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Sometimes I feel Wikipedia is a funny place... In the article you quote they provide a calculation from our patent application (see, e.g., http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ) proving that a homogeneous shell made of any existing material cannot be both light enough to float in air and strong enough to withstand ...


3

The black hole event horizon is not a thing i.e. not a physical object. It is just a surface in spacetime from which light can never escape to infinity. Also, if we take the Schwarzschild description of a (non-rotating) black hole then it is a point mass hidden away behind the event horizon. You can't spaghettify a point mass. When two black holes merge, ...


3

I've just remembered that there was once a suggestion to use a mixture of xenon and oxygen under high pressure to allow people to float/fly/swim in it. It was also stated that water could be lighter than such a mixture. According to Smithsonian Physical Tables the critical point for xenon is $16.6\,\text{C}^{\circ},\quad ...


2

Since you have not specified the "real world" size of the ship, let's take a 74-gun ship of the line https://en.wikipedia.org/wiki/Seventy-four_(ship) as the desired type, firing a 36-pound cannon. The bore on such a cannon https://en.wikipedia.org/wiki/36-pounder_long_gun was about 175 mm, with a shot weight of about 39 lb. Shrinking this cannon to a bore ...


1

The formula is $$R = \rho \frac{l}{A},$$ where $R$ is the resistance, $l$ the length of the medium current is flowing in and $A$ its cross-sectional area. $\rho$ is the resistivity, a property of the material. An intuitive way of understanding the dependence on $l$ and $A$ is the following. The longer the wire (increase $l$), the more collisions electrons ...


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You're basically looking for a smoothing algorithm, something that takes a collection of points and turns it into a density. This can be done with the help of a kernel, a weighting function $K(u)$ that satisfies the following two conditions: \begin{align} \int_{-\infty}^{+\infty} K(u)\,du=1 \\ K(u)=K(-u)\quad\forall u \end{align} The Wikipedia article I link ...


1

Short: No Because: Assume material SG ~= 7 Assume ~= non-sonic shot. Say 500 m/S based on this superb reference R = 0.05mm = 5E-5m E= 1/2 mV^2 m kg = 4/3.Pi.R^3 x sg x 1000 kg/m^3 = 4/3x 3.14 x (5E-5)^3 x 1000 x 7 = 3.E-9 kg At say V = 500 m/s E = 0.5 x 3.5E-9 x 500^2 ~~= 500 micro Joule. (438 uJ calc) A VERY solid teaspoon weighs 50 gram. ...



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