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0

experiments reveal that the pressure at a point acts in all directions.the pressure does not depend on the shape size or area of the container.


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Lots of poor info here. Ribs are not an impenetrable barrier, compression happens between them. Fat can also be compressed so is not bouyant. Equalization of bouyancy for humans happens at approx 4m below the surface. So jumping off a medium height bridge into water 'just to cool off' means you won't get to the surface unless you swim upwards.


1

The Venturi effect is not specific to fluid flow in pipes but rather flow of fluid in general. The basis of the effect comes from the Bernoulli equation which accounts for the energy in a flowstream. Fluid which moves with a high velocity has high kinetic energy derived from a potential (pressure) energy. It's both the conservation and conversion of energy ...


2

This is a classic example, often used in fluid dynamics classes, of Bernoulli's principle. This is the principle which underlies the Venturi effect: increasing the flow speed leads to a drop in pressure. The governing equation for an flow of an incompressible fluid such as water is $$\frac{V^2}{2} + gh + \frac{P}{\rho} = \mathrm{constant}$$ To answer your ...


1

A 2 piece fiberglass pool filter is plenty large and can withstand the pressure. A stainless steel belt goes around the sections after you put the RUT in there. Add a hose and use a bicycle pump if you cant achieve 40 foot hose head. It is only 18 psi.


1

Think about it like this: friction happens at the boundary layer (in the vicinity of where the water touches the pipe). So doubling the diameter will double the circumference (double the friction) but the quadruple the cross sectional area (quadruple the flow). Will you get a lot more pressure by installing a bigger pipe in that one section? No. Using an ...


4

As several people have advocated using pressure vessels, I'll add my own take on this. Pressure vessel safety Compressed air in an uncertified vessel can be extremely dangerous. You need a pressure of about 1.2atm gauge to simulate 40ft of water. That's about 2.2atm absolute. I'm not going to perform an integration here, but air at these pressures is ...


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I was going to suggest building a centrifuge. I am guessing you don't need to attach cables to your robot. And yes, a centrifuge can be dangerous. But Paul has less complicated solutions.


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Fit a valve onto the body of your submarine, and use a vacuum pump to pump out all the air. 1 atmosphere is equivalent to 10.3 meters (33.8ft) of water. Then you only need 6-7 feet of water to reach the pressure difference you want to test.


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Find a rigid 40ft-long water hose, attach it to your tank vertically and fill it with water.


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I think John Rennie's is a bit misleading. You don't actually need 12 tons of weight. You can get the same 18 psi from a 40 feet garden hose hung vertically. Attach it to the lid of a pressure cooker, and you'll have a "pool" that's 40 feet deep. The pool doesn't need to have the same diameter throughout. The two challenges are (1) attaching the garden hose ...


-2

How much space do you need, and what shape does it need to be in? If it's a stationary bot, call an excavating company and ask them for a quote on drilling a 24 inch hole 40 feet deep. They will bring a humongous machine out to your place, punch a rather large hole in the ground, and be gone before lunchtime. Visit a home center for some 6mil plastic rolls, ...


4

Fill the cavities of your vehicle with an environmentally friendly liquid like glycerin. This will take care of a possible water leak, at least to the extent necessary to survive for a short amount of time in a low pressure environment. Before you do that you have to ask the event organizer if that is an allowed design strategy (what "big oil" can do is not ...


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It's easy to work out what weight you need, because it's the weight of 40 feet of water. The pressure at a depth of 40 feet is simply due to the weight of the 40 feet of water above. Let's work in SI units, so 40 feet is 12.2 metres. Suppose the top of your tank has an area of one square metre, then the amount of water above it would be 12.2 cubic metres ...


0

I'm not quite sure of your level, so apologies if any of the following is telling you how to suck eggs. When people say informally that there is a vacuum inside a vessel, they most often mean that the pressure exerted by the gas on the inside of the vessel is less than that exerted by the surrounding gas on the outside. The latter is most often the Earth's ...


2

An equation of state is a relation of state variables: $$ p=p(\rho,\,T,\,\mu,\,\alpha) $$ where $\mu$ is chemical composition and $\alpha$ the acentricity (itself dependent on $\mu$), the other variables take their normal meaning. There are some equations of state where the dependencies on these variables is non-linear (e.g., the Peng-Robinson eos), so ...


8

The maximum speed of sound is the speed of light - the maximum speed at which "information" can be propagated. This will occur for an equation of state that satisfies $P = \rho c^2$, where $P$ is the pressure and $\rho$ the density. Such an incompressible equation of state may be approached in the cores of neutron stars due to the strong nuclear force ...


