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2

This page quotes the pressure inside a soap bubble as $\frac {4\gamma }R$, where $\gamma$ is the surface tension, about $25\text { dyne}/\text{cm }$ for soapy water, and $R$ is the radius of the bubble. For $R=1$ cm, the pressure is then $100 \text { dyne}/\text{cm}^2 = 10 \text{ Pa}$. This is released when the bubble pops. It doesn't seem like much with ...


10

The air pressure inside the (intact) bubble is larger than in the surrounding. This pressure difference is called Laplace pressure and is caused by the surface tension between the soap film and the air. When the bubble pops the compressed air expands, thus creating a pressure wave, which you ultimately hear as the typical popping sound.


0

Cavitation is the formation of bubbles in a liquid when a sufficiently strong negative pressure is applied. A point in the liquid experiences a “negative pressure” if the local pressure goes below the average pressure in the liquid. This can happen when water in a pipe has a very abrupt turn, near the propellers of ships and submarines, in presence of a ...


0

The rubber is not linearly elastic. Initially there is some plastic deformation of the thick rubber, which requires more pressure. When you deflate it, and inflate for a second time, it is easy from the start because the rubber is thinner already. Even if the rubber were linearly elastic, the reduction in curvature as the balloon inflates more than ...


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Almost there. Since this looks like a homework question I'll just give a hint. The force balance equation is due to the buoyancy force which is due to the difference in density inside the ballon vs. outside the ballon. Once you work that out, the weight that you missing will come back into the equation.


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The air inside the balloon is less dense than the air outside; this difference is what causes the lift for the balloon. When you heat the air in the balloon, it expands until the balloon is full. At this point the balloon is still on the ground since there is not enough lift. You need to heat the air more, which expands the air more and causes some of the ...


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It seems like the dyson fan uses a regular fan to pull air into its base creating its air multiplying effect. If that's the case a single fan and motor could drive multiple dyson structures at once, which could result in an overall lower net weight. Not sure how well these things would perform outdoors in air turbulence, but they might make for nifty indoor ...


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Intuitively, you can think of it this way: Pressure is measured as Force / Area, e.g. Pounds per Square Inch (PSI), or Newtons per Square Meter (Pascals). Initially, the surface area inside the deflated balloon is small, meaning more pressure will be required to inflate it. Once you begin inflating the balloon, the surface area inside grows larger and ...


9

What is probably being referred to here indirectly is the fact that air with moisture in it is less dense that dry air. The question becomes, is the buoyancy force of an empty egg with the optimal moisture content of air sufficient to overcome it's weight? Searching around I see that water vapor has a density of 0.804g/Land dry air has a density of 1.27 ...


1

No. For the same reason that you can't pick yourself up by your shoelaces. However, if you were to fill a container with water and close off the escaping steam with the eggshell, it would be possible to "levitate" the eggshell with the rising steam.


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My guess is, this is related to low pressure on your mouth (and much less noticeably - high pressure on your chest) which makes it harder for you to inhale. -When you lower your head, you prevent your mouth from being drained of air by making it into sort of a bubble free from the airflow. This could be related to a RollerCoaster - Have you ever tried not ...


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Actually it depends on the material of balloon. If the balloon is made of less elastic material then it will become even more difficult to blow it even after little inflation cause it will reach its elasticity limit very early and this might not be true in case of more elastic balloons.


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The elastic nature of rubber is varying on pressure/temperature inversely, which makes it harder when cool and softer when applied pressure. So to inflate a balloon in a normal condition we need to put more pressure initially. As when it expands, the pressure inside increases which reduces the elasticity of rubber making it easier to blow later on. The ...


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Answer to question (1) is NO: Using the equations of the comments $(P_1 V_1=P_2 V_2$), you get: $P_2=P_1/(1+.06666)$ (in this case of a small drop in pressure the answer will be approximately correct though, because $1/(1+x) \simeq 1-x$ for $x<<1$) Answer to question (2) is YES: Using $n_2RT_1=(1-0.06666)n_1RT_1$ you get $P_2=(1-.0666)P_1$


2

This phenomenon is probably related to the cold shock response, a set of physiological changes that come about in response to rapid temperature change, such as that experienced by a human whose face is immersed in a cold fluid. The response is accompanied by respiratory changes, including an initial gasp (see here). It is related to the dive reflex. This ...


