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


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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.


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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 ...


2

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 ...


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 ...


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 ...


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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. ...


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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 ...


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" ...


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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 ...


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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: $$ ...


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 ...


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

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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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

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


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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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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

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 ...


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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(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

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 ...


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

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.


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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 ...


1

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 ...



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