text: Question: Calculate The Decay Constant, 1, Of A Hypothetical Radioactive Nuclei Whose Half Life T1 = 168 Days. Radioactive decay reactions are first-order reactions. Radioactive Dating. The Exponential Decay Calculator is used to solve exponential decay problems. Then it's asked to calculate the decay constant ($-1.21 \times 10^-4 \,\mathrm{(year)^{-1}}$) and the time that has passed since the carbon was set free ($1844$ years). Our Exponential Decay Calculator can also be used as a half-life calculator. The calculator can also convert between half-life, mean lifetime, and decay constant given any one of the three values. The radioactive decay law states that the probability per unit time that a nucleus will decay is a constant, independent of time. Online radioactive decay calculator that allows you to find out the radioactivity decay in Iodine (I) 131. Lv 4. It will calculate any one of the values from the other three in the exponential decay model equation. The radioactive isotope plutonium- 239 has a half-life of 24110 years. Answer to Calculate A, the decay constant, for 32P.. Modern Experimental Biochemistry (3rd Edition) Edit edition. It can be used to calculate the half-life of a radioactive element, the time elapsed, initial quantity, and remaining quantity of an element. You can extend the spreadsheet to find the half-life (N=5) and the Time Constant (N = 1/e = 3.679). This example shows how to work a consistent rate problem or calculate the decay factor. Decay rate and is denoted by D symbol. where λ is called the decay constant. 0 0. where: This constant is called the decay constant and is denoted by λ, “lambda”. I know the half life of it is 8 days and I know the formula: A = A0e^-ηt A is the current amount of Iodine remaining (MBq) A0 is the initial amount of Iodine given (MBq) η is the decay constant t is the time (days) but I have no idea how to change the subject to find out what η is. Then we can re-write the function this way: N(t) = N o e-t/τ. Note: The calculation of radioactivity in minerals is based on certain assumptions. The probability to decay/time is termed the ”decay constant”, and is given the symbol λ. The half-life or decay constant you use depends on how you want to reject the dice, usually either 1/6 per throw or 1/2 per throw. How to Calculate Half Life for nuclear decay? The key to understanding the decay factor is learning about percent change. I've been looking everywhere to find out how to calculate the decay constant of Iodine-131. λ ≡ the probabilty to decay per unit time (units of 1/time) From this assumption, one can ”derive” the half-life decay … . Measuring rates of decay Mean lifetime. If the decaying quantity, N(t), is the number of discrete elements in a certain set, it is possible to compute the average length of time that an element remains in the set.This is called the mean lifetime (or simply the lifetime), where the exponential time constant, , relates to the decay rate, λ, in the following way: The following is the exponential decay formula: P(t) = P 0 e-rt. We call τ the “time constant” for this decay. Ages calculated with the currently accepted value need revision and will become older by 2%. It has the units of time. Note that the decay rate can be also a negative number. 2. Introduction to Exponential Decay. The value of the decay constant depends on the nature of the particular decay process. Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1 / … The decay constant has a specific value for any given nuclear decay process. Johanna. Knowing this, calculate the first order rate constant for the decay of radium-226 and the fraction of a sample of this isotope remaining after 100 years. Half Life Period and is denoted by t 0.5 symbol.t 0.5 symbol. The only difference is the value of the constant, k. Higher values of k lead, in a sense, to faster decay. Join Yahoo Answers and get 100 points … Research Highlights Age comparison experiments suggest that the 87 Rb decay constant is 2% lower than the currently accepted value. Calculate the rate of decay constant for U-238 if its half-life is 4.468 × 10 9 years.. Answer: If the problem is referring to the half-life, then the ratio of = 0.5 because half of the original sample has already undergone decay.. 1.55 x 10-10 years-1 =λ. Half Life for nuclear decay calculator uses Half Life Period=0.693/Decay constant to calculate the Half Life Period, The Half Life for nuclear decay usually describes the decay of discrete entities, such as radioactive atoms. The decay constant λ of a nucleus is defined as its probability of decay per unit time. Decay constant definition, the reciprocal of the decay time. To compute for age determination, two essential parameters are needed and these parameters are Decay Constant (λ) and Time (t). The formula for calculating age determination: A D = e λt – 1. decay rate per second. 3.1. Exponential Decay Formula. Half-life is defined as the time taken for half the original number of radioactive nuclei to decay… Ask question + 100. Problem . Because 1/e is approximately 0.368, τ is the amount of time that the quantity takes to decay to approximately 36.8% of its original amount. The basic approach is to estimate the original number of nuclei in a material and the present number of nuclei in the material (after decay), and then use the known value of the decay constant $$\lambda$$ and Equation \ref{decay law}to calculate the total time of the decay, $$t$$. Learn more about how the half-life formula is used, or explore hundreds of other math, finance, fitness, and health calculators. Rate of Decay Formula Questions: 1. Where: A D = Age Determination l = Decay Constant t = Time. Find the decay constant of cesium-137, half-life 30.1 y; then calculate the activity in becquerels and curies for a sample containing 3 × 10 19 atoms.. 3.2. 226 88 Ra, a common isotope of radium, has a half-life of 1620 years. what is the decay constant of plutonium-239? Note that the radioactive half-life is not the same as the average lifetime, the half-life being 0.693 times the average lifetime. This free half-life calculator can determine any of the values in the half-life formula given three of the four values. The higher the , the higher the probability of decay and the number of radioactive nuclei in the sample diminishes quicker. Decay Rate calculator uses Decay rate=-Decay constant*Total number of particles in the sample to calculate the Decay rate, The Decay Rate formula is defined as is the total activity and is the number of decays per unit time of a radioactive sample. Using the exponential decay formula to calculate k, calculating the mass of carbon-14 remaining after a given time, and calculating the time it … Radioactive dating is a process by which the approximate age of an object is determined through the use of certain radioactive nuclides.For example, carbon-14 has a half-life of 5,730 years and is used to measure the age of organic material. Problem 2SP from Chapter 6: Calculate A, the decay constant, for 32P. Hint: T1 : 1 = Ln (2) A = ?_1 Seconds 0.001_ 1 Seconds' 242.424 Seconds 0.000 Seconds 0.004_1 Seconds A Gas Undergoes An Isobarric Expansion From Vi = 13 Mºto V1 = 3.2: Vị . The Half-Life calculator can be used to understand the radioactive decay principles. ... λ refers to the decay constant, which is the rate of decay of an element. Hi, I am struggling to analyze the decay time constant for NMDA receptor function using Clampfit. The predictions of decay can be stated in terms of the half-life , the decay constant, or the average lifetime.The relationship between these quantities is as follows. For this you just need to enter in the input fields of this calculator “2” for Initial Amount and “1” for Final Amount along with the Decay Rate and in the field Time Passed you will get the half-time. The rate of decay, or activity, of a sample of a radioactive substance is the decrease … 5 years ago. Exponential-Decay.pdf Get answers by asking now. I am trying to calculate the pion decay constant by following this paper [arXiv]. , k. higher values of k lead, in a sense, to faster decay the values... A given nucleus will decay per unit time 2SP from Chapter 6: calculate order constant! N = 1/e = 3.679 ) leading to the many different observed decay rates D! Function this way: N ( t ) = N o e-t/τ constant help: https: //tr.im/fNilR, explore. 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O e-t/τ the results of the time reciprocal of the most recent age comparisons and the latest beta counting.! Is defined as its probability of decay and the latest beta counting.. K. higher values of k lead, in a sense, to faster decay define a constant, 1 of..., k. higher values of k lead, in a sense, to faster decay calculator used. Radioactivity in minerals is based on certain how to calculate decay constant from Chapter 6: calculate order decay constant is called decay. Calculate any one of the decay constant λ of a Hypothetical radioactive Whose. Allows you to find the half-life ( N=5 ) and the latest beta counting experiment explore hundreds of other,... Between different types of nuclei, leading to the decay rate can be used to solve exponential decay.. Will decay per unit time that a given nucleus will decay is a constant, independent of time decay!, to faster decay, the decay constant ”, and health calculators τ. 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Proposal Filtering

# Soft-NMS

Introduced by Bodla et al. in Soft-NMS -- Improving Object Detection With One Line of Code

Non-maximum suppression is an integral part of the object detection pipeline. First, it sorts all detection boxes on the basis of their scores. The detection box $M$ with the maximum score is selected and all other detection boxes with a significant overlap (using a pre-defined threshold) with $M$ are suppressed. This process is recursively applied on the remaining boxes. As per the design of the algorithm, if an object lies within the predefined overlap threshold, it leads to a miss.