2

Sound travels fastest in less compressible materials. But it is also affected by the state of the material, specifically its temperature. As a mechanical wave, sound must overcome the inertia of the material in which it travels. Higher temperatures mean greater kinetic energy in the molecules which carry the compression wave, and therefore less inertia ...


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The speed of sound is a function of the compressibility of materials and their density: $$c=\sqrt{\frac{E}{\rho}}$$ Where $E$ is the bulk modulus (sometimes written as $K$) and $\rho$ the density. Compressibility itself depends on the material; for instance diamond, with relatively low density (3.52 g/cm3) and very stiff covalent bonds, has a high speed of ...


1

It's a good question . I also got confused when i studied it in fluid mechanics and static tube . I think you will get the answer if you visit https://en.wikipedia.org/wiki/Pressure#Liquid_pressure the part of (Direction of liquid pressure)


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If you take liquid water for example and start heating it at atmospheric pressure, the temperature increases. It corresponds to following a horizontal line along $p=p_{atmospheric}$ on the above graph. Under liquid state, the volume is small hence it corresponds to the left region on the graph. If you keep heating, the liquid will start turning into ...


0

Let us assume that the volume of the heavier fluid remains constant (as we always do), and that the drilling pipe and the accompanying setup is vertical. In the absence of the packer, buoyant forces would take over and push up on your pump. But in the presence of your packer, the heavier fluid is restricted from creating vacuum under the packer due to the ...


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The Question is basically "how Quickly"? It's definetly not "instantly", and there is some delay, which causes movement inside the fluid and it must take some time before viscous forces kills these movements. And as they are, by nature exponential, means the lower the velocity, the smaller the losses and thus the time can be considerably different according ...


0

Let’s assume that by “swimming in space”, it is meant that there is no gravity whatsoever. A (basic) microscopic view of the pressure as the force exerted by the particles (e.g., molecules) of a fluid on a surface implies that still there will be a pressure on a macroscopic body in the fluid, regardless of external forces, like gravity — although in a static ...


0

The reason for why you do not need to supply any energy is because there is no net work done. Lets assume for a second your container exists in a vacuum. This is essentially an isolated system, not considering black body radiation. This means that none of the energy contained in the box ever escapes and is doomed to stay in the box for all eternity. If ...


0

The real reason is that the gas and container are assumed to be in thermal equilibrium. If the container could radiate into void, then the pressure would slowly decline as the temperature fell. The ideal gas law ignores inter-molecular forces and the finite size of molecules, so eventually the gas will condense and other forces, other than elastic ...


0

Pressure of a gas comes from the average, collective change of momentum as its particles bounce against each other and the walls of anything encountering it. What gravity does is provide a counter-pressure - the weight of the gas itself - which stops it from expanding freely into space. The weight of the air above Earth's surface averages about 101,325 ...


0

Atmospheric pressure will just act as an extra "weight" for your object, such that $$ \left| N \right| = \left| mg \right| + \left| P_0 A_{top} \right| $$ Think of the pressure from above as just being the weight of all the air stacked on top of the box. From the sides you have just got $$ \left| P_0 A_{left} \right| = \left| P_0 A_{right} \right| $$ so ...


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The free body diagram is accepted but one should remember to make lines of forces close to each other for forces acting on same direction.As usually it can be seen in books like HRW.


9

You should ignore the sides of the pool when calculating the pressure. This is because weight of the water must be fully balanced by the upwards force exerted by the pool's bottom. The isotropy of the water then ensures that an exactly equal amount of force is exerted regardless of the direction. (Newton's third law is satisfied because the force on one wall ...


0

lets be simple latent heat refers to the heat required to overcome molecular bonds. latent heat of vapourisation of water at 1 bar ,100c is 2257kJ/kg which means, that much heat is required to break inter-molecular forces and turn into gasoeus phase as pressure on molecules increases they require more heat to overcome the pressure force acting or to escape ...


2

Just assume the pressure at the center applies over the whole floodlight. It's actually less at the top and more at the bottom, but since the pressure is linear in depth, and the floodlight is vertically symmetric, the pressure differences between the top and bottom should exactly cancel.


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What i think is that since the floodlight is 25cm in diameter a distance of 12.25 Could also be subtracted to get a better approximaion since 25+30=55 is almost half of a meter and can make a difference.……………………………………………………… .You are going right though.


