# Tag Info

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Here is a bit of basic physics. If you drop a mass $m$ with area $A$ from a height $h$ onto snow, and it penetrates the snow to a depth $d$, then the average pressure on the snow during the fall is calculated as follows: Total distance dropped: $D = h+d$. Total gravitational energy: $E = mg(h+d)$ Retarding force $F$ acted over distance $d$ to do the ...

0

Your roof won't rise in this case. Your roof will simply crush. If tornado wants to lift your roof, its outer air would exert an equal downward force on ground. If its of exact diameter that force would be exerted on your walls that support roof and crush down your roof.

0

Your force will transmit through the cube to the bottom and it will spread like a projection from your pressure point. having sand underneath it will give away a pression mark the size of your cube showing how your force transmitted. the height will no matter, being a solid cube and your force will not deformed it and lose itself in the process.

3

Like tides on the ocean, it is a consequence of the tidal forces (if the claim is correct at all which I am verifying now). Tidal forces result from inhomogeneities of the gravitational acceleration. If we approximated the external gravitational acceleration $\vec g$ caused by the Moon or the Sun to be constant all over the Earth and its vicinity, the Earth ...

3

Due to the cylindrycal shape of the lamp, pressure forces act compresing the glass. Since the compressive strength of glass is quite high, $\approx 10\,000$ bar, atmospheric pressure is not enough to break it.

2

Well, let's look at the equations for a flow initially at rest: $$\frac{\partial \rho u}{\partial t} = -\frac{\partial p}{\partial x}$$ So, if we assume density is constant in the flow, you get: $$\frac{\partial u}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x}$$ and so you can see that in both your cases, the gradient in pressure is the ...

1

You are right to be skeptical of your answer. The technical specs show you should get something on the order of 100 m$^2$ . Your problem lies in basic trig: the force of the pressure is normal to the surface, and the surface is tilted by 5° - this means that the vertical component of force will be $\cos 5°$ times the normal force - you used the $\sin$ which ...

0

The downward force exerted by a column of water due to the gravitational attraction of the Earth is called the weight of the column of water. If the column is of unit cross sectional area the mass of the column is height X density and its weight is mass X gravitational field strength (g) = height X density X g and because this weight is acting on unit area, ...

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$P = h \rho g$ ($\rho$ is density) is an approximation that can only be used when both $\rho$ and $g$ are constant over the height change $h$. The accurate relationship is proven in most standard texts on the subject and indeed on the wikipedia page for hydrostatic equilbrium, by considering the equilibrium of a thin slab of thickness $\Delta h$ and area ...

1

The pressure of the tires would decrease because when the car is lifted, the force on the tires from the mass of the car decrease, causing the area of the tyre to increase. Since the formula for pressure is PRESSURE=FORCE/AREA, the bigger the area of the tyre, the less the pressure of the tyre.

1

Well, I don't know if this is as specific as you asking, but the pressure in the tire has to balance the external forces acting on the tire. In this situation, there are two to consider: Air pressure pushing in on the tire and the amount the earth pushes up on the tire (3rd law equivalent of the weight of the car). When the car is up on a lift, the only ...

1

Air beds and water mattresses are similar examples. The person standing on the hose will need a force of about 700 N to be exert upwards. This force will be transmitted over the contact area between the person's soles and the hose. If you know that area you can equate this extra pressure exerted by the soles to the extra pressure that the water in the ...

2

This is not limited to hydrofoil hulls. You can sail across the wind ("reaching") on other low-drag shapes such as windsurfers at speeds greater than the windspeed. The fastest such craft, so far as I've heard, are iceboats -- which are sort of an ultimate "hydrofoil," in that they skate along the frozen lake surface with almost no friction at all. ...

2

Just look at it as a wedge. To a 0th approximation, a sail or wing through fluid is just like a knife through jello. It can only move along a line. A sailboat has two of those. It has a sail cutting through the air, and it has a keel (centerboard, leeboard, hull) cutting through the water. Neither one of those can move sideways in its fluid, so if the air ...

1

Think of it like this: Assume that you want to make pasta. For a piece of pasta to be "cooked" a certain amount of energy has to be given to it (depends on the type of pasta) (also as anna v mentioned It is the energy that changes the molecular bonds that results in cooking and the higher the temperature the faster). The way to transfer this energy to it is ...

