Wikipedia:Reference desk/Archives/Science/2023 January 8

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January 8

What % of glucose undergoes glycolysis, what % of pyruvate undergoes Krebs cycle?

The difference is whether we are jogging/exercising or not. What's the 2nd largest fate for glucose and pyruvate? If pyruvate does not go to Krebs cycle, is it urinated out? I'm guessing the next-common for glucose is glucogenesis.

Also, amylase breaks down starch, while sucrase breaks down sucrose. What happens if sucrose meets amylase or starch meets sucrase, does nothing happen? How does it know the correct 1, like is it a lock-and-key? Thanks. 67.165.185.178 (talk) 01:11, 8 January 2023 (UTC).

Glucolysis occurs within cells and Krebs cycles within mitochondria, so how is it going to be urinated out? It seems to me that once sugar is inside the cells, 100% will be metabolized. All these systems are balanced, unless one has diabetes. As for your second question, an enzyme that meets a chemical that isn't one of it substrates might bind momentarily, but it will soon let go. Abductive (reasoning) 15:14, 8 January 2023 (UTC)

For the 1st part, after pyruvate undergoes Krebs cycle, it becomes CO2. So it is breathed out. You are right it is not urinated or excreted out but I think it is ultimately breathed out even if it gets metabolized. But on your 2nd part on enzymes, I am curious to how that works. I previously asked and looked into, if I have a bowl of water with salt and sugar, is it possible to separate the salt from sugar or sugar from salt, I have those answers. But the biology answer is, enzymes eat out the sugar in the blood vessels and stomach. What happens if you threw in the enzymes (amylase and sucrase) into a bowl of water with salt and sugar, will it know to break down the sugars? Or will nothing happen? 67.165.185.178 (talk) 16:21, 8 January 2023 (UTC).

Enzymes are highly specific as to what chemical they work on. "Lock and key" was formerly used as a metaphor for this, but it is actually somewhat more complicated. Usually only one particular substance will really fit into the enzyme's active site well enough for the enzyme to be able to work. If two molecules are extremely similar, an enzyme might be able to work on the second one to a limited degree. But salt is completely different from any sugar, so there is no chance of enzymes that breakdown sugars working on it. If anything, the ions of the salt might bond to various parts of the enzyme and make it less active. --User:Khajidha (talk) (contributions) 15:00, 11 January 2023 (UTC)

I'm also hearing that sucrose dissolves in your stomach to become chyme. So that's reacting with HCl? So, if you have a bowl of water with sugar and salt, probably nothing will happen if you throw in sucrase, though maybe if you throw in HCl and then throw in sucrase, the sucrase will eat the (chyme). 67.165.185.178 (talk) 03:20, 12 January 2023 (UTC).

I think you are misunderstanding what chyme is. --User:Khajidha (talk) (contributions) 03:33, 12 January 2023 (UTC)

Turning soil

The second paragraph of Plough says, unsurprisingly, The prime purpose of ploughing is to turn over the uppermost soil, bringing fresh nutrients to the surface while burying weeds and crop remains to decay. But the section Mould-board ploughing says To grow crops regularly in less-fertile areas, it was once believed that the soil must be turned to bring nutrients to the surface. Why does it say it was once believed? Is it not in fact necessary to turn the soil? Is this a contradiction?  Card Zero  (talk) 09:09, 8 January 2023 (UTC)

@Card Zero There are many factors a farmer is juggling when deciding to plough or not. The article on no-till farming gives a reasonable account of most of them. For example, what benefit is achieved by bringing nutrients to the surface may be outweighed by the soil erosion that it can encourage. Mike Turnbull (talk) 13:39, 8 January 2023 (UTC)

Dust Bowl shows what can happen with traditional ploughing in a dry area. And it can lead to bad erosion on sloping ground if there is a lot of rain. So farmers are being a lot more careful about what they do when preparing the ground so the surface is protected. NadVolum (talk) 12:42, 12 January 2023 (UTC)

Surviving free fall

If I were in free fall at terminal velocity and hit the top of a frictionless parabolic ramp my kinetic energy in the vertical direction would be gradually converted to horizontal. Assuming a comfortable survivable acceleration of say 2g, what size would the ramp need to be?

Sounds like homework. But why are you doing homework at this time of year? Graeme Bartlett (talk) 11:49, 8 January 2023 (UTC)

@Graeme Bartlett Schools and colleges where I live re-opened for learning since 4 January. Bazza (talk) 11:53, 8 January 2023 (UTC)

A quarter-circular ramp may give you a constant acceleration. The parabola gets tighter as you get to a low point. Graeme Bartlett (talk) 11:55, 8 January 2023 (UTC)

Not at all homework; it was something I was thinking about on the drive home last night and worked it out for a circular ramp setting omega squared r to 2g. The parabolic ramp seemed to make more sense at the time but I guess it doesn't. 2A01:E34:EF5E:4640:B0B4:C298:AF41:C711 (talk) 18:01, 8 January 2023 (UTC)

As we can read in Circular motion § Formulas, the acceleration due to change of direction of an object moving at constant speed along a circular trajectory is constant (in absolute size). However, the object, not being in free fall, will also experience the gravitational force.  --Lambiam 13:13, 8 January 2023 (UTC)

I agree with Graeme Bartlett - the parabola is inappropriate and may even lead to a multitude of possible answers. Instead, the question would be better based on the Tautochrone curve. Dolphin (t) 13:30, 8 January 2023 (UTC)

