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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 ...


30

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...


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: ...


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 ...


16

The reason is electron degeneracy pressure. The cores of giant planets are dense enough that the electrons in the gas occupy about $h^3$ of phase space each. The Pauli exclusion principle means that they cannot all occupy low energy/momentum states. This means that even at relatively cool temperatures the gas can still exert considerable pressure due to the ...


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 ...


12

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 $70\text{ kg}$, ...


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

Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity. One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning). ...


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 ...


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 ...


8

Ok, trying my luck with a physics answer. Let's first look at the boundary conditions given in the movie, since we're particularly talking about that here. The water planet is said to have $130\%$ of earth's gravitational acceleration on the surface. So we have \begin{equation} g_W = 1.3 g_E \end{equation} This is a given and not to be violated. And in fact ...


7

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

Basically, it has to do with the density of the material as a function of temperature. The density of iron increases as it cools, that is, solid iron is more 'packed tight' than when it is melted. This is understandable, since the kinetic energy of the iron atoms decreases as the temperature drops (ie: the average velocity of the atoms decreases), allowing ...


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 ...


4

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


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

You would be wise to somehow determine the exact fluid used by the original manufacturer. Consider that each of the floats has a fixed density, and has a temperature marked on its hanging tag. So you need a liquid which will have the correct, different density at each temperature marked on a tag. In short, the liquid you choose must match both the ...


4

This is an instrument that measures fog density and has an experimental plot, figure 9 . Once you have the relative humidity at the fog appearance at a temperature and pressure , one can use known equations to get the density. This link gives a calculator.


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

There more sides to this scenario that you're considering. Firstly, if we are assuming that the temperature is the same at sea level and on the high mountains, then the speed of sound doesn't actually change, as a constant temperature will take care of the air pressure-density ratio. $$c = \sqrt{\kappa \frac{p}{\rho}} $$ Where $p$: static air pressure, ...


3

I'll use this answer to provide some information that's mostly orthogonal to what Phonon said. As Phonon pointed out, the speed of sound depends on temperature, not pressure. It's cold on the top of high mountains, so the speed of sound would tend to be lower. Some mechanisms for sound production have a frequency that depends on the speed of sound, and ...


3

This answer will not make me popular because it gets people up to speed fast on protecting themselves from thieves and levels the playing field for people who like to maintain their advantage over others. There are a few methods we use to determine if gold is bunk or real...Methods that test if your gold is hollow, filled, alloyed (and the alloy percentages ...


3

Assuming you are talking about exoplanets, I'll offer this. To obtain a density you need a mass and radius. Masses come via two methods - either measuring the radial velocity variations of the star it orbits (the bigger the RV variations, the bigger the planet mass), or so-called transit timing variations. This latter works in multiple "transiting planet" ...


3

Far away from a black hole, spacetime is curved only a little bit, and many different things could curve it like that out there. It's like if you had a dollar in your pocket, and it's been there for a long time, and you can't remember if you got it from your boss or from your friend. But a dollar is a dollar. So you could have a massive star, or a black ...



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