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August 31

Winds picking up at night

I know the winds are powered by the sun, therefore the winds can only pick up during the day as the presence of the sun add energy (heat) to the air, causing pressure differences and triggering the wind. But sometimes the winds can pick up in the middle of the night, even though the sun has long been set. How can the winds pick up significantly at night? Maybe the winds must've powered by something other than the sun, but by what? Planet Star 05:57, 31 August 2017 (UTC)[reply]

See Wind#Causes of wind. The Sun heats the atmosphere and ground during the day and the ground temperature sinks during the night as the ground loses more heat to the atmosphere. Large-scale air pressure differences cannot equalize immediately and are the cause of Prevailing winds, including local gusts that arise both by day and by night. All wind is powered by the Sun. Blooteuth (talk) 10:05, 31 August 2017 (UTC)[reply]

Bucket-spoon for cement

The masons working with trowels use a tool that is shaped like a small bucket (or a big ladle) and having a handle - for taking the cement close to where they work. In Romanian it's called "cancioc" but I can't find the translation. Can anyone help me to find it's name please? Here's some photos with this tool.

Thanks. — Ark25 (talk) 08:24, 31 August 2017 (UTC)[reply]

Initial googled dictionary searches merely suggest "ladle". In the UK I've not seen such a tool used, and various British online catalogues of bricklaying tools do not include one, so its use may be a non-UK practice, with no English-language term specific to this builder's tool.

British builders sometimes use a different tool for carrying small quantities of mortar/cement, a flat square plate with a vertical handle underneath called a hawk: this is held in one hand/arm with a small quantity of mortar piled on the plate, while the other hand uses the trowel (c.f. Hamlet's remark "I am but mad north-north-west. When the wind is southerly, I know a hawk from a handsaw.") Another name for this tool was a mortar board, but as the link suggests, this name has now mostly transferred to the academic headgear originally so-named as a joke. {The poster formerly known as 87.81.230.195} 90.204.180.96 (talk) 11:19, 31 August 2017 (UTC)[reply]

It's a mortar ladle.[1] More Eastern European than British. Used with wetter mortars than British bricklaying mortar, such as those for rendering. Some have a four foot long wooden handle, so that they can be used as a dipper for lifting mortar up onto staging. In Britain we'd use a bucket, and an assistant to carry it up. Andy Dingley (talk) 12:35, 31 August 2017 (UTC)[reply]

The version with the long wooden handle is a brick hod, used for carrying bricks or mortar. Dolphin (t) 13:27, 31 August 2017 (UTC)[reply]

A hod is quite a different beast. They aren't waterproof, for one thing (no one wants a rain-filled hod). Andy Dingley (talk) 13:34, 31 August 2017 (UTC)[reply]

Thanks for all the answers. Searching for "mortar ladle" on Google images returns images with "Mortar and pestle". Only found one image - named "Construction steel ladle". Does anyone know the name of this tool in any other languages? And maybe there is a Wikipedia article on some other language about this tool? — Ark25 (talk) 14:05, 31 August 2017 (UTC)[reply]

2:56 Have you heard, it's in the stars! See Big Dipper in Ursa Major. [2] [3]. The Western asterism is now known as the "Northern Dipper" (北斗) or the "Seven Stars of the Northern Dipper" (Chinese and Japanese: 北斗七星; pinyin: Běidǒu Qīxīng; Cantonese Yale: Bak¹-dau² Cat¹-sing¹; rōmaji: Hokutō Shichisei; Korean: 북두칠성; romaja: Bukdu Chilseong; Vietnamese: Sao Bắc Đẩu). The personification of the Big Dipper itself is also known as "Doumu" (斗母) in Chinese folk religion and Taoism, and Marici in Buddhism. Blooteuth (talk) 16:47, 31 August 2017 (UTC)[reply]

Au contrair User:Blooteuth, surely a dipper is one of these, used for scooping water out of a bigger container such as a barrel. You can get platic ones now. Alansplodge (talk) 17:30, 31 August 2017 (UTC)[reply]

