules are arranged in a regular pattern with well defined distances between ...
-1
Why does a pierced compressed air can move? One answer is: when it is not pierced at 'the target point', there is a pressure acting on the target point (I should say area, but nevermind) on the right from the inside, as well as on its mirror point on the left, and so for every pair of points, so that the total force is zero. When you pierce the can, there ...
4
This answer is a bit of a long story, but I have split it up for the different statements for your convenience. Having thought about it a bit more after the discussion with @Mephisto I actually believe that Bernoulli's equation is not applicable in points B and C, because it is based on conservation of energy and therefore only applies if wall friction is ...
2
Warning: Another user has given a better answer. This one was chosen as the best one, before the other answer was written.
First, a simplified approach based on Bernouilli's equation for incompressible fluids:
Points B and C are directly in contact with the surface of the tube, thus they are nearly at zero speed with respect to it. But the fluid in A and D ...
0
Ok, thank you all for your comments. That's how I solved it, but I'm not quite sure if my reasoning is correct. Any feedback would be great!
Searching for materials on pressure vessels theory, I found this:
http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect05.d/IAST.Lect05.pdf
In the end, there's an example on inflating ballons where ...
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