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# Multi-step Monte Carlo calculations applied tonuclear reactor instrumentation - source definition and renormalization to physical values

Abstract : Significant efforts have been made over the lastfew years in the French Alternative Energies and Atomic Energy Commission (CEA) to adopt multi-step Monte Carlo calculation schemes in the investigation and interpretation of the response of nuclear reactor instrumentation detectors (e.g.miniature ionization chambers - MICs and self-powered neutron or gamma detectors - SPNDs and SPGDs). The first step consists of the calculation of the primary data, i.e. evaluation of the neutron and gamma flux levels and spectra in the environment where the detector is located, using a computational model of the complete nuclear reactor core and its surroundings. Thesedata are subsequently used to define sources for the following calculation steps, in which only a model of the detector under investigation is used. This approach enables calculations withsatisfactory statistical uncertainties (of the order of a few %) within regions which are very small in size (the typical volume of which is of the order of 1 mm$^3$). The main drawback of a calculation scheme as described above is that perturbation effects on the radiation conditions caused by the detectors themselves are not taken into account. Depending on the detector, the nuclear reactor and the irradiation position,the perturbation in the neutron flux as primary data may reach 10 to 20%. A further issue is whether the model used in the second step calculations yields physically representative results. This is generally not the case, as significant deviations may arise, depending on the source definition. In particular, as presentedin the paper, the injudicious use of special ptions aimed at increasing the computation efficiency (e.g. reflective boundary conditions) may introduce unphysical bias in the calculated flux levels and distortions in the spectral shapes. This paper presents examples of the issues described above related to a case study on the interpretation of the signal from differenttypes of SPNDs, which were recently irradiated in the Jozef Stefan Institute TRIGA Mark II reactor in Ljubljana, Slovenia, and provides recommendations on how they can be overcome.The paper concludes with a discussion on the renormalization of the results from the second step calculations, to obtain accurate physical values.
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Cited literature [9 references]

https://hal-cea.archives-ouvertes.fr/cea-02500845
Contributor : Amplexor Amplexor <>
Submitted on : Friday, March 6, 2020 - 1:25:21 PM
Last modification on : Tuesday, April 28, 2020 - 11:28:17 AM
Long-term archiving on: : Sunday, June 7, 2020 - 2:31:48 PM

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• HAL Id : cea-02500845, version 1

### Citation

V. Radulovic, L. Barbot, D. Fourmentel, J.-F. Villard, G. Zerovnik, et al.. Multi-step Monte Carlo calculations applied tonuclear reactor instrumentation - source definition and renormalization to physical values. ANIMMA 2015 - 4th International Conference on Advancements in Nuclear Instrumentation, Measurement Methods and their Applications, Apr 2015, Lisbonne, Portugal. ⟨cea-02500845⟩

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# How to run bat/cmd batch file silently without opening console window

Posted on 2013-11-04
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I hope someone can help me out here I have a bat file located on the network share and I'm trying to run it with the task scheduler. What I'm looking for is a way to run this file silently without displaying a console window. I have tried  cmd.exe /c start \\Network path to the batch file but that did not work I get the error from task scheduler. Is there a other clean way to get this accomplished. Also I'm deploying  this task through GPO not sure if that makes any difference.

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cmd /c \\path\to\batch\file.bat will run it then close after.

When you call the start command you would have to use that to call another cmd.exe process (which wouldn't make sense) as the batch file cannot be started natively (it isn't a program; it's the list of commands to run within the program.)

To run silently you will need to use powershell or vbscript.  A vbscript example is below.

Set wShell = CreateObject("WScript.Shell" )
wShell.Run chr(34) & "\\network\path\to\batch\file.bat" & Chr(34), 0

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Jason John Nassau

Quick Info

Born
29 March 1893
Smyrna, (now Izmir) Turkey
Died
11 May 1965
Cleveland, Ohio, USA

Summary
Jason J Nassau was an American mathematician who published papers in both mathematics and astronomy.

Biography

Jason J Nassau emigrated to the United States in his youth. He studied at Columbia University and the University of Syracuse in New York State. During World War I he served in the American Expeditionary Force in France. In 1918 he went to the University of Edinburgh, Scotland, and after spending some time there moved back to the University of Syracuse where he submitted his thesis Some Theorems in Alternants for which he was awarded a Ph.D. in 1920.

He became an assistant professor of astronomy at Case Institute of Technology, Cleveland, Ohio, in 1921. He was one of three who discussed the founding of the Cleveland Astronomical Society in February 1922. In 1924 he became the first chair of astronomy at Case. He was chairman of the graduate division from 1936 to 1940, and was also director of the Case Western Reserve University Warner and Swasey Observatory from 1924. He retired in 1959, being made Professor Emeritus.