Bien sûr, it's what Swedes call a Vannøse and equally useful for water or cement. I looked up "platic" and am no wiser. Blooteuth (talk) 18:41, 31 August 2017 (UTC)[reply]

Thanks but "Vannøse" seems to be a word used mostly for frying pan - check the first image in no:Øse - also Google images contain mostly cooking items when searching for "Vannøse". I think I'll ask on russian.stackexchange.com - german.stackexchange.com etc. Too bad there's no Polish SE yet. — Ark25 (talk) 21:44, 31 August 2017 (UTC)[reply]

The water dipper seems to have a longer handle and it's not used for masonry. — Ark25 (talk) 21:47, 31 August 2017 (UTC)[reply]

Feynman Lectures. Exercises. Exercise 12-12 JPG. Lecture 12

. .

...
12-12. An object rests at the base of a frictionless 20° incline 1.00 m long (slant). If the incline is accelerated along the table with an acceleration a = 4.00 m sec ^-2 , how long does it take the object to slide to the top of the slope?

— R. B. Leighton , Feynman Lectures on Physics. Exercises

I have solved the exercise by means of reference frame connected with the incline . But I would like to know how to solve it in inertial frame of reference, connected with the table. I suppose there is only 2 forces on the object png . But I have no idea, how to find R. We have 2 equations of motion for each axis (3 unknowns) and we cannot use the principle of virtual work because the system is not balanced... Username160611000000 (talk) 15:28, 31 August 2017 (UTC)[reply]

I didn't go diving in your bazillion links to see your answer, but I'm thinking that in an inertial frame of reference, the inclined plane is being accelerated upward at 9.80665 m/s^2 and, say, to the right at 4.0 m/s^2. That should give a resultant of 10.59105 m/s^2 that is directed 67.81013 degrees up, 22.18989 degrees right. But the plane itself is tilted right 20 degrees, so the angle is just 2.18989 degrees to the right. Now the object on the plane receives a force that is its mass times the acceleration times the cosine of that; but it receives no force along the plane, so it lags behind according to sin 2.18989 * 10.59105 m/s^2 = 0.4046998 m/s^2. since d = 1/2 (a*t) * t, t = sqrt(2d/a) = 2.223046, I think. Wnt (talk) 03:04, 2 September 2017 (UTC)[reply]

I'm thinking that in an inertial frame of reference, the inclined plane is being accelerated upward at 9.80665 m/s^2 and, say, to the right at 4.0 m/s^2 -- Why? According to my image png to the body was at the top of the incline, the incline should be accelerated to the left. In this ex. we have the table (inertial reference frame or lab frame), the incline and the body.Username160611000000 (talk) 10:12, 2 September 2017 (UTC)[reply]

Oh, I just went by the text above, and the question didn't say which way the incline inclines. (I have no patience to go through half a dozen links, enable scripts, download what may be pirated textbooks etc. in the process of trying to figure out what the question is) Whichever way, we automatically assume the acceleration is in the direction that would make the object tend to go up it. My mental image was mirror image of yours.

Ok, but then the incline can't be accelerated upward, because the table acts on it with reaction force equal to the weight. Username160611000000 (talk) 03:51, 3 September 2017 (UTC)[reply]

dozen links, enable scripts, download what may be pirated textbooks -- When I type jpg or png, these are direct links to the images , no need scripts or to download something (actually all images in browser windows are downloaded. If you will turn off javascripts globally, it's safe). Username160611000000 (talk) 03:55, 3 September 2017 (UTC)[reply]

Unless the ramp is in free fall, it is being accelerated. The 9.8 m/s^2 force per mass downward does not exist in an inertial frame of reference. So the ramp is being accelerated 9.8 m/s^2 upward by the force provided by the table, just as it is being accelerated 4.0 m/s^2 to one side by some other force. The two should be considered equivalent in an inertial frame of reference, which is what you requested. Wnt (talk) 16:04, 4 September 2017 (UTC)[reply]

Why was Hurricane Adolph added to the name list?