Assume we have a smooth symmetric convex curve whose tangent is vertical at the point of entry, whose depth to the bottom (counted from the point of entry) is $d$, and whose curvature at the bottom is $\kappa .$ Let $v_{0}$ denote the speed of the infalling acronaut at the point of entry, and $g$ the gravitational acceleration. We ignore the effect of friction (even though the concept of "terminal velocity" assumes friction) and model the acronaut as a point particle. At entry, the particle is in free fall. Unless the change of direction of the curve is insufficiently gradual, the maximum force experienced will be at the bottom of the curve. There the initially vertical velocity is entirely horizontal; from the law of conservation of energy we know that the speed $v$ is determined by ${\tfrac {1}{2}}v^{2}={\tfrac {1}{2}}v_{0}^{2}+gd.$ The vertical acceleration due to change of direction equals $v^{2}/r$, in which $r=\kappa ^{{-}1}$ is the radius of curvature. To this we must add the gravitational acceleration $g$. Putting things together, this results in

$\kappa ({\tfrac {1}{2}}v_{0}^{2}+gd)+g\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\text{----------------------}}$    $\kappa (v_{0}^{2}+2gd)+g.$

So, ideally, the curve should be flatter at the bottom than the tautochrone curve. If $\kappa =0,$ the formula simplifies to just the least possible value, $g.$ The precise shape of the curve is irrelevant.  --Lambiam 23:30, 8 January 2023 (UTC)

Its unclear whether or not the air resistance about their body remains constant or is reduced any by the OP's "frictionless ramp" (which seems unlikely since air still has to be displaced). Considering then that the air resistance of their terminal velocity remains constant (in practice these vary a lot), their tangential velocity would also remain constant (so there would be no increase in their kinetic energy) until they almost completely stopped falling and then their velocity decreases thereafter upon the level terrain. In a nutshell, with the air resistance a smaller ramp is required than without it. Modocc (talk) 03:25, 9 January 2023 (UTC)

@Lambiam Solving for $v^{2}$ requires $v_{0}^{2}+2gd$ not ${\tfrac {1}{2}}v_{0}^{2}+gd$ if I'm following your derivation of the total acceleration correctly. I just now noticed this. Modocc (talk) 00:33, 12 January 2023 (UTC)

Thanks for spotting this; now corrected.  --Lambiam 01:53, 12 January 2023 (UTC)

I have used the free fall slide at Questacon, so they do work.[1][2] (but not tested at terminal velocity) Graeme Bartlett (talk) 21:35, 8 January 2023 (UTC)

This all seems much more complicated than I originally thought. When the faller hits the top, vertical, part of the ramp her acceleration is the combination of g downwards and D/m (drag/mass) upwards (net magnitude zero). When the faller reaches the bottom, horizontal, part of the ramp her acceleration is still g downwards and D/m horizontally (net magnitude less than 2g, but the drag force will be lower since her speed has been reduced by air resistance). At any point inbetween her acceleration is the combination of g downwards, D/m tangential to the local curve and omega squared r towards the local center of curvature. Seems like this is more of maths question than a science one after all. 2A01:E34:EF5E:4640:EB51:3C5F:A88E:8F35 (talk) 09:56, 9 January 2023 (UTC)

Likewise, let the heroine simply grab hold of a taught rope swing that is long enough to gradually oppose g until its insignificant, but the drag persists hence the deceleration. Modocc (talk) 19:17, 9 January 2023 (UTC)

• In terms of shape, you want neither a semi-circular nor parabolic ramp for this. I believe the ideal shape for such a ramp is almost always a tautochrone curve, which, IIRC, allows for an ideal deceleration. These curves are also called brachistochrone curves or cycloids, depending on how they are created and/or utilized, but they're all functionally identical. --Jayron32 16:21, 10 January 2023 (UTC)

  I think that for the tautochrone curve to work its miracles, the sliding object must have zero velocity somewhere along the (finite) curve,  --Lambiam 17:19, 11 January 2023 (UTC)

  Could be. I was spitballing a bit, but from my memory "tautochrone" usually ends up being relevant for optimizing behaviors. It shows up in all kinds of optimized models from Fermat's principle to Geodesics to calculating the ideal path of a pendulum when measuring out periods of equal time. Though, doing some more looking, it appears the ideal curve may be a clothoid rather than a cycloid. Clothoid loops are used in roller coaster design for smoother accelerations. See Here. Regardless of which shape I am mis-remembering, I'm pretty sure it isn't a semi-circle. --Jayron32 19:00, 11 January 2023 (UTC)

  With air drag it can be modeled as a damped oscillator. Thus if the frictionless ramp is sufficiently large such that it has a fairly long "flat" and then it ramps upward at its lowest point their speed could be brought to zero by the air drag alone. But it would be far larger than what the OP required which was simply to bring the vertical velocity component to 0. Modocc (talk) 19:12, 11 January 2023 (UTC)

  With a sufficiently long flat in a uniform gravitational field, the speed in the idealized mathematical model will eventually get arbitrarily low, but in the end some additional minute gravitational pull is needed to bring the object to a full stop. I think the whole operation can be done without getting the g-force ever above 1 g.  --Lambiam 02:04, 12 January 2023 (UTC)

  Right. I wasn't very clear on this, but by "flat" in quotes I meant a long relatively flat or almost flat length with a stationary point about which they will oscillate back and forth and settle into. Modocc (talk) 02:27, 12 January 2023 (UTC)

OBAFGKMRNS to OBAFGKMLTY

Why did the extended Stellar sequence change from OBAFGKMRNS to OBAFGKMLTY?? Georgia guy (talk) 16:05, 8 January 2023 (UTC)

The R, N, and S types are (or were) for stars with non-standard compositions. The rest of the sequence (OBAFGKM) are in order of decreasing temperature, the L, T, and Y classes continue the pattern of temperature decrease. --User:Khajidha (talk) (contributions) 18:39, 8 January 2023 (UTC)

Discovery of brown dwarfs caused the revision. Graeme Bartlett (talk) 21:23, 8 January 2023 (UTC)