Шта радите када вам је досадно ?

DRGeorgy · 1日前

Не брини, већ сам му дао савет!

And btw, while browsing this subreddit in order to find informations related to my question, I found one redditor advise this : https://www.youtube.com/user/Kocayine/videos.

Was it a joke ? Because it's a little bit terrifying…

DRGeorgy · 1日前

Тролујем на Редиту.

FriendlySceptic · 1日前

Ultimately the question comes down to statistics. The overriding concern for bodies regulating the rules of blood donation is usually preventing new infections via blood transfusions, particularly of pathogens such as the HIV virus. Ideally one could achieve this by screening all batches of blood in order to exclude any that show the presence of any pathogens. However no blood test is 100% effective. Every test will exhibit both false positives (a result indicating that the blood was infected when in fact it wasn't) as well as false negatives (a result indicating the the blood was not infected, when in fact it was). Of these two possibilities, false negatives are the bigger concern because it could mean that an infected sample of blood could pass the screening, in which case there is above a 90% probability of the recipient being infected. Such false positives are especially common shortly after infection when the donor's blood is infected by HIV but the body has not yet produced a sufficient number of antibodies to be reliably detected.

Currently the risk of blood infected by HIV passing the screening and posing the possibility of infecting the recipient (the so-called residual risk) is about 1 in 2 million (using US data). This number may sound small, but keep in mind that roughly 30 million blood transfusions are carried out per year in the US alone. Putting the two numbers together, roughly 15 people per year would on average risk being infected by HIV via a blood transfusion in the US. This number, while unfortunate, is rather small. However in the past it was much higher. Before better screening techniques, both in terms of the physical screening of the blood as well as via risk assessment of the donors, infection rates from transfusions were much higher. Between 1978-1984 alone, up to 12,000 people were infected in the US!

Finally, coming directly to the question of excluding gay men from donating blood, the reason is because as a group men who have sex with men (MSM) have a much higher rate of HIV infection than most other risk groups. For example, in the US while MSM constitute roughly 6% of the population, they account for over 60% of new HIV infections and the rate of infection of MSM is 60 times higher than that of other men. The fact that the prevalence rate is so much higher for MSM is so much higher than in the general population, combined with the imperfect screening techniques for infected blood means that lifting the restriction on the donation of blood by gay men could (and statistically would) raise the risk of new HIV infections via blood transfusions. More detailed questionnaires taking more specific categories of risk behavior into account as well as increasingly better screening techniques may change this situation in the future.

XenonBG · 2日前

Ovo je sjajno!

ark2010 · 3日前

Such a scheme already exists and is called concentrated solar power. However in this system it's not the water that's heated directly, but rather there is an intermediate step where a different working fluid, usually a liquid salt first collects the heat before transferring the energy to water. The basic scheme is as follows. 1) First you use an array of mirrors, to collect the light as shown here and to then concentrate it down to a smaller area filled with the working fluid. 2) The hot salts then flow towards a heat engine such as a steam turbine, which converts the thermal energy into electrical energy.

While concentrated solar power is practical, it nevertheless is less common that photovolatic energy (i.e. solar cells). One reason is that the efficiency of CSP systems tends to be lower than that of photovoltaics. Even with the cheapest "good material," namely crystalline silicon (the material based on which 90% of the solar cells in use are based), module efficiencies (i.e. the efficiency of converting solar energy to electricity) are now approaching 20%, but with fancier systems the number can double. In contrast, many CSP plants operate at efficiencies at about 20% or even a bit less. Moreover, the efficiency of photovolatic cells generally rises with the concentration of light, so if you were to couple a high efficiency photovoltaic cell with an optical concentrator (a scheme called concentrated photovoltaics), you can obtain efficiencies over 40%.