Sagittarian Milky Way (talk) 16:35, 31 August 2017 (UTC)[reply]

Do the articles Tropical cyclone naming and List of retired Pacific hurricane names help? Blooteuth (talk) 16:55, 31 August 2017 (UTC)[reply]

I can see why it was retired. Why was it on the name list and for so long? Sagittarian Milky Way (talk) 17:07, 31 August 2017 (UTC)[reply]

They try to use a variety of names to avoid picking from one cultural group. Adolph is a common name. Once a storm is given a name, the name is commonly retired to avoid confusion by having a second Hurricane Adolph. Similarly, Katrina has been retired. So has Hugo, and many others. 209.149.113.5 (talk) 17:14, 31 August 2017 (UTC)[reply]

It's unreasonable to blame Adolf for Adolph. Blooteuth (talk) 18:48, 31 August 2017 (UTC)[reply]

This is more for the language desk, but I note in passing that it seems to be more acceptable in British publications, than in American ones, to anglicize names, for example to write "Adolph Hitler". The one that particularly jangles my nerves is when people refer to "George" Cantor. On the other side, I think Americans are more likely to write "Josef" Stalin, which can be seen as a hyperforeignism, given that "Josef" is not particularly closer to "Ioseb" than "Joseph" is. --Trovatore (talk) 19:28, 31 August 2017 (UTC) [reply]

I thought that the question may be wondering why the name of a bad person was used. I assumed that the question was based in intelligence, even though I know that the United States is currently in a wildly fascist frenzy of censoring history. That is very new. Not long ago, liberal groups were fighting all the way to the Supreme Court to allow idiots to say and do things that any reasonable person would find offensive. Therefore, I expect the current spasm to pass as soon as the media has trouble selling advertising with it. 209.149.113.5 (talk) 19:53, 31 August 2017 (UTC)[reply]

It's perhaps worth noting what actually happened here. While this isn't explicitly noted anywhere that I noticed if you look at our articles and the sources [4] and understand how things wrong, the name Adolph was I guess used since naming started in 1960. None of the retired ones would suggest Adolph was a replacement. Then in 1995 Ismael was retired due to the damage and deaths and replaced with Israel. When people noticed this (I'm not sure if the name Israel was only decided in 2001, or it had been decided after 1995 but no one paid attention) some people raised concerns over Israel especially since there was a minor risk of "Israel destroys ....." headlines. So Israel was never actually used and replace instead with Ivo which didn't actually do anything. People also noticed Adolph at the same time as it was in the same season and so Adolph was retired after that season [5]. In other words, if Ismael hadn't done all that damage or Israel hadn't be chosen to replace it, there's a fair chance we'd still have Adolph. Nil Einne (talk) 20:21, 31 August 2017 (UTC)[reply]

Adolph was retired because it was a major hurricane. All major hurricanes have their names retired. 71.85.51.150 (talk) 00:37, 1 September 2017 (UTC)[reply]

Except that all sources presented thus far disagree with you. All including NOAA say that the reason it was retired was due to political reasons, and NOAA and IIRC some other sources (I think this includes at least one of our articles) also say hhurricanes which don't make landfall and don't cause any notable damage nor deaths or other significant impact to human populations (the worst seems to have been closure of two ports to small ships) despite some warnings and fears it would, are not necessarily retired even if they were very strong. If you have other sources which contradict these sources, please present them. To be fair, Adolph was new record for the basin in May since recording begun in the mid 1960s which in itself is not necessarily that significant since there could have been a lot of those, but was also the first category 4 in May since recording begun, so there are reasons it may have been retired even without making landfall (although I'm not sure if records 'in May' are really enough to push them into retiring in the absence of something else). So perhaps there was discussion over whether to do so and the political reasons are what pushed them over the edge. Or maybe they actually decided to retire the name due to the significance of the last Adolph, but then said they did it for political sensitivity reasons. Maybe there are blogs or records which talk about these issues. But none of them have been presented this far and they don't seem to be in our articles so we can only go by what the sources actually say namely that it was retired for political reasons, and with a strong hint the reason this came about was because the plan to name a source Israel drew attention to the issue. Nil Einne (talk) 09:10, 1 September 2017 (UTC)[reply]