1

First consider a pot with a loose lid over it. As the temperature rises, the vapor pressure of the steam rises as well (even before boiling) and the steam will push its way out under the lid. You'll get cooking temperatures over 100C, but not by much, and a lot of the energy you expend on cooking just goes out into the air instead of into your food. And to ...

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Also, though this might not be an actual use for this, you could make steak very rare with it. In the case of a very rare steak, you need it to be cooked just enough to make it safe, kill bacteria, and make it easier to digest, but no more than that. If you put it high pressure, you could heat it up to a high temperature without cooking it. That way, you ...

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Let me start off with something that should be obvious but I feel the huge need to say it -- I am not trained in the biomedical field and what I put in my answer should not be construed as definitive and ready-for-use in actual patients. I will do my best and present what I think is correct, but I am not a doctor and don't even pretend to play one on TV. ...

2

An egg has to reach an inner temperature of 100C to cook and in water the egg shell is kept at 100C for five minutes in a heat bath. Thus your question is answered by "can the egg shell be heated to 100C by the much hotter thin gas in the thermosphere" In vacuum the egg will radiate away and go close to 0 kelvin, so in the thermosphere it will be a fight ...

1

EDIT : The egg I am considering doesn't have a shell Temperature is a measure of the average kinetic energy of particles in a gas. In the thermosphere, the highly energetic solar radiation collide with the (having very less pressure) air, and are thus given very high velocities and so have a high temperature. However, the temperature shouldn't play a big ...

1

For the question, can a soap bubble exist in a vacuum? Your answer places certain assumptions on the soap bubble. Asump 1: There is pressure inside the bubble. Asump 2: There is no gravity. Asump 3: The structure of the bubble has the strength to exert force against the pressure inside the bubble. 1 A total vacuum exerts the same pressure inside the ...

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i think ,heat energy is required to change the temperature of the gas as well as for the the expansion of the gas against external pressure hence,change in temperature at constant pressure ,therefore Cp>Cv

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The answer you are looking for is that $$\rho \propto 1/V$$ and $$m/\rho = 1/V$$ with this knowledge you should be able to see how your final equation relates to $$pV = nRT$$

-1

This is a side view in a high pressure area and low pressure area: The high pressure air flows outward from its center (moving from higher pressure to lower pressure) and to the ground, whereas the low pressure air flows towards its center (from low pressure air to even lower pressure air). The low pressure air forms clouds when it flows towards its ...

2

There are three points to be noticed: If you just blow without closing the lips, you would change the boundary condition. The trumpet waveguide is not "nicely predictible", the approximation of an open tube does not work cause the bore variations $S(x)$. You need to solve this kind of beasts for reasonable 1D propagating pressure approximation:  ...

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I agree with don_Gunner94's answer. If the fluid come out from the constricted passage to atmosphere,it will experience atmospheric pressure,which is same as the pressure acting at top of the container. Even according to Bernoulli's principle, Static pressure + Dynamic Pressure = Constant Therefore, the pressure acting on the fluid when it is inside the ...

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Apparently this is only an empirical formula, obtained from measurements of sound velocity versus porosity in a single type of material, published in: Krief et al, A Petrophysical Interpretation Using The Velocities Of P And S Waves (full-waveform Sonic) (1990).

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I suppose that the best way to answer this question is by using an analogy. Take a glass of water(at room temperature) and place it in a refrigerator. What happens to the water? It cools down. Now you take this glass of cold water and keep it back outside. What happens now? The temperature of the water comes back to the room temperature. Why? Because the ...

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This is not really a hard question...it's just that the proper assumptions haven't been enumerated! You have two tires. Purchased from the same lot, identical in every respect that matters. First Assumption: Tires are mounted identically, on identical rims, prepared identically, with identical UNLOADED inflation pressure. Second Assumption: The tires are ...

2

There are two factors at play: the first is surface tension, which tends to generalize the action of forces across an area, instead of those forces localizing at one infinitismal point. Second, atmospheric pressure is still in effect. As gravity pulls on the liquid in the pipette, pressure in the bulb diminishes, so external atmospheric pressure pushes ...

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If the pipette is long and thin, then the force of the liquid exerted by gravity is not enough to overcome the force exerted by the air as the liquid tries to fight its way out. Because of the short radius of the pipette, the fluid would have to travel very far to get out of the tube. If the pipette were short and fat, then the liquid has a much shorter ...

