studiot

But that does not prevent there being limitations. In fact there are a whole variety of them for many different purposes, made more complicated by differences in criminal and civil law. Coincidentally I had a conversation last night with an old school friend, now retired, but a solicitor by profession. He was telling me about a some grit in a piece of bread which damaged a tooth. I asked if he was going to sue and he said "probably not, but he had three years to think about it "Me "3 years ??" Yes he said, you have to issue proceedings within 3 years for this type of thing. If only Imatfaal was still here to explain.

The subject being the takeup of CO2 in desert conditions. In fact I devoted 7 paragraphs in my last post to that as against the four lines to an aside, which I maintain offers a balanced view of that aside, before putting it fully aside.

As with so many things in the world, comparisons can be made to fit many different and often opposing profiles. Here are some examples:- It depends who's buying. Perhaps solar or other renewable energy is more expensive per unit than traditional mains power. Especially if you buy a large amount. But suppose you want to put in 500 miles of new illuminated notices to motorists along a rural motorway? Do you include the cost of digging and backfilling the 500 mile trench to lay the cable when comparing the cost with pole mounted solar panels? And what about the maintenance of this cable over the years v simply fitting a new panel, as you would a light bulb? And who pays for the cable itself, since you are the only user? Or suppose you want to keep the batteries in your sailing yacht topped up. Is not a small wind turbine mounted on the mast an obvious solution? These have been available for many years. Both of these applications demonstrate a particular point. Applications generally are low density power users at the point of use whether domestic or light industrial. No one or two users combined can sink the output of a modern power station, that takes many tens of thousands or even hundreds of thousands. So you have a place of very high density power, which has to be scaled down for each user, as well as distributed. Who pays for this very expensive system? Now compare this with renewable. Renewable tends to be low density power. Yet we still try to collect it all back together to power station densities. Why? Surely it makes more sense to use it at low density.

Out in my garden this afternnon, picking giant peaches for Mr Dahl, I spotted this little fellow flitting about amongst the 2 foot butterflies. He came to rest for a moment on a gardens seat so I snapped him quickly.

Thank you for this comment on my question, I will follow it up. Before I came across the China article I had not given any consideration to the thought of getting carbon dioxide into groundwater but thinking about it I realise that it must happen and, more to the point, the uptake rate will likely increase in any water desert areas. Carbon dioxide is mildly soluble in water, obviously more so in conditions where the resulting acid can be withdrawn eg by reaction with carbonate rocks. The equilibrium is a dynamic one and over the open ocean the water (at the top at least) will be fully saturated with carbon so the net solution rate will be low. The oceans remove so much CO2 simply because they are so vast and of course because of the photosynthesising life in them. In the ground we have vadose water (vadose = german for wandering) above the phreatic surface ( = water table) https://www.google.com/search?q=vadose+water&ie=utf-8&oe=utf-8&client=firefox-b In this region the saturated ground below the water table is in gaseous equilibrium with the air in the soil or rock pores. So it will take up CO2. Here is your mechanism. I suppose that in desert areas there is a mostly dearth of groundwater, so if there is a large subsurface reservoir it will be working pretty hard extracting CO2. Thank you for your reply, I am sorry that you received 2 downvotes as a result of answering my thread, perhaps your last line was too strident and perhps was taken to suggest they are all like this, though I am sure that some have chosen the easy route to obtain grant money on a bandwagon subject. But certainly not all, there are definitely seriousnplayers out there as well. So here is a +1 as partial compensation. I would welcome any further conversation on the subject.

Whilst I can think of many legal applications of your film, such a device would be illegal in the UK and no doubt the US and many other places. Interestingly you are asking a similar question to the one I asked decades ago (not here). Electronic devices that had a control electrode to switch on or increase current or voltage go back to the 1920s (Valves, thyratrons, thyristors, fets, transistors etc) Although corresponding switch off devices were invented in the early 1960s they were not made practical until the 1990s. (GTO devices)

I thought it was the other way round.

I already noted that Wolfram Alfa will solve the exponential equation for you. Type in solve 16^x=x^2 WA will automatically choose the graphs with appropriate limits. edit cross posted with strange! But the is is the point of a discussion site. To extend one's knowledge and capability. I certainly cleared some rust out.

