pavelcherepan

Pantheism fits the description. Taoism might work too.

Then this discussion should be in Speculations forum. Also there's been drilling, rock sampling, spectral analysis and whole heap of other studies done on Mars to confirm the presence of water. Also, would you please attach a Google map image showing the supposed volcanic crater or give coordinates of it? What picture are you talking about? Lava formations coming onto the ocean floor are generally referred to as "pillow lavas", because they do look like pillows. So your discussion of volcanic eruption onto the ocean floor must be supported by examples of known pillow lavas in the region with thicknesses of several kilometers as you assume. There are some pillow lavas in the northern part of India, but at most those are several meters thick and discontinuous.

Just a thought here. Say, humanity goes to colonize Titan, which is awesome, much cooler than Mars both literally and figuratively. So, given that Titan has a nice and dense atmosphere in order to explore it faster using planes makes sense. So a normal jet engine uses atmospheric oxygen as oxidizer and carries fuel with it. Atmosphere of Titan on the other hand has about 5% of methane (in the troposphere) and no oxygen so is there any roadblocks from engineering perspective in creating a jet engine that uses atmospheric fuel rather than oxidizer and carry oxidezer on board? It it possible to create an engine that would work on a 5% concentration of fuel in the air?

So with this logic an american teenager, for example, has free will to go either for Democrats or Republicans, but then if later in life he/she makes a choice and sticks with it for the rest of the life, does it mean that he/she had lost free will? Or is it maybe because a person is very stubborn, or maybe the original choice was just so good, or maybe the choice was not good, but he/she is too stupid to understand it?

Hi there, I'm slightly confused. There are different interpretations of Satan, but for now let's talk about the fallen angel, who had rebelled against God after God created men and after gaining knowledge of upcoming incarnation of God as a man and refusing to bow to inferior creatures. So in this version he has free will (1) and exists eternally (2) being an angel. Since the reason for his rebellion didn't go away, there's no reason for him to repent, even though he retains free will, and then there's no salvation, so (not 5). Using the same logic I see no contradiction between Satan being forever "evil" from our perspective and still retaining free will. In his own perspective he'd still believe that he's doing the right thing. I see no reason why he'd change his mind. And then (7) breaks down.

In my studies I found that what really helps improve memorizing a lot of information before exams is making cheat sheets. I made them nice and tiny and laminated and all handwritten. I didn't actually use those at exams, but the process of making those helps immensely. First of all, as cheat sheet is generally small, you can only put so much information on one, and also you should try to limit the number of them to a minimum, as a result you have to carefully process all the information that you're going to write down and prioritize. This alone stimulates your memory a lot since long-term memory works best by association and you've already done such a work on carefully ranking all the data by the matter of priority. Then comes the writing part, where the very mechanical process of scribbling words on a piece of paper and obviously reading them at the same time improves your memorizing. Most of the time, when I was thinking about some question on the exam I could picture in my head which particular cheat sheet the answer was on and where on the sheet it was located. After that remembering is extremely easy. Also you could try mnemonic methods as other posters have suggested.

One thing sort of fits what's described in the paper are Deccan Traps in south-west India, but those were a long way away from Tibet.

You haven't shown what reaction is used to synthesize PbBr2. I would assume, that it's the most common one: 2KBr(aq.)+Pb(NO3)2 = PbBr2(s) + 2KNO3(aq.) 4.32g of lead(II) bromide is 0.0117 mols as you have calculated, then 0.0117 mols of lead(II) nitrate is required for the reaction. Lead nitrate is created from metallic lead via the following reaction: Pb + 2HNO3 = Pb(NO3)2 + H2(g) Since we have 0.0117 mols of lead nitrate, the same amount of metallic lead is required, 0.0117 mols or 2.48 grams.

First of all, creation of Tibetan plateau through a collision of continental plates and volcanism are not mutually exclusive. It is a normal situation for converging plate boundaries to develop volcanism specifically on the side in which direction the oceanic plate is being subducted. As subduction goes, oceanic plate partially melts and the molten magma raises up due to its buoyancy and then normally finds its way to the surface through faults created as a result of stress. Although, in the situation with Himalayas and Tibet, volcanism precluded formation of the mountains and overlapped with orogeny only slightly, because Tethys ocean was completely closed about 50 mya and a couple million years after that the last bits of oceanic crust would have sunk into the mantle and no longer contribute to volcanism. Secondly, the paper attached doesn't specify the location of presumed "giant volcanic crater" and I've been looking on Google maps and can't find one. Thirdly, plate tectonics on this planet is not very supportive of forming long-term giant volcanoes. Due to plate moving relative to one another and varying angle of subduction over time (it normally starts shallow and finishes very steeply) the location of partial melting zone moves with respect the overlaying continental crust from further inland towards the converging boundary. You can see this in modern day Kamchatka, Russia. Older volcanic ranges (dormant) are further west (inland) and all the active ones are along the eastern coast. And lastly, the paper says "The lava had flown through the sides of the crater except through Himalayan ranges", but large-scale volcanism could not happen AFTER the Himalayan range has formed, because by that time all remainder of the ocean plate is long gone and there is no source for magma. Then volcanism would have to mostly happen BEFORE the range had formed and then it would overlay other deposits and then we'd see extrusive rocks on top of Mt. Everest, rather than limestone, dolomite and marble. EDIT: Oh, and two more points. What the paper or rather the abstract describes can mostly be understood as a shield volcano which is highly unlikely to form on a continent in the absence of "hot spot". Subductive volcanism on a continental plate generally has magma ranging from andesitic to rhyolitic which results in explosive eruptions that produce little lava, but a lot of tuff and breccia. Also, what the paper describes is not a theory but a hypothesis. If it were a theory it would have to include a lot more of important independently verified data, such as correlated stratigraphy columns, that show the extent of presumed lava flow, extrusive rock samples from various areas that have been dated using radiometric methods and much more.

