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My problem with the assumption is that sound is quite poorly absorbed in water. The $$30 cm = 1/4 \lambda$$ size means you'd look at waves of about 120 cm = 12 Hz. Absorption at those frequencies is measured in deciBels per kilometer. If we'd model the bottle as a cylinder, we might get a standing wave pattern that could persist for several kilometers (i.e. ...


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TL;DR: The load does not significantly increase the pressure in the tire, but not inflating the tire more will increase friction. This will heat up the tire. Correct pressure ensures correct contact area - preventing wear on the tire, and keeping rolling friction low. Full answer: Going to use simple math, round numbers (no calculator): 1000 kilo car, 4 ...


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If one uses a fan to blow air on a fixed disk, the air pressure on the side of the disk facing the fan will be higher than on the reverse. The relationship between the movement and pressure will depend upon a variety of factors including the size of the disk, angle at which the air is blowing, etc. The net force applied to the disk, however, will be ...


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I will attempt to go more in-depth into the physics here. In practice, the unloaded weight of the car is non-zero, but I'll consider two states where the first state refers to zero weight on the tire. In order to make a practical calculation for the car, then, you'll use the numbers for the second state, subtracting one vehicle weight from another. There ...


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This is actually hinted at in an existing answer (on re-reading) but (and this is a rather handwaving take on the question as we don't have much data on tyre deformation, expanded from an intended comment): There's a further explanation which I believe to be closer to the correct one, as the pressure change shown can be large compared to the mass change, ...


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For an ideal gas at constant temperature, then $PV$ is a constant. So they key issue is whether the volume of air in the tyres is changed by the load. I would think the specifics of this are down to the particular tyre, and the operating pressure and temperature. To first order one could imagine the unladen tyre to be represented by a cylinder with maximal ...


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The first explanation is correct: tire pressure increases with an increasing load. The second explanation has a problem in that the phrase "the amount of air inside the tire is constant" is being too vague as to what "the amount of air" means. The number of molecules of air inside the tire remains constant. But the volume of air does not remain constant; ...


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The question Does the pressure inside a tire equal to its average ground pressure? is related. If we can ignore the rigidity of the tyres then the air pressure in the tyres multiplied by the four tyre contact patches must be equal to the car weight, so the pressure would be given by: $$ P = \frac{Mg}{A} $$ where $M$ is the car mass and $A$ is the total ...


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The data manufacturers provide assume the tyre pressure is adjusted with tyres cold (ambient temperature) prior to loading. Though the small error introduced by making adjustments after loading will not be significant, the difference between cold and hot tyres is greater. Recommended pressure is higher for a 'full laden' car. That is why a spare tyre has to ...


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Assumptions: The hole is in the region below the air pocket (so water, not air, is leaking) Air pocket volume is $V_p$ when the pressure is $P$ Isothermal process (slow expansion: temperature constant) Volume of container doesn't change with pressure (probably not true… - this will underestimate the leakage rate) you can write the rate of change of the ...


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Everything depends on the size of the air pocket, since you can treat the water as incompressible. As water is lost, the air pocket expands, lowering the pressure. If the air pocket is large, it takes a lot of water loss to lower the pressure a certain amount. If the air pocket is small, the pressure will be very sensitive to loss of water. Check out ...


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The area of the manometer tube makes no difference. All that matters is the difference in the heights of the two ends (labelled $x$ in your diagram). That's why pressure units like the torr exist that are (or rather were) defined as the pressure difference when the difference in height of a mercury manometer is 1mm. All that matters is the height difference. ...


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Interesting question. I suppose one should compare several scenarios Lie still Go forward - either straight, or hard to port, or hard to starboard Go in reverse The rate at which water enters the ship is (to first order) proportional to the pressure differential - lower the pressure and live longer. Maybe even long enough for the Carpathia to come and ...


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breathing: Since the tyre is a about 2 atm at the surface, at 50m it will have a pressure of 6 atm if the tyre is thin (membrane) or less if is "partially rigid" (down to 2 atm if the tyre is totally rigid). If it behaves like a membrane, air will come out if you push the valve (with the tongue, I guess it would require practising a lot :D) because the ...


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The pressure at 50 m is about 6 atm. Since the pressure in the tyre has a pressure of about 2 atm, there are 2 possibilities when the valve is opened. The tyre (which has been compressed by the water pressure) is rigid enough to return to it's previous shape and water enters the tyre (very likely). Or the air stays in the tyre and the water stays out. The ...


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I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is $$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$ all in the horizontal direction. I assume that a fraction $(1-\cos\theta)$ of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of ...


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Assuming: the problem is static ($\rightarrow$ no velocity gradient, so no shear, only the spheric part remains non-zero), the density is homogeneous ($\rightarrow$ $\rho$ is a constant), the cross-sectional area of the pipe is constant, (neglecting the variations of $g$ (!)) the pressure $p$ various linearly with the height $z$: ...


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If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark: $$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$ Where the $g$ subscripts pertain to the gas and the $a$ subscript to the ambient. The gas density $\rho_g \equiv M / V$ is the ratio of the mass of gas (M) in the tank to the volume ...


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The pressure at the bottom of a pipe, or any other column of liquid, for that matter, is directly proportional to the height of the liquid. The pressure of the liquid is equal to: $\rho h$, where $\rho$ represents the density of the liquid, and h represents the height. So, assuming that the cross-sectional area of the pipe is constant, then the pressure ...


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The name of the property is itself a clue here : enthalpy of vaporization. By nature, enthalpy does take into account the work required to push the atmosphere. You can see the impact of increasing the pressure on the enthalpy of vaporization on a Mollier diagram. Increasing the pressure has the overall the effect of reducing the enthalpy of vaporization, ...


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When a pipe is bent the outside curve - what would be the longest path through the curve - has the highest pressure and the lowest speed. The inside curve - the shortest path through the curve - has the lowest pressure and the highest speed. In short, when the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always ...


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The simple answer to "why is the pressure inside a soap bubble higher than outside," is that a higher pressure than the local atmosphere is required to make the bubble in the first place! This requirement comes from the need to counterbalance the surface tension force. For stable conditions, $$F_i = F_o + F_s $$ Where $F_i$ is the force due to inside ...


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I'll give brief answers to your questions. If you need more detail, you should ask your questions separately. What's the difference between heat and work at the atomic level? Isn't heat simply work between particles colliding with different momentum against each other? Treating a substance semi-classically, one can say that at the atomic level, the ...


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Heat is energy in transit. It only comes into play when one body is transferring heat to other. Feynman best explained in a video on youtube, it is just the vibrations of atoms at an atomic scale. So, if a more vibrating solid object comes in contact with solid with relatively low "jiggling", we say heat is transferred. [Ref]: Feynman lectures on physics ...


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Working out how much force is required to move through a fluid, whether it's air or water, is surprisingly difficult because there are two effects you need to take into account. However neither of those effects is pressure (or at least only indirectly related to pressure). The first effect is viscosity, which is simply how thick the fluid is. Obviously it's ...


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The air in lungs, ears, stomach and intestine has the same pressure as the air pressure outside the human body, ensures that you don't get crushed by the air pressure. If you have already managed living almost compensating the air pressure on you, moving in that air is simple.


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Once boiling starts, there is an equilibrium between the liquid and vapor phase. The pressure, temperature and partial molar Gibbs energy are equal for each phase so that water molecules have no preference for one phase or the other. That's for intensive variables. However, the total enthalpy of the liquid and vapor is not fixed : it keeps increasing as more ...


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When you heat water on the stove top, you see bubbles forming on the bottom of the pot. The bubbles are created where the heat applied (if you move the pot, you see the bubbles forming in a different spot) and is sufficient to convert the liquid into a vapor (less heat would just heat the water). These bubbles form even though the water is below its boiling ...


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An answer to v1 of the question (that didn't emphasize the sides): According to this cooking website, the pressure cooker valves are designed to be naturally "open." As the water heats up, the pressure inside will be slightly higher than outside (which is why you see steam escaping). At some point the valve is designed to "close" once a certain minimum ...



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