BTW, I'm not sure what you meant by major hurricane, but if if you simply meant it was a category 4, you may want to take a look at List of Category 4 Pacific hurricanes. It lists 118 hurricanes. Unfortunately it doesn't separate between the eastern and central Pacific although I believe eastern Pacific hurricanes may be more common and in any case I'm fairly sure you'll find other category 4sin the eastern Pacific which weren't retired because they too didn't have any significant impact. As the earlier list shows, the retired names is a lot smaller and not all of these are even category 4 or higher. In fact I'm fairly sure that means you'll find some List of Category 5 Pacific hurricanes which weren't retired. Edit: Actually I just found Hurricane Dora (1999) which was category 4, if you look at Tropical Storm Dora, you'll see the name is still being used.Nil Einne (talk) 10:13, 1 September 2017 (UTC)[reply]

Are foot orgasms real?

Do they truly exist? — Preceding unsigned comment added by Uncle dan is home (talk • contribs) 21:53, 31 August 2017 (UTC)[reply]

Well, I only found this reference. It appears that foot orgasm does exist for one woman in the world. 50.4.236.254 (talk) 23:31, 31 August 2017 (UTC)[reply]

Did'nt Coca Cola build an world empire on that? When we are heated up and very thirsty the drinking of cold sweet water with a fruity taste can get so intense that we may even be stunned and have "involuntary muscular contractions". That looks pretty close to "orgasms". --Kharon (talk) 02:36, 1 September 2017 (UTC)[reply]

I'd like whatever Kharon is getting, please. This study describes how to induce in animals an itch paroxysm comparable to that which occurs in man. Blooteuth (talk) 12:33, 1 September 2017 (UTC)[reply]

Guinea worm. μηδείς (talk) 21:36, 1 September 2017 (UTC)[reply]

You're definitely not one of the Knights who say Ni in organism. Dmcq (talk) 21:46, 1 September 2017 (UTC)[reply]

I just bought reading glasses today. μηδείς (talk) 18:29, 2 September 2017 (UTC)[reply]

Whatever floats your boat, some have spontaneous orgasms for neurologic reasons or through intense fantasy ideas, some have them while dancing or kissing... some even have them from pain or the sensation of submission (and others via sadism). I remember some tabloid news mentioning someone having orgasms from being thrown mellons at. For more information: orgasm, BDSM, fetishism. Feet are also quite sensitive, it's possible that for some they are like an erogenous zone. —Paleo Neonate – 21:48, 1 September 2017 (UTC)[reply]

Look at the sensory homunculus. The big toes are next to the genitals, as far as the brain is concerned. Have your lover suck your big toes while stimulating you genitally. Any good hetaira knows this. μηδείς (talk) 18:27, 2 September 2017 (UTC)[reply]

Rocket consumption and energy

A curiosity: but the rocket consumption (assuming that the weigth of propellent is very small compared to the weight of the accelerated object and an efficiency of 100%) is proportional to the impulse given to the object or to the kinetic energy given to the object? Because in the first case, for example, with a dobule velocity given the consumption would be double, but in the second case the consumption wolud be four times higher. The more logical solution would be the first (because according to the relativity an object can be considered ever stationary in his frame of reference, so the consumption can only be a function of the time multiplied for the force and even because a force produces a consumption even if doesn't move an object) , but how could this conciliate with the kinetic energy theory? Thanks for answering, I know that this isn't an easy question, Francis. — Preceding unsigned comment added by 82.48.80.146 (talk) 22:03, 31 August 2017 (UTC)[reply]

You may be interested in our article on the Oberth effect. -- ToE 22:20, 31 August 2017 (UTC)[reply]

By conservation of energy, we can say that the initial potential energy of the rocket (the fuel) equals the final kinetic energy of the payload and propellant, plus the gravitational potential energy, plus energy lost to friction and heat. Typically, the kinetic energy is much greater than the gravitational potential or heat because rockets go very fast. So in a simplified analysis, you need four times the fuel to make your payload go twice as fast. This is true for cars as well: it takes four times as much energy to accelerate a car to 20 m/s compared to 10 m/s (ignoring friction).