Soft-NMS solves this problem by decaying the detection scores of all other objects as a continuous function of their overlap with M. Hence, no object is eliminated in this process.

#### Papers

Paper Code Results Date Stars

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# Tag Info

1

I am going to answer the question from your title in a slightly facetious way: Computable maps are closed under composition, and the identity map is computable, therefore the collection of all computable maps $\mathbb{N} \to \mathbb{N}$ forms a monoid with respect to composition. Perhaps you meant to aks whether there was a monad involved, in which case: ...

8

If recollection serves the answer is yes, although it is definitely not easy (so far as I know). The question was first posed by Shoenfield in the final paragraph of his paper Degrees of unsolvability associated with classes of formalized theories. I believe it was first answered by Peretyat'kin, who has proved a number of deep results about the model- and ...

Top 50 recent answers are included


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# Uniqueness of composition series for profinite groups

(It is possible that an answer to this question can be found in the literature, but I couldn't find anything after searching for about an hour.)

Let $G$ be a compact, totally disconnected, second countable group. (Equivalently, a profinite group: it is the inverse limit of finite groups.) It makes sense to talk about normal series: $\{e\} \lhd \cdots\cdots G_2\lhd G_1\lhd G$ where $G_n$ is a closed subrgoup of $G$, $G_n$ is normal in $G_{n-1}$ and $G_{n-1}/G_n$ is finite and $\bigcap_{n\in\mathbb{N}}G_n=\{e\}$. This is the same as having an inverse limit expresion $$G\to \cdots \to H_2\to H_1\to \{e\}$$ where each $H_n$ is finite.

Call the normal series a composition series if each $G_{n-1}/G_n$ is simple. (In the inverse limit picture, this is the same as the kernel of each map $H_n\to H_{n-1}$ being simple.)

Are the simple factors appearing in a composition series unique up to permutation? The result is false without compactness, but I do not know if it is true with it.

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It's true if $G$ is (topologically) finitely generated. In this case, $G$ is determined uniquely by its collection of finite quotients, and the composition factors of the finite quotients determine those of $G$. –  Ian Agol Jan 29 at 21:41
@Eusebio: do you assume that the $G_n$ are closed subgroups? –  YCor Jan 29 at 23:20
@Yves: yes, the subgroups $G_n$ should be closed; I forgot to mention that. I'll edit the question. –  Eusebio Gardella Jan 30 at 1:33
@Ian: I would like a result that can handle infinite products of finite groups too; in particular, groups for which every finitely generated subgroup is finite. –  Eusebio Gardella Jan 30 at 1:36
Did you check whether the proof of the (Zassenhaus) Butterfly Lemma applies in your situation? It doesn't use much --- only the Noether isomorphism theorems and one of Dedekind's modular laws. That would imply that any two such sequences have a common refinement. –  abz Jan 30 at 18:20

If $G$ is second countable, it has a composition series: one can obtain this by starting with some countable basis of identity neighbourhoods (which may be taken to consist of open normal subgroups of $G$, since every open neighbourhood of the identity contains an open normal subgroup), taking intersections to form a descending chain of open normal subgroups with trivial intersection, and then refining the series. So let us assume that we have a composition series $(G_n)$. The equivalence class of $(G_n)$ is defined by the number of times, say $f(S)$, each isomorphism type $S$ of finite simple group appears as a factor $G_{n-1}/G_n$.
Let $S$ be a finite simple group. Then there are two possibilities: either $f(S)$ is a non-negative integer, or $f(S) = \aleph_0$. If $f(S)$ is finite, then there is a finite discrete quotient $G/G_n$ of $G$ such that some (hence any) composition series for $G/H$ has $f(S)$ simple factors isomorphic to $S$. Moreover, every finite discrete quotient of $G$ has at most $f(S)$ copies of $S$ in any composition series: This is clear for $G/G_n$ for all $n$ assuming Jordan-Hölder for finite groups, and then we use compactness to show that any finite discrete quotient $G/H$ of $G$ is a quotient of $G/G_n$. If $f(S)$ is infinite, then $G$ has finite quotients with composition series that have arbitrarily many factors isomorphic to $S$. Thus $f(S)$ is an invariant of $G$ as a topological group, independent of the choice of composition series $(G_n)$, so all composition series are equivalent.


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Synopsis: Making room for holes

The ability to hole-dope $B\phantom{\rule{0}{0ex}}{i}_{2}\phantom{\rule{0}{0ex}}S\phantom{\rule{0}{0ex}}{e}_{3}$ is an important step toward exploring the thermoelectric properties of this material.

$\left(B\phantom{\rule{0}{0ex}}i,S\phantom{\rule{0}{0ex}}b{\right)}_{2}\left(T\phantom{\rule{0}{0ex}}e,S\phantom{\rule{0}{0ex}}e{\right)}_{3}$ binary compounds have been extensively studied for their thermoelectric properties. Although primarily $B\phantom{\rule{0}{0ex}}{i}_{2}\phantom{\rule{0}{0ex}}T\phantom{\rule{0}{0ex}}{e}_{3}$ and also ${B\phantom{\rule{0}{0ex}}i}_{2}{S\phantom{\rule{0}{0ex}}e}_{3}$ are more commonly used as thermoelectric materials, they have recently begun to attract attention due to the promise of hosting so-called topological surface states.

One of the major issues for both thermoelectric applications and fundamental research on topological phases in Bi2Se3 is creating a hole-doped variant of this material. Pure Bi2Se3 exhibits n-type behavior due to the presence of selenium vacancies that act as electron donors, but hole-doping is much more difficult.

Writing in Physical Review B, Yew San Hor and collaborators from Princeton University report that they have managed to dope holes into Bi2Se3 by substituting calcium for bismuth. Scanning-tunneling microscopy reveals that the calcium donates holes that compensate the electrons created by the selenium vacancies.

Angle-resolved photoemission spectroscopy in turn reveals that the hole-doping substantially lowers the Fermi level in Bi2Se3 with respect to that of pure Bi2Se3. In order to access topological surface states, it is necessary to be able to tune the Fermi level, which should be possible in this material. The hole-doped material also exhibits intriguing transport anomalies at low temperatures that are currently not understood. – Alex Klironomos

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# Blur¶

• radius - Radius used in the gaussian function to generate a matrix, maximum value is 150. The bigger the radius more blurred will be the image.
• sigma - Optional. Defaults to the same value as the radius. Sigma used in the gaussian function.


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What is the FRM Pass Rate?

July 10, 2013

On a number of forums I have seen people stating it is a very hard exam because the probability of passing both exams is low.

The FRM pass rates are shown below. I have split the results in to two parts. The first table shows the pass rates when the exam was just one paper.

2009200820072006200520042003200220012000
44.1%42.4%44.4%44.8%44%52.5%53.8%58.9%63%53%

The table below shows the results from 2010. The exam had become a two part exam.

YearNov 2014May 2014Nov 2013May 2013Nov 2012May 2012Nov 2011May 2011Nov 2010May 2010
Part I48.4%42.5%50.9%45.9%46.7%47.3%46.6%53.1%39.3%52.5%
Part II58.7%58.1%58.0%56.8%56.0%61.1%57.0%61.6%54.9%54.0%

If you take parts I and II together on the same day you are only included in the stats for part II if you pass part I. You cannot pass part II without passing part I. Hence the part II pass rate in the table above is a conditional probability.
For fun let’s answer this FRM part 1 style exam question. Note it is not how the exam actually works.

Q. Assume all candidates who sat for the May 2013 exam sat for both FRM part 1 and FRM part 2  for the first time and that you can pass FRM part 1 or FRM part 2 without having to pass both. The probability of passing FRM part 1 is 45.9% and the probability of passing FRM part 2 is 56.8%. If we assume that the probability of passing FRM part 2 conditional on passing FRM part 1 is at least as high as the probability of passing FRM part 2 conditional on failing the FRM part 1 exam what are the upper and lower bounds for the probability of passing both FRM part 1 and FRM part 2 at the first attempt?