Well originally said this because this separates exponents and other functions of x on one side of the equation and pure numbers on the other, I also looked at something similar to what OldChemE said, +1 , but inExcel. Isn't is so easy these days to get a graph of your 16x times table? [math]{x^2} = {16^x}[/math] [math]{x^2}{x^{ - 2}} = {x^{ - 2}}{16^x} = 1[/math] Not just any old numbers but 1 in particular. 1 has two advantages A) 1 is an easy square number B) the log of 1 to any base = zero, always a useful number to have in an analysis for a solution. anyway I thinking harder I realise that this extra line does not add anything useful although it does lead to a solution, so [math]{x^2}{x^{ - 2}} = {x^{ - 2}}{16^x} = 1[/math] This can be recognised as the difference of two squares so [math]\left( {{{16}^{\frac{x}{2}}} - x} \right)\left( {{{16}^{\frac{x}{2}}} + x} \right) = 0[/math] removing the square root [math]\left( {{4^x} - x} \right)\left( {{4^x} + x} \right) = 0[/math] It is now clear that either the first bracket or the second or both are zero. But it is also clear that the bracket cannot be zero for any [math]x \ge [/math] Of course we can do the difference of squares again to simplify further [math]\left( {{4^{\frac{x}{2}}} - \sqrt x } \right)\left( {{4^{\frac{x}{2}}} + \sqrt x } \right) = 0[/math] [math]\left( {{2^x} - \sqrt x } \right)\left( {{2^x} + \sqrt x } \right) = 0[/math] But I am still in a quandary. Was the OP meant as a puzzle (challenge) or was this really homework, in which case I can't proceed further.

Would a Kerr cell do you? https://www.google.co.uk/search?source=hp&ei=tYajXdHUBYvIwQKSlL7AAw&q=kerr+cell&oq=kerr+cell&gs_l=psy-ab.3..0l8.196.2072..2706...0.0..1.748.3248.2-1j0j2j2j1......0....1..gws-wiz.......0i131.IRR3KjHGLV0&ved=0ahUKEwiRg-Oxh5rlAhULZFAKHRKKDzgQ4dUDCAc&uact=5

Interestingly one logarithmic method does indeed gets you stuck. [math]{16^x} = {x^2}[/math] take logs to the base 16 since log16(16)=1 [math]x{\log _{16}}\left( {16} \right) = 2{\log _{16}}\left( x \right)[/math] [math]x = 2{\log _{16}}\left( x \right)[/math] put this into wolfram alpha equations solver and you get no real result. but it does give [math]{\log _{16}}\left( { - 0.5} \right) = - 0.25[/math] but WA can solve the equation using the algebraic form directly and algebraic continuation. What I don't understand is why this is in homework help, if it is called a 'challenge' and what you want to achieve ? There is another section for puzzles and brain teasers.

Well I made x = minus one half Does this help?

The point of school science projects is to find out what you can and can't do. It is not like woodwork where the object is to show you can accurately cut a piece of wood to a certain shape and size, and this is a measure of your success. So the point of the ac v dc is that the generated magnetic field will be different. Can you think why? So do you really want an inverter to convert dc to ac ? And where would you get mains voltage dc at that power level? Rewinding coils is a serious undertaking. It can be done but there are many opportunities for failure along the way. And what would you achieve by rewinding? Yes I'm very happy to discuss your ideas and try to make you think. So don't get the idea I'm being negative or avoiding the issue. I'd hoped your teacher would also perform this function, which is why I suggested talking to him/her.

The General Philisophy section is a good place to ask then. And the philosophical answer (after at least 5 whiskys) is that unless there was both something and nothing you would not be able to distinguish between them.

They have calculators these days. I wasn't even allowed a slide rule. But what is wrong with my method? It allows you to almost dispense with taking logs at all.

I can't imagine why you haven't tried what was suggested. You have 2 perfectly good suggestions to be going on with. These days logs are not taught so you may be wary of using them. If in doubt ask for more help.