6

Let us first summarize what do we actually experience when inflating the balloon. For the very first bit of volume, we have to exert a lot of energy. Or alternatively, we have to apply a lot of pressure coming from our lungs because for the change of energy $\delta E$, change of volume $\delta V$ and extra pressure $\Delta P$ (that is the difference between ...


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The volume of a balloon grows linear, while the surface (which you actually stretch) doesn't. So although you're blowing the same amount of air into a balloon, you don't stretch the surface as much as in the beginning.


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When in doubt, use mathematics. Imagine the balloon as a sphere (close enough for this answer) of initial radius $r_0$ and thickness $t$. Let's inflate it just a little bit (to radius $r_0 + \Delta r$). Now we can take a look at what happens by taking a cut through the equator of the sphere. The total circumference at the equator is $2\pi r$; with the ...


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I think that most of the answers here are incorrect since it has nothing to do with decreasing resistance of rubber. In fact, the force required to stretch the balloon increases, not decreases while inflating. It's similar to stretching a string, ie. the reaction force is proportional to the increase in length of the string - this is why there is a point ...


0

Because air is a cheap, light engineering material. The wheels perform several functions: Lift the car off the ground Provide a low rolling resistance Transfer energy from the engine to the vehicle body Provide friction to the surface so energy transfer (accel or deccel) is efficient and the handling is sufficient Provide some compliance on rough road ...


2

The material of an unstretched baloon does resist stretching; this can largly be overcome by manually pre-stretching it (something I learned from my mom when I was about 5 years old). The other thing I'd consider (and that I don't see mentioned so far) is the cumulitive force exerted on the inner surface of the baloon as it's inflated, calculated by Pressure ...


31

Take a strip of balloon rubber and pull it. It will get harder the more you pull. So why is it that inflating the balloon gets easier (at least long before the breaking point)? The balloon starts with very high curvature, so the air pressure is distorting each spot on its surface a lot relative to its 1cm neighbors for example. All the rubber's tension ...


1

Because the rubber is thicker initially. Thick rubber is harder to stretch than thin rubber, in proportion to its thickness. And the thickness of the rubber in a balloon is inversely proportional to its surface area.


15

Like Dev said above, the material your typical round balloon is made from has a non-linear stress strain curve. When just starting to inflate it is fairly stiff, but then as it starts to blow up the stiffness goes down somewhat until it approaches its maximum size. We measured this in my undergraduate advanced lab class, and while I don't have the data ...


1

(I am refering only to the updated question) This is a question which has been asked numerous times by physicists after the formulation and popularization of Maxwell's equations. They have previously known the wave equations on strings where the matter of the string was vibrating, or on water, where the water surface was the one carrying the wave. How ...


1

This is a question that was drastically changed and the other answer and the comments to the question and the answer are discordant with the edited question. I already advised in the comment that you read a simplified article in wikipedia on electromagnetic radiation. Classical electromagnetic radiation cannot be simplified easily by analogies. I will try ...


4

You can rearrange the terms to have any constant as the base of the exponent: $D = 1.25 e^{(-0.0001h)}$ $= 1.25 (e^{0.0001})^{-h}$ $= 1.25 (2^{\frac{0.0001}{ln 2}})^{-h}$ $\approx 1.25 (2^{0.00014})^{-h}$ $= 1.25 \times 2^{(-0.00014h)}$


10

It's actually a surprisingly straightforward differential equation. If you assume that the acceleration due to gravity $g$ doesn't change with altitude (a good approximation if the atmosphere is thin compared to the radius of the earth), Bernoulli's relation tells you the change in the pressure $P$ with height $h$: $$ \frac{dP}{dh} = -\rho g$$ Meanwhile the ...


3

Euler's constant appears naturally in phenomena where the spatial gradient of a quantity (or rate of change with time) is proportional to the quantity itself: $$\frac{\mathrm{d}X}{\mathrm{d}x} = X/x_0$$ ($x_0$ determines the strength of the proportionality, and keeps units straight.) The solution of this differential equation is $$X=X_o e^{x/x_0}$$ $X_0$ ...



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