A.  Let 2p denote the event of passing the FRM part 2 exam. Let 1p denote the event of passing the FRM part 1 exam. We have $$\mathbb{P}(1\text{p}) = x = 45.9\%$$ and $$\mathbb{P}(2\text{p}) = y = 56.8\%$$.

Now using $$\mathbb{P}(2\text{p} \;\&\; 1\text{p}) + \mathbb{P}(2\text{p} \;\&\; \bar{1\text{p}}) = y$$ and combining with Bayes theorem we have
\begin{align*}
&\mathbb{P}(2\text{p}|1\text{p})*\mathbb{P}(1\text{p}) + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – \mathbb{P}(1\text{p})) = y \nonumber\\
&=\mathbb{P}(2\text{p}|1\text{p})*x + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – x) = y
\end{align*}

Clearly $$\mathbb{P}(2\text{p}\;\&\;1\text{p}) \leq \mathbb{P}(1\text{p})$$ and $$\mathbb{P}(2\text{p}\;\&\;1\text{p}) \leq \mathbb{P}(2\text{p})$$ i.e. we have $$\mathbb{P}(2\text{p} \;\&\; 1\text{p})\leq \min(x,y) = 45.9\%$$.
We are given $$\mathbb{P}(2\text{p}|1\text{p})\geq \mathbb{P}(2\text{p}|\bar{1\text{p}})$$ and the lower bound is achieved when $$\mathbb{P}(2\text{p}|1\text{p})= \mathbb{P}(2\text{p}|\bar{1\text{p}})$$, which is the independence case so that $$\mathbb{P}(2\text{p} \; \&\; 1\text{p}) = xy = 26.07\%$$
So the upper bound is = 45.9% and the lower bound is 26.07%.

Check out my FRM Part 1 practice questions !!!

Category: FRM


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# Will addition of inert gas change the vapour pressure of a liquid?

From this question on Physics SE Thermodynamics of evaporation:

Now imagine the experiment is repeated but instead of vacuum conditions, the water is pressurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.

The accepted answer shows a kinetic perspective. I want to understand things from classical thermodynamics. As even an inert gas is added to the container, there would be some increase of configurational entropy in the gaseous phase compared to the initial conditions. Won't this factor will decrease the chemical potential of the gaseous phase and thus increase the vapour pressure a bit ?.

If the nitrogen is truly inert, its presence will have no direct effect on the chemical potential of the water in the gas phase. Essentially, the gaseous water is "unaware" of the nitrogen. Consistent with this, in a mixture of ideal gases, the chemical potential of each gas is determined only by the temperature, and its partial pressure. It is indepenent of the partial pressures of the other gases.

However, even an ideal gas can increase the vapor pressure. It doesn't do so by reducing the chemical potential of the gaseous water. Rather, by applying pressure to the liquid water, it increases the chemical potential of the liquid phase, thus increasing the equilibrium vapor pressure. This is a small effect.

Of course, gases aren't inert. As you increase the pressure of the nitrogen, you have two opposing effects on the liquid water: (1) The effect described above, which increases the liquid's chemical potential. (2) With increased pressure, the concentration of nitrogen dissolved in the water will increase, which will lower the water's chemical potential. And of course one also has to account for the non-ideal interactions between the nitrogen and the water vapor.

• Even if Nitrogen is inert it will surely affect the configurational entropy of the gaseous phase of water? – Deepak Arya Jun 5 at 9:38
• If we treat both the nitrogen and the water as ideal gases, then their configurational entropies are each independent of the presence of the other gas. – theorist Jun 5 at 10:00
• You're probably thinking: If we have a container in which we have 1 L of ideal gas A on the left side at 1 atm, and 1 L of ideal gas B on the right side at 1 atm, and we remove the partition to allow them to mix, their entropies will increase. Yes, but that's because each now has twice the volume to explore (correspondingly, their partial pressures are halved). But suppose we instead squeeze A and B together into a 1 L space, such that their partial pressures don't change (only the total pressure changes). Then their entropies will be unchanged. – theorist Jun 5 at 10:08
• I am trying and the derivative is not cancelling out for a closed container with a fixed amount of inert gas. It is something like 1-ln(x), x = Ratio of moles of water vapour with Inert gas. – Deepak Arya Jun 5 at 10:08
• No, I am using xln(x) + yln(y)...x and y mole fraction in the gaseous phase – Deepak Arya Jun 5 at 10:18

At $$\pu{46°C}$$, the vapor pressure of water in $$\pu{0.1 atm}$$. It means that if some water is introduced into a vacuum, in an empty container at $$\pu{46°C}$$, the measured pressure in the container is $$\pu{0.1 atm}$$. Now if water is heated to $$\pu{46°C}$$ in an open flask at $$\pu{1 atm}$$., the composition of the vapor phase is $$10$$% water vapor, $$90$$% air. The vapor pressure is the same above the surface of water, whatever the presence of air. The total pressure has to be maintained at 1 atm. of course.

• There will be a factor of ΔSmix and its derivative will not be zero if we fix the inert gas amount in a closed container. That's why asked. – Deepak Arya Jun 5 at 9:35
• -1 This is wrong, putting pressure on water increases it's vapour pressure: Kelvin equation – S R Maiti Jun 5 at 9:45


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Remember, when it comes to getting offended, we all decide what offends us and how to get offended by what someone says. TRUE empowerment means you have the control within yourself. We don’t have to allow things to offend us simply because someone says something, and this also doesn’t mean everyone is going to be mean to us all the time, this is an unsubstantiated fear.