1

You can take the integral of the pressure, and add up all the pressures on this disc, to find the total force. However, the integral really does work out to be very, very close to the assumption of uniform pressure for water at over one diameter deep. The integral works out to: $$\rho g\pi r^2(h + \frac{r^2}{4h})$$ The last term is pretty small when r ...


3

The physical absurdity - or at least highly hyperbolic situations - of most of Roald Dahl's scenarios is the essential Dahl - it's wholesale a part of his humor and his lack compliance with physical laws is, in this respect, quite deliberate. Having said this, the "Great Big Greedy Nincompoop" disappearing up the tube is wholly possible, given the right ...


0

Will the humidity of the air inside the bell change or remain the same? Assuming there is no change in temperature, the relative humidity will decrease. If it changes, that would seem to suggest that for some reason the water molecules are preferentially being sucked out by the vacuum pump. Why would that be? You may have a misunderstanding of ...


0

Check Newton's third law: For every action there is an equal and opposite reaction. You push on the right, and the fluid pushes on you. Of course, in a control volume, fluid can move across the surface, but there is still equal and opposite pressure on both sides. Unless there isn't, in which case the fluid accelerates (changes velocity), but you're not ...


0

Are you saying the fluid is pulled on right side and pushed from left side! Pressure is always compressive. The flow is taking place from left to right because compressive pressure on left side is more than that at right.


1

Take for example, a submarine. The pressure against the walls of the submarine increments as this goes deeper. For another example, take an airplane, as it goes higher the pressure against the walls decrease. So yes, the pressure changes as the altitude (or depth) changes (if you are submerged on a fluid in a gravitational field). Even small amounts of ...


1

Hydrostatic pressure inside a pool on earth is given by: $p=ρgh +p_{atm}$, (g: gravity, h: depth, ρ: fluid density, $p_{atm}$: atmospheric pressure) Assuming 0 gravity and no atmosphere, there would be no pressure. You would feel no pressure at all whether you are on the surface or in the center of the sphere. Also, you wouldn't be able to float to the ...


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Best answer is given from xkcd's what-if: https://what-if.xkcd.com/124/. It's not space, but it describes the fluid flow in lower gravity - such as how you could jump out of the pool just by performing aquadynamic maneuvers, or walk on the water. It is really a cool read. As mentioned in the xkcd article - diving and floating, being primarily about ...


0

Other answers are right, but let me put it without math: Water can't come out of the bottle if air can't go in. (Except: if the water in the bottle is so tall that water can come out even if air can't go in.)


2

Water stops draining from the jar into the dispenser once it forms an interface as draining of more water would result into the formation of a vacuum in the jar because no air can rush into the jar to displace the water as it has an interfacial-lock. Consider the water level above interface $= h$, water level below interface $= x$ now $$P_{surface}= ...


1

For submarines, it depends on the depth. In the worst case, the one you always see in the movies, the submarine is at the bottom of the Atlantic and the water will always rush in, the only way you could stop it is to increase the air pressure inside the submarine to match the outside pressure of water, which would make life more than a little uncomfortable ...


0

Your drawing shows an empty box . If it is at room temperature and not air tight the inside and outside air pressure are in equilibrium, the arrows balance against the rigidity of the wallss If the box is air tight and the inside is a vacuum then it will be crushed. Nobody makes vacuum tubes from polysterine. Why we can have vacuum tubes ( incuding ...


1

It looks like you don't take into account elastic forces in the rigid cube. The pressure on the cube's faces will tend to bend them inwards, the force will be passed to the ribs, so the faces and the ribs will get compressed, and they will resist compression. Perhaps it is easier to understand what happens if you consider an evacuated spherical shell, rather ...


0

I am reading your question as wondering why the total force on the (smaller) inside surface area of the cube is not crushed by the (larger) outside surface area. Perhaps exaggerating this will make it a bit clearer. Yes the total force on the outside is larger, but much of that force does not reach the interior of the cube. If you imagine the force from ...


1

I think you might have actually touched on something interesting here. One explanation for the difference in temperature is simply that part of the energy removed from the balloon + air system comes from the air, so the balloon will cool more slowly when filled with air. But there might be more to the story. I haven't done any calculations on this, so it's ...


2

Let me quote this line which says that: I touch it that it's temperature did not drop down Its better to use thermometer to check the readings as it gives you accurate reading. Please check this link as it shows what you did wrong: Why does cold metal seem colder than cold air? The process of touching and determining its temperature is wrong. There ...


1

At height h/5, the distance to the bottom is h/5, and the distance to the top is 4h/5. The note means the pressure is proportional to the distance from your position to the top. So depth = h, and height = 0 at the bottom.



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