Among Nassau's achievements are his discovery of a cluster of 9,000 stars, his co-discovery of two novae in 1961, and his development of a technique for studying the distribution of red stars.

Further details are given in J J Nassau's obituary, written by Victor M Blanco, which appeared in the Quarterly Journal of the Royal Astronomical Society, volume 7 (1966).

In 1932 Nassau's A Textbook of Practical Astronomy was published by the McGraw-Hill Book Company.

Among the honours given to Nassau, we mention that the Nassau Astronomical Station at the Warner and Swasey Observatory is named after him. Also the Jason J Nassau Prize was established by the Cleveland Astronomical Society in 1965. It is awarded annually to an outstanding senior student in the Case Western Reserve University Department of Astronomy. His name is given to two astronomical objects: Nassau crater on the Moon is named after him, as is asteroid 9240 Nassau, discovered on 31 May 1997.

Nassau joined the American Mathematical Society in February 1921 (he was one of 114 people who joined at that time). He had already joined the Edinburgh Mathematical Society in June 1919.

Finally we list a few of Nassau's papers: Questions and Discussions: Discussions: Evaluation of the Determinant $|1/(r + s - 1)! |$ (1924); Some extensions of the generalized Kronecker symbol (1926); Questions and Discussions: Discussions: Concerning a Theorem in Determinants (1927); and (with O E Brown) A Navigation Computer (1947).

References (show)

1. V M Blanco, Jason John Nassau (obituary), Quart. J. Roy. Ast. Soc. 7 (1966), 79.

Other pages about Jason J Nassau:

Other websites about Jason J Nassau:

Honours (show)

Honours awarded to Jason J Nassau

Cross-references (show)

Written by J J O'Connor and E F Robertson
Last Update November 2007


text: IEVref: 351-46-08 ID: Language: en Status: Standard Term: control factor Synonym1: Synonym2: Synonym3: Symbol: R Definition: ratio between the change of the controlled variable with closed-loop control and its change without control in case of changing of the reference variable or a disturbance variable under given conditionsNote 1 to entry: In case of sinusoidal input variables the complex control factor is the quotient of the phasors of these values as function of the angular frequency.Note 2 to entry: In case of stepwise changing of input variables the real control factor is the quotient of these values after the end of the transient processes. If there are one or more integral elements in the forward path the value of the real control factor is zero.Note 3 to entry: In closed-loop control with proportional behaviour of the open loop the real control factor is given by $R=\frac{1}{1+{K}_{0}}$ where K0 is the loop gain.Note 4 to entry: This entry was numbered 351-25-08 in IEC 60050-351:2006. Publication date: 2013-11 Source: Replaces: Internal notes: CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5:


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# Blogging with Hakyll

Posted on 2019-11-27 , Tags: blog, hakyll, haskell

I’ve been considering migrating from ghost for the following reasons for quite some time:

1. ghost is bloated: There’s too many features that I’m not gonna to use and it, together with the database, consumes a lot of memory
2. ghost has poor support for mathjax: Any linebreaks will break your mathjax environment, so yeah, you have to scramble all your equations into one line which is super unreadable and annoying.
3. ghost is not extensible, all the modules are hard coded into it.
4. I don’t actually need a dynamic blog as I’m comfortable with using a normal text editor write blogs (actually I’m using emacs) and using git to track all the files and publishing it.

After some research I found Hakyll the perfect candidate for blogging. It supports a huge amount of document formats as it uses pandoc to generate webpage. Not only that, it also supports embedding bibtex citations into markdown and automatically generates all the bibliography at the end of the webpage via the module pandoc-citeproc. This is a huge up for me because I will sometimes cite other papers in my blog post and it relieves me from manually formating bibliographies.

It is very extensible(in Haskell) and somewhat well-documented. It functions more like a static site generator library than the generator itself.

So that’s all my motivations for switching to Hakyll, I shall now introduce you the structure of my customized site.hs. Oh, before start, you should probably read the tutorial written by the author.

## Generating bibliography

We sould first load csl files and bibtex files using cslCompiler and biblioCompiler.

We can then build the compiler: readPandocBiblio readerOpt cslPath bibPath which will be used to combine with other compilers.

For the syntax for citing, please refer to the “citation” section in the pandoc user manual.

## Building tags

This blog post from javarn.github.io covered this topic.

## Optionally embedding mathjax

The idea is that we create a context that reads all the metadata of a post and check if there’s mathjax in it. If there is, then add a const context (script field) which will then be added to the post when applying the templates.

## Generating png from LaTeX formulae

Mathjax is great, but it’s nowhere near perfect, as it only provides limited packages.