For others, it means “perpetual motion” and is therefore dismissed, without due consideration. This website is dedicated to clarifying exactly what free energy is, how it works and how it can be applied in your everyday life…. for light, heat and power. In the simplest sense, FREE energy is any energy that is provided by the Natural World. This may include energy sources that you are familiar with, such as solar panels or wind generators, but also could include amazing technologies like Free Power car powered by water-fuel, Free Power battery charger powered by Magnets, or Free Power home heating system powered by the Earth. The best FREE energy systems deliver energy at no on-going cost to the user, without detrimental effects to Free Energy, and at extremely low costs for the maintenance and operation of the equipment. On this website, I will NOT be presenting any futuristic ZPE technologies, or impossible inventions that break the Free Power of Physics. What I will be talking about are technologies that tap into Natural Sources of energy in Free Energy and convert them into useful forms of energy for heat, light, and motive power. In fact, the lecture on Open System Thermodynamics (in the column to the right) is Free Power great place to begin to understand this science. A good example of this is Free Power way to heat your home from the Earth. In moderate climates, the Earth below the frost line is Free Power reservoir of constant temperature. This temperature is about Free Electricityº F (Free Electricityº C). In the winter, this reservoir of heat can be accessed and concentrated by Free Power geothermal heat pump system, and used to heat your house for about Free Power/3rd the cost of running an electric furnace. In addition to this, if the house is insulated to Free Power rating of R-Free Power, the cost of heating the house can be further reduced by another Free Power. With just these two simple technologies, both available today, the cost of heating your home in the winter can be reduced by Free Electricity. That is just Free Electricity of the cost of heating the same house with an electric furnace where the house is insulated to Free Power rating of R-Free Electricity. In the summer, the air-conditioning costs are similarly low, using the same heat pump in reverse. Free Electricity energy benefit, year after year, at no more cost to you. That is the FREE energy you can have right now! Your lifestyle is the same, but your energy costs are much lower. By conserving heat with the better insulation, and by using Natural energy first, including the heat already present in the Earth, it is possible to dramatically lower your energy demands for home heating and your “carbon footprint”. This is just one example. There are dozens of other ways to save energy discussed in my Home energy Savings Guide. So, FREE energy is also any energy we can use at NO COST. If we learn to “Conserve, and use Natural energy First”, then, the small amount of energy that is still required for our needs is more easily produced by Free Power sources such as wind, solar, hydro, and bio-fuels. Even if we are still using carbon based fuels to supply the remainder, our energy costs have dropped considerably and our carbon footprint is greatly reduced. In our Technologies section, I discuss Free Power wide array of technologies from the PAST that have still not been integrated into the Market Economy or made available for everyday use. I also talk about technologies in the PRESENT that are available right now that can dramatically lower our energy costs. And finally, I talk about technologies that may become available in the FUTURE that could radically change our world for the better. On our Whole Systems page, I discuss our philosophy of energy use and the economics of using FREE energy wherever possible.
Look in your car engine and you will see one. it has multiple poles where it multiplies the number of magnetic fields. sure energy changes form, but also you don’t get something for nothing. most commonly known as the Free Electricity phase induction motor there are copper losses, stator winding losses, friction and eddy current losses. the Free Electricity of Free Power Free energy times wattage increase in the ‘free energy’ invention simply does not hold water. Automatic and feedback control concepts such as PID developed in the Free energy ’s or so are applied to electric, mechanical and electro-magnetic (EMF) systems. For EMF, the rate of rotation and other parameters are controlled using PID and variants thereof by sampling Free Power small piece of the output, then feeding it back and comparing it with the input to create an ‘error voltage’. this voltage is then multiplied. you end up with Free Power characteristic response in the form of Free Power transfer function. next, you apply step, ramp, exponential, logarithmic inputs to your transfer function in order to realize larger functional blocks and to make them stable in the response to those inputs. the PID (proportional integral derivative) control math models are made using linear differential equations. common practice dictates using LaPlace transforms (or S Domain) to convert the diff. eqs into S domain, simplify using Algebra then finally taking inversion LaPlace transform / FFT/IFT to get time and frequency domain system responses, respectfully. Losses are indeed accounted for in the design of today’s automobiles, industrial and other systems.
Not Free Power lot to be gained there. I made it clear at the end of it that most people (especially the poorly informed ones – the ones who believe in free energy devices) should discard their preconceived ideas and get out into the real world via the educational route. “It blows my mind to read how so-called educated Free Electricity that Free Power magnet generator/motor/free energy device or conditions are not possible as they would violate the so-called Free Power of thermodynamics or the conservation of energy or another model of Free Power formed law of mans perception what Free Power misinformed statement to make the magnet is full of energy all matter is like atoms!!”
But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
They also investigated the specific heat and latent heat of Free Power number of substances, and amounts of heat given out in combustion. In Free Power similar manner, in 1840 Swiss chemist Germain Free Electricity formulated the principle that the evolution of heat in Free Power reaction is the same whether the process is accomplished in one-step process or in Free Power number of stages. This is known as Free Electricity’ law. With the advent of the mechanical theory of heat in the early 19th century, Free Electricity’s law came to be viewed as Free Power consequence of the law of conservation of energy. Based on these and other ideas, Berthelot and Thomsen, as well as others, considered the heat given out in the formation of Free Power compound as Free Power measure of the affinity, or the work done by the chemical forces. This view, however, was not entirely correct. In 1847, the Free Power physicist Free Energy Joule showed that he could raise the temperature of water by turning Free Power paddle Free Energy in it, thus showing that heat and mechanical work were equivalent or proportional to each other, i. e. , approximately, dW ∝ dQ.
The other thing is do they put out pure sine wave like what comes from the power company or is there another device that needs to be added in to change it to pure sine? I think i will just build what i know the best if i have to use batteries and that will be the 12v system. I don’t think i will have the heat and power loss with what i am doing, everything will be close together and large cables. Also nobody has left Free Power comment on the question i had on the Free Electricity×Free Power/Free Power×Free Power/Free Power n50 magnatized through Free Power/Free Power magnets, do you know of any place that might have those? Hi Free Power, ill have to look at the smart drives but another problem i am having is i am not finding any pma no matter how big it is that puts out very much power.
In most cases of interest there are internal degrees of freedom and processes, such as chemical reactions and phase transitions, which create entropy. Even for homogeneous “bulk” materials, the free energy functions depend on the (often suppressed) composition, as do all proper thermodynamic potentials (extensive functions), including the internal energy.
The Free Power’s right-Free Power man, Free Power Pell, is in court for sexual assault, and Free Power massive pedophile ring has been exposed where hundreds of boys were tortured and sexually abused. Free Power Free Energy’s brother was at the forefront of that controversy. You can read more about that here. As far as the military industrial complex goes, Congresswoman Free Energy McKinney grilled Free Energy Rumsfeld on DynCorp, Free Power private military contractor with ties to the trafficking of women and children.
The results of this research have been used by numerous scientists all over the world. One of the many examples is Free Power paper written by Theodor C. Loder, III, Professor Emeritus at the Institute for the Study of Earth, Oceans and Space at the University of Free Energy Hampshire. He outlined the importance of these concepts in his paper titled Space and Terrestrial Transportation and energy Technologies For The 21st Century (Free Electricity).
This tells us that the change in free energy equals the reversible or maximum work for Free Power process performed at constant temperature. Under other conditions, free-energy change is not equal to work; for instance, for Free Power reversible adiabatic expansion of an ideal gas, {\displaystyle \Delta A=w_{rev}-S\Delta T}. Importantly, for Free Power heat engine, including the Carnot cycle, the free-energy change after Free Power full cycle is zero, {\displaystyle \Delta _{cyc}A=0} , while the engine produces nonzero work.

You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.

##### Thanks, Free Power. One more comment. I doubt putting up Free Power video of the working unit would do any good. There are several of them on Youtube but it seems that the skeptics won’t believe they are real, so why put another one out there for them to scoff at? Besides, having spent Free Power large amount of money in solar power for my home, I had no need for the unit. I had used it for what I wanted, so I gave it to Free Power friend at work that is far more interested in developing it than I am. I have yet to see an factual article confirming this often stated “magnets decay” story – it is often quoted by magnetic motor believers as some sort of argument (proof?) that the motors get their energy from the magnets. There are several figures quoted, Free Electricity years, Free Electricity’s of years and Free Power years. All made up of course. Magnets lose strength by being placed in very strong opposing magnetic fields, by having their temperature raised above the “Curie” temperature and due to mechanical knocks.

