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5

Take A to be a cylindrical fluid element of height $h_A$ and cross sectional area $A_A$, as the entire portion of the fluid above the section marked $A$. Take B to be another cylindrical element of the height $h_B$, with cross sectional area $A_B$, as the entire portion of the fluid about the section marked $B$. As you have noticed, if the sections marked ...


4

The height of the puddle I will use the common definition of puddle in the field of capillarity (which I believe you refer to) which is: a droplet on a flat horizontal surface flattened substantially by gravity as shown in the schematic below, coming from the book by De Gennes (2003). The droplet on the left is just a droplet (with contact angle ...


4

I don't know what you define as too large, but soap bubbles can reach very large sizes. https://www.youtube.com/watch?v=5bjggctu3kw It's a balance between the surface tension being weak enough to allow the film to stretch to large size but strong enough to not tear as it flexes. The soap added to water is for weakening the surface tension.


4

Sometimes a picture tells a thousand words... It all depends what question you ask Wolfram Alpha: Floating point arithmetic leads to rounding errors. Non SI units are rarely defined precisely (an exception is the inch which is exactly 25.4 mm - and thus other derived units of length). But getting back to the "what is the value" - we should go with ...


4

It looks like a bit of artistic licence was taken (gosh :-) ). You are correct that, the bags would not float that high up -- at least if I'm correctly viewing them as having only a tiny air pocket at the top. The physics is simple: take the total mass of the bag+fish+bagwater, and calculate the equivalent volume of seawater for the same mass. That's ...


3

The flaw in your reasoning is that you believe that pressure and velocity are said to be inversely proportional, $p=K/v$, by the law. Instead, Bernoulli's law says that these two variables are in inverse relationship (but not "proportionality") which means that one of them is a decreasing function of the other, and you wrote what the function is. $p=K/v$ is ...


3

The main considerations are grip and rolling efficiency. A tyre dissipates energy as it flexes, and any energy dissipated in the tyre means extra effort from the rider or motor and therefore fewer miles per gallon. The more you pump up the tyre the harder it becomes and the less it flexes, so higher pressures are more fuel efficient. However harder tyres ...


2

To first order, the speed of sound is not affected by pressure. Pressure waves can be shown to fulfill the D'Alembert wave equation $(c_S^2\,\nabla^2 - \partial_t^2)\psi=0$ where the wavespeed $c_S$ is given by: $$c_S = \sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus of the medium in question and $\rho$ its density. Now, for an ideal gas, the bulk ...


2

How can I calculate the gas pressure given particles per cubic centimeter, and its temperature in Kelvin? as pointed out in comment by KyleKanos $PV=Nk_BT$ where $P$ is pressure, $V$ is volume (in $m^3$), $N$ is the number of particles, $k_B$ is Botzmann's constant and $T$ is temperature in Kelvin. If you rearrange it $P= {N \over V}~k_BT$ so ...


2

At the lower limit, if the bubble is very small the pressure inside will be so large that the gas inside can dissolve into the shell of the bubble, and from there diffuse out to the atmosphere. That limits the life time of small bubbles. On the large side, huge bubbles (several meters diameter) are certainly possible. These tend to be unstable because they ...


2

The net pressure on the liquid is just the atmospheric pressure. Pressure in a fluid acts in every direction, but as the point is on the surface, $\text{P}_{water}=h\rho g=0$ as $h=0$. So only atmospheric pressure will be acting on point A. The height of the liquid column doesn't affect the pressure on top. Pressure in a liquid is affected by the weight of ...


1

1 PSI is 0.0689475728 bar which means that 1 bar is 14.50377397 PSI. Thus, $$ \frac{6894.7573\,{\rm bar}}{1}\times\frac{14.50377397 \,\rm PSI}{1\,\rm bar}=100000.0002900755\,\rm PSI $$ which is slightly off from both sources. NIST says that 1 Bar = $10^5$ Pascal 1 PSI = $6.894757\times10^3$ Pascal Thus, $$ 6.894757\times10^{-2}\,{\rm ...


1

Sticking to SI units for now, the unit of force is the Newton while the unit of mass is the kilogram. To convert kilograms to Newtons you multiply by $g$. So if we express a pressure in Newtons per square metre we multiply the mass by $g$. But if we were going to express the force in kilograms per square metre we wouldn't multiply by $g$. However in SI ...


1

As John Rennie explained, in the American Engineering System, force is expressed in lb$_f$, mass in lb$_m$, and acceleration in ft/sec$^2$. This system is not coherent. Hence, a conversion factor other than one must be used in the equation for force; that is, $F=\frac{ma}{g_c}$, where $g_c=32.174 \frac{lbm \cdot ft}{sec^2 \cdot lbf}$ is a constant, known ...


1

The issue is not "equilibrium" - the issue is boiling. During boiling, there is explicitly NO equilibrium: the water wants to get out of the liquid phase, and into the vapor phase. The temperature of the liquid is sufficiently high that liquid can evaporate below the surface (strictly speaking the temperature must be slightly above boiling for that, as the ...


1

When you press the plunger down, it forces air into the drain and increases the atmospheric pressure on it. If the item is dislodged, the pressurized air is free to travel throughout the rest of the piping. When you then pull back up on the plunger, the vacuum created will force anything inside the tube to be forced upwards. Boyle's law is: $$ p_1 V_1=p_2 ...


1

Q1: How does one interpret tensile strength, yield strength, etc.? The answer is to interpret them as the result of a test that tells you what the material can withstand in an engineering application. The type of machine used to measure tensile strength is popularly called an Instron machine (the most famous manufacturer is Instron; kind of like how tissue ...


1

I'm not a fluid mechanics expert, but my mechanical systems knowledge suggests it might be simply a natural oscillatory behavior, which is always present but in this case is more noticeable due to the aggressive initial response (i.e fast influx of air) your chamber experiences. So what is causing this inexplicable pressure drop? Once the chamber has ...


1

Hetrzian calculations assume infinite width for the parts and in real life tires have a finite width. What that means is the if the contact is line contact (like a cylinder on a plane) as opposed to a point contact (like a football on a plane) the pressure distribution is going to be abruptly interrupted at the ends, compared to an infinitely long line ...



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