Secondly, you need a lot of upfront capital to even create a CSP plant because of its large scale integrated design, while it's easier to distribute solar cells either in solar farms or even on residential rooftops. Moreover maintaining these plants can be fairly expensive. One reason is that because the relative position of the sun in the sky changes, tracking systems must be used to keep the mirrors oriented at the optimal angle. Moreover running hot salts through metal pipes often causes corrosion imposing additional costs. Still, concentrated solar power does have one key advantage over photovoltaics, namely the possibility of thermal energy storage so that you can get energy out even when the sun isn't shining (such as at night). For this reason, the capacity of CSP is projected to increase significantly in the next decades, especially in regions with a low population density and a high level of insolation (i.e. lots of sunlight), such as deserts.

tilluminati · 3日前

Within the Standard Model elementary particles (e.g. electrons) are generally treated as point particles that lack an internal structure or an intrinsic size. However, this does not mean that in practice these particles behaves as though all its density was concentrated at a point in space. Because of the uncertainty principle, we know that there is an intrinsic tradeoff between how well we can know the position and momentum of a particle simultaneously. As a result, particles are treated like so-called wave packets, which defines their spatial location probabilistically, giving an envelope for their position that looks something like this. If the particle is found in an isotropic environment and wasn't prepared in a well-defined state, then this envelope will on average reflect the symmetry of the environment, thus making it spherical. This is analogous to say how the electron distribution in a hydrogen atom is spherically symmetrical.

poinc · 3日前

I believe that the great divide between northern and southern Italy helps to explain this duality, namely how Italy can shine according to some macroeconomic metrics such as its strong manufacturing and high level of exports while still under-performing according to other criteria such as its overall competitiveness. There was an interesting article in The Economist recently, which highlighted how pronounced this geographic asymmetry is and how it in fact continues to grow. For example, while Italy overall sends out a large volume of exports, only about 10% of these come from the Mezzogiorno (Southern Italy). This means that Italy can have some economic sectors which are highly productive, but it also has many that have chronically underperformed and when taking the weighted average, the success of the former is often hidden by the failure of the latter.

Infu-P · 3日前

The color of clouds is almost entirely due to scattering of light by the constituent water droplets. Specifically, the type of scattering at play is called Mie scattering, which occurs when the scattering center (in this case the water droplets) are comparable in size to the wavelength of the incident light. A key property of Mie scattering is that it is fairly spectrally flat, in other words all parts of the visible spectrum are scattered approximately by the same amount. This is the reason why thin clouds appear white, when you look at the clouds what you are seeing is the diffuse light that they scatter and transmit as shown in this simple cartoon. This is actually the same reason why paper looks white.

As for the difference in color between different clouds, this is mostly due to how much water they contain. The amount of scattered light will depend on how many water droplets the initial beam of sunlight must pass through. For thin clouds, a large fraction of the scattered light will reach the ground and those clouds will look white. As the clouds get thicker and thicker, however, there will be an increasing likelihood for additional scattering events, which will serve to reduce the fraction of light that will make it through. Increasingly the cloud will look darker and darker, going from white to increasingly dark shades of grey.

casebash · 4日前

Grey is not a color, it is just white light of a reduced intensity. So basically a beam with a spectral profile that at high intensities stimulates the three types of cone cells to give a response that we interpret at white light, at lower intensities will give a response that we see as shades of grey.

casebash · 4日前

The Wikipedia answer is right about the big picture, but a bit wrong about the details. It's true that our eyes use information from both the blue and red cones in order to distinguish between blue and violet. However the phrasing here:

L-cone (red cone) in the eye is uniquely sensitive to two different discontinuous regions in the visible spectrum

is rather misleading. The absoprtion spectrum of the L cells don't have another band in the blue range, but rather only have a weak tail as you point out.

casebash · 4日前

This is a great point! I got a bit too caught up in emphasizing why you can't simply think of magenta in terms of its wavelength since no spectral color will have that appearance, but I completely glossed over the reason that the purples are similar in appearance to spectral violet. That bit form Wikipedia nails the answer. Just as spectral yellow light can be simulated by a mixture of red and green light, so spectral violet can be partially simulated by creating a properly weighted mixture of blue and red light that would approximate the response elicited by violet light.

velidb · 4日前

I ovo bi kao trebalo da bude smešno? :(

casebash · 4日前

To understand the answer, it helps to step back and consider how color vision works in humans. Under sufficiently intense lighting when color vision is possible (formally called photopic vision), it is mediated by specific biological dyes located in cone cells in the retina of the eye. There are three types of cone cells, called S, M, and L cells (which is why our vision is called trichromatic), which have approximately the following spectra. The weighted sum of these three spectra (weighted by the relative abundance of the individual cone cells and their responsivity) gives the overall shape of the spectral sensitivity of our eyes.