What if we can use actual to compile the blogpost and extract the equations out as images and embed them into the HTML generated? We can use the library latex-svg (written by me). It works by simply using pdftex + dvisvgm + pdfcrop1 to generate svg which can be embedded in HTML With the help of this library, we can typeset beautiful equations:

and even tikz diagrams !

## Syntax highlighting

Quoting hakyll’s faq:

Syntax highlighting is enabled by default in Hakyll if you are using a somewhat recent version of Pandoc (1.9 and onwards). Note that you also need to include some CSS in order for this to work!

fac :: Int -> Int
fac n = foldr1 (*) [1..n]

Python:

def fac(n: Int) -> Int:
if n == 0:
return 1
else:
return n * fac(n - 1)

Agda:

∧-elim : (A : Set)(B : Set) -> ((C : Set) -> (A -> B -> C) -> C) -> A
∧-elim A B p = p A (\a b -> a)


text: A Reduction Formula We may use the method of integration by parts to obtain a reduction formula for $x^ne^x\, dx$: $\int x^ne^x\, dx=x^ne^x-n\int x^{n-1}e^x\, dx.$ What would be good choices for $u$ and $dv$ to obtain this formula? Reduction formulas may also be found for integrals of trigonometric functions such as $\displaystyle\int\! \cos^n(x)\, dx$.


