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July 16

Losing streams

Can someone help me find a reliable source that includes the Nile as an example of a losing stream? It definitely fits the description (its largest water level is at Khartoum, after which it loses a lot to evaporation and irrigation), but all I'm finding is false positives and publications in fields other than geology/hydrology. Nyttend (talk) 13:27, 16 July 2017 (UTC)[reply]

I'm not sure it should be counted as a losing stream by the definition there if it is losing water due to irrigation and evaporation rather than to an underground river or aquifer. Dmcq (talk) 14:37, 16 July 2017 (UTC)[reply]

The question wades into the muddied waters of political contest about irrigation along the river Nile. For millenia, traditional agriculture in Egypt was based on the seasonal filling of the Nile that tends to flood over its east bank. Major dams have been built along the Nile's course in Egypt to restrain the outflow to the delta and to retain a head of water to feed irrigation channels, see Video about Nile Irrigation (1940-1949). The net result is more loss of water by Evaporation. See Water resources management in modern Egypt. Colonial-era agreements between Great Britain, Egypt and Sudan that gave Egypt the right to veto dams in the upstream nations are now disputed. The upper basin nations Ethiopia, Kenya, Rwanda, Tanzania, Uganda and Burundi signed an agreement in 2011 that was intended to reapportion Nile water without a significant affect on Egypt and Sudan, who both strongly objected. In 2011 Ethiopia began work on a massive hydroelectric dam on the Blue Nile near the border with Sudan. Egypt objected to its construction, but in 2015 Egypt, Ethiopia, and Sudan signed an agreement on principles concerning the dam. Blooteuth (talk) 15:51, 16 July 2017 (UTC)[reply]

Take a look at the Colorado River[1]. Scary to look at it. --AboutFace 22 (talk) 00:52, 17 July 2017 (UTC)[reply]

GPS, Gravity affecting time on earth and space, and so on

How does it work? I would like to understand in the simple manner please. 116.58.200.103 (talk) 16:27, 16 July 2017 (UTC)[reply]

Maybe you could narrow your focus a bit? We've got articles on Global Positioning System, gravity, time dilation, and gravitational time dilation for your perusal. There's also a simple English version of the time dilation article located here. This outside site might also be helpful for the basics. Matt Deres (talk) 16:39, 16 July 2017 (UTC)[reply]

I believe time dilation can be ignored for GPS calcs. Just use the speed of light delay to various satellites to find the distance from each, then combine that info to triangulate in on your location. StuRat (talk) 03:28, 17 July 2017 (UTC)[reply]

Yes, it can be ignored for instantaneous positional calculations because atmospheric effects introduce greater errors, but it needs to be taken into account to keep each satellite's clock synchronised. Dbfirs 06:26, 17 July 2017 (UTC)[reply]

Simply put, a place near to a gravity source is subject to a different progression of time compared to a place under less gravity influence. This is a very minimal, barley measurable difference near low gravity sources but extreme near the famous black holes. Since GPS works with extreme short time differenc measurings between multiple satelite signals this can causes a theoretical measure error. However it is far to small to have any practical relevance. --Kharon (talk) 10:30, 17 July 2017 (UTC)[reply]

To expand on why this is: general relativity says that space and time are really just one interlinked thing, spacetime. General relativity also says that gravity is a warping of spacetime caused by things with mass-energy. So, if you're near a massive object, like the Earth, time passes more slowly for you than it does for someone far away. For more, see introduction to general relativity, and I highly recommend PBS Space Time's videos on relativity. --47.138.161.183 (talk) 21:55, 18 July 2017 (UTC)[reply]

Feynman Lectures. Exercises. Exercise 9-8 JPG. Lecture 9

. .

I have solved this exercise by conservation of energy and by dynamic eq. But I need explanation of more deep thing. E.g. we have a dynamometer tied to the wall and a mass 1 kg pulling the free end of the dynamometer horizontally (by pulley). Why does dynamometer show 10 N, but not 20? There are 2 forces applied to the dynamometer : weight of mass 1 kg = 10 N and a reaction of the wall = 10 N. It turns out that all dynamometers are graduated in such a way as to show half the force. Is it correct? Username160611000000 (talk) 21:13, 16 July 2017 (UTC)[reply]

There is no halving or doubling. When you stand on the ground the force that resists you from it is your weight. The force between your feet and the ground is simply that, it is not double your weight because of adding up the force from you weight and the ground resisting. Dmcq (talk) 21:35, 16 July 2017 (UTC)[reply]

Then what happens when the tie is burned and the dynamometer flies horizontally (at first moments )? Or better the massless dynamometer pulls out a brick from the wall and all system (brick - tie - dynamometer - tie - load ) accelerates. From conservation of energy the acceleration of the system =

^{g{\frac {m_{\text{load}}}{m_{\text{load}}+m_{\text{brick}}}}}

. Then the force on the brick =

^{F_{\text{d. on br.}}=m_{\text{brick}}g{\frac {m_{\text{load}}}{m_{\text{load}}+m_{\text{brick}}}}}

. The force from the brick on dynamometer =

^{|F_{\text{br. on d.}}|=|F_{\text{d. on br.}}|}

. The force from the load on dynamometer =

^{F_{\text{l. on d.}}=m_{\text{load}}g}

. If

^{m_{\text{dyn.}}=0}

then

^{F_{\text{br. on d.}}-F_{\text{l. on d.}}=0}

. But we saw that

^{F_{\text{br. on d.}}\neq F_{\text{l. on d.}}}

How is it possible? Username160611000000 (talk) 08:54, 17 July 2017 (UTC)[reply]

It seems in a static system the thread tension differs from that in the moving (accelerating) system ... Username160611000000 (talk) 13:41, 17 July 2017 (UTC)[reply]

I do not know why you point at exercise 9.8. It does not have a dynamometer or a wall or anything going horizontally. It does have a ceiling or roof and two equal masses instead of one plus an extra mass. There is no force on the dynamometer from a burnt tie. Dmcq (talk) 13:21, 17 July 2017 (UTC)[reply]

All these questions are connected. To solve next ex. 9-9 we need to know tension in ropes, to know the tension we must understand all the machinery in ex. 9-8. Username160611000000 (talk) 13:49, 17 July 2017 (UTC)[reply]

Fine so whatever it is you are talking about has some relevance to exercise 9.9. That doesn't answer the question. Where is the exercise with the dynamometer and a wall and a horizontal tie? There is nothing like that in exercise 9.8. There is nothing like that in exercise 9.9. Dmcq (talk) 14:26, 17 July 2017 (UTC)[reply]

There is no an exercise with dynamometer. I introduced dynamometer because I didn't understand how to calculate a tension in the tie. The dynamometer is simplification of a task of finding the tension. The next step was to find the force from the pulley on the ceiling. And in Ex. 9-8 there is just a pulley and a ceiling. It seems that force in kgf on the ceiling ≠ sum of masses in Atwood's Machine [ graph for masses 1 kg and k kg ]. Username160611000000 (talk) 14:41, 17 July 2017 (UTC)[reply]

The tension in the tie is 10 N. It is not double that or half that. In 9.8 there was an acceleration because one side was heavier than the other. There is no static equivalent if the weights are different, only if you are thinking of putting the average weight on both sides. If you put all the weight on one side there would be no tension in the rope even though the total weight is still the same. Dmcq (talk) 15:14, 17 July 2017 (UTC)[reply]

So from 2nd Newton's law the forces on the ends of massless dynamometer are equal. But the tension is not a usual force. It's something like pressure -- force per unit of something. And by convention it is equal the magnitude of one of two forces applied, but not the sum.Username160611000000 (talk) 18:02, 17 July 2017 (UTC)[reply]

It is just a force, nothing special. The dynamometer does not move because the force is resisted by an equal force. When the rope attached to the wall is pulled the wall resists with an equal force otherwise the rope would move. The rope is pulled with 10N. The wall resists with 10N. The rope does not move. There is no 20N. The total force applied to a bit of the rope is 10N - 10N = 0N and therefore it does not move. Dmcq (talk) 21:09, 17 July 2017 (UTC)[reply]

I haven't gone into scribd to try to get your illustration, but this is a familiar confusion. For some reason, where tension is concerned, it is tremendously tempting to add the upward pull and the downward pull. But they are just action and reaction. You rarely feel the urge to do the same thing when a 10 kg weight sits on a bench, even though the bench is putting an "additional" 10kg upward - the compressive force is simply 10 kg, and it remains 10 kg even if you stick a penny under the weight "receiving compressive force from both sides". But with tension it all seems somehow more confusing, until you think about it. Wnt (talk) 00:22, 18 July 2017 (UTC)[reply]

How visible/noticeable is the Aurora borealis expected to be tonight in the northern US?

This article on USA Today, Aurora borealis possible Sunday night across northern U.S., Canada, says that the Aurora borealis is expected to visible tonight in the northern US. I live in suburbs of Chicago and I'm trying to figure out if I will be able to see it. Will I be able to see it over the light pollution? And if so, will it be a steady glow or is it something like watching a meteor shower where I have to be patient and look at the right spot in the sky to see it? A Quest For Knowledge (talk) 23:33, 16 July 2017 (UTC)[reply]

University of Alaska forecast is pessimistic. A more detailed forecast from Space Weather says the best chances are from sunset to around midnight. The aurora is fairly hard to forecast so it could be anything from a bust to a good display. In any case you'll want to get as far from the city glow as possible. Shock Brigade Harvester Boris (talk) 23:41, 16 July 2017 (UTC)[reply]

Also, head towards the north, find a place with a low northern horizon and look north and probably low. Maunder Minimum 2 might be coming so if you haven't seen an aurora this solar cycle or the even weaker one after that might be your last chance (besides going to the Arctic) Sagittarian Milky Way (talk) 01:26, 17 July 2017 (UTC)[reply]

And wait till about 10:30pm to look cause twilight lasts longer in the country than in the city. Sagittarian Milky Way (talk) 01:49, 17 July 2017 (UTC)[reply]

Thanks for all the help/tips. A Quest For Knowledge (talk) 02:09, 17 July 2017 (UTC)[reply]

There are lots of free webcams all over the world online "24/7". You can find multiple that will show you real time simply by googling "borealis cam canada". --Kharon (talk) 10:04, 17 July 2017 (UTC)[reply]

If you look at our article Aurora you will see photographs of the phenomenon. They look like green curtains flickering across the sky. I saw one over a back garden and they are quite unmistakeable. 92.8.217.19 (talk) 12:04, 17 July 2017 (UTC)[reply]