True here as well. My original conditions were that Vsmall = V, Vmiddle = 2V, Vouter = 9/4V, probably imatfaal misread it or i didn't explain too well, but what matters are not numbers themselves, but the method of solution. I've done calculations using the proper volumes, it's too easy once the method of arriving to the result has been shown so well.

Sensei, you're right here, but using full heights of all three cones makes it easier to get to the final result. In my original calculations (which were obviously incorrect but only by a difference between square root and cubic root of the same number) I used partial heights and getting to the result was very tedious. It's much simpler with full heights as imatfaal has done and I can still get ratios I need.

I would still argue that technology did advance at a higher rate during wars. Maybe not science, but technology and engineering definitely. During a war you have a lot of pressing issues that you have to develop solutions for quickly and a lot of stimuli as at time the very survival is at stake and also the time is of the essence. As a result there is a great push to solve myriads of problems as fast as possible and using as little resources as possible. The science proper on the other hand would suffer a lot given that international collaboration and availability of resources of multiple international scientific institutions are important in order for the science to advance faster. All or most of these are not available during wars and financial/human resources are taken from scientific front towards technology.

Wow! Thanks a lot guys! I thought that using triangular sections and areas of those might not be the correct way but for some reason couldn't figure out a way to do it with volumes. Imatfaal, you're a legend! I should be fine to finish it from here.

Here are a couple of really good videos from two of the most popular science channels on Youtube. The first video explains how permanent magnets work and the second one explains how special relativity helps understand electromagnetism. Check them out, hopefully they will answer most of your questions: https://www.youtube.com/watch?v=hFAOXdXZ5TM https://www.youtube.com/watch?v=1TKSfAkWWN0

Howdy all! I have yet another question, this time in the realm of mathematics. For a project I'm doing at work I need to create an optimal sampling method for a drill cone and so the basics of the problem are like this: We have 3 cones nesting inside one another or a cone with 3 "layers". They all have the same base angle [latex]\alpha[/latex] but different radii and heights. If we use V as the volume of the bottom cone, then middle cone is 2V and the outer (largest) cone is 9/4V. Based on this I need to calculate percentages of the total height of the entire cone for the following: h3 - from the top of the outer cone to the top of middle cone h2 - from the top of middle cone to the top of bottom one h1 - from the top of inner cone to the ground Probably it will be easier with a picture. I've done calculations but for the ease of it I just used a section through the middle of the cone, so that I only have to deal with areas of triangles and in the end I got the following: h1 = 0.632*h, where h is the total height of the cone h2 = 0.262*h h3 = 0.106*h Does this seem plausible? I could attach my calculations but those are pretty messy. Thanks for the help!

Oy vey! I knew it was some very stupid mistake. [latex]\frac{gt^2}{2}[/latex] is the distance traveled by free-falling body, not sure why I used it here. Thanks for the help guys!