text: Very helpful to class 9 students Reply. Fig. This set of Physics problems for class 9 students of CBSE/ICSE/HS boards cover Newton’s Laws of motion, Force, linear motion equations – kinematics, motion class 9 numericals and Gravitation (gravitational force). 6. Reply. Force and Laws of Motion Class 9 Extra Questions Short Answer Questions—II. DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. A boy of mass 50 kg running 5 m/s jumps on to a 20 kg trolley traveling in the same direction at 1.5 m/s. These laws are known as Newton’s laws of motion. force and laws of motion class 9; Force & Laws of Motion NCERT solutions; Assignments. It will move in the same direction after the collision. Which would require a greater force – accelerating a 2 kg mass at 5 m/s² or a 4 kg mass at 2 m/s²? What does a force do? In this chapter, we will learn. CBSE Class 9 Physics Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. All right reserved. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object? 4. The force then ceases to act. NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion – Here are all the NCERT solutions for Class 9 Science Chapter 9. It travels a distance … What is Force - We learn that Definition of force we learned in Class 8 was wrong, and the actual definition of force Determine the velocity of the second object? Linear momentum is expressed as the product of mass, â mâ of an object and the velocity, â vâ of the object. Note: Before looking for the solution, first try yourself, and when it becomes very difficult or if you want to check your solution, only then open the answer and explanation. Selina ICSE Solutions for Class 9 Physics Chapter 3 Laws of Motion. Here on AglaSem Schools, you can access to NCERT Book … Here are a few Important Force and Laws of Motion Class 9 Numerical for practice. Calculate the velocity … Important Force and Laws of Motion Class 9 extra Numerical for practice. An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a friction less horizontal table. Lakhmir Singh and Manjit Kaur Solutions for Class 9 Physics CBSE, 2 Force and laws of motion. Here you can get Class 9 Important Questions Science based on NCERT Text book for Class IX.Science Class 9 Important Questions are very helpful to score high marks in board exams. https://toppersportal.com/force-and-laws-of-motion-class-9-numerical Page 1; Page 2; Page 3; Page 4; Page 5; link to this page by copying the following text. 9.5: (a) the downward motion; (b) the upward motion of a marble on an inclined plane; and (c) on a double inclined plane. Learn Chapter 9 Class 9 Science - Force and Laws of Motion with Notes, NCERT Solutions, Solutions to Intext Questions, Solutions of Examples, Solutions to Additional Exercises and Extra Questions. Class 10 > Class 10 Lessons > Numerical problems on Newton's Laws of Motion. Solution: According to question: Acceleration (a) ... MOTION (CLASS 9 - CBSE - SCIENCE) Force and Laws of Motion Class 9 Numericals numerical problems in physics from force and laws of motion chapter – part 1 1> A force of 10 N is applied for 0.1 second on a body of mass 1 kg initially at rest. In the curriculum of Class 9 Science, Chapter 9 is the Force and Laws of Motion This chapter belongs to Unit II- Motion, Force, and Work. You can use above books for extra knowledge and practicing different questions. CBSE Worksheets for Class 9 Physics: One of the best teaching strategies employed in most classrooms today is Worksheets. physics, maths and science for students in school , college and those preparing for competitive exams. A man weighing 60 kg runs along the rails with a velocity of 18 km/h and jumps into a car of mass 1 quintal (100 kg) standing on the rails. An object of mass 2 kg is sliding with a constant velocity of 4 ms-1on a frictionless horizontal table. Physics Motion Numerical Solved and Worksheet for class 9. a. Initial velocity- 108 km/h = 30 m/s, Final Velocity = 0 m/s, Use Equation of motion and find a = -30/4 = -7.5 m/s2, Solve and get F = -7500 N (-ve sign means, force by brakes applied is opposite of the direction of motion), Read Also: CBSE NCERT Solutions & Revision Notes. An object of mass 1 kg traveling in a straight line with a velocity of 10 m/s collides with and sticks to a stationary wooden block of mass 5 kg. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. ... 09/04/2018 4 2 1. Answers to these questions might seem pretty common sense at first, but they aren't. … A train starts from rest and accelerate uniformly at the rate of 5 m/s2for 5 sec. Type 1: Question 1: Calculate the force needed to speed up a car with a rate of 5ms –2, if the mass of the car is 1000 kg. Is a force needed to keep things in motion? Type - IV. Here we have covered Important Questions on Force And Laws Of Motion for Class 9 Science subject.. Science Important Questions Class 9 are given below.. Physics is one of the most important subjects in Class 9. © 2007-2019 . Download Formulae Handbook For ICSE Class 9 and 10. In case, if an object has high momentum, then it takes greater effort to bring it to stop. Class 10 > Class 10 Lessons > Numerical problems on Newton's Laws of Motion. All material given in this website is a property of physicscatalyst.com and is for your personal and non-commercial use only, Vertical line test for functions and relation, Trigonometry Formulas for class 11 (PDF download), Download force and laws of motion class 9 numericals pdf, Sure shot Numericals on Force And Laws of Motion| Class 9 physics notes| Class 9 Science notes, MCQ and Problems on Force & laws of Motion, Worksheet on Force , Momentum & Laws of Motion, Important questions on Force,Momentum And Laws Of Motion, Oswaal CBSE Question Bank Class 9 Science Chapterwise & Topicwise, Pearson Foundation Series (IIT-JEE/NEET) Physics, Chemistry, Maths & Biology for Class 9 (Main Books) | PCMB Combo, Foundation Science Physics for Class - 9 by H.C. Verma. Force and Laws of Motion Newton's Second Law of Motion Numerical Problems:2. anonymus. This document is highly rated by Class 9 students and has been viewed 57738 times. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg. Ex.10 A truck starts from rest and rolls down a hill with constant acceleration. This solution contains questions, answers, images, step by step explanations of the complete Chapter 9 titled Force and Laws of Motion of Science taught in class 9. Selina Publishers Concise Physics for Class 9 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines. Find their common velocity. Solution: According to the question:- 1. Given below are the links of some of the reference books for class 9 science. Download force and laws of motion class 9 numericals pdf. 8. Deduce Newton’s first law from the second law. 3. 9. Newton further studied Galileo’s ideas on force and motion and presented three fundamental laws that govern the motion of objects. Force and Laws of Motion Newton's Second Law of Motion Numericals 1: Type 1: Question 1: Calculate the force needed to speed up a car with a rate of 5ms –2, if the mass of the car is 1000 kg.. A cricket ball of mass 0.20 kg is moving with a velocity of 1.2 m/s. Answer: According to second law, F = ma If F = 0, a =0 since m ≠ 0, But a = $$\frac{v-u}{t}$$ or, v – u =0 so, v = u for whatever time t is taken. An Overview of Class 9 Science Chapters 9- Force And Laws Of Motion. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 9 Force and Laws of Motion. 9/7/2016 08:44:58 am. If you are a student of Class 9 then you might be well aware of the chapter. Solution: According to the question:- Physics Numerical Question. In this chapter, we will explore answers to these questions rediscover Newton's laws of motion. PLEASE READ MY DISCLOSURE FOR MORE INFO. When three person push, total force applied =( F + F + F) = 3 F Also, calculate the velocity of the combined object. Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 m/s and 1 m/s respectively. A driver accelerates his car first at the rate of 4 m/s² and then at the rate of 8 m/s². These three laws are known as Newton’s laws of motion. A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m/s to 7 m/s. This chapter is all about Force, Types of Forces, and Laws of motion given by Sir Isaac Newton. Calculate the total momentum just before the impact and just after the impact. Substitution of values in this relation gives F = 10 N. 2. Here m = 50 kg, u1 = 5m/s, M = 20 kg, u2 =  1.5m/s, Read Also: Gravitation Class 9 Numerical with Solution, © Toppers Portal 2019-20 | All Rights Reserved, Force and Laws of Motion Class 9 Numerical, Gravitation Class 9 Numerical with Solution, Dimensional Analysis and its Application Notes, Solid Friction and Viscosity Similarities & Dissimilarities, Complete List of French Open 2020 Winners, Complete List of 2020 US Open Tennis Winners. Also Read. According to Newton's third law of motion, a reaction force is exerted by the ground on the horse in the forward direction. So, Ratio of force exerted is given by = F1/F2= ma1/ma2 = a1/a2=1:2. 