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11

The biggest, immediate problem with "openning the door" of a spacecraft is not that you would die immediately from exposure to the vacuum of space: you don't - you have of the order of minutes to do something about it. The problem is the violent outrush of air. User rob offers this answer to the Physics SE question Do airlocks in space decompress violently ...


6

if the pressure is the same, there in no net force on the piston, so it will remain at rest, so neither gas will expand, and this will stay this way because there is no heat exchange that could change the pressure on either side. Ask your teacher to return his diploma.


5

people in spaceships opening doors and closing them again with no suits on. Is it possible in "real life" No, it is not. Any sane engineer will build doors that open inward, or have latches that over-center when closed so it is simply impossible to open an airlock in a pressurized vessel. An aircraft, for example, has about 6-8 tons of pressure holding ...


3

I will first elaborate what wind-gusts are not, for mental clarification: Thermodynamics as you may have encountered it in a physics course, is a static theory, as opposed to dynamic theories. More specificly that what is taught at university is usually equilibrium thermodynamics - You calculate the equilibrium state of a system where it will end up, but ...


3

Yes, there is a force, because the air pressure inside the tire is larger than the pressure outside, so air will try to leave from any hole it finds, however small. It might be through some micro cracks on the tire or, more often, through leaks from the air valve.


3

A reasonable sub woofer at sound power level of 130 dB would produce pressure fluctuations of 60 Pa. Compare this to the ambient pressure of 100'000 Pa and you will see that related temperature fluctuations would be negligible. It extinguishes fire because it pushes the air back and forth. For the small fire in this video you could take a small air blower ...


2

I thnk the answer to this lies in 'Chaos Theory', which can also be described as non-linear dynamics. To give an example of a simple linear system - if you want to travel from the West Coast of the USA to the East Coast of the USA by car travelling at an average speed of 60 miles per hour for 12 hours driving a day you can make a prediction of how long the ...


2

A gas (like air) will tend to diffuse through any material, and the rate of diffusion will be roughly proportional to the pressure difference between the inside and the outside. It so happens that it tends to diffuse through the rubber of ordinary bike inner tubes fairly rapidly. In addition, the inner tube wall is quite thin, and the thinner the material ...


2

You could probably get a negative pressure in polymer physics, so you could view a big block of rubber as behaving this way. Basically: negative pressures happen when an increase in volume causes a decrease in entropy. Polymers might be a good example because you have these molecules which "want" to be tangled up and kinked ("want" in the sense of "it is ...


2

Pressure is the (outwardly directed) force normal to any area. This definition most naturally fits hydrostatic pressure, e.g. in gases and liquids. In ideal media, this kind of pressure is never negative. In real media, that is not necessarily true. The most obvious example occurs at the boundary of just about any liquid: There a negative pressure acts on ...


2

The force exerted by an airflow is given to a good first approximation by the classic drag equation: Force = 0.5 x Rho x Cd x A x V^2 Rho = gas density Cd = drag coefficient wrt flat plate drag ( 0 < Cd <= 1) A = projected area relative to flow V = velocity Mainly opinion: I am unaware of the particle size of Martian dust, but this information will ...


2

Your teacher is wrong... in a way. What he probably meant was same mass, different temperture.


2

I actually added this to the answer to your other question but will repeat it here... If the balloon is bigger, the time that the toy can hover will increase - by a surprisingly large amount. Using the result from my other answer that pressure (and thus flow rate) scales with $1/r^2$, and volume scales with $r^3$, then time (which is the time it takes for ...


1

The ideal gas law states $PV=nRT$. If you change $n$ (the amount of air) or $T$ (the temperature), you will change the product $PV$. So the answer is "no".


1

The exact answer to your question can be found at http://hyperphysics.phy-astr.gsu.edu/hbase/ppois2.html#tub There it shows that the flow rate goes as the 4th power of the radius of a pipe - so if you double the diameter, air will flow 16x faster. The reason behind this is something I described in some detail in this earlier answer.


1

Bad Things Happen The air, as it has no pressure or enough gravity to keep it in the ship, will attempt to expand. Air, in fact, attempts to expand to fill the container it is placed in. If there is no walls to the container, like on a planet, it will only be stopped by gravity. When the airlock is unsafely open or a hole is made in a spaceship, the air ...


1

I will focus on just a little bit of one of your questions - the relationship between compressibility, density and pressure - and per my comment, recommend that you narrow down the scope of your question. As you know, in a gas we experience "pressure" because molecules hit the walls of the containing vessel. When I double the number of molecules in the same ...


1

You would have quite a problem to keep your water liquid. Normally, the water will evaporate when pumping. So you should go to low temperatures, but it freezes there. I thin you should thing about some other material to make bubbles in vacuum.


1

Yes. When you do stress testing of materials (for example, the Brazilian test of a disk shaped test object) you apply stress along a single axis (using for example an Instron machine). This is a good way to measure elastic properties of materials. On the other hand if you have a pressurized container (for example the hydraulics in your car brake system), ...


1

The formula you cite refers to the work done when the change of volume is made at constant pressure, which is not the case here. Let's say I start with the balloon in the air with volume $V_0$ at pressure $p_0$. The state I want to end up with is the balloon underwater at a depth with pressure $p_1$ and the balloon having volume $V_1$. I cannot directly ...


1

It doesn't. They'd have to exchange heat through the piston.


1

Since the pressure is dependent on depth, and your depth will vary if the structure (like a dam) is vertical, you will need to take the derivative of a vertical slice of the structure, and from there you will be able to multiply it by the horizontal (XY plane) area. your dx will be the change in depth making up for the "height" of the area for that ...


1

The first equation you have just comes from strength of materials. It's equivalent to $\epsilon=\frac{\sigma}{Y}$, where $\epsilon$ is strain which is $\frac{\Delta l}{l_0}$. This is valid as long as the material is in the range of small deformations. The $C_0$ and $l_0$ should be the values at equilibrium with no additional external pressure. The reason ...


1

It has to be the same, for a still pond of water, as the pressure of the air immediately above it, i.e. approximately $100{\rm kPa}$. To think carefully about this, you would think of a layered fluid (in this case a column of air above a column of water) and calculate the compressive stress at each position along the column. Imagine a balloon of air in ...



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