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(well as far as i could understand your question) according to the law of continuity which says $A\times v = k$ where k is a constant, $A$ is the area of cross section of the pipe through which the fluid is flowing and the $v$ velocity of the fluid through that cross section. so let is consider two holes, hole 1 (h1) and hole 2 (h2) of $A$ as $A_1$ and ...

1

If the fluid is incompressible and/or the pipe is flowing at steady state, then the mass inflow will equal the mass flow out.

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Conservation of energy works pretty well. Assume that instead of a squirting fountain and a paddle wheel, you had a cylinder and a piston to allow the compressed water to do work. The work done would be $\int PdV$. At the end of the expansion the water in the lower container is at atmospheric pressure. Now we can either repressurize by opening the red ...

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I'm guessing that your problem lies in your determination of the force exerted on the slanted walls. Pressure always acts perpendicular to a surface, so this perpendicular pressure force must be resolved into horizontal and vertical components. You can then integrate that over the slanted walls (taking into account the hydrostatic changes in pressure with ...

0

Yes! The explanation is very simple . Frm first law of motion if the net force wasn't zero either the fluid or bucket would move and accelerate.

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You will not be able to produce any useful energy with this system. The previous answers have picked up on a number of the issues... Water is nearly incompressible, so the gain in pressure that you are counting on is too small to do any significant work. Don't forget that the work required to fill a 1km tall pipe with water is part of the system. The ...

0

In order to send it back to the container you have to do work against gravity. The energy obtained by rotating wheels would be from the kinetic energy of the fluid which came into existence because of the work we did. The energy from wheel would equal the work we did to pour the fluid back in. So energy gain would be 0. Moreover the efficiency of the wheel ...

2

There are two possibilities: water (or whatever fluid you're using) is compressible or it's incompressible. If it is incompressible, when the red valve is closed the tank pressure drops to zero, and opening the yellow valve produces no effect on the generator. If it is compressible, its pressure remains constant when the red valve is closed, and it ...

3

Some energy would come from the part where "the water would be collected and sent back into the container", which you somewhat glossed over.

-1

According to my understanding the balloon pops because its "surface tension is disrupted. The particles in the balloon experience some mutual attraction holding them together. Nonetheless the rubber stores potential elastic energy. When the needle is poked into the balloon the tension is disrupted and the rubber can tear resulting in the pop.

1

How about neither of them? The balloon pops because of surface tension of the balloon! Surface tension is what keeps the balloon in shape. Now when you puncture the balloon with a needle, the force along the edge will be imbalanced and the tension will pull the surface without anything cancelling it. Because of the rubbery nature of the latex, it will ...

2

The 'balloon in a jar' demonstration just illustrates the principle of lung ventilation, but really shouldn't accepted too closely in modeling the dynamics of breathing. In a real lung the only gas-filled space is the space that communicates with the upper ways, and terminates at the other extreme with sac-like structures called alveoli. The space outside ...

0

I think the referenced experiment is interesting but not an exact model of the lungs. But in the "close enough" realm, what's happening is that the movement of the diaphragm (and the ribcage) happens as various muscles act against the external air pressure. Once there's additional volume inside (and assuming no collapsed lung, the interior of your lungs, ...

0

Since the water is accelerating, the pressure distribution within the water must change to accommodate Newton's 2nd law locally. The net result of this is that the vertical pressure gradient becomes $\rho (g+a)$ rather than $\rho g$. It is as if gravity had increased.

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If you do a force balance on a small element of surface on a balloon or drop (say, "window shaped"), there is curvature of the surface, and the surface tension forces acting at the edge of the element (or the elastic forces in the case of the balloon) have a net resultant in the radial direction. This is why the pressure inside is higher than the pressure ...

2

From an engineering POV you could use sound, either in the audible or ultrasonic range to disrupt the adhesion. A transducer could be placed in the liquid at one end of the tube and arranged so some sound travels along the tube. If that is not possible them maybe attach a transducer to the outside. All experimental though. One of those "try it and see" ...

3

Probably, yes. 13.56MHz was the frequency used and it was a 300W field. http://www.rsc.org/chemistryworld/Issues/2008/April/ColumnThecrucible.asp The effect doesn't appear to be well described. So while the salt is necessary, the exact role it plays isn't detailed. It could be that some complex with the salt is resonating.

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The stuck bubble problem happens when there is static equilibrium between the circumferential contact forces of surface tension and the forces due to the pressure difference upstream and downstream of the bubble. The gas pressure inside the bubble is uniform and between the upstream and downstream liquid pressures. In that equilibrium state there is a liquid ...

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