13

Pressure is actually $$P=\frac{F_\bot}{A}$$ where $F_\bot$ is the force component perpendicular to the surface in question, and $A$ is the area of the surface. Therefore, there is no "division by a vector" here. Certainly, the area vector is used in various areas of physics; this is not one of those areas (pun always intended). I suppose if you ...


8

The pressure is the same, whether it is a sheet of paper or piece of rubber. The object doesn't alter the applied pressure (stress). What is different is the effect of the pressure (stress) on the object. That is, the resulting strain (change in thickness) that occurs due to the applied pressure (stress). Can you add the formula for strain? Stress σ is ...


7

Let's just use the ideal gas law. Let $P_{ab}$ = absolute pressure, $P_g$ = gauge pressure and $P_a$ = atmospheric pressure. $$P_{ab} V = NkT$$ From this is it obvious if I double Pab and keep the volume the same, then N must double. Now write Pab as $$(P_g + P_a)V = NkT$$ If I only double $P_g$, then N does not double. Your question about pressure being ...


4

Yes we are assuming this. The gas laws deal with gases in equilibrium, and with macroscopic samples of gas (say of linear dimensions thousands of times greater than the mean separation of the molecules). On this scale any pressure differences will cause bulk movements in the gas so that the pressures quickly equalise.


4

Yes, their is a compressive stress throughout the material of the pipe. However, since this stress is the same on both sides of the pipe, it is isotropic (the same in all directions) throughout the pipe and will not generate any tensile shear stresses in the pipe. Since most bulk materials are much stronger in compression than in tension, compressive stress ...


3

The ideal gas law takes no account of weather patterns over land and water; it takes no account of atmospheric circulation. So the connection you attempt to draw between the gas law and weather reporting is invalid. Furthermore, the gas law's relationship between temperature and pressure you cite requires that the volume be held constant. No such rule ...


3

The reason that pressure acts normal to all surfaces is that, in reality, it is not actually a scalar. At static equilibrium, the (2nd order) stress tensor is isotropic, and the parameter we call pressure is the magnitude of this stress tensor: $$\boldsymbol{\sigma}=-p\mathbf{I}$$where I is the so-called (isotropic) identity tensor or metric tensor. The ...


2

Your textbook doesn't seem very understandable! This might help: First, sound waves in air consist of small amounts of back-and-forth air movement accompanied by a small amount of rising-and-falling air pressure. The characteristics of air (its mass per unit volume and its elasticity) cause a pressure change to result in air movement, and air movement to ...


2

The downward pressure of the elephant foot is done on the contact surface and is $$p_d = \frac{F_e}{A}$$ For a thin sheet of paper, the upward reaction of the ground on the other paper surface is equal, because the weight of the paper itself is very small compared to the elephant force. But if the thickness is too big, the weight of the upper layers of paper ...


2

It depends on the sound source. Obviously, if you hit a drum in vacuum, there will not be any sound emitted even if you hit the drum with the same impact. If you somehow manage to build a sound source that emits the same acoustic power (watts) at 1013 hPa and at 500 hPa, and your dB meter is designed to work properly across that pressure range as well, then ...


2

Neither paper nor rubber would significantly reduce the pressure on the ground due to the elephant's feet. This is because the paper or rubber would bend easily, so not spreading the force exerted by each foot over the 'nominal' area of the paper. But if a rigid board or sheet of metal were used instead of paper, you could then use your formula. Provided the ...


2

Gravity holds the atmospheric gases to the earth. Very light gases like helium near the top of the atmosphere can gain enough energy from a collision with a heavier molecule to exceed the escape velocity. The pressure at any level is caused by the weight of the gasses above.


2

If a body is moving very slowly, the air can easily move around it, resulting in the negligible drag terms. In practice, that void gets filled by air near the void, and the atmosphere takes the least effort required to populate the empty space. When you move the box, the air in front of the box is displaced, creating a slightly higher pressure region, ...


2

Yes. See Mantle and Earth's Inner Core. Large parts of the interior of the earth are solid when you look at them for a short time. If you bent a rock harder and harder over a period of say 100 years, it would break. This is why we get earthquakes. Mountains and other structures on opposite sides of a geologic fault move slowly past each other. They are ...