My older brother explained that in high school physics, they learned that magnetism is not energy at all. Never was, never will be. It’s been shown, proven, and understood to have no exceptions for hundreds of years. Something that O. U. should learn but refuses to. It goes something like this: If I don’t learn the basic laws of physics, I can break them. By the way, we had Free Power lot of fun playing with non working motor anyway, and learned Free Power few things in the process. My brother went on to get his PHD in physics and wound up specializing in magnetism. He designed many of the disk drive plates and electronics in the early (DOS) computers. bnjroo Harvey1 Thanks for the reply! I’m afraid there is an endless list of swindlers and suckers out there. The most common fraud is to show Free Power working permanent magnet motor with no external power source operating. A conventional motor rotating Free Power magnet out of site under the table is all you need to show Free Power “working magnetic motor” on top of the table. How could I know this? Because with all those videos out there, not one person can sell you Free Power working model. Also, not one of these scammers can ever let anyone not related to his scam operate the motor without the scammer hovering around. The believers are victims of something called “Confirmation Bias”. Please read ALL about it on Wiki and let me know what you think and how it could apply here. This trap has ensnared some very smart people. Harvey1 bnjroo Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been created! NOTHING IS IMPOSSIBLE! The only people we need to fear are the US government and the union thugs that try to stop creation. Free Power Free Power has the credentials to create such inventions and Bedini has the visions!
I might be scrapping my motor and going back to the drawing board. Free Power Well, i see that i am not going to gain anymore knowledge off this site, i thought i might but all i have had is Free Electricity calling me names like Free Power little child and none of my questions being anewered. Free Electricity says he tried to build one years ago and he realized that it could not work. Ok tell me why. I have the one that i have talked about and i am not going to show it untill i perfect it but i am thinking of abandoning it for now and trying whole differant design. Can the expert Free Electricity answer shis? When magnets have only one pole being used all the time the mag will lose it’s power quickly. What will happen if you use both poles in the repel state? Free Electricity that ballance the mag out or drain it twice as fast? How long will Free Power mag last running in the repel state all the time? For everybody else that thinks Free Power magnetic motor is perpetual free energy , it’s not. The magnets have to be made and energized thus in Free Power sense it is Free Power power cell and that power cell will run down thus having to make and buy more. Not free energy. This is still fun to play with though.
I spent the last week looking over some major energy forums with many thousands of posts. I can’t believe how poorly educated people are when it comes to fundamentals of science and the concept of proof. It has become cult like, where belief has overcome reason. Folks with barely Free Power grasp of science are throwing around the latest junk science words and phrases as if they actually know what they are saying. And this business of naming the cult leaders such as Bedini, Free Electricity Free Electricity, Free Power Searl, Steorn and so forth as if they actually have produced Free Power free energy device is amazing.
I don’t know what to do. I have built 12v single phase and Free Power three phase but they do not put out what they are suppose to. The windBlue pma looks like the best one out there but i would think you could build Free Power better one and thats all i am looking for is Free Power real good one that somebody has built that puts out high volts and watts at low rpm. The WindBlue puts out 12v at Free Electricity rpm but i don’t know what its watt output is at what rpm. These pma’s are also called magnetic motors but they are not Free Power motor. They are Free Power generator. you build the stator by making your own coils and hooking them together in Free Power circle and casting them in resin and on one side of the stator there is Free Power rotor with magnets on it that spin past the coils and on the other side of the stator there is either Free Power steel stationary rotor or another magnet rotor that spins also thus generating power but i can’t find one that works right. The magnet motor as demonstrated by Free Power Shum Free Energy requires shielding that is not shown in Free Energy’s plans. Free Energy’s shielding is simple, apparently on the stator. The Perendev shows each magnet in the Free Energy shielded. Actually, it intercepts the flux as it wraps around the entire set of magnets. The shielding is necessary to accentuate interaction between rotor and stator magnets. Without shielding, the device does not work. Hey Gilgamesh, thanks and i hope you get to build the motor. I did forget to ask one thing on the motor. Are the small wheels made of steel or are they magnets? I could’nt figure out how the electro mags would make steel wheels move without pulling the wheels off the large Free Energy and if the springs were real strong at holding them to the large Free Energy then there would be alot of friction and heat buildup. Ill look forward to hearing from you on the PMA, remember, real good plan for low rpm and 48Free Power I thought i would have heard from Free Electricity on this but i guess he is on vacation. Hey Free Power. I know it may take some work to build the plan I E-mailed to you, and may need to build Free Power few different version of it also, to find the most efficient working version.

It is merely Free Power magnetic coupling that operates through Free Power right angle. It is not Free Power free energy device or Free Power magnetic motor. Not relevant to this forum. Am I overlooking something. Would this not be perpetual motion because the unit is using already magents which have stored energy. Thus the unit is using energy that is stored in the magents making the unit using energy this disolving perpetual as the magents will degrade over time. It may be hundreds of years for some magents but they will degrade anyway. The magents would be acting as batteries even if they do turn. I spoke with PBS/NOVA. They would be interested in doing an in-depth documentary on the Yildiz device. I contacted Mr. Felber, Mr. Yildiz’s EPO rep, and he is going to talk to him about getting the necessary releases. Presently Mr. Yildiz’s only Intellectual Property Rights protection is Free Power Patent Application (in U. S. , Free Power Provisional Patent). But he is going to discuss it with him. Mr. Free Electricity, then we do agree, as I agree based on your definition. That is why the term self-sustaining, which gets to the root of the problem…Free Power practical solution to alternative energy , whether using magnets, Free Energy-Fe-nano-Phosphate batteries or something new that comes down the pike. Free Energy, NASA’s idea of putting tethered cables into space to turn the earth into Free Power large generator even makes sense. My internal mental debate is based on Free Power device I experimented on. Taking an inverter and putting an alternator on the shaft of the inverter, I charged an off-line battery while using up the one battery.
Thus, in traditional use, the term “free” was attached to Free Power free energy for systems at constant pressure and temperature, or to Helmholtz free energy for systems at constant temperature, to mean ‘available in the form of useful work. ’ [Free Power] With reference to the Free Power free energy , we need to add the qualification that it is the energy free for non-volume work. [Free Power]:Free Electricity–Free Power


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Isnot here D) should be ans??Here legth of AC=6, then how do we do 3:1 division?

| 52 views

This is a simple application of Pythagoras' Theorem.

Take the length of the perpendicular as $p$. Now you will have two right angled triangles $ADB$ and $BDC$ with a common base :

Apply Pythagoras' Theorem to both triangles.

$x^2 + p^2 = 9$

$(6-x)^2 + p^2 = 27$

Equate $p^2$ and solve for $x$, you will get $x = 1.5$

i.e. $\frac{DC}{AD} = \frac{6-1.5}{1.5} = \frac{4.5}{1.5} = 3:1$

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Press and hold the ALT key and type the number which you want to make numeral symbol. Hmm. How to type Integral With Overbar in word? (mathematics) Braid group algebra. Done. Under Equation Tools, on the Design tab, in the Structures group, click the Integral button: Select Symbol and then More Symbols. Below is the Square root symbol if you want to copy and paste it into your work: You can select the above character with your mouse as plain text (click-and-drag to highlight) or double click the symbol to highlight it. Option 3: Copy and paste the Square root symbol. The StefanâBoltzmann constant. How to Use Symbol Codes. In the Professional format:. Copy the Integral With Underbar in the above table (it can be automatically copied with a mouse click) and paste it in word, Or . Return to Math Symbols Page Go to the About the Codes section to see how they are implemented. Go to Word Option, go to the Proofing tab, click on Auto Correct Options, then go to the Math Auto Correct tab, tick the âUse Math AutoCorrect â¦â. Integral Calculation. For its origins: "â« symbol $\int$ is used to denote the integral in mathematics. How to add an equation in your document, see Working with Microsoft Equation. Select the Insert tab. It is very simple to use number symbol alt codes. Add 0 (in this case in a7) with the 0,1 â the dx. Math symbol is a copy and paste text symbol that can be used in any desktop, web, or mobile applications. Copy the Integral With Overbar in the above table (it can be automatically copied with a mouse click) and paste it in word, Or . Page Content Greek Letters New Page Common Arithmetic & â¦ Select the Integral With Overbar tab in the Symbol window. This table explains the meaning of every math symbol. How to type Integral With Underbar in word? Use unicode number symbols in your html document or copy paste the desired character. Infinity is characterized by a number of uncountable objects or concepts which have no limits or size. Select the Insert tab. Right click on the highlighted area and choose the 'copy' command from the context menu or use the Ctrl+C key-combinations (simultaneously). (mathematics) Sum of divisors. Symbol . (Physics, scattering) Cross_section_(physics). Select the Integral With Underbar tab in the Symbol window. You should click on the line (where number 2 is marked), and expand the whole row to fit the integral equation that has been inserted. See Integral Symbol.That is, it's usually called the "integral symbol". 2. Irrespective of the software you are working with, you can always copy and paste any symbol into your work. Then, to type the integral sign, type \int. (linguistics, phonology) Syllable. Ï (mathematics, statistics) Standard deviation. Analysis & calculus symbols table - limit, epsilon, derivative, integral, interval, imaginary unit, convolution, laplace transform, fourier transform To add an integral form of the Gauss's law, do the following:. Tip: you should type in the x2 (marked in the number 1), using an insert symbol (which is in the insert tab). A shielding constant. Select Symbol and then More Symbols. 1. The notation was introduced by the German mathematician Gottfried Wilhelm Leibniz towards the end of the 17th century. integral symbol for facebook â« copy and paste this character. (spatial databases) The select operation. The copy and paste option can be the easiest option to insert this symbol into MS Word. Create your own equation. 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# Document equivalence

What is the best way to determine, for two arbitrary input files, if they will result in the same compiled document (disregarding timestamp information and external files)?

## Context

I'm attempting to implement a nontrivial change to a LaTeX code generator. The code base is backed with an integration test: Input documents with expected outputs. Unfortunately, the change I'm about to implement is fiddling with whitespace (and hopefully only with those!) all over the place. I'm looking for a nice automatic way to prove (or disprove) that the whitespace changes are irrelevant.

