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13

I don't think this sounds unreasonable as an estimate at all. Let's check it. One designs a building as a compromise between two competing factors: One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ...


1

Young's modulus is the ratio of tensile stress to tensile strain for a material: E = (F/A)/(∆L/L) = (F * L) / (A * ∆L) F/A is force per area, and (∆L/L) is change in length per original length For structural steel, Young's modulus is 200 gigapascals. This quantity can be used to predict how much the steel will compress under a given weight per unit area. ...


1

Let us make an estimate. Let the skyscraper be 400 m tall, each storey 4 m high, 500 people per storey, 20 m2 per person, 10000 m2 per storey, let us assume that the building is a 100x100 m2 square in the plan and that it only has 10 cm thick structural walls in a 25x25m2 grid. So the cross-section area of the structural walls is 2x5x100x0.1 m2=100 m2. Let ...


1

According to the wiki page the 747 has a max takeoff "weight" of about 350,000 kg (depending on the model) and a wing surface of about 500 m2. That means that a force of 3.5 MN must be carried by 5 million square cm, or 0.7 N per square cm. If you can somehow split this evenly between the top and bottom surface, then you need to come up with a paper surface ...


1

I noticed that no direct answers were provided to this question, so I decided to answer it from my own experience. In general, alloy additives can shift the density of an alloy either up or down. I don't know prescisely what the mechanics of the density shift are, but high-speed steels are a good example because of their tungsten content. An extreme example ...


2

It's hard to do this one analytically, but that's why we have computers. I used a commercial FEM package to set up your problem and simulate steel with some dummy loading conditions. Your intuition is basically correct away from the corner. Indeed, the stress is uniform far from the corner, and it only has interesting behavior close to the feature. However, ...


1

Such a material (apart from conductive ones) typically doesn't exist. The reason: when a material is interrupting bluetooth, it's either reflecting or absorbing the radio wave. Reflective materials are also absorptive materials, so this comes down to finding a material that absorbs radio waves. For metals and other conductive materials, electrons are very ...


0

The problem is basically that radio waves are big and tend to be able to "curve around" and "penetrate" obstacles. Metals and conductors will be the hardest things to get around/through. So maybe your problem is that you're just putting, say, one concrete block between two bluetooth devices, rather than a whole concrete wall. You could try metal and ceramic ...


1

The absolute cheapest thing could be to wire a variable resistor in series to try to get the discharge voltage down: it's possible that you could reach a regime where you're only ionizing some of the gases in the air, without melting the resistor. Since ozone's "badness" comes from being hyper-reactive you might be able to remove it chemically by putting ...


1

This may be cheating - but ozone is generated by the interaction of the high voltage discharges and the oxygen in the air. Why not flush the system with nitrogen - if there is no oxygen, no ozone will be produced. And a few mL per minute is not a lot of nitrogen. A 70 L bottle of nitrogen at 200 bar should expand into 14000 liter of nitrogen at STP - at 3 ...


3

Gray is not a color - it is a shade. A shade of white. We perceive something as gray when it has no obvious "color" (that is, the amount of red, green, blue stimulus that the cones in our retina receive are roughly the same), and Something else in our field of view is brighter Our eyes adjust our perception of gray or white relative to "something else". ...


1

Simple answers: When you cut diagonally, less material is being moved aside during a given bit of time. The time of the cut is longer, the force the same, therefore you are applying the same force to less material in a given moment. Therefore it's easier to cut the longer arc, even though the force is unchanged. Also, gravity is assisting your diagonal ...


12

The moon appears white from the Earth for two reasons. The first is that the reflected spectrum of sunlight is very broad and contains no very significant features. On this basis, the spectrum of the moon could be considered pinkish, as the reflectance of sunlight (which appears almost white to the human eye) is twice as effective at red wavelengths than ...


5

It is not white, see this link for photos: http://www.mikeoates.org/mas/projects/mooncolour/intro.htm See also this image I took, it does not appear white, one of the seas looks blue. See this photo with increased saturation. It appears that on average the Moon is grey with light coloration aligned with the terrain ranging from yellow to blue. The ...


15

The moon is actually grey. You can see this if you look at images taken in space, or, preferably, on the moon itself. For example, this one, of Buzz Aldrin: (Courtesy of NASA) But, seeing as how at night you compare it to a black sky, it appears white.


14

For emphasis: The Moon is grey, but it looks white because of scattering along with sunlight. The dark parts - which are less common - are maria, plains of volcanic rock (basalt). They are relatively old, as there has been no recent volcanic activity on the Moon. Most are on the near side of the Moon. By contrast, the white parts are often referred to as ...


2

The dominant load that a large structure has to face it's its own weight. The weight load scales as volume, which means that it goes as $\lambda^3$ The canonical strain $\epsilon$ upon a given load $F$ is $$ \epsilon = \frac{F}{k} = \frac{ F L }{ A E} $$ so the strain will scale as $\lambda^3 \lambda^{-1} = \lambda^2$, which means that scaling up the ...


3

The flexural rigidity of a vertical beam will be greater with a diagonal cut. This allows more of the force to be channeled into cutting the target as opposed to bending the target. The determinants of flexural rigidity affected by a diagonal as opposed to a horizontal cut are the cross section of the beam and distance from the supported end of the beam. ...



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