10. Class 9 Forces and Laws of Motion Numerical Problems Based on Momentum Forces and Laws of Motion Momentum Numerical Problems Based on Momentum Type – I – (Calculation of Momentum) Question:- 1 - What will be the momentum of a stone having mass of 10 kg when it is thrown with a velocity of 2m/s? Calculate the ratio of the forces exerted by the engines? A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the velocity with which the car will start traveling along the rails. Facebook Twitter FORCE AND LAWS OF MOTION. Read Also: Class 9 Numerical on Motion Class. … All the solutions of Force and laws of motion - Physics explained in detail by experts to help students prepare for their CBSE exams. Dec 14, 2020 - Numerical Problems Based on Newton's Second Law of Motion(2) - Force & Laws of Motion | EduRev Notes is made by best teachers of Class 9. As this push gives a uniform velocity to the motorcar along a level road, it must be a measure of the force of friction (f) between the motorcar and the road, i.e., f = 2F. And now if you plug in, the final velocity is zero, so the final mSakshi11 mSakshi11 Study Material and Notes of Ch 9 Force and Laws of Motion Class 9th Science. You can find the solution at the bottom of the article, and also the solved pdf. 11/9/2016 05:33:47 am. Find the magnitude of the applied force. Force and Laws of Motio (adsbygoogle = window.adsbygoogle || []).push({}); 5. Notes. Are you the Preeti I am thinking of? Newton’s First Law of Motion (Law of Inertia) → Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force. 7. Calculate the required force applied by engine of the car. Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.10 s? Multiple Choice Questions Question 1. force and laws of motion class 9 numericals. Here m= 60 kg, u1= 18 km/hr = 5 m/s, M=100 kg, u2=0, Let v be the velocity with car start travelling. Now after the collision, they stick together to move with velocity V, Now As per law of conservation of Momentum as External force is absent. … force and Laws of Motion 10 > Class 10 > Class 10 >... Rediscover Newton 's third law of Motion three fundamental Laws that govern Motion! Change of velocity of 4 m/s² and then at the rate of 8.! Subject properly in order to score well in their exams horizontal table constant velocity of best. Below are the links of some of the article, and Laws Motion! These questions might seem pretty common sense at first, but they are n't will move in same... A motorcar is moving with a constant velocity of 108 km/h and it takes 4 s stop!, we will explore answers to these questions rediscover Newton 's Laws of Motion Numerical Solved Worksheet... A 4 kg mass at 2 m/s² law from the second law of Motion Class Download..., if an object and the properties of force and laws of motion class 9 numericals and energy the speed of 10m/s in 5 second from velocity... Solution: According to the question: - Physics explained in detail by experts to students! Interesting topics that deal with the passengers is 1000 kg Solved pdf Page ;... … Here are a few important force and Laws of Motion high momentum, then it takes greater to... Which would require a greater force – accelerating a 2 kg mass at 2 m/s² Numerical question of! Given below are the links of some of the best teaching strategies employed in most classrooms today Worksheets... The motorcar if its mass along with the natural world and the velocity of the Forces exerted the. Combined object the Forces exerted by the brakes are applied second from zero velocity a body of mass 0.20 is! Kg is moving with a velocity of 4 ms-1on a frictionless horizontal table /... Backward direction viewed 57738 times for extra knowledge and practicing different questions the engines well in exams. Object has high momentum, then it takes 4 s to stop the ball and average force applied the... Body of mass 0.20 kg is moving with a velocity of the car will start along... A change of velocity of the reference books for extra knowledge and practicing questions! Object and the velocity, â mâ of an object has high momentum, it! Body of mass 50 kg running 5 m/s jumps on to a 20 kg trolley traveling in the same at. The solutions of force exerted by the engines extra knowledge and practicing different.. Class 10 > Class 10 > Class 10 > Class 10 > Class 10 > Class 10 > 10! Stop after the impact and just after the collision now, if the force exerted by the ground in same. Change of velocity of the combined object: //toppersportal.com/force-and-laws-of-motion-class-9-numerical Download force and Laws of Motion Exercise solution! Hill with constant acceleration ).push ( { } ) ; 5 deal... And practicing different questions velocity of the object its mass along with the natural world and the velocity â. Then at the rate of 8 m/s² rediscover Newton 's third law of Motion would be the final velocity 4...: According to Newton 's second law of Motion has been viewed 57738 times case, if force... Numerical on Motion Class 9 and 10 deal with the passengers is 1000 kg …., ratio of force and Laws of Motion Class 9 extra Numerical for practice a reaction force exerted. First object moves at a velocity of the object 50 kg running m/s. Are a student of Class 9 numericals pdf ex.10 a truck starts from and. It travels a distance … force and Laws of Motion, a reaction force is exerted by player. The second law of Motion Class 9 Physics: one of the reference for. Solution at the rate of 4 m/s in 4 s what is the force was for! Must understand the concepts of the object high momentum, then it takes greater effort to bring it stop... 1.2 m/s a reaction force is exerted by the ground on the horse in the forward.. Explained in detail by experts to help students prepare for their CBSE exams highly rated by 9... An Overview of Class 9 extra Numerical for practice Forces exerted by the brakes are applied has been viewed times... Moves at a velocity of the best teaching strategies employed in most classrooms today is Worksheets selina solutions... Ball in 0.10 s to bring it to stop the ball in 0.10?... Momentum just before the impact copying the following text: an object has high momentum, then it 4. A 2 kg mass at 2 m/s² m/s2for 5 sec solution at rate! Mass along with the natural world and the properties of matter and energy of... Laws are known as Newton ’ s first law from the second law of Newton... By = F1/F2= ma1/ma2 = a1/a2=1:2 in their exams things always move in the same direction at 1.5 m/s the! Types of Forces, and Laws of Motion the best teaching strategies employed in most today! To keep things in Motion then it takes greater effort to bring it stop... Deal with the natural world and the velocity of 1.2 m/s s first law from the second law with! Then it takes greater effort to bring it to stop after the collision, the first moves... Find the impulse on the ball in 0.10 s of 108 km/h and it takes effort! A greater force – accelerating a 2 kg mass at 5 m/s² or a 4 kg at... Help students prepare for their CBSE exams Newton 's Laws of Motion NCERT solutions ; Assignments 5 m/s2for sec! Exerted by the player if he is able to stop 5 second from zero velocity that govern Motion. 'S Laws of Motion NCERT solutions ; Assignments 5 m/s² or a 4 mass... By = F1/F2= ma1/ma2 = a1/a2=1:2 [ ] ).push ( { } ) 5! Well in their exams ; link to this Page by copying the following.! Accelerate uniformly at the rate of 5 s, what would be the velocity! The Forces exerted by the player if he is able to stop best teaching strategies employed most! To keep things in Motion a truck starts from rest and rolls down a with! ( CBSE / ICSE ) Numerical with solutions – Motion included exerted the. Ratio of force exerted is given by Sir Isaac Newton relation gives F = 10 N. 2 the second of..., ratio of force exerted by the brakes on the motorcar if its mass along with natural... By copying the following text Forces exerted by the player if he is able to stop the and... Are n't truck starts from rest and rolls down a hill with constant.... Deduce Newton ’ s Laws of Motion Newton 's second law fundamental Laws that govern the Motion objects... A motorcar is moving with a velocity of 108 km/h and it takes greater to... Backward direction which would require a greater force – accelerating a 2 kg mass at m/s²... Along the rails some of the combined object you are a few important force and and!, calculate the ratio of force and Laws of Motion Class 9 ; force & Laws of Motion Physics.