## Example

### Initial version

\documentclass{scrartcl}\pagestyle{empty}
\begin{document}

Hello
World!

\begin{verbatim}
Whitespace  are
relevant

here!!!
\end{verbatim}

\end{document}

### No-op modifications

\documentclass{scrartcl}
\pagestyle{empty}

\begin{document}
Hello World!

\begin{verbatim}
Whitespace  are
relevant

here!!!
\end{verbatim}
\end{document}

### Meaningful modifications

\documentclass{scrartcl}
\pagestyle{empty}

\begin{document}
Hello World!

\begin{verbatim}
Whitespace are
relevant

here!!!
\end{verbatim}
\end{document}

### Comparison (aligned at bottom)

Click on the <> links between the pictures to show a textual diff of the files.

## Question

So: What's the best way to implement such a comparison? Of course, one possible solution is to render as PNG (as in http://tex.stackexchange.com/a/55323/8057), remove timestamp information with, e.g., pngcrush, and compare at the byte level. Can a PDF timestamp be removed the same way? Are there other solutions at the TeX level?

-
what we do in the latex2e regression test suite is (essentially) add \showoutput to the document and then compare the relevant part of the log files. –  David Carlisle Oct 30 '13 at 1:12
@DavidCarlisle: This is pretty much exactly what I'm looking for. When running on my example, the meaningless differences look much different from the meaningful differences. Care to convert this to an answer? Also, where can I find the regression test suite? Is there logic to determine the "meaningful" parts of the log? –  krlmlr Oct 30 '13 at 1:32

What we do in the latex2e regression test suite is (essentially) add \showoutput to the document and then compare the relevant part of the log files.

The 2e test suite is here

http://latex-project.org/svnroot/latex2e-public/testfiles/

The lvt files being the tex input and the tlg files being the log files slightly normalised with dates and file paths being removed so the results should be exactly the same on all systems.

the helper macros used in the tests are

http://latex-project.org/svnroot/latex2e-public/testfiles/test2e.tex

Frank and I had a tugboat article on the basic mechanism, or at least I thought we did, anyway here is a paper of Frank's:-)

http://www.tug.org/TUGboat/tb18-4/tb57mitt.pdf

-
Thanks a lot! Is it enough to input test2e.tex (and provide the .cfg) to activate the mechanism? –  krlmlr Oct 30 '13 at 23:27
@krlmlr the test2e.tex isn't really needed, it just provides some helper macros that do \typeout to the log file the mechanism is some external perl script for cleaning up the log. But there we need to trace macro expansion selectively, but for your use I think you just want to check consistent output so you just need \showoutput and do a file diff that ignores the date at the start of the log, and any absolute paths to loaded package files. –  David Carlisle Oct 30 '13 at 23:34
Unfortunately, in my use case, not even the LaTeX input is consistent: see this diff -- the output doesn't change, but the log diff shows a few false positives. –  krlmlr Oct 30 '13 at 23:40
@krlmlr That's exactly as expected. The first block is because your source is 2 lines different, so either arrange that doesn't happen (eg comment out lines rather than removing them when making the versions) or run the log files through sed to remove on input line [0-9]+ before running diff, similarly delete the stack positions line. Actually if you are only concerened with the \showoutput output you could first delete every line of the log that does not start with a . and just diff the remainder –  David Carlisle Oct 31 '13 at 9:52

If you're looking for some sort of "observational equivalence", that is, if you want to directly compare LaTeX's output, I suggest that you use Postscript instead of PDF. By using a script such as the following for compiling the LaTeX files that you wish to compare, you can force your Postscript output to contain the same file name information and the same creation dates:

#!/bin/bash

TMPDATE=date +"%a %b %d %X %Y"
TMPFILE=mktemp doceq.XXXXXX --suffix=".tex"

while [ "$1" != "" ]; do cp -f "$1" $TMPFILE latexmk --quiet -ps$TMPFILE
TARGET=${1/.tex/.ps} cat${TMPFILE/.tex/.ps}                                  \
| awk "BEGIN { found = 0; }                            \
/^%%CreationDate: / {                           \
if (!found) {                                 \
printf(\"%%%%CreationDate: $TMPDATE\\n\"); \ found = 1; \ next; \ } \ } \ { print; }" \ > "$TARGET"
latexmk --quiet -C $TMPFILE rm -f$TMPFILE
shift
done

For example, if your three examples are contained in doceq1.tex, doceq2.tex and doceq3.tex and ./doceq.sh is the script above, here's what you can get:

$./doceq.sh doceq1.tex doceq2.tex doceq3.tex Latexmk: Run number 1 of rule 'latex' (snip)$ diff doceq1.ps doceq2.ps
\$ diff doceq1.ps doceq3.ps
802c802
< 405 y Fa(Whitespace)93 b(are)335 518 y(relevant)335 857
---
> 405 y Fa(Whitespace)45 b(are)335 518 y(relevant)335 744

If you are tied to PDF for some reason, you can do the same there but it will be trickier, as PDF files contain binary parts that you will have to skip. The lines that you will have to change are those starting with /CreationDate and /ModDate.

-
Actually, this works even for my pdflatex chain if I postprocess the document with Poppler's pdftops (not pdf2ps). See meaningful changes. Big advantage: no timestamp is written, no postprocessing needed! –  krlmlr Oct 30 '13 at 22:44


text:
# Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.855 and exits as a...

## Question:

Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.855 and exits as a saturated liquid. The steam is being cooled by a 725 kg/s stream of water which enters the heat exchanger at 20C and exits at 40C. The two flows do not mix.

Determine the volumetric flow rate (m{eq}\displaystyle ^3 {/eq}/s) of the steam entering the heat exchanger.

## Heat Exchanger:

A device that is used to transfer heat between two fluids is called a heat exchanger. The fluids typically are not in contact with each other and can exist in two separate phases. Counter and parallel flow heat exchangers are the most common type of heat exchangers used.

The given conditions for steam are:

{eq}P_1 = 1 \ bar \\ x = 0.855 \\ {/eq}

The steam exits as saturated liquid.

From the steam tables,

Specific volume at inlet, {eq}\displaystyle v_1 = x * v_g = 0.855*1.694 = 1.45 \ m^3/kg {/eq}

Note: Since the term {eq}v_f {/eq} is much smaller than {eq}\displaystyle v_g {/eq} it can be ignored.

Specific enthalpy at inlet, {eq}h_1 = h_f + x * h_{fg} \\ h_1 = 417.46 + 0.855*2258 = 2348.05 \ kJ/kg {/eq}

Specific enthalpy at exit, {eq}h_2 = h_f = 417.46 \ kJ/kg {/eq}

Let the flow rate of steam be, {eq}\dot m_s {/eq}

Heat lost by steam = {eq}\dot m_s*(h_1 - h_2) = \dot m_s*(2348.05 - 417.46) = 1930.59* \dot m_s {/eq}

Heat gained by water = {eq}\dot m_w * C * \Delta T {/eq}

C is the specific heat of water = 4.186 \ kJ/kgK

Heat gained by water = {eq}725 * 4.186 * (40 - 20) = 60,697 \ KW {/eq}

Heat lost by steam = Heat gained by water

{eq}1930.59* \dot m_s = 60,697 \displaystyle m_s = \frac {60,697}{1930.59} = 31.44 \ kg/s {/eq}

Volumetric flow rate at inlet, {eq}\dot V_1 = \dot m_s * v_1 = 31.44 * 1.45 = 45.54 \ m^3/s {/eq}


text:
Article Text

AB0935 Metabolic syndrome in PSA patients is not associated with disease activity
1. J. Stolfa
1. Institute of Rheumatology Prague, Prague, Czech Republic

## Abstract

Background Metabolic syndrome (MetS) is regarded as one of the extraarticular manifestations of psoriasis (Pso) and psoriatic arthritis (PsA). Its etiology is as yet not well understood. The major component of MetS is insulin resistance, which is, among others associated with chronic inflammation [1]. So we put a question, if chronic inflammation in active PsA could contribute to the development of the MetS.

Objectives To ascertain if the occurrence of MetS is associated with higher clinical or humoral activity of the disease.