Well it's a good thought, but will probably turn out differently than you expect. Induction hobs are ac devices so the mag field generated is alternating or oscillating. Ordinary electromagnets are dc. Here are a couple of picture from google. 700 × 601 250 × 184 These show the flat spiral winding of the coils and the resultant field with its harmonics. Have a think about what this means talk to your teacher and come back to us if you want more advice.

Wrong It is not a good idea to try to mess with basic definitions. The period of any vibration or oscillation is the time it takes for one vibration to complete. (not 9192631770 of them!) The physical quantity known as period has a unit of seconds , nanoseconds, femtoseconds etc. Here is a downloadable pdf that may help you now and in the future It contains some useful tables. https://i1.dainikbhaskar.com/web2images/education/phy_unt_13659_13897.pdf frequency has units [math]\frac{1}{{{\rm{seconds}}}}[/math] or seconds-1 or s-1 or dimensions T-1 in the scheme outlined in the pdf. Note the distinction between units and dimensions.

No. This is the question that exercised Cantor and Dedekind so much and boiled down to What do you want the properties of 'point' and 'line' to be ? An infinitesimal was regarded as a function, although the modern view of a function as a type of mapping was just then arriving. This requires that one can (theoretically) separate off any point (or subset or aggregate of points) of a 'point set' and perform an oepration (function) on it (them). Changing the court from clay to grass or even concrete has implications in tennis and changing the underlying ground in continuum maths has similar repercussions.

How can something be generalised Newtonian and non Newtonian ? I did ask you to state Newton's Law of Viscoscity. Using your symbols it is [math]\tau = A\frac{{du}}{{dy}}[/math] or in words shear stress = a constant * rate of shear It is also a straight line through the origin. So Newton's law has B = 0 What about the other constants in your test equation? Can you now allocate these for Newton's Law? You should sketch this graph, plotting shear strees against on the y axis rate of shear on the x axis. So to generalise Newton.s Law What will the introduction of a constant B do to any such plot? What about the values of n and A ?

Does anyone have more up to date information on the underground water system in the Tarim Basin China which is alleged to operate as a giant carbon sink ? Full 2015 article here https://www.scmp.com/tech/science-research/article/1845192/huge-hidden-ocean-under-xinjiangs-tarim-basin-larger-all-great

That is the question. You surely must have some ideas of some of them, especially the value of B for parts (I) and (II) So what have you done so far? What is Newtons law of viscosity?

As the football coach said "It is a game of two halves" So perhaps you should be considering half periods? You have dashed off a lot of formulae/calculations but what about the Physics? Can you describe the Physics of what happens? Remember this is a mechanical system with no dissipative forces so it obeys the principle of conservation of (mechanical) energy. I suggest you say something like Let the bob be drawn a small distance sideways to the left and then released. Then describe what happens at each significant point in the travel and the Physics of why. I think the significant points are 1) The initial starting point 2) The bottom of the travel as the string becomes vertical. 3) The bottom of the travel as the string passes vertical. 4) The top of the travel as the string reaches its max rightward excursion. 5) The bottom of the travel as the string again arrives at the vertical. 6) The bottom of the travel as the string passes vertical now travelling leftwards. 7) The top of the leftward travel when the string is back at the initial starting point.

w/q = p/I is a long way round to get to Ohm's law, but yes ok to here [math]E = IR = B\frac{{dA}}{{dt}}[/math] However you will not get the right answer if you just follow the example in Res & H . Theirs is simpler. Remember that R depends upon the length of the perimeter of the triangle and this is increasing with time and the motion of the blue wire. Backalong I suggested to get the geometry done first. This was because in R&H the blue wire has a constant length whereas here it is increasing. In fact all linear dimesions increase in the asame ratio, which is why I wanted you to draw the line velocity vector through the origin at 45o. So you could see for youself that the length of the base of the triangle (ie the blue line) increases linearly with time by the area of the triangle increase as the square if this since it is the product of two dimensions, both subject to the same increase in time. The geometry of similar triangles, as mentioned, confirms this by another way.

Well done you have learned how to do superscript. But as a member since 2008 you surely know we don't do your homework for you. However here is a hint to start you off. multiply the equation through by x-2 Now tell us how you got +10 upvotes on one post and I will