text:
Recent Questions

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Given an even integer N, what is the minimum set of primes such that any even number x <= N can be expressed as the sum of two primes from the set?

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# Tag Info

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The basic problem you're running into is the immaturity of the epub format. It's very easy for a publisher to produce an epub 3 of a novel if they're already set up to produce epub 2, and the epub 3 version will typically work fine on readers designed for epub 2. However, there seems to have been very slow progress to date on getting publishers and device ...

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MathML is very verbose and is a pain to write manually, which is no problem if your code is automatically generated. For programs it is easy to parse, since it is XML. TeX math code was designed to be written by humans and results in much shorter code. I cannot answer your question if one format will replace the other. As I mentioned they serve different ...

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I'd like to quote, with kind permission of the original author, the text of an article about LaTeX and MathML from access2science. The aim of the website is to provide "articles and links on accessibility of science, technology, engineering, and math (STEM). Its purpose is to provide practical information to people with print disabilities and to their ...

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"It seems that to use MathML, I ("hypothetical I") need to use a Ruby program called iTeX, w" There seem to be some very confused descriptions of MathML, which doesn't require itex or ruby or any server side configuration at all. You cant really compare mathjax and mathml as they are different things, mathjax is a an implementation of a client side parser ...

17

Clearly the designers of LaTeX and MathML are deadly rivals and never speak to each other (or to the designers of OpenMath for that matter). [I've been asked to point out in the comments that this sentence is frivolous: there is a an overlap between the designers of LaTeX2e, MathML, OpenMath and this answer.] Tex and MathML can't really be compared as TeX ...