2

The "absolute" pressure of the fluid blood in a scuba diver's blood vessels will rise to the ambient pressure level under water. As the heart beats the pressure in any part of the circulatory system will rise and fall leading to the flow of blood throughout the body. The fluctuations in pressure due to the beating of the heart will be small ...


1

Yes. However, it usually works out the same as if you ignored air pressure. Homework problems in beginning physics courses are often simplified so you can concentrate on the part of physics in the current chapter. So it is certainly true for your problem. So let's see why you can ignore it. Consider an object not sitting on a surface. It has a top and bottom ...


1

It is possible to calculate but it is very unreliable. The reason being the density of atmosphere at different places and heights varies thereby changing the buoyant force on the apparatus. However to calculate the height we can do this. Measure the total mass of the apparatus( you may include the balloon and string too). Find out total volume it occupies. ...


1

When you write down $Q = vA$, it's implicit that the velocity profile is uniform over the cross-section A (and purely perpendicular to it). In general, $$Q = \int \mathbf{v} \cdot \mathrm{d}\mathbf{A}$$ This no longer implies that $v \propto \frac{1}{R^2}$. If we assume $\mathbf{v}$ has purely radial dependence and is aligned with $\mathrm{d}\mathbf{A}$ as ...


1

There are general equations for drops on a surface but they are different equations and require other/more information. When a fluid interacts with a surface, the hydro -phobicity or -philicity also needs to be taken into account, this is usually expressed via a contact angle. The equation talks about drops that are non-spherical in the sense that they aren'...


1

In the case of a pipe where sudden contractions (or expansions), regulating valves, bends etc occur, Bernoulli's Equation (here for negligible gravity) is written as: $$p_1+\frac12 \rho v_1^2=p_2+\frac12 \rho v_2^2+\Delta E$$ Where: $$\Delta E=\Delta\Big(p+\frac12 \rho v^2\Big)$$ For sudden contractions ('vena contracta'), the coefficient of contraction is: $...


1

It is due to the pressure difference between inside the can and outside. When you blow up a balloon you are pushing more air particles inside it than would like to be there. As soon as you pop it they all rush out. This is because they are getting pushed more from behind (from the many more air particles inside the balloon) than from the outside (the ...


1

The answer to your first question is contained in the phase diagram for $H_2O$ shown here. Note how the line separating ice from water, line A to D, is negatively sloped. This means that if you start in the solid phase and apply enough pressure (move upward on the Y axis), you will incur a phase change to liquid. So to answer you question, one just needs ...


1

If the plate and cylinder are rigid, there is not enough information to answer. The container below the plate is basically a pressure vessel. The water within that vessel could be at virtually any pressure. It is disconnected from the upper container by the strength of the iron plate. This strength means the upper and lower pressures don't have a fixed ...


1

To a good approximation yes. Pascal's law won't hold exactly for an emulsion because the interface between the two fluids will have a non-zero interfacial tension, and there will be small excess pressure inside the emulsion droplets given by the well known formula: $$ \Delta P = \frac{2S}{r} $$ where $S$ is the interfacial tension and $r$ is the radius of ...


1

Pressure is only the measure of the force an area experiences. How it responds to has nothing to do with the pressure. It's response is measured by the strain: $$strain=(L-L_0)/(L_0)$$ or the elongation/ the original length. Here is the difference: An elephant steps on a square of fabric, so the fabric is stretched quite a bit. Measure the amount it's length ...


1

Let's consider a "light detector" first. A photo detector measures the light intensity, which is proportional to the square of the electric field, $I \propto |E|^2$. Therefore, if the electric field is maximal, we obtain a "large" signal and if at its minimum, we obtain a "small" signal. Now, let's consider a sound detector. If ...


1

The problem with the argument comes in the line "However, pressure in a fluid is caused by nothing other than the collision of water molecules against other things." Pressure in an ideal gas is caused by nothing other than the collision of gas molecules. Pressure in condensed matter comes from interatomic repulsion. It's true that when you ...


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