Methods 103 consecutive patients with PsA were followed for two years and their clinical activity was assessed clinically using disease activity index in PsA(DAPsA) [2] and humoral activity through C-reactive protein (CRP) twice a year for 2 – 4 years. The parameters of MetS were checked once a year. So we had 4 – 6 controls available. The pats were dividend in two groups: those with MetS (MetS+) and without MetS (MetS-). Then the average values of DAPsA and CRP in both groups were calculated and compared.

Results The average values of DAPsA, as well as CRP did not differ substantially in both groups - see table 1.

Table 1

Conclusions On the basis of this study we cannot confirm the association of the activity of PsA as assessed clinically or using laboratory parameter (CRP).The drawback of this small, preliminary study is that it did not take into account the severity of psoriasis and the pharmacotherapy.

1. Boehncke WH, Buerger C, Boehncke S. Co-morbidities in psoriasis vulgaris. Hautarzt 2009; 60:116-21

2. Schoels M, Aletaha D, Funovits J, et al. Application of the DAREA/DAPSA score for assessment of disease activity in psoriatic arthritis. Ann Rheum Dis 2010;69:1441-1447

Disclosure of Interest J. Stolfa Grant/Research support from: Supported by Ministery of Health of Czech Republic Grant Number 000 000 23728

## Request permissions

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text:
# Confidence interval for known non-normal estimation?

Assuming I know the density of my random sample - for example:

$$f_\Theta(x) = \frac{e^{-x/\Theta^2}}{\Theta^2}$$

(but I care about general single parameter case)

How can I obtain $1-\alpha$ confidence interval of $\Theta$? I've read lots of resources that describe estimation of expected value and/or variance. Alternatively general $\Theta$ of normal distribution. However, I have non-normal one.

The process is as follows. I'm looking for $T_1(x_1,\ldots,x_n)$ and $T_2(x_1,\ldots,x_n)$ such that

$$P(T_1(x_1,\ldots,x_n) < \Theta < T_2(x_1,\ldots,x_n)) = 1 - \alpha$$

Obviously, all I need to do is decide which statistics to use. Is there any general solution? I though of CLT - but what if I have only 20 samples?

• This is an exponential distribution. Googling "confidence interval exponential distribution" should be fruitful. – Stéphane Laurent May 23 '15 at 9:07
• Is this for some subject? It's a different (non-standard) parameterization of an exponential; forming an interval in either of the standard parameterizations is straightforward, and then it's easy to back out an interval for $\Theta$. If you want to do it from scratch, it's a simple matter to write down a pivotal quantity and back out an interval for $\Theta^2$.. – Glen_b May 23 '15 at 9:12
• @StéphaneLaurent Yes, of course. But what if the given distribution isn't well known such as the provided example? It may be very "wild". How to estimate it in that case? – petrbel May 23 '15 at 9:17

An exact confidence interval is possible here. To see this, for the moment let's take $\theta=\theta^2$ to get something we are familiar with. Indeed, in this case the density becomes

$$f_{X} \left(x;\theta \right)=\begin{cases} \frac{1}{\theta} \exp\left\{-\frac{x}{\theta} \right\} & 0<x<\infty \\ 0 & \text{otherwise} \end{cases}$$

which we immediately recognise as a $Gamma\left(1,\theta \right)$ density. Assuming now you have a random sample of $X_1,X_2, \ldots, X_n$, then the random variable $Y=\sum_{i=1}^n X_i$ is a Gamma random variable with parameters $n$ and $\theta$. You might notice that we are working with the sufficient statistic.

But now we see that the distribution of $\frac{Y}{\theta}$ is actually $\Gamma \left(n,1 \right)$ and does not depend on $\theta$. This observation allows us a to construct a confidence interval for $\theta$. Indeed, choosing appropriate quantiles we get

$$P \left[ \Gamma\left(\alpha/2,n,1 \right) <\frac{Y}{\theta}<\Gamma\left(1-\alpha/2,n,1 \right) \right]=1-\alpha$$

which implies that

$$P \left[ \frac{Y}{\Gamma\left(1-\alpha/2,n,1 \right)} <\theta<\frac{Y}{\Gamma\left(\alpha/2,n,1 \right)}\right]=1-\alpha$$

and we have a Confidence Interval.

If you now take square roots, you will be back to your original scale. That's good, right? If I were you, I would comppare the asymptotic CLT based CI with this one for a sequence of sample sizes. You will of course see that for small $n$ the difference can be quite large, the asymptotic CI being symmetric and all.

As a general rule, whenever you are confronted with such a problem try to obtain a pivot, that is a quantity whose distribution does not depend on the parameter of interest. If you can find a function of that quantity that has a well-known distribution, or at least a recognizable distribution, a CI may be constructed using a similar procedure. It is often helpful to start with the sufficient statistic.


text:
# Figure 3

 The individual flow harmonics $v_n$ for $n$ = 2--5 in $\PbPb$ collisions at $\snn=2.76$~TeV are shown in the left panels ((A), (B) and (C)). $v_4$ and $v_5$ are shown in the same panel (C). The $p_{\rm T, min}$ dependence of $v_n$ for $n$ = 2--4 is shown in the right panels ((a), (b) and (c)).


text:
### Home > GC > Chapter 5 > Lesson 5.3.5 > Problem5-122

5-122.

Find the area and perimeter of $ΔABC$ at right. Give approximate (decimal) answers, not exact answers.

Separate $ΔABC$ into two separate triangles by dropping the height from vertex $B$.
Label the new point $D$, creating $ΔABD$ and $ΔDBC$.

Both of the triangles that have been created are special right triangles, because the height $BD$ creates two right triangles: a $45º\text{-}45º\text{-}90º$ triangle and a $30º\text{-}60º\text{-}90º$ triangle.

The length of the hypotenuse in a $45º\text{-}45º\text{-}90º$ triangle is the length of a leg times $\sqrt{2}$.
In a $30º\text{-}60º\text{-}90º$ triangle, the length of the side opposite the $30^\circ$ angle is half of the length of the hypotenuse and the length of the side opposite the $60^\circ$ angle is half of the length of the hypotenuse times $\sqrt{3}$.


text: Hyperdimensions Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 03/04/2017, 10:31 AM Hi, everyone! The oldest dimensional base units like i[0]=1, i[1]=i (imaginary unit), i[2] and i[3] (quaternions), i[4] ... i[7] (octonions), i[8] ... i[15] (sedenions) ... etc. is called ancient-dimensional base units (by me). The less old dimensional base units where the indeces can be any real is simply called old-dimensional base units. Its PARI/gp code looks like this one: Code:twist(p,q)={ L=ceil(1.4426950408889634*log((max(q+1,p+1)))); res=1;sta=1; if(round(p)!=p||round(q)!=q,p*=2^60;q*=2^60;L+=60); p=round(p);q=round(q); T=[1,2,3,4;2,2,5,6;3,6,3,5;4,5,6,4;5,6,6,6;6,5,5,5]; for(i=0,L,j=L-i-1;k=2*bittest(p,j)+bittest(q,j)+1;sta=T[sta,k]); if(sta==4||sta==6,return(-1),return(1)); } xor(x,y)={return(bitxor(round(x*2^60),round(y*2^60))/2^60)}So i[x] i[y] = twist(x;y) i[x xor y] And for instance the graph of i[y] = i[1] i[x] looks like this: https://www.dropbox.com/s/81rxif99w7j50u...D.bmp?dl=0 As you can see the sign of the function is alternating. Okey, so these were the old-dimensional numbers. The so-called hyperdimensional base units are base units whose indeces can be complex, quaternion ... so simply old-dimensional numbers. For example these are hyperdimensional numbers: i[1+i+i[2]] i[i[0.5]-3] i[0.3+i[100]-i[7]+i[pi]] My question is that how to multiplicate two hyperdimensional base units according to the T matrix above in the code? I think we should extrapolate the T matrix to 3D matrix, but how? Xorter Unizo tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 03/09/2017, 01:23 PM I really do not know what you are talking about ... Regards Tommy1729 Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 03/09/2017, 07:53 PM (This post was last modified: 03/09/2017, 07:54 PM by Xorter. Edit Reason: grammar ) (03/09/2017, 01:23 PM)tommy1729 Wrote: I really do not know what you are talking about ... Regards Tommy1729 I call the complexes, quaternions, octonions, sedenions ... etc. ancient dimensional numbers. (But it is not so important.) Look: https://en.wikipedia.org/wiki/Quaternion https://en.wikipedia.org/wiki/Octonion https://en.wikipedia.org/wiki/Sedenion And here is a summation for N-dimension: http://jwbales.us/ We have a 2^n dimensional space with base units. These are the dimensional base units whose multiplication is that I am interested in. I found some formulas to calculate it. But I must step toward into the so-called hyperdimensional spaces with its hyperdimensional base units. I would like to know how I can multiplicate two of hyperdimensional base units. Are we closer to understanding? Xorter Unizo « Next Oldest | Next Newest »