The theory of relativity says that the laws of physics are the same in all stationary reference frames. It does not apply to the payload being accelerated in a rocket. If the velocity of an object changes, then the object's reference frame is not stationary. You might be interested in the twin paradox, which is resolved in a similar way.

For rockets going into space, our quadratic approximation is very optimistic, because propellant takes up 85% or more of the mass of real rockets. That means that adding more propellant increases the weight of the rocket significantly, and you need to add more propellant to propel that extra weight, and so on. Your velocity increase from adding propellant diminishes, or phrased another way, you need an exponential amount of propellant to keep increasing the velocity of a rocket. NASA has a good explanation with real numbers. C0617470r (talk) 00:05, 1 September 2017 (UTC)[reply]

Our article Tsiolkovsky rocket equation may be of interest. {The poster formerly known as 87.81.230.95} 90.204.180.96 (talk) 16:45, 1 September 2017 (UTC)[reply]

Thanks to all of you for answering. But there is one thing that I don't realize. I give you an example: the Earth moves respect to the Sun at a speed of 35 km/s. If I launch a 2 kg object on the Earth in the same direction of orbitation movement at the speed of 5 m/s the object gains respect to the Sun an accretion of kinetic energy of 350.025 Joules (35.005^ equals to 1.225.350.025 respect to the quadratic of 35.000 that is 1.225.000.000) with a consumption of energy of only 25 Joules. And in other, if for example is given to a 100 kg object a thrust (in the opposite direction of gravitational force) of only 50 kg of force, there is howewer a propellent consumption even if there isn't any movement. How colud be this explanable without thinking to the more logical solution: the consumption is proportional to the force applied, not to the kinetic energy? Francis — Preceding unsigned comment added by 87.3.17.123 (talk) 22:33, 1 September 2017 (UTC)[reply]

What may be bothering you is that the change in kinetic energy of an individual object is not invariant under Galilean relativity, whereas change in momentum is invariant. Consider a 2 kg object R (for Rocket) which changes its velocity by 1 m/s. In the reference frame where the object was initially at rest, its momentum changed from 0 kg⋅m/s to 2 kg⋅m/s, and it's kinetic energy changed from 0 J to 1 J. But in the reference frame where it was initially traveling at 1000 m/s and finally traveling at 1001 m/s, its momentum changed from 2000 kg⋅m/s to 2002 kg⋅m/s (the same Δ of 2 kg⋅m/s), and its kinetic energy changed from 1,000,000 J to 1,002,001 J. At fist glance it may not make sense that in one frame the object gained only 1 J, but in another it gained 2001 J. How can that be consistent with conservation of energy? Part of the answer is Newton's first law. Your object wouldn't have changed velocity unless it was acted upon by a force, and you must consider the entire system, including those other objects from which the force originated, in order to apply conservation of energy. So, lets give that object a reason for changing velocity.

Consider a second object with an attached spring, together called F (for Fuel), total mass (object and spring) 2 kg, such that if F is connected to R with the spring compressed between them, and then the connection is broken, the two objects will fly apart at a relative velocity of 2 m/s. In the frame where the objects were initially at rest, after the catch is released each is traveling at 1 m/s, so the kinetic energy of each object went from 0 J to 1 J, and that 2 J total increase came from the potential energy of the spring. Now consider the frame where they are initially traveling at 1000 m/s, with final velocities of 1001 m/s and 999 m/s for R and F, respectively. R's kinetic energy changed from 1,000,000 J to 1,002,001 J, the same 2001 J increase we calculated earlier, and F's kinetic energy change from 1,000,000 J to 998,001, a loss of 1999 J. Combining the two, we have that same 2 J increase in total kinetic energy. Energy is conserved.

