What is the advantage of having a Q cycle in the ETC in spite of its complexity?
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Good question.  I presume that "Q cycle" is in reference to "coenzyme q/CoQ/ubiquinone/UQ"'s activity.\n\nMy guess is that it evolved to help stabilize NADH; otherwise, if NADH were to build up, it could cause some serious havoc in the cell by reducing everything.
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This weblink will explain the differance between the 2\nhttp://www.erg.abdn.ac.uk/users/gorry/eg2069/async.html
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http://en.wikipedia.org/wiki/Kreb%27s_cycle
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To provide a simple answer that does not involve a lot of mathematics, differential equations are used in almost every scientific field to provide a mathematical description of a process.  They are useful because most processes can be described in terms of changes that result from small differences in elements of the process.  When these small changes are represented in mathematical form the result is a differential equation.  Because the changes are small, it is possible to make assumptions that simplify the expression of the process.\n\nFor example, consider the discharge of a capacitor into a resistor.  We know that the charge stored in the capacitor is Q = V*C, where V=voltage on the capacitor, and C is the capacitance.  A small change in charge, dQ will cause a small change in voltage, dV.  The voltage across the resitor will be V, and the current in the resitor V/R.  The current is the amount of charge divided by the time, or change in current is change in charge divided by change in time, or dI = dQ/dt.  (Note: here is the simplifying assumption:) For very small changes in Q, we can assume that the voltage does not change, so that the current is constant in that interval.  Therefore dV=(1/C)dQ = idt = (V/C*R)dt; rewriting this as a differential equation gives\n\ndV/dt = V/(C*R),\n\nwhich describes the discharge of a capacitor into a resistor.  This is a particularly simple example, and differential equations of more complex processes can get very messy, and many cannot be solved explicitly.
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the retention times are longer for everything. The gas flow rate is the only variable decider in the speed of transit through the columns. Slower flow means slower movemnt independent of column interactions\n\nThe separation should theoretically be better for everything besides the components that don't interact at all with the column. They come out at dead/void volume anyway. The other components are better separated because the slower flow rate means they have more time to interact with the stationary phase and so the differences in elution rate become more pronounced.
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Good day:\n\nCircular buffer is very important to store analog data before transfering to digital format. Circular buffer is effective to swicth to many devices simultaneously. Intel webiste provide lots of DSP information!\n\nThank you, bye!
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The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
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Time constants in electronic circuits.
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Ok, so the relative co-efficient of x is 1, so the unrelative co-hedrent homeomorphic must obviously be 2 to the power of Y-as Y is homeomorphic to X's relative undescribed co-efficient.\nUnderstand my friend. I hope I helped.
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Easy to use, logical, does what it is supposed to, and doesn't cause problems that it is not intended to.  It would also be nice for it to be aesthetically pleasing.\n\nThis is a very generic description, but then yours is a very general question.
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This again is a very vague question, I can only guess that the output stage shows the timing graphs. But unless you provide us with a link to a particular site or reference of what you are trying to ask, then we can not know the question exactly.\n\nPlease bear in mind you lose two points every time you ask a question. So make you answer as much as you ask !\n\nAnhar Hussain Miah
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independent: voltage, length of wire, number of coils\ndependent: magnetic effect (paper clips)
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Advantages, you only have to hit the 1 key.\ndisadvatages, no one knows what the heck your asking.\nDUH
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In compiler theory, loop transformations play an important role in improving cache performance and effective use of parallel processing capabilities. Most execution time of a scientific program is spent on loops. Thus a lot of compiler analysis and optimization techniques have been developed to make the execution of loops faster.
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one immediate advantage comes from trying to answer your question without mathematics font on my computer!! \n\n but I would say the main advantage occurs when mutiplication of bases or radicands (things under the root) is reqiured ie:\n** ill have to use root(m,x,n) to mean "the mth root of  \n   x to the nth power** \n\nsuppose you must multiply root(3,x,2)*root(4,x,3)\n\nusing rational exponents we have x^(2/3)*x^(3/4). now since base x is the same problem reduces to merely adding 2 fractions w/ different denominators.\n 2/3+3/4 => 4*2/4*3+3*3/3*4 => 8/12+9/12 => 17/12\n\nso x^(2/3)*x^(3/4) = x^(17/12) = x^(12/12+5/12) = x^(12/12)*x^(5/12) = x*x^(5/12) or x*root(12,x,5) or x times the twelveth root of x to the 5th power.\n\nnow attempt this multiplication using only radicals to verify my conclusion
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I don't really understand what your question is.  In the first place, heat always flows from a warmer object to a cooler one.  That maximizes the number of possible arrangements of energy quanta, leading to an increased entropy.
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Imagine a block of memory from 0 to N. A cyclic buffer fills from 0 towards N, and when it reaches N it starts back at 0 again.\nThis kind of buffer needs to be "serviced" (have the data removed) before new data overwrites data that hasn't been removed yet.\nThere is normally an "in" pointer for incoming data, and an "out" pointer for outgoing data, If the "out" pointer reaches the "in" pointer, the buffer is empty again. If the "in" pointer reaches the "out" pointer, you've overflowed the cyclic buffer!