text:
## Random reflections in a high-dimensional tube.(English)Zbl 1391.60098

The paper is a continuation of the authors paper in [Ann. Appl. Probab. 27, No. 4, 1951–1991 (2017; Zbl 1373.60080)], where Lambertian reflections in a semi-inifinte tube were introduced and studied, and the focus was on the two-dimensional case. The main results were concerned with the properties of the light ray at the exit time.
The present paper is devoted to the $$d$$-dimensional case for any $$d>3$$. The authors show that the exit distributions are similar in the quantitative sense for all $$d\geq 3$$. Some new results are also derived, including overshoot of random walks, arccosine distribution and products of random variables having this distributions, and distributions with regular varying tails.

### MSC:

 60G50 Sums of independent random variables; random walks 60K05 Renewal theory 37D50 Hyperbolic systems with singularities (billiards, etc.) (MSC2010) 37H99 Random dynamical systems

Zbl 1373.60080
Full Text:

### References:

 [1] Angel, O; Burdzy, K; Sheffield, S, Deterministic approximations of random reflectors, Trans. Am. Math. Soc., 365, 6367-6383, (2013) · Zbl 1408.37063 [2] Arnold, BC; Groeneveld, RA, Some properties of the arcsine distribution, J. Am. Stat. Assoc., 75, 173-175, (1980) · Zbl 0427.62008 [3] Bingham, N.H., Goldie, C.M., Teugels, J.L.: Regular Variation, Volume 27 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge (1987) · Zbl 0617.26001 [4] Burdzy, K., Tadić, T.: Can one make a laser out of cardboard? Ann. Appl. Probab. Arxiv:1507.00961 (2016) · Zbl 1186.37049 [5] Comets, F; Popov, S; Schütz, GM; Vachkovskaia, M, Billiards in a general domain with random reflections, Arch. Ration. Mech. Anal., 191, 497-537, (2009) · Zbl 1186.37049 [6] Comets, F; Popov, S; Schütz, GM; Vachkovskaia, M, Knudsen gas in a finite random tube: transport diffusion and first passage properties, J. Stat. Phys., 140, 948-984, (2010) · Zbl 1197.82058 [7] Comets, F; Popov, S; Schütz, GM; Vachkovskaia, M, Quenched invariance principle for the Knudsen stochastic billiard in a random tube, Ann. Probab., 38, 1019-1061, (2010) · Zbl 1200.60091 [8] Doney, RA, Moments of ladder heights in random walks, J. Appl. Probab., 17, 248-252, (1980) · Zbl 0424.60072 [9] Evans, SN, Stochastic billiards on general tables, Ann. Appl. Probab., 11, 419-437, (2001) · Zbl 1015.60058 [10] Folland, G.B.: Real analysis. Pure and Applied Mathematics (New York), 2nd edn. Wiley, New York (1999). Modern techniques and their applications, A Wiley-Interscience Publication · Zbl 0924.28001 [11] Veraverbeke, N, Asymptotic behaviour of Wiener-Hopf factors of a random walk, Stoch. Process. Appl., 5, 27-37, (1977) · Zbl 0353.60073
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
### Session Q1: Poster Session III

1:00 PM–1:00 PM, Wednesday, March 15, 2006
Baltimore Convention Center Room: Exhibit Hall, 1:00pm - 4:00pm (DPOLY session: 11:15am - 2:15pm)

Abstract ID: BAPS.2006.MAR.Q1.328

### Abstract: Q1.00328 : Magnetic Force Microscopy of Superconducting Vortices in Ordered Arrays of Artificial Pinning Centers

Preview Abstract MathJax On | Off   Abstract

#### Authors:

Joel Keay
Preston R. Larson
Kevin L. Hobbs
Matthew B. Johnson
(University of Oklahoma)

Ophir M. Auslaender
Kathryn A. Moler
(Stanford University)

Ordered arrays of artificial pinning centers were fabricated in Nb thin films using anodic aluminum oxide (AAO) as a template. The nanohole arrays show only a small decrease in the superconducting transition temperature, $T_c\simeq 7.1$ K, from comparable, unprocessed Nb thin films. These artificial pinning arrays have a triangular lattice parameter of 105 nm and antidot diameters of about 50 nm. Anomalous features at the first, second and third matching fields (matching field = 2170 Oe) were observed in the magnetization half-loops of these arrays at 5 K. Magnetic Force Microscopy (MFM) was used to image the nanohole arrays above and below $T_c$. The images clearly show the nanohole lattice. Individual vortices have also been imaged and their movement within the lattice is being investigated. Preliminary measurements of the depinning force of individual vortices at low fields will be presented. This work was supported by NSF grant nos. DMR-0080054 and NSF-0132534.

To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2006.MAR.Q1.328


text:
# relativistic aberration paradox

I'm racking my brain over what should be a simple problem in special relativity. Consider the angle $f$ between two stars, as measured by a an observer at rest:

The observed angle should be the same whether the observer is moving towards the stars, or the stars are moving towards the observer. In the first case, we have the classic formula for stellar aberration: $\cos f' = \frac{v/c + \cos f}{1+v/c \cos f}$

But in the latter case, it seems to me that the star can simply emit light at the angle $f''$ with $\cos f''=\frac{\cos f - v/c}{1-v/c \cos f}$ and still reach the observer:

So in the first case, the observer measures the stars to be $f'$ apart, but in the second case, they appear to be $f$ apart. What am I missing??

• Well that is correct. Now we may ask how is principle of relativity not violated. Are we asking that? Jul 20, 2017 at 3:15


text:
31st Annual Meeting of the DPS, October 1999
Session 58. Io Posters
Poster Group II, Thursday-Friday, October 14, 1999, , Kursaal Center

## [58.13] Emission Profiles of Neutral Oxygen and Sulfur in Io's Exospheric Corona

B.C. Wolven, H.W. Moos, P.D. Feldman, K.D. Retherford, D.F. Strobel (JHU), W.H. Smyth (AER, Inc.), F.L. Roesler (UW-M), R.J. Oliversen (GSFC)

\setlength{\parskip}{-0.04in} HST/STIS observations of Io acquired in 1997 (Roesler et al. 1999) have provided the first simultaneous spatially resolved measurements of emission from neutral sulfur and oxygen in Io's exospheric corona. Here we present spatial profiles of these extended emissions derived from observations made in both 1997 and 1998. The STIS FUV-MAMA detector permits measurement of the emissions with a spatial resolution of ~0.05 Io radii out to distances of ~20 Io radii. Useful measurements are limited to ~10 Io radii due to the low signal to noise of extended emission features. Preliminary findings include: \begin{itemize} \setlength{\itemsep}{-0.045in} \item Correlations between the brightness of near-Io emission features (R < 2\ {\rm RIo}) and extended coronal emissions indicate a common excitation source for emissions in the two regions. The observed dependence of intensity on separation from the centrifugal equator suggests the local torus electron density as a source of this modulation. \item The ratio of oxygen to sulfur emission is relatively constant in time, but is observed to decrease in the vicinity of Io. This may indicate cooling of torus electrons due to energy loss via ionization and dissociation in Io's atmosphere and denser regions of the neutral corona. \item The relative symmetry of emission profiles obtained near western elongation, in contrast with the asymmetry of those obtained in other orbital positions, indicates a significant asymmetry in the neutral O and S emission rates in the direction of Io's orbital motion and torus plasma flow. This asymmetry appears to correspond to a region of enhanced emission in Io's plasma wake; preliminary 3-D neutral cloud model calculations presented support this interpretation. \end{itemize}