Now look at it just a little differently and consider two of these 2 kg rockets which started from rest, one loaded with only one unit of fuel and the other loaded with a huge amount of fuel so that after expending all but one unit of fuel it is traveling at 1000 m/s. Then they both expend their last unit of fuel, and the first rocket accelerates from 0 m/s to 1 m/s gaining 1 J, and the second accelerates from 1000 m/s to 1001 m/s gaining 2001 J. If you think if unfair that the same unit of fuel did so much more work on the moving rocket than on the stationary one, remember that half of the huge amount of fuel loaded onto the second rocket was spent getting that last unit of fuel up to speed, and it is giving up some of its kinetic energy which was gained at that fuel's expense. You could even say that the last bit of fuel was supercharged beyond its 2 J of potential energy with the 1,000,000 J of kinetic it had as well, though it only manages to give up 0.2% of that. But change your reference frame and that supercharging seems to disappear. -- ToE 23:44, 1 September 2017 (UTC)[reply]

Is global sea level drop due to heavy rainfall events measurable?

We can read here:

"Harvey has unloaded 24.5 trillion gallons of water on Texas and Louisiana"

If all that water is on land at the same time, then that amounts to roughly 0.3 mm sea level drop. Now it may be that a lot of the water has flowed back to sea by the time the last part of the 24.5 trillion gallons fell from the sky, so the sea level drop may be a bit less. Count Iblis (talk) 22:35, 31 August 2017 (UTC)[reply]

I concur with your calculation: 24.5E12 us_gal / 360E6 km^2^Ocean = 0.26 mm. -- ToE 23:16, 31 August 2017 (UTC)[reply]

Do you mean measurable as in within the level of significant figures or greater than noise level? Or do you mean measurable as in "I can see that distance on a ruler"? The answer is different depending on which you are asking about. --Jayron 32 01:26, 1 September 2017 (UTC)[reply]

No, you have to include all water in the atmosphere, all water in the underground, in rivers and inland seas, all water bound by Biomass and all water bound by ice, because all that is part of the Water cycle, as is the Harvey rain volume and the oceans you picked out. Additionally water is not a static mass and therefor you should include Ocean currents, moon position, local Atmospheric pressure(weather) and last not least the local Physical geodesy to conclude a sea level rise or drop. --Kharon (talk) 02:11, 1 September 2017 (UTC)[reply]

It might be worth considering that 'Harvey' has not been the only catastrophic event in the world recently. Richard Avery (talk) 07:27, 1 September 2017 (UTC)[reply]

Just to be explicit, that is not a discernible change. The precision of annual, global average sea level is 5-10 mm. Local sea level can be discerned more precisely when the weather is very calm, but most of the time it is not. Dragons flight (talk) 08:44, 1 September 2017 (UTC)[reply]

The first bit sounds reasonable (not to say unimpressive!), any ref for that? SemanticMantis (talk) 14:17, 1 September 2017 (UTC)[reply]

It would seem trivial to construct a pseudoharbour to emulate "Very calm" weather. All the best: Rich Farmbrough, 12:52, 2 September 2017 (UTC).[reply]

If something useful seems trivial, but is not widely done, there are is usually at least one explanation. It is not as trivial as it seems ;-). In this case, you would need a body of water that communicates with the sea (so as to reflect the actual sea level), but which excludes not only surface waves, but also dynamic changes in pressure (e.g. those caused by waves colliding against an underwater inlet). --Stephan Schulz (talk) 13:55, 2 September 2017 (UTC)[reply]

It's a big assumption that the amount of water that fell was an addition to the amount in the air anyway or all came from the sea. And shouldn't the sea level actually rise nearby if the land is heavier with water on it though it might fall elsewhere in the world? If so that makes the concept of sea-level just a little murkier.Dmcq (talk) 17:17, 1 September 2017 (UTC)[reply]

I see, so an order of magnitude below what can be detected. Count Iblis (talk) 05:28, 2 September 2017 (UTC)[reply]

Also, consider that the land itself is plastic, so when it is depressed in one area, it may rise in another. Think about what happens when you press into dough or clay with your hand. The handprint is depressed, but the rest of the soft matter rises around the depression. If it were sea-water causing this to happen, the effect would pretty much balance out. μηδείς (talk) 18:23, 2 September 2017 (UTC)[reply]