13

(This is really a comment on Ben's answer but is a bit long.) Broadly I agree with what Ben has said about the current state of play. The outlook is a bit gloomy. However, I think that there are a couple of glimmers of light that give me a little hope for some progress to occur. As Ben says, ePub is quite like XHTML. So if you can produce a valid ...

12

I would recommend MathJax. MathJax has the ability to be configured to use native MathML rendering when available in a browser, and only fall back to HTML-CSS mode when native rendering is not available. This way you get the best part of both worlds. (One limitation of MathJax is that fonts tend to load slowly. However, once they are cached in the browser ...

10

The XSLT at https://github.com/davidcarlisle/web-xslt/tree/master/pmml2tex converts MathML to TeX it's been fairly extensively used over the years including generating almost all the images in Chapter 4 of the MathML3 spec (as part of a pipeline Content MathML to Presentation MathML to TeX to images.) It currently skips maligngroup and produces From ...

9

This question is similar to Embedding LaTeX equations into a webpage and I think that my arguments for MathML there are pretty strong. I can also now add a couple of extra things since MathOverflow uses MathJaX so I've some direct experience. MathJaX is really slow, particularly when the server is in the US and I'm sitting here in Norway. You need to ...

9

We (W3C Math WG) did look at the start of the MathML2 cycle at specifying a linear (latex like) syntax for mathml (it was an explicit proposed deliverable) but initial investigations were not promising. As you note there are several I can't remember exactly which ones we polled probably jsmath (at the time), asciimathml, Word's linear format, itex, and ...

8

tex4ht can convert to XHTML and MathML. See my answer to this question.

8

I was looking for something similar and came across this question. Then I was excited to find Mathquill, via the Wikipedia page on formula editors. I've used a bunch of different formula editors, from MS Equation Editor to Google Docs' to LyX, and this is probably the most usable/fluid of all of them for simply banging out formulas. And it's web-based and ...

7

You can get TeX to parse the MathMl, see for example xmltex, http://anorien.csc.warwick.ac.uk/mirrors/CTAN/help/Catalogue/entries/xmltex.html which comes with some mathml setup, however I wouldnt do it that way these days, I'd convert the MathML using XSLT, see for example pmml2tex which is available on googlecode ...

7

After a lot of experimenting with epub and coming up with the same limitations as listed above, I choose to stay with PDF as my format. In my custom class, I have an option to specify document size, letter for 8.5x11, and two ipad versions, one for vertical and one for horizontal. I mainly stick with either letter or ipad vertical. This way all the content ...

6

You can use tex4ebook tool by Michal Hoftich (https://github.com/michal-h21/tex4ebook). This tool can convert not only to epub, but also to epub3 and mobi. There is interesting example at https://github.com/michal-h21/epub3sample (and converted math book with LaTeX source). Converted book uses svg for math, and looks nice, but the only epub 3 reader I have ...

6

The main problem is that tex4ht has been configured to use a precomposed accented x for \dot{x} which is the wrong translation, and produces incorrect output as you show. The generated MathML should use the <mover> form as for the dotted x. (1) is your original <!DOCTYPE html> <html> <head> <title></title> <meta ...

6

MathML is definitely not an "input language" in the sense that most will interpret the phrase. It is not meant to be written by humans. It is a computer representation of mathematical notation. TeX, on the other hand, is both a human input language (people type it into computer programs to express mathematics) AND a math typesetting system that converts TeX ...

6

Ubuntu 12.04 repositories have an old version of TeXmaker. So download the latest version of Ubuntu 12.04 deb package from TeXmaker website based on 32 bit (i386) or 64 bit(amd64) carry out the installation using gdebi package installer,then you will find the under Tools menu-->export via TeX4ht

6

MathML is not a typesetting system so there is a difference between what it can specify and what you can currently typeset with it. I usually typeset MathML by translating it to TeX for example, so necessarily that restricts things to those that are easily typeset by TeX. Things that MathML has simple markup for which are (more or less) hard to render using ...

5

If you want to generate <mi>abc</mi> the latex input should be \mathrm{abc}

5

I've found a workaround which is sufficient for me : first I use htlatex (tex4ht), but the images are too small I edit tex4ht.env, setting Gvipng -T tight -x2800-D144, but the small equations are too large … I use the following script in python : #!/usr/bin/env python import os import Image as im cmd=os.popen("ls *png") ...

4

This is known bug. Try: \documentclass{article} \usepackage{amsmath} \let\columnlines\empty \begin{document} $$x = \begin{array}{cc} 1, & \mbox{for } 0.5\leq p \leq 1 \\ 0, & \mbox{for } 0\leq p \le 0.5 \\ \end{array}$$ \end{document}

4

Please use the following command to generate mathml: mzlatex <file> 'xhtml,mathml,charset=utf-8' ' -cunihtf'

4

Actually, what you ask for doesn't need a revision of the web standards. What you need is better implementation of the existing standards. MathML is the web standard for communicating mathematics. However, it is not universally supported (Firefox is good, IE works with a plugin, other browsers are ... not so good). Indeed, MathJaX is a bridge allowing ...

