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32

The speed of sound in a liquid is given by: $$ v = \sqrt{\frac{K}{\rho}} $$ where $K$ is the bulk modulus and $\rho$ is the density. The bulk modulus of mercury is $2.85 \times 10^{10}$ Pa and the density is $13534$ kg/m$^3$, so the equation gives $v = 1451$ m/sec. The speed of sound in solids is given by: $$ v = \sqrt{\frac{K + \tfrac{4}{3}G}{\rho}} $$ ...


31

When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform ...


23

It's not the falling that's fatal, it's the deceleration at the end that kills you. Something like water or concrete does this on a sub-meter distance (which requires extremely high forces). On the other hand a gas is much less dense, so it cannot decelerate a falling object nearly as quick. Sometimes inflatable cushions are used as safety nets (think: ...


21

If gas A and gas B are of different density, then the situation sketched is not in equilibrium: the water level on the side of the light gas will be higher. There, the containers are moving down, and you have to push your containers through this net difference in level. You do need to put in energy here, which is probably the piece that you are trying to ...


16

Squeezing the bottle does decrease its volume. Rather than a bottle, it may be more helpful to think of a full toothpaste tube; the mechanics will be the same. If you squeeze the middle of the tube, the middle will collapse, the back will expand, and the front will expand and squirt out some toothpaste. Treating the toothpaste in the tube (or the water in ...


16

The density does increase with depth, but only to a tiny extent. At the bottom of the deepest ocean the density is only increased by about 5% so the change can be ignored in most situations. If you're dealing with these sorts of depths you also need to take temperature into account because the water temperature changes with depth and the density also ...


12

Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is ...


11

Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly: $$ F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right) $$ You can imagine that ...


11

The ocean surface is not as hard as the ground but if you drop from a plane, you would hit it with such a high velocity that the pressure would most likely kill you or cause very serious damage. Considering air resistance, the terminal velocity of a human, right before reaching the water, would be at most some $150\text{ m/s}$. If you weigh $70\text{ kg}$, ...


10

The surface area of the bottle is conserved, but the volume is not. Squeezing the bottle deforms it into a shape whose volume to surface area ratio is lower than it was previously. As an example consider a bottle whose cross-section is initially a circle. The volume of the bottle will be $V_0=\pi r^2h$ where $h$ is the height of the bottle, and the ...


10

It's actually a surprisingly straightforward differential equation. If you assume that the acceleration due to gravity $g$ doesn't change with altitude (a good approximation if the atmosphere is thin compared to the radius of the earth), Bernoulli's relation tells you the change in the pressure $P$ with height $h$: $$ \frac{dP}{dh} = -\rho g$$ Meanwhile the ...


8

The answer is No and the reason is the equivalence principle which says that there exist natural units in which the gravitational mass (the mass $m$ in $F=GMm/r^2$) is equal to the inertial mass (the mass $m$ in $F=ma$) for all objects in the Universe. This is equivalent to the statement that all objects, regardless of their composition, density, and other ...


7

Due to the crystal structure of the solid phase of water, the molecules arrange themselves in a rigid, ordered fashion and end up being, on average, farther apart from each other (than they are in the liquid phase), and thus less dense. Less dense things float because of buoyancy.


7

"The speed of sound is variable and depends on the properties of the substance through which the wave is traveling. In solids, the speed of transverse (or shear) waves depend on the shear deformation under shear stress (called the shear modulus), and the density of the medium. Longitudinal (or compression) waves in solids depend on the same two factors with ...


6

The density of mercury (13.534 g/cm^3) does not imply high intermolecular forces. It simply reflects that the mercury atom is much more massive than a water molecule. The atomic weight of mercury is 200.6, while the molecular weight of water is about 18, so mercury atoms take up $\frac {200.6} {18\cdot 13.534}=0.823$ as much volume as a water molecule. ...


6

I'm not a physicist. So I am treading very carefully attempting to answer a question here... :) A physical example that may help explain this is rock skipping. When you skip a rock, it will 'bounce' off of the water when at high speeds. Eventually it slows enough to no longer bounce but 'sink' into the water. Picture your body doing the same thing. Your ...


6

as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it? No, it doesn't - or at least only negligibly so. At normal pressures, liquids are essentially incompressible. This table gives the compressibility of some liquids, including water. Note that the units are to be multiplied by ...


5

jdj081's answer is good. I just want to address where I think you originally went wrong. Your confusion lies in using the word "volume" in two different ways. You should differentiate between volume of the container (capacity is a better term, as jdj081 states) and volume of the liquid. The liquid's volume doesn't change. The container's volume does. ...


5

In the moist air of the clouds, the water condenses on dust particles. At the altitude where this happens, it is usually below the freezing point of water, so it quickly freezes. If winds and updrafts keep these particles of ice in the moist air, they collect more water. Eventually, the weight of the ice particles overcome the updrafts and fall to the ...


5

You said the right word: liquid! $P=\rho g h$ holds only if the fluid that you are considering is not compressible, that is a liquid. Try to fill a syringe (without the needle) with some water, then close the hole and try to compress it: you will notice that you cannot do much, indeed liquids are not compressible, this mean that the density $\rho=m/V$ does ...


5

Hint: It seems the substitution $z \longrightarrow z^{\prime}=r-z$ will work to prove the second equality of eq. (8). (Geometrically, this corresponds to a reflection of the $z$-axis around $z=\frac{r}{2}$.)


5

In addition to Bernhard's answer, just because three gases (Gas A,B and air - which is itself a mixture of nitrogen, oxygen, and other gases) have different densities, it does not mean they will remain seperated when in a container. In fact, as entropy of the system increases over time, Gas A, B and air will make an even (if heterogeneous) mixture.


4

As @Qmechanic points out in his answer, making the substitution $z \rightarrow z - r$ (going from the first version to the second in the identity) works. That's the math answer. The physical intuition is that, in order for the system to have perfect cylindrical symmetry, it must extend to $z = \pm \infty$. That corresponds to integrating your $z$ variable ...


4

I don't think that this question is still fully resolved, water is a fascinating molecule! But here are some thoughts. Clearly, if ice is lighter than liquid water it is because it doesn't pack as well. Its an example of how a random-ish packing can be more efficient than an ordered packing of a "weirdly" shaped molecule. Imagine throwing LEGOs into a box, ...


4

I think you're asking if there is some special cutoff density after which spacetime "collapses" and forms a black hole. If this is your question then the answer is no, there is no specific cutoff. Density unites are $\frac{\mathrm{mass}}{\mathrm{volume}}$ but the size of black holes is dependent on the mass and the size is not proportional to the volume ...


4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...


4

John Rennie has provided an exact mathematical treatment of the equations behind the calculation of the speed of sound. I don't want to detract from that treatment, and of course the Wikipedia articles we both draw from provide a broader treatment; but an intuitive understanding of the 'why' has been equally helpful for me, in the past. The following is my ...


4

Air is lighter because there are fewer molecules per unit volume compared with a unit volume of liquid water. A mole of water is 18 grams, so a liter of water contains about 55 moles (1000 grams). A mole of air at standard temperature and pressure, however, occupies a volume of 22.4 liters, much more. Dividing a mole of 02 (32 grams) by 22.4, you have ...


4

The trouble is that your table, or whatever object it is, will act as a waveguide. That's because the sound waves will (partially) reflect of the wood/air surface then travel back into the table and interfere with other waves. The result is going to be hideously complicated to calculate. As LuboŇ° says in a comment, if the thickness of the table is much less ...


4

You can rearrange the terms to have any constant as the base of the exponent: $D = 1.25 e^{(-0.0001h)}$ $= 1.25 (e^{0.0001})^{-h}$ $= 1.25 (2^{\frac{0.0001}{ln 2}})^{-h}$ $\approx 1.25 (2^{0.00014})^{-h}$ $= 1.25 \times 2^{(-0.00014h)}$



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