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32

I drew an image to illustrate the forces at play. For any curved surface of the bubble, the tension pulls parallel to the surface. These forces mostly cancel out, but create a net force inward. This compresses the gas inside the bubble, until the pressure inside is large enough to counteract both the outside pressure, as well as this additional force from ...


18

The increased pressure is caused by the surface tension between the soap and the surrounding air. This can be seen by a simple equilibrium energy argument. The total energy of the system reads $$ E = E_i + E_o + E_s \;, $$ where $E_i$ is the energy associated with the air inside the bubble, $E_s$ is the interfacial energy, and $E_o$ denotes the energy ...


16

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.


13

Is there a particular way that you think this scheme will fail? Rather than have an airlock with that particular portion of ship, you can simply pressurize the entire vessel. There are practical reasons why you would not want to do this at great depths (related to how much gas you use and toxicity), but the problems are not related to how the access works. ...


5

This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to ...


5

For the case that you have drawn, the behavior of the drop is actually the exact opposite of what you mention: it will move from right to left. This is caused by surface tension and the curvature of the droplet caps which creates a larger pressure in the drop at side B than at side A. To make it more quantitative. Let's assume that the funnel is ...


5

No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ...


4

Water vapor is invisible. I think you mean fog - fine water droplets condensed from water vapor. Pressurized planes fly with an 8000 foot equivalent altitude and humidity in the cabin is low. But, at 35,000 feet (called Flight Level 35) it is likely one would get a brief fog. If loss of pressure is fast, you would only get to watch for a few seconds. At ...


3

The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth. Correct x 2. You have 2 choices with an underwater habitat - build it really strong to take the pressure, or just pressurize it and you can make the whole thing out of plastic. Submarines chose the former, as they may ...


3

Realistically? No. For this to happen, your body's density would need to increase faster than the density of water increases as you descend. This does happen because the bulk modulus of water is extremely high (about $2.2 \times 10^9\, \mathrm{Pa}$ compared to that of air (about $1.4 \times 10^5\, \mathrm{Pa}$). So ignoring the human body part of the ...


3

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 ...


3

It is like a balloon. The pressures of the inner and outer air tend to equilibrate, creating a force over the balloon surface from the higher pressure to the lower one, trying to make them equal (the force goes from inside to outside, when you inflate it, from outside to inside when you deflate it). That's why it changes its size, because the gas pressures ...


2

If the flow is laminar, i.e. not turbulent, then the relationship between flow rate and pressure is given by the Hagen–Poiseuille equation: $$\text{Flow rate} = \frac{\pi r^4 (P - P_0)}{8 \eta l}$$ where $r$ is the radius of the pipe or tube, $P_0$ is the fluid pressure at one end of the pipe, $P$ is the fluid pressure at the other end of the pipe, ...


2

I think I see your question, if the surface of the air bubble were perfectly flat, and the air + cup didn't float, then the surface would have an equal pressure across it and it would not move. The system, however, is in an unstable equilibrium, the slightest perturbation will cause the bubble to rise out of the cup. Consider this, since the pressure of the ...


2

The diver would cease to be buoyant when the their average density matches that of the surrounding water; where density is weight (or mass) divided by volume. The density of the surrounding water is not going to change much with such depths as water is very hard to compress. However the diver's density will increase as the air in their lungs compresses; ...


2

Some of these responses are hilarious, yet some have a bit of accuracy. For the man who asked the question... never believe anything seen on TV. Ever. Have divers been to depths greater than 100m, 200m? Absolutely. Does it require a monumental amount of time, money, planning, and training? You bet. Not just the decompression, but the acclimation to the ...


2

No. the whole weight will not act on the base of the container 2. If the whole weight had acted on the base of container 2, then the pressure on the base of container 2 would be equal to that of container 1 i.e., mg/A, where mg is the whole weight of the fluid and A is area of the base. But as you know the pressure at a depth depends on the height of the ...


2

Basic idea Picture in your mind a deep ocean of water. Imagine a column of the water, going from the surface down to a depth $d$. That column of water has some weight $W$. Therefore, there is a downward force of magnitude $W$ on that column of water. However, you know the column of water is not accelerating, so there must be an upward force of magnitude $W$ ...


2

Consider this diagram showing the three columns you describe all connected to the same body of water: Your question asks whether the three pressures $P_1$, $P_2$ and $P_3$ will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if $P_1 > P_2$ then water would flow from the base of ...


2

Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as $\sigma$. You can pick a Cartesian basis $\{ e_1, e_2, e_3 \}$ and express the components of the tensor in that basis $$ \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & ...


1

As Rahulgarg mentioned, the pressure does not depend on the shape but on the depth. However, the direction of the force caused by pressure can be approximated as being normal to the surface, hence the total force on the sides will depend on the shape. For a fluid at rest like the one I think you are assuming, the pressure at any depth will be $p=p_0+\rho g ...


1

If you replace the fluid scenarios with solids with matching have the same shapes, then indeed the pressure at bottom would be the weight of each solid divided by base area, i.e., they would be different! Then what's different about the fluid case? I presume in scenario 1 you have a prism, in scenario 2 you have a bottom-heavy frustum, and in scenario 3 ...


1

There is no quick answer, except if the droplet is completely non-wetting or if it is at least partly wetting. If it is completely non-wetting, it will be move towards the wide side of the funnel until it is a spherical drop touching only its wall. If it is at least partly wetting, it will move to the narrow side until it reaches its apex (if air is ...


1

The piezo electric effect is caused by the distortion of, normally, a crystalline solid. By applying pressure, you are creating an electromechanical interaction between the crystals electric and mechanical states. It is also a reversible process, i.e. applying a current to an piezo electric material will distort it (this is how most buzzers work or the ...


1

As a scuba diver you know that pressure increases when you go deeper. Imagine a cylinder held vertically under water. The force on the top of the cylinder is pressure times area (by definition of pressure). On the bottom of the cylinder the area is the same but the force is greater (deeper, more pressure). The difference between the two is the buoyancy ...


1

Consider this container in pressurized air but zero-gravity (and ignore surface tension, which would make the liquid ball up). If your guess were right, the liquid would squirt out the small hole on the right, but that ignores the role of the wall on the right, which counters the pressure on the left. Think of the liquid as a collection of horizontal ...


1

When the cup is tilted up, the water wants to flow into the cup. That is what water does - it attempts to flow downstream. In doing so it displaces the air. Now the air experiences the force of the water (pressure below bubble > pressure above)


1

It depends on the direction in which you choose the $y$-axis. If a positive $y$ means a displacement to the left, when this figure is accurate. But you right, that choice is at least unconventional and should have been indicated in the figure.


1

If you take a one meter long section of your cylinder and imagine cutting it in half, the force separating the halves is pressure*diameter*1m This is resisted by 2*wall thickness*1m of wall. So the stress in the wall is the ratio of these:$$\text{stress}=\frac {\text{pressure*diameter}}{2* \text{wall}}\\ \text{wall}=\frac {\text{pressure*diameter}}{2* ...


1

Below the yield stress your fluid is behaving like an elastic solid. Imagine putting your tank in zero g, so there are no forces, and then removing the base. The result would look like the left hand figure in the diagram below: Now turn on gravity, or apply an external force and the result will be the middle diagram. As long as the stress is below the ...



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