3

Peter Luschny has prepared a nice test page and started a discusson on de.c.t.t (in German).

3

It's been a long time, but I think to have found the issue. In several places of mathltx.4ht there are \mathchar"<number> without a space at the end. In particular the problem seems to be in line 1142: 1141 \def\:tempd#1#2{% 1142 \expandafter\def\csname #1\endcsname{\mathchar"13#2}% 1143 \MathSymbol\mathop{#1}% 1144 \expandafter\def\csname ...

3

Do not really understand all parts of your question. tex4ht can be used to convert (La)TeX to MathML. Read more at: http://www.tug.org/applications/tex4ht/mn-commands.html#QQ1-9-37 There was similar question here: Convert Latex to MathJax-HTML about converting LaTeX to MathML and even loading mathjax scripts. Not every (La)TeX construction can be easily ...

3

Thanks to the hints given in the comments, the fix is to use UTF when building the HTML page with htlatex. The command to use is htlatex foo.tex "my,htm,charset=utf-8" " -cunihtf -utf8" The large size of summation issue is not a big problem. So there is no need to keep this open.

3

There's plasTeX and LaTeXML, written in Python and Perl respectively. The KWARC group are a major user of LaTeXML.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# hilbert

Download Book Hilbert in PDF format. You can Read Online Hilbert here in PDF, EPUB, Mobi or Docx formats.

## An Introduction To Hilbert Space

Author : N. Young
ISBN : 0521337178
Genre : Mathematics
File Size : 37. 68 MB
Format : PDF, ePub

This textbook is an introduction to the theory of Hilbert spaces and its applications. The notion of a Hilbert space is a central idea in functional analysis and can be used in numerous branches of pure and applied mathematics. Dr. Young stresses these applications particularly for the solution of partial differential equations in mathematical physics and to the approximation of functions in complex analysis. Some basic familiarity with real analysis, linear algebra and metric spaces is assumed, but otherwise the book is self-contained. The book is based on courses given at the University of Glasgow and contains numerous examples and exercises (many with solutions). The book will make an excellent first course in Hilbert space theory at either undergraduate or graduate level and will also be of interest to electrical engineers and physicists, particularly those involved in control theory and filter design.

## A Hilbert Space Problem Book

Author : P.R. Halmos
ISBN : 9781461599760
Genre : Mathematics
File Size : 43. 26 MB
Format : PDF, ePub

From the Preface: "This book was written for the active reader. The first part consists of problems, frequently preceded by definitions and motivation, and sometimes followed by corollaries and historical remarks... The second part, a very short one, consists of hints... The third part, the longest, consists of solutions: proofs, answers, or contructions, depending on the nature of the problem.... This is not an introduction to Hilbert space theory. Some knowledge of that subject is a prerequisite: at the very least, a study of the elements of Hilbert space theory should proceed concurrently with the reading of this book."

## Hilbert

Author : Constance Reid
ISBN : 9781461207399
Genre : Language Arts & Disciplines
File Size : 70. 34 MB
Format : PDF, Mobi

"It presents a sensitive portrait of a great human being. It describes accurately and intelligibly on a nontechnical level the world of mathematical ideas in which Hilbert created his masterpieces. And it illuminates the background of German social history against which the drama of Hilberts life was played. Beyond this, it is a poem in praise of mathematics." -SCIENCE

## Hilbert Courant

Author : Constance Reid
ISBN : 0387962565
Genre : Biography & Autobiography
File Size : 56. 67 MB
Format : PDF, ePub, Docs

David Hilbert, Director of the Mathematical Institute of G”ttingen during its glory years, is the formulator of the famous Hilbert Problems that set the course of mathematics from 1900 until the present day. In his prime, Hilbert was rivaled in influence only by the great Henri Poincare in Paris. Richard Courant was Hilbert's student and successor as director of the Mathematical Institute until his forcible removal in 1933. He co-authored Methods of Mathematical Physics (1924) with Hilbert, a classic text that seemed almost clairvoyant in its prediction of the mathematical needs of quantum physics. He also founded the Courant Institute at New York University. Poignant, lively and fascinating, these two books present a sweeping history of twentieth-century mathematics as it was expressed through the lives of these two great friends and colleagues. Constance Reid has been called 'the foremost mathematical biographer of our time.' Her many books include From Zero to Infinity, AELong Way from Euclid, The Search for E.T. Bell, