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What do you catabolize first during starvation: muscle, fat, or both in equal measure? by strong_grey_hero in askscience

[–]incognito_dkMuscle Biology | Sports Science 942 points943 points x2 (0 children)

Finally something in r/askscience where my degree can be of use (PhD in muscle biology)

Whenever you stop eating, your substrate preference will be about 2/3 fat and 1/3 carbohydrates. Those carbohydrates will come from stored glycogen in your liver and muscles.

When those glycogen stores run out, the liver will try to defend the blood glucose through gluconeogenesis, synthesizing glucose from amino acids from protein broken down elsewhere in the body and glycerol from triglycerides. This metabolic phase is characterized often by decreases in blood sugar and associated tiredness and hunger. It is also the phase in which muscle catabolism progresses at the fastest pace.

However, 12-24 hours after running out of glycogen, the body will gradually go into ketosis, in which the liver synthesizes ketone bodies from fatty acids. These ketone bodies can substitute and/or replace glucose in the metabolism, reducing the need for breakdown of protein for amino acids for gluconeogenesis. After a couple of days the substrate preference will have changed to 90% fat and 10% carbohydrates, thereby reducing muscle catabolism strongly. This state can be maintained for as long as there is enough fat. The longest documented therapeutic fast was 385 days during 100+ kg weight loss in an obese patient. Mind you that a kg of bodyfat contains enough energy to go for 3-6 days depending on body size and activity level.

Ketosis and relying predominantly on fats will continue until only the essential bodyfat stores are left at approximately 5-7% in men and 10-14% in women. At this level the substrate preference for fats disappear and muscle catabolism increase sharply again. At this point death will usually occur within very few weeks.

Is the Monty Hall problem the same even if the door opened by the host is chosen at random? by JooJoona in askscience

[–]Gankro 57 points58 points  (0 children)

For those who trust science more than math, here's a simulation of both versions of the game (push the big red Run button in the top left):

http://play.integer32.com/?gist=77751c1b3ac7e2c0387deda53924b979

(apologies for the sloppy code, copy-pasting was easiest)

As you can see, science and math agree:

  • if the host knows and avoids the car, it's 66/33 (in favour of changing)

  • if the the host guesses and the universe explodes when he picks the car, it's 50/50

Edit: the best explanation I can give is that half of your wins are being cannabalized by the universe exploding. Whenever universe-death would have happened in the "host knows" case, the host would pick the other door and you would win instead.

Edit2: or to put it another way: picking the car first makes it impossible for the universe to explode, so the universe not exploding is evidence that you picked the car, and so switching becomes less-excellent.

Can a Mars Colony be built so deep underground that it's pressure and temp is equal to Earth? by 2Mobile in askscience

[–]Astromike23Astronomy | Planetary Science | Giant Planet Atmospheres 2755 points2756 points  (0 children)

Short answer: If you wanted to dig on Mars to reach a depth where the pressure would be 1 atmosphere, i.e. equivalent to sea level pressure on Earth, it would most likely be much too warm.

Long answer: Consider the case of Death Valley on Earth. Since it lies below sea level, the atmospheric pressure there is actually greater than what's found at sea level, roughly 1.1 1.01 atmospheres. Similarly, we could dig below the surface of Mars so that the weight of the overlying atmosphere would be the equivalent of 1 atmosphere.

We can calculate how deep a hole one must dig by using the "scale height" - this is the difference in altitude needed to produce a factor of e = 2.718x increase in pressure. In Mars' case, this is equal to 11.1 km.

Now, the pressure at the surface of Mars is a measly 0.006 atmospheres, while we want to go to 1 atmosphere. The number of scale heights we want to dig is then:

ln (1.0 / .006) = 5.12 scale heights

...which, for a 11.1 km scale height means we want to dig 5.12 * 11.1km = 56.8 km. Note that this is over 4 times deeper than the deepest hole ever dug on Earth, so this is already a pretty tough technological achievement.

Now, how warm would it be when we get there? For this, we need to consider the adiabatic lapse rate; this tells us how much the temperature drops as we ascend in the atmosphere, or similarly how much the temperature increases as we descend. (It's also for this reason that Death Valley has the highest temperatures recorded on Earth.)

In the case of Mars, the adiabatic lapse rate is 4.4K/km. In other words, for every kilometer we descend, the temperature increases by 4.4 K.

Thus by descending 56.8 km, we're increasing the temperature by 56.8 * 4.4 = 250K. Since Mars' average temperature is 223 K (= -50 C, -58 F), that means the final temperature at 1 atmosphere of pressure would be 473K (= 200 C, 391 F).

EDIT: Since a lot of people are asking:

  • This is unrelated to whether Mars has a "dead core" or not. This temperature increase is not due to geothermal (or in this case, areothermal) energy. Rather, it's a simple consequence of taking the current atmosphere and compressing it adiabatically as it fills up our hole. A similar transformation would be suddenly opening the doors on a pressurized jet at 33,000 feet...the air would quickly expand to the thin ambient pressure and cool down in the process by 65o - 98o C, depending on how humid the air inside the airplane was.

  • You can't generate electricity from this temperature change. It seems counter-intuitive, but even though the temperature has increased, there's no extra energy added to the system - this is the definition of an adiabatic transformation.

If light is "sucked" into a black hole due to its gravity, does this mean that light can accelerate? How does this work with special relativity? by tylercwats in askscience

[–]rantonelsHigh Energy Physics | Superstring Theory 19 points20 points  (0 children)

For any coordinate system (t,x) you can build on spacetime with a black hole, the speed of a light ray defined as dx/dt will not be in general c as it travels.

However, these are not inertial coordinates, so there's no reason for dx/dt to be c. I can make up any sort of weird coordinates I can think of, and dx/dt is completely arbitrary. However2, it turns out that I cannot build coordinates that are everywhere inertial, so that's the best I can manage.

At any given point in spacetime, and with your thrusters off, you can whip out your ruler and clock and make measurements. This defines a very very small coordinate patch around you which is inertial and in which therefore special relativity holds. dx/dt is c but only for light rays lucky enough to pass through your little bubble at this moment.

You can extend your coordinates into a global coordinate system throughout spacetime. But you cannot make it inertial. Curvature messes things up. In particular, if you have a bunch of friends scattered in spacetime, each with an inertial frame, it's not in general possible to create a global coordinate system matching all of those little coordinate patches.

So everyone which is at that point and that moment in which the light ray passes, at any point along its path, will measure light going at c. But any given observer with an associated global coordinate system (and a way to receive that information, possibly with a delay) will measure light's coordinate velocity (how light's space coord changes with the time coord) to be sometimes faster or slower than c.

If there was a planet with the same orbit and orbital period as the Earth but on the opposite side of the Sun would we ever detect it from Earth? by AladdinSane1690 in askscience

[–]lolfunctionspace 92 points93 points  (0 children)

Try it out for yourself :) Here's the source code.

https://github.com/mjsprengel/GravitySim/blob/master/GravitySim.py

If you want to try out a 1 part in 1 million difference, go into the globalreset() function and change 400 to 400.001, then when you run the program, hit the start button and time it. The orbit decays into chaos pretty quickly.

Joey Chestnut ate 70 hotdogs in 10 minute today. What is your bodies reaction to 19,600 calories in that short of timespan? by Villyfresh in askscience

[–]Mc9306 35 points36 points  (0 children)

What is happening?

Joey Chestnut ate 70 hotdogs in 10 minute today. What is your bodies reaction to 19,600 calories in that short of timespan? by Villyfresh in askscience

[–]yungdutchoven 108 points109 points  (0 children)

It bothers me more than it should that this got gilded instead of the OP.

Joey Chestnut ate 70 hotdogs in 10 minute today. What is your bodies reaction to 19,600 calories in that short of timespan? by Villyfresh in askscience

[–]zk3033 3080 points3081 points  (0 children)

TL;DR - He most likely won't absorb a good deal of that 20k Calories.

The ultimate way the calories end up in our blood (and then the rest of our system) is via solubilized fats (help from the bile), broken down simpler sugars (saliva, stomach acid, and bile enzymes), and simple amino acids (stomach acid, and bile enzymes). I'm going to walk down the GI tract.

They're not chewing it that much, so most of the food is in solid chunks without much saliva (that's why they need water after all). This will prevent all future steps from accessing all the nutrients, and is probably the biggest block to the bulk Calories.

It probably won't be broken down that much in the stomach. The pylorus, the sphincter at the end of the stomach that leads to the rest of the intestines, is controlled by pressure, concentration (acidity), and a few other things. 70 dogs is a lot of volume and pressure. Also, the body's stomach can't produce that much acid to accommodate that much food - so the pylorus will have to let a lot of un-broken down food through. This isn't to mention that food is unbroken - it'd be a lot easier to dissolve crushed up chalk in vinegar than a whole piece.

Once in the early small intestines (the duodenum), the pancreas can't produce that much EDIT: bile digestive enzymes, the liver can't produce that much bile, and the gallbladder couldn't have stored enough bile to break down that much food. A lot of the fat won't become emulsified, and thus can't be absorbed. Much of the proteins and carbs won't be broken down into the smaller pieces that can be absorbed. And, of course, you can't absorb large chunks of food.

The rest of the small intestines is where the Calorie absorption goes on. The transit time through the intestine is longer than a normal meal, but a lot of the food is not in body-absorbable forms at this point.

The large intestines is where it gets interesting. The GI's natural peristaltic rhythm moves things through (somewhat), but all that undigested nutrition is going to meet up with the large intestinal bacteria. There will be gas (the a main bacterial byproduct), but all this "digestion" doesn't really go into the human body. Plus, the large intestines don't have the same nutrient absorption machinery the small intestines do.

In the colon, the large undigested chunks will coalesce with dead bacteria, dead red blood cells (it's what gives the stuff most of the brown color). The large intestines and colon serve to dehydrate the remains. The un-emulsified fats will serve as...let's say...barriers to coalescence. This will make looser stools. Large chunks, slightly smaller than the bites that were taken way up top, will still remain. And don't forget the gas.

The parts that DO get absorbed go to the liver, and are processed as normal. The sugars and amino acids are converted to fats or carbohydrates (glycogen, etc.) - but there's going to be a lot of excess nitrogen compounds in the blood (amino acids, unlike carbs and fats, have a lot of nitrogen) that will be eliminated as urea in the urine or can exacerbate gout. The sodium will easily be taken care of by the kidney. If they were to donate blood the next day, it would probably be a milky red color (versus the deep red we are familiar with). This is due to the solubilized and re-packaged fats. Some are from the intestines to the liver, and some are from the liver to the rest of the body (the familiar HDL, LDL etc.) These will get absorbed by the body eventually - either by other tissues, or back to the liver.

I should point out a lot of the unabsorbed calories are because the food was whole protein and carbs and fats. If 20k Calories of sugar (sucrose) were ingested, the body could likely absorb almost all of it. Edit: I provided the answer to this hypothetical here

Edit 2: I underestimated what 20k calories of sugar is like. Some rough stuff will happen before that is absorbed - please see comments below the link for better explanations.

Joey Chestnut ate 70 hotdogs in 10 minute today. What is your bodies reaction to 19,600 calories in that short of timespan? by Villyfresh in askscience

[–]leadchipmunk 1553 points1554 points  (0 children)

That was, without a doubt, the most informative journey through the GI tract that I have ever experienced.

Is it possible to raise a number to the power of any real 2x2 matrix? by chunkylubber54 in askscience

[–]functor7Number Theory 10 points11 points  (0 children)

The answer is "Yes", but this is a roundabout way to do it. When you raise a number to a complex number, you get a complex number. When you raise a number to a split-complex number you get a split-complex number. If you were to do this in the way you suggest, then we would be raising numbers to a combination of both and the result would be a combination of both, which isn't too helpful. Furthermore, if we were to want to raise numbers to 3x3 or 4x4 or nxn matrix powers, then this would not give any hint on how to generalize to these dimensions.

For any number, ex=1+x+x2/2+x3/3!+x4/4!+... If M is any nxn real or complex matrix, then it makes sense to talk about the series

  • eM := Id+M+M2/2+M3/6!+M4/4!+...

so we say that this is what eM is. We'll have a matrix in the end, and it will behave like an exponent would. In particular, if M is a 2x2 matrix that represents a complex number, z, then eM will be the 2x2 matrix representation of the ez.

AskScience AMA: I’m Professor Brian Hare, a pioneer of canine cognition research, here to discuss the inner workings of a dog’s brain, including how they see the world and the cognitive skills that influence your dog's personality and behavior. AMA! by Dr_Brian_HareProfessor | Duke University | Dognition in askscience

[–]Dr_Brian_HareProfessor | Duke University | Dognition[S] 789 points790 points  (0 children)

So sorry for your loss and what a fascinating observation. I would love to say there's an experiment or systematic study to cite for this one, but there is not. Scientist have written papers about other animals - like primates - that they interpret as grieving the loss of their offspring. That said dogs showing behaviors that can be interpreted as 'grief' is something that has been recorded throughout the ages. The best one I know of is of Napoleon Bonaparte. At the end of a battle in Italy, Napoleon came across a dog sitting beside the body of a fallen soldier, licking his hand. Later, during Napoleon’s final years in exile, he would write; ‘This soldier, I realized, must have had friends at home and in his regiment; yet he lay there deserted by all except his dog… I had looked on, unmoved, at battles which decided the future of nations. Tearless, I had given orders which brought death to thousands. Yet, here I was… moved to tears. And by what? By the grief of one dog.’

Why do people go bald on the top of their heads, but not the sides? by bofshc in askscience

[–]Ellocomotive 780 points781 points  (0 children)

I'm sure someone with a more detailed answer will respond, but the hairs on the top of the head are structurally different from the ones on the side. When you start to get older, your body starts converting testosterone in your body to dihydrotestosterone (DHT) at a steadily increasing rate (until you hit your developmental peak, which for males is somewhere between late twenties to mid thirties).

EDIT: /u/TheTousler and /u/hatetheloss contributed this correction: As we age, our hair follicles become more sensitive to DHT. Amount of DHT doesn't increase, but sensitivity in the hair does.

The increased amount of DHT (which is essentially more potent than testosterone alone) affects the hair follicles with the certain protein makeup on the top of the head, effectively killing them. The ones of the side are spared of this horror.

It's why those who use steroids typically use Propecia in concert, and also why it happens in middle age. It's why most elite level athletes are in their late 20's when they hit their peak-more DHT in the body. Hair transplants work by using hairs that don't have the faulty genetic make up and moving them to the top.

EDIT: I'm adding some more information from a reply I made below regarding genetics:

Well that's where those of us who lost our beautiful flowing manes lost out on winning the genetic lottery:/. Genetics play a big part-I have the gene for the weaker hair on the top of the head, whereas my significantly older father in law does not.

Is there any biological reason why some people sing better than others? by ArthurCurryAquaman in askscience

[–]pneruda 362 points363 points  (0 children)

Not necessarily singing. Singing requires precise vocal and respiratory control and is in large part due to training.

However, pitch is something different, and quite fascinating. The phenomenon of 'absolute pitch' (or 'perfect pitch') is one whereby the individual can instantly identify any note sounded, and often recreate it vocally.

A study in 2001 (full text) found that a region of the brain called the planum temporale is different in those with absolute pitch. The planum temporale is an interesting region in the brain -- it's one of the few areas which is asymmetrical from the left side to the right side (in general, structures on one side of the brain are mirrored in size and location in the other). It's located in the region of our brain primarily responsible for understanding speech (Wernicke's Area), and is typically up to about 10 times larger in the left hemisphere than the right (especially in right-handed individuals, whereas the larger side is more random in left-handed individuals).

One of the most interesting roles of the PT (along with other aspects of Wernicke's area) is in Prosody, which simply put is the emotional content of speech. Someone with damage to the right hemisphere's PT may have impaired prosody -- they can understand the semantic content of speech, but may be impaired in their ability to interpret nuance, subtlety, sarcasm, etc.

How are the two linked? Well, interestingly, when you throw someone in an MRI and check which bits of their brain are "lighting up" when they listen to speech, you predominantly get the left hemisphere. When you have them listen to music, you get the right hemisphere. This is evident even in newborns. The idea seems to be that our ability to interpret emotional valence in speech and our understanding and appreciation of music are linked.

Back to the neurobiology of absolute pitch, though. The asymmetry is interesting in one particular aspect -- it appears to be the result of "pruning". Remember, musicians with absolute pitch tend to have PTs that are more asymmetrical than others. That is, their left PT is even bigger than normal compared to their right PT. It isn't due to an increase in the size of their left PT, but rather a reduction in the right. These results, in general, are fascinating but not particularly well replicated (some studies confirm the difference in PT asymmetry, others find it only between musicians and non-musicians). Regardless, there's a lot of evidence for differences in brain anatomy between those with absolute pitch and those without, suggesting that there is a strong physiological basis for at least some aspects of musical talent.

On a final note, I should mention that the above deals almost entirely with the ability to hear music, rather than produce it vocally. Understanding speech and producing it are two tasks which are handled by almost completely separate brain regions, and there's little overlap. The same is very likely true of something like pitch, and to some extent then, vocal singing ability.

Is atmospheric pressure affected by terrain? by Netrilix in askscience

[–]AugustusFink-nottleBiophysics | Statistical Mechanics 0 points1 point  (0 children)

It depends if you measure height from the Earth's center or if you measure height from sea level. Sea level is a trickier concept than most people realize.

The presence of dense rock underneath you means that the force of gravity will be a little higher at the top of a mountain compared to a point that is (1) at the same latitude and (2) the same distance from the Earth's core but located over the ocean. And more gravity means a little more air pressure.

When we calculate sea level, we try to establish how high the sea would be if it could flow to that position on the Earth's surface. This equipotential surface is called a geoid. Here is a picture that helps visualize how it deviates from the ellipsoid shape a smooth but spinning Earth would adopt. The geoid bulges away from the ellipsoid a bit where the mountains are and sinks down where the ocean is.

Ignoring air currents and weather, then air pressure should just scale with your height above the geoid. That is what we mean when we talk about your altitude above sea level. So using this measure, the influence of terrain on air pressure has already been factored in.

edit: Thanks for the gold!

Hi Reddit, I’m Margaret Leinen, here to talk about the world’s oceans and how we observe them. Ask Me Anything! by AmGeophysicalU-AMA in askscience

[–]JumalOnSurnud 153 points154 points  (0 children)

We concluded that it is not possible for us to determine unequivocally whether ExxonMobil is participating in misinformation about science currently, either directly or indirectly, and that AGU’s acceptance of sponsorship of the 2015 Student Breakfast does not constitute a threat to AGU’s reputation.

Sorry to be a jerk, but this is insane and there is no way that you or the other members of AGU believe this. It's clear that you've made this decision from political and financial reasons, not based on the reality of Exxon's actions.

To be clear, Exxon IS directly and indirectly funding and disseminating misinformation. If you and AGU truly believe otherwise it makes the validity of your credentials incredibly suspect, and if not it makes your convictions incredibly suspect.

Is it possible to create 100% vacuum? by rick_dick_ulous in askscience

[–]gawdybaubles 348 points349 points  (0 children)

"about a millionth of a millionth as much pressure as in space around earth" ... what a cool statement! :) In these places of extremely low pressure, about how far apart are the remaining molecules from each touching each other? On the order of inches, feet, or nano-meters?? I really have no idea how to think about the atoms and molecules at those extremely low pressures. A millionth of a millionth makes me think that for every one molecule in the super-low pressure areas there are 1 trillion molecules in the higher pressure areas.

Is it possible to create 100% vacuum? by rick_dick_ulous in askscience

[–]PA2SK 3549 points3550 points  (0 children)

Heh, finally my job is useful. I design vacuum chambers for a living. The short answer is no, we cannot achieve a perfect vacuum. Even the best pumps, like cryopumps, which actually condense gas molecules onto the pump surface, cannot get every single molecule. Even if you could capture all molecules all materials outgas slightly, meaning they emit gas particles. So you will constantly have molecules entering the chamber. We use special cleaning methods, low outgas materials, and even bake hardware to try and reduce this but it still happens. We cannot even create a vacuum equal to that in outer space. Hope this helps.

Can we calculate how long it would take an object, moving in space, to stop on its own? by NathanielGarro- in askscience

[–]ColossalMistake 34 points35 points  (0 children)

It would depend heavily in what sort of space you're talking about. Space in the solar system vs. interstellar space, vs. intergalactic space. Each will have different (average) densities.

The highest density I can find for any of these is about 1000 atoms/cubic centimeter near our galactic core. 1000 is a nice round number. The levels elsewhere are much lower, probably closer to 1 atom per cubic centimeter and in certain places, like intergalactic space, probably .1 atom per cubic centimeter.

What we're looking for here is a drag equation fine tuned for the level of atoms per cubic centimeter in space. The equation for viscous resistance is F(d)=-bv. Where b is a constant that depends on the properties of the fluid and v is the velocity of the object.

I admit this is where I lost my footing. We're talking about a VERY tiny force acting on your hypothetical object. So tiny that it is almost negligible. My best guess is that it would take more time than the universe has existed to stop your object using only the viscous resistance from ordinary space.

Can we calculate how long it would take an object, moving in space, to stop on its own? by NathanielGarro- in askscience

[–]KnowsAboutMath 47 points48 points  (0 children)

What we're looking for here is a drag equation fine tuned for the level of atoms per cubic centimeter in space. The equation for viscous resistance is F(d)=-bv. Where b is a constant that depends on the properties of the fluid and v is the velocity of the object.

It's not going to be a linear drag. At these low gas/dust densities, the Knudsen Number will be extremely high, leading to non-hydrodynamic behavior. Drag will be collision-dominated, not flow-dominated.

There should be nonlinear drag, with a force law that looks like:

F ~ -b v2

This comes about because the frequency of particle-sphere collisions scales linearly with the sphere's velocity, but the average momentum transferred to our sphere per collision also scales linearly with velocity, leading to the v2 dependence.

We can get a VERY rough estimate by 1) assuming that the sphere's velocity is always measured with respect to the average rest frame of the gas, 2) and ignoring the "bouncing" of gas particles off the surface of the sphere. 2) means we pretend that 100% of each gas atom's momentum is transferred to the sphere, as if it were absorbing them. This is clearly not true, but it'll allow us to get a ballpark estimate.

Putting all this together (with Newton's 2nd law) yields the following for the rate of change of the sphere's velocity:

dv/dt = -(m n pi R2 / M) v2

Here m is the average gas atom mass, n is the gas number density (how many per unit volume), R is the sphere radius and M is the sphere mass.

Define c = m n pi R2 / M, and let the initial sphere velocity be v(0) = v0. The above differential equation has the solution:

v(t) = v0 / (1 + v0 c t)

Note that this object never "stops", it just gets slower and slower forever. To get an idea of the time scales, we note that it takes an amount of time t = 1/(v0 c) to get to half of its original velocity.

Let's assume these are hydrogen atoms it is hitting, at a density of 1000 atoms/cubic centimeter. Suppose this is an iron sphere of radius 10 cm. Then c ~ 1.6 x 10-23 cm-1 . If we assume this thing starts out at a speed (relative to the gas) of 100 mph (= 4470 cm/s), then the "speed-halving time" turns out to be about 1.4 x 1019 seconds, or about 45 billion years.

That's just as a ballpark.

ETA: In fact, this kind of thing is a very interesting problem that pops up a lot in rarefied gas dynamics (which I used to do for a living). For this simplified case, you can actually solve this problem nearly exactly, and get not only the average drag component, but the deviations from that average, which will act more or less like Brownian motion... but there's no room here for all of that.

Can we calculate how long it would take an object, moving in space, to stop on its own? by NathanielGarro- in askscience

[–]nairebis 11 points12 points  (0 children)

Note that this object never "stops", it just gets slower and slower forever.

Wouldn't that mean your answer must be wrong, then? The density is so low, it seems to me you can figure it out purely from a standpoint of Newtonian mass and collision physics. With every atom collision, the object would transfer some momentum to the atom. If the velocity is low enough, then a collision with an atom ought to eventually bounce it back the other way. Wouldn't it then just bounce back and forth among the atoms, very very slowly?

Or to put it another way, if I put a bowling ball in space and pushed it toward a big cloud of marbles, I don't think it would just slow forever. Wouldn't it just eventually bounce randomly inside the marble cloud?

Edit: I'm wrong and misremembering classical mechanics. See below in the thread. I'll still take the gold for at least raising a question many people may have had. :)

Is it possible to block out the sun from Earth using a dinner plate? by st3richards in askscience

[–]Afinkawan 23 points24 points  (0 children)

Am I correct?

You are correct that something much larger than the sun could block out the sun, yes.

Edit: thank you kind internet stranger!

Do bees socialize with bees from other hives? by TheTedd in askscience

[–]Satoyama_Will 265 points266 points  (0 children)

Nah, honeybees are sweethearts. Just stay away from the front of their nests.

I've never had a bad encounter with any of the bumble/carpenter bees. Wasps on the other hand... they are an important part of an ecosystem (eating rotting flesh, pollinating, etc) but man, they can be little bastards.

I think the main thing to remember is that, if you're not allergic, being stung means you've just participated in one of nature's miracles: the ability for a bug to take nectar, pollen, whatever, from the environment, and turn it into a searing, painful venom. Hahha!

Do bees socialize with bees from other hives? by TheTedd in askscience

[–]Satoyama_Will 4887 points4888 points x2 (0 children)

Beekeeper here. I assume you mean honeybees (apis mellifera ligustica, carnica, etc)

Hives are known to experience "drift", where bees change colonies. This can happen to both workers (female bees) as well as drones (male bees), moreso with the latter.

This happens for a number of reasons. Bees have a range of two miles and do much of their extracolonial travel visually, using landmarks, so it's often speculated that they become lost or confused and end up someplace else. They are often exhausted from their travels, so we can speculate that they may just collapse to another hive in an act of self preservation, unusual for a "superorganism" that relies on strict cooperation!

So let's talk about socializing. It's well known that a honeybee can make an "offering" to the guards at the entrance of another hive, by regurgitating nectar from their crop, thus bribing the guards who would normally stop other bees. They are then tagged with the scent profile of that hive - like a lot of the pheromone properties in the hive, this part is loosely understood.

Often, honeybees going from one hive to another are robbing. This happens when there's enough honey for another colony's scouts to smell. The attacking colony will send scouts who attempt to overwhelm the guards and rob out the honey stores of the other hive. This usually happens when the robbed colony is weakened, either from queenlessness or disease.

Drifting, however, is different, and represents an interesting behavior among bees. Some people believe it's a vector for disease, which it probably is, but the below study indicates that it's probably not that big of a factor. I think it's just a species preservation behavior. Think human travelers moving from town to town and getting in good with the guards at the gate before being let in. We do it because it just makes sense. Why not invite in some extra hands (or in the case of bees, extra claws)?

An interesting note besides socialization: the Cape honeybee (Apis mellifera capensis) is a subspecies from far southern Africa, originating in a specific coastal region. They have a special faculty, thelytoky, which allows them to lay an egg that is essentially a clone of themselves. This allows them to perform much more effective supercedures (queen raising ) than, say, the european honeybees, who need to raise a queen from an egg already deposited in the hive if there's an emergency or they lose a queen. European honeybees can only produce unfertilized drones, meaning a colony of the european bees without a Queen or eggs will perish.

Well this Cape honeybee, should it infiltrate another hive, such as those of African honeybees (apis mellifera scutelleta), can wreak havoc by laying clones of itself. Those clones hatch and then make clones of themselves. In short order, they overcome the latent genetics in the hive. This has been a menace for many beekeepers in Africa, because Cape honeybees are ill suited to life outside the Cape region.

/u/niandralades2 brings up a great point here, that species preservation is not a good explanation for drifting behavior: https://www.reddit.com/r/askscience/comments/4odwzk/do_bees_socialize_with_bees_from_other_hives/d4cxpwv We talk about it a little bit there.

/u/RosesFernando brings in some information about the wax (a lipid) retaining the scent of the hive, as well as giving some sources on relative aggressiveness of guards: https://www.reddit.com/r/askscience/comments/4odwzk/do_bees_socialize_with_bees_from_other_hives/d4cptzd?context=3

Why is that when you subtract a number from its reverse, the difference is a product of 9? by kingkiller_123 in askscience

[–]functor7Number Theory 2115 points2116 points x2 (0 children)

A number is divisible by 9 exactly when its digits sum to a multiple of 9. So 1836 is divisible by 9, 1936 is not. More exactly, a number and the sum of its digits both have the same remainder when we divide by 9. For example:

  • 3823 divided by 9 is 424 remainder 7. Also, 3+8+2+3 = 16 divided by 9 is 1 remainder 7

  • 92838845 divided by 9 is 10315427 remainder 2. Also, 9+2+8+3+8+8+4+5 = 47 divided by 9 is 5 remainder 2.

So if I want to know the remainder of a number after dividing by 9, I can just look at the remainder of the sum of its digits. They'll always be the same.

Now, remainders work nice under addition, subtraction and multiplication. If the remainder of N divided by 9 is 3 and the remainder of M divided by 9 is 2, then the remainder of N+M divided by 9 will be 2+3=5. If I add numbers, I can just add remainders. Same thing with subtraction and multiplication. Sometimes you might have to take an additional remainder, but it's good. For example

  • The remainder of 40 divided by 9 is 4, and the remainder of 55 divided by 9 is 1. The remainder of 40+55 will be 4+1=5. So 40+55=95 has remainder 5 after dividing by 9.

  • The remainder of 43 after dividing by 9 is 7 and the remainder of 80 after dividing by 9 is 8, so the remainder of 80-43 after dividing by 9 is going to be 8-7=1. In fact, 80-43 = 37, which has remainder 1 after dividing by 9.

  • 3823+92838845 will be divisible by 9 because 3823 divided by 9 has remainder 7 and 92838845 divided by 9 has remainder 2. So 3823+92838845 divided by 9 will have remainder 2+7=9. But 9 can't be a remainder, and 9 divided by 9 has remainder 0 so 3823+92838845 will have remainder 0, which means that it is a multiple of 9.

In particular, if N and M have the same remainder after dividing by 9, then N-M will be divisible by 9. Their remainders are the same, so subtracting them will give remainder zero, which means that it must be divisible by 9.

If you have a number, then reordering it's digits won't change the remainder after dividing by 9, because this does not change the sum of the digits. The remainder of 435 after dividing by 9 is the same as the remainder of 4+3+5=12, which is 3. The remainder of 354 after dividing by 9 is the same as the remainder of 3+5+4 = 12, which is 3. This means that 435-354 will have remainder 3-3=0, so it is divisible by 9.

So, because reversing the digits does not change the sum of the digits, the difference will always be divisible by 9.

The field of math dealing with remainders is Modular Arithmetic.

EDIT: It works for 11,22,33 etc and palindromes, 313,2332 etc, because 0=0*9, which means zero is a multiple of 9.

Would a field that reduces your inertia violate any important laws of physics? by zimirken in askscience

[–]ididnoteatyourcatExperimental Particle Physics 2 points3 points  (0 children)

I'm not sure what you mean by "separated." Photons are ripples in the electromagnetic field. Quarks are ripples in quark fields, and so on. Ripples in photon fields don't interact with much, so they just keep going until they encounter an electron ripple (for example), which might cause the electron and photon to scatter. Ripples in quark fields on the other hand cause ripples in the gluon field, and quark ripples are very strongly attracted to gluon ripples, so they both get sort of glued together, the ripples basically orbiting each other.

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