The key point is that color differentiation is only mediated by the three distinct channels associated with each type of cone cell in this (S,M,L) space. To get a more intuitive picture, imagine that if for instance if your eye is illuminated by deep blue light (say around 440nm), your optical system would record a signal like (1,0,0) but if it's illuminated by deep red light, you would get something like (0,0,1). I.e. in the first case the full signal is mediated by the blue channel, in the last only by the red channel. When you get a mixture of the two colors, your brain would get a signal like (1,0,1). Our visual processing system maps such spectral profiles to the so-called purples, which have no direct spectral analogue (i.e. there is no light of a pure wavelength that would illicit such a response). While the appearance may be approximately similar to violet, this similarity is misleading since violet actually exists as a true spectral color with a wavelength in the range of 400nm.

A somewhat better example to illustrate how our trichromatic vision works is when cosidering the case of yellow light. Yellow light exist as a true spectral color with a wavelength of about 570nm. However, a mixture of red and green light can also be perceived as yellow. To see why, take a look again at the spectra of the three cone cells . Notice that yellow light falls into a spectral range where it can activate both the M and L channels, so light from a yellow laser would give a signal like (0,1,1). However, a similar effect can be achieved by using say a combination of a green laser and a red laser if you weight the intensities of the two just right that you get 1*(0,1,0) + 1*(0,0,1) = (0,1,1). As far as your eyes are concerned, the signal from the pure yellow color and the properly weighted mixture of red and green will be indistinguishable!

too_many_cheetos · 4日前

When a current flows through a conductor, it produces a magnetic field as described by the Biot-Savart Law. The coiled wire through which the alternating current(AC) current flows acts as a so-called solenoid, which in the center has parallel magnetic field lines as shown here. While the direction of the magnetic field lines is horizontal, this direction depends on the direction in which the current is flowing. Because the electrical current is alternating, the direction of the magnetic field will switch accordingly.

This alternating magnetic field in turn produces so-called eddy currents in the metal suspended. Because the metal has a finite resistance, these currents end up depositing heat through Joule heating (i.e. resistive heating). This is basically the same mechanism as to how any old fashioned resistor like the coil in a space heater produces heat. Eventually the heat that is deposited in the metal becomes sufficient to raise the temperature of the metal above its melting temperature. As an aside, the way in which the coil heats the chunk of metal is essentially the same operating mechanism as that used in induction heating.

jackoboy9 · 5日前

Yes, the total intensity of the light will be significantly reduced by using this filtering scheme. Unfortunately, however, it's not easy to change the wavelength of the photons making up the beam to change the effective color from red to orange. The reason is that this process would require taking lower energy red photons and converting them to higher energy orange photons.

Going the other way (say from orange to red) is much easier and can be achieved by using a material that absorbs the orange light and then re-emits the light in the red. Such a scheme is in fact used quite often. Many LED-lit displays use a blue LED as the original source of light and then use red and green emitters to convert the blue light to red and green in order to get all three RGB cmponents.

However, going from red to orange would require a more complicated scheme where the energy of multiple lower energy photons would have to be combined to create one higher energy photon. Note that there are ways to achieve this process called photon upconversion, however this process is generally rather inefficient.

Buckfost · 5日前

Removed: Editorialized Title

If you wish to resubmit this story, please do so using the original title.

jackoboy9 · 5日前

I want to know what colour filter I need to put over a red LED to make it appear orange.

It makes more sense to think of what spectral characteristics you need for this specific applications rather than how the filter will appear, since filters with different effective colors can achieve this effect. To start off, realize that a simple filter does not actually change the wavelength of the incoming photons, but rather only attenuates (reduces the intensity of) the components of the light in specific spectral ranges. For example, if you take white light and a filter removes the blue and green components, the light will now effectively look yellow/orange. Filters achieve this effect by preferentially absorbing and/or reflecting light in specific spectral ranges and transmitting the rest.

