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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 ...


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 ...


2

As an experimental answer, for 12" latex balloons, I could lift about 5 grams (in addition to the balloon). It of course will depend on how full you fill the balloons.


2

It depends on the fluid. Consider, for example, an ideal gas at fixed temperature near the surface of the Earth. Does the density vary in such a column? Yes. Let's investigate as follows. Imagine that the column is in the $z$-direction and has cross-sectional area $A$. Let $z=0$ at the ground. Consider a small, vertical "piece" of the column between ...


2

First and foremost, a primer over what "buoyancy" is is needed. Pressure decreases with altitude. The atmospheric pressure at the top of the balloon is a tiny bit less than the atmospheric pressure at the bottom of the balloon. This pressure difference results in a tiny net upward force on the balloon. The balloon rises if this tiny net upward force exceeds ...


2

You should be able to start with methylated spirits - ethanol with a bit of methanol mixed in to make it toxic and cheap (or ethanol if you can get your hands on it - but it will be expensive because of excise taxes unless you can prove "scientific exemption".) It is much lighter than water and highly miscible with it. Once calibrated you do need to seal it ...


1

Density is mass divided by volume. $$D=m/v$$ Most often in physics, we use the symbol Rho to represent density, so that $\rho=m/v$. With the density you mentioned of Uranium-238, the answer to your question is in isotopes. An isotope is when you have an atom of an element (meaning all your samples have the same number of protons. If you had a hydrogen ...


1

ice is less denser than water because in ice the molecules arrange themselves in a rigid tetrahedral structure due to which cage like spaces remain in their bonding. But water molecules remain in linear bonding form. As the volume of ice becomes greater, it is less denser.



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