Howdy all! I've ran into a slight issue and I hope you guys can clarify where I went wrong. Basically there I was today sitting at some boring meeting barely listening and out of boredom was trying to derive formulas for ballistic projectile trajectory. Unfortunately, when meeting finally finished and I went into wiki and compared what I have to what the actual formulas are mine were incorrect and I've ran some numbers and I definitely get wrong results. It's all very embarrassing, but I've checked my scribbles and all seems reasonable. Anyways, I went like this: At point 1 on the surface level we have a projectile launched with a velocity [latex]v_1[/latex] at an angle [latex]\alpha[/latex]. At point 2 the projectile has reached maximum height and about to start falling back down. By energy conservation we have [latex]E_{k1} + E_{p1} = E_{k2} + E_{p2}[/latex]. Obviously, at point 1 projectile being at surface level the potential energy is 0, then: [latex]\frac{mv_1^2}{2} = \frac{mv_2^2}{2} + mgh[/latex] Dividing both sides by m and multiplying by 2 we get [latex]v_1^2 = v_2^2 + 2gh[/latex], but since the vertical component of the original [latex]v_1[/latex] vector is now 0, [latex]v_2[/latex] is essentially the horizontal component of the starting velocity or [latex]v_1*cos(\alpha)[/latex]. Then, [latex]v_1^2 = v_1^2*cos^2(\alpha)+2gh[/latex] Simplifying that I got: [latex]h = \frac{v_1^2(1-cos^2(\alpha))}{2g}[/latex], or [latex]h = \frac{v_1^2sin^2(\alpha)}{2g}[/latex] After that I went to derive the range of projectile. Since two parts of trajectory, namely, surface to highest point and highest point back to surface are essentially mirror images of one another I got: [latex]R = v_{1h}*2t [/latex], where [latex]v_{1h}[/latex] is the horizontal component of the starting velocity and [latex]t[/latex] is the time required for projectile to get to the highest point of trajectory. At point 2 vertical velocity is 0, then: [latex]v_{1v} - gt^2/2 = 0[/latex] simplifying for t, [latex]t = \sqrt{\frac{2v_{1v}}{g}}[/latex], but since [latex]v_{1v} = v_1*sin(\alpha)[/latex] [latex]t = \sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex] And then range becomes: [latex]R=v_{1h}*2*\sqrt{\frac{2v_{1}*sin(\alpha)}{g}} = 2v_{1}cos(\alpha)\sqrt{\frac{2v_{1}*sin(\alpha)}{g}}[/latex] And this is all wrong. I've definitely done some very stupid mistake but for the life of me can't find where exactly. Any ideas?

I guess I have to re-post it. "Normal magnetite is attracted to a magnet but some specimens are automagnetized and have the ability to attract small pieces of iron, small pieces of magnetite and other magnetic objects. This form of magnetite, known as "lodestone", was man's first encounter with the property of magnetism. Lodestone is easily identified because it is usually covered with small particles of magnetite and other magnetic minerals (see photo). " This. http://geology.com/minerals/hematite.shtml

"Normal magnetite is attracted to a magnet but some specimens are automagnetized and have the ability to attract small pieces of iron, small pieces of magnetite and other magnetic objects. This form of magnetite, known as "lodestone", was man's first encounter with the property of magnetism. Lodestone is easily identified because it is usually covered with small particles of magnetite and other magnetic minerals (see photo). " http://geology.com/minerals/magnetite.shtml Of course, why wouldn't I? Mineral Composition Magnetic Order Tc(°C) σs (Am2/kg) Magnetite Fe3O4 ferrimagnetic 575-585 90-92 Hematite αFe2O3 canted AFM 675 0.4 Where [latex]\sigma[/latex] is saturation magnetisation at room temperature. http://www.irm.umn.edu/hg2m/hg2m_b/hg2m_b.html

I think that if the velocity vector is along the x-axis and observers are also on the x-axis then they wouldn't observe any contraction.

Actually, you can also use magnet for separation of those. For example in iron ore beneficiation, for magnetite magnets are used to separate it from waste rock, but for hematite is so weakly magnetic that magnets are not used for separation, instead density flotation separation is used.

Yes, John, I do. That's why I said "piece of iron", not a magnet. Maybe I should've made it more clear. Black dust will stick to the iron, red dust shouldn't, as it's very weakly magnetic.

Fe3O4, or more precisely Fe2+Fe23+O4 is magnetite and the "red dust" is hematite. The former is ferromagnetic and the latter is anti-ferromagnetic, hence a piece of iron would work fine to separate those.

Thanks! I didn't think of Oberth effect. Although I think in the movie when landing crafts got separated main engines were off already. I might be wrong though.

I was re-watching Interstellar last week (my wife hasn't seen it before) and had a question about yet another physics inaccuracy in the movie... or so it seems to me. In the end in order for Dr. Amelia Brand to get to the last planet in the system they haven't visited yet her and Coop fly Endurance and perform a powered gravity assist using central black hole for that purpose. First issue is that you can't use central body of the system containing pretty much all the mass as a gravity assist, because in that stellar system FoR is would have zero momentum and hence you can't get any out of it. Although they did briefly mention a neutron star in the same system, so it's possible that the barycentre of the system is somewhere outside of black hole. But let's forget this for a moment. The other issue is the way gravity assist itself was done. They fired engines of the spaceship and landing crafts at the closest approach to BH and then as the fuel ran out detached landing crafts... to get more momentum. This doesn't make sense to me. In the movie it looked like crafts were simply detached, not fired away and in that case why would Endurance get extra velocity from it? [latex](m_e+m_c)*v=m_ev+m_cv[/latex] where e is fore Endurance, and c is for landing crafts. I can't see why they had to drop Matthew MacConaughey into black hole apart from the fact that without it the plot would fail even to a greater extent than it had as it is.