So now let's look at your specific example. A red LED lamp has a profile that looks roughly like this. To put this in context, here is what the visible spectrum looks like. Comparing the two, you can see that to get light that looks orange from the red LED, you need to filter it in such a way that you only let the short wavelength tail (say everything below 620nm through).

You could achieve this by using a 620 short-pass filter, which will block off all light redder than 620nm, but leave the rest through. It's spectral profile will look something like this (this graph shows a short pass filter at 600nm). If you were to look at the sun through this filter will look like white light with part of the red component removed, so it will have a blueish tint. On the other hand, the same effect could be achieved using a bandpass filter, which only transmits a band of light (e.g. only light from 580-620nm), and which would have a profile that would look like this. Now if you were to look at the sun through this filter, it would look orange, because only orange light is transmitted!

Tl;Dr You need a filter that preferentially absorbs light redder than about 620nm, but the exact appearance of the filter will depend on how exactly it achieves this effect.

Berthelmaster · 5日前

While a photon has no rest mass, it nevertheless still possesses a momentum. This is a direct result of the energy-momentum relation, which states that the energy can be expressed as

E² = (pc)² + (mc² )^2,

where E is the energy, p is the momentum, m is the rest mass and c is the speed of light. Because light has a finite energy and a zero rest mass (which makes the term on the right equal to zero), it immediately follows that it must have a finite momentum. Reflection of light causes this momentum to change direction, which results in a force due to Newton's second law. This force is called radiation pressure, which makes things like solar sails possible.

fnsv · 6日前

Submission delisted until a source is provided.

Please provide a source and the post will be re-approved.

crnaruka · 6日前

Removed: Editorialized title

Panonia · 6日前

Možda i nisam sa ove planete...

SpamFilterHatesMe · 6日前

Sve najbolje zelim novom moderatoru /r/europe

Hvala!

crnaruka · 6日前

Hi. Nine months ago, you (the user named "crnaruka") started trolling r/croatia with pro-Ustaša posts. After I removed your posts and banned you...

This I can actually directly and unambiguously refute. I was not banned from /r/croatia for trolling with pro-Ustaša posts. What happened was that a user was asking about the "futuristic monuments" that dotted former Yugoslavia and I explained to that user where they were located and that many of them were memorials to the atrocities committed by the the Ustaše. At that point you banned me even though I had not violated the rules of /r/croatia, and when I appealed the ban, you clarified that it wasn't a matter of me breaking the rules, but simply of you not wanting me around. As proof, here is our conversation over modmail.

Therefore, since I had already been banned for that rather innocuous comment, I obviously could not have posted any trollish songs (or anything else for that matter), and it would hardly have been possible for any other user to repost them then...

rogerology · 7日前

No, in fact this happens surprisingly often. Many people view the boundary between life and death as a sharp line, however, physiologically this is not so. After your heart stops, many biological processes continue almost normally until the lack of fresh blood and the drop in blood pressure start to cause the body to gradually shut down. At the end of the day, as intricate as the body is, ultimately it's just a very complicated biochemical/mechanical system. The heart is just a muscle, which can be jump-started using an electrical shock, not unlike how a car can be jump-started, setting all the machinery back in action. The biggest problem is that irreversible brain damage can result in the interim as the brain is starved for blood and oxygen (hypoxia).

khangsta · 7日前

The reason is that water by virtue of being polar has a strong dipole moment (or more accurately the two properties are synonymous), as shown here. This dipole moment can act as handle on which the electric field created by the charged object can act (this is called the charge-dipole interaction). The field will then preferentially align the molecule so that its dipole is oriented parallel to the field and at the same time the molecules will feel an attraction in the direction of the charged object, creating a macroscopic deflection.

The same effect will exist for n-octanol, which also possesses a dipole moment because of the presence of the hydroxyl group (OH). However, the deflection induced by the field will be weaker, both because the dipole moment of n-octanol is weaker, but also because the total mass of the molecule is larger than that of water (basically, you have additional CH3 groups that add to the mass but don't contribute to the attraction.).

Finally, this interaction would be significantly weaker still if instead of water of n-octanol you had a very non-polar molecule like CCl4, which does not possess any permanent dipole moment.

crnaruka

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