# Tag Info

24

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...

5

The front of the train compresses air which can blow you away, while at the back of the train air rushes back in after the train has displaced it. This backdraft is especially troublesome in closed areas such as subways, where a train exits a small tunnel near a platform and the displaced air rushes back into the vacated tunnel. Next time you see a big truck ...

5

As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ...

5

You need to take density of the air into the question. And weight of the balloon itself. The Archimedes' law says: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. Therefore a balloon can support the same weight as equal volume of air has (reducing gravitational ...

4

No, there is no contribution to the pressure from the gravitational attraction between the particles. To see this you need to appreciate that the pressure is an ensemble property, and look at the stress-energy tensor for a single point particle. This is: $$T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta \delta\left( x - x_p(t) \right)$$ where $v$ ...

3

Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$. Suppose the pressure is too low, such that ...

3

Yes you are right. The 10kg piston acts as a force over the area of the piston, increasing the pressure and decreasing the volume of the gas inside. When the set up is tilted, the force no longer acts on the gas, but sideways, so the pressure equalizes.

3

Pressure is a scalar and does not have a direction. This is discussed in some detail in the answers to Define Pressure at A point. Why is it a Scalar?, though this might be a bit technical. When you measure a pressure you are actually measuring the force applied to a surface. For some small bit of surface $\delta {\bf A}$, the force produced on that surface ...

3

I would say pressure is better defined by $$\vec{F} = P \vec{A}.$$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ...

3

The net upward force is, according to Wiki buoancy: $$F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-m_\mathrm{balloon} \cdot g$$ For helium, the $m_\mathrm{balloon}=\rho_\mathrm{helium}V_\mathrm{disp} + m_{shell}$, thus F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-\left(\rho_\mathrm{helium}V_\mathrm{disp} + m_{shell} \right)\cdot ...

2

Your approach is pretty much correct. First you calculate the buoyancy generated by 1L of Helium. This can be calculated by the difference in the mass of 1L of Helium and 1L of air, which you can look up on the internet. The answer is ~0.001kg, as you got. Next you estimate the volume of a balloon. A sensible assumption, followed by some questionable ...

2

I believe the confusion is that you believe pressure will always increase as temperature increases. This is only the case in a closed environment such as inside the tire. In an environment such as the atmosphere which is, essentially, in an unconfined environment, the density will decrease with temperature as well. This does not happen inside of a closed ...

2

The point where this pressure law brakes down is when the surface tension of the water starts to play a leading role. You can try this yourself with a narrow tube (I would say make it considerably smaller in inner diameter than a sixteenth of an inch). If you submerge the tube in water and you pull it out, the molecular forces between the water molecules and ...

2

Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation: $dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$ If you insulate your air cylinder well enough, $\delta Q = 0$. Assuming that your air cylinder does not deform, $\delta W = 0$. Since you are filling your ...

2

Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? That's not exactly true. Deviations ...

2

Sorry for the enormous delay, I was caught with more work than I thought, then, here it is. @DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system. My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the ...

2

Like any object moving through a fluid, a high-speed train distorts the air as it moves through it. Broadly speaking, there are three main regions of flow structure around a high-speed train: the upstream distortion, boundary layer and wake. These can be collectively referred to as the slipstream. The effects of the slipstream on a static observer (e.g. a ...

2

The rotation is part of the key to the storm itself. Primarily the pressure and temperature differences are what causes these systems to take the shape and forms that they do. Once a tropical depression starts to form you can already see rotation in the moisture around the low pressure zone, even through it typically looks nothing like a hurricane. Not ...

1

How do slight changes in these properties result in a large change in pressure, microscopically? Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the ...

1

The ideal shape for the water to just 'hang' is actually the cube that you've proposed. In this case all of the forces on the water are uniform across the interface. For one region to slip down, another needs to move up (as you've indicated in your diagram). But since every location is experiencing the same forces, there's no reason any spot to start ...

1

What you observed is known as the Hydrostatic Paradox. You are confused that the force exerted onto the tank wall corresponds to the weight that is larger than the weight of water contained in the tank. You are probably familiar with levers: you can put a small weight onto the long arm of the lever and thereby exert a large force at the short arm. The gain ...

1

Since you know the force on the bottle (roughly 200N), you would have to get an approximate area over which this force is distributed. You could try by spreading some ink on either the bottle or the weight to estimate the contact area. While this is not completely correct (the wall of the bottle does redistribute the pressure on the outside to a larger area ...

1

Yes, area is a vector, which is the normal to the surface. ($\vec{A}=A\vec{n}$) $\vec{F} = -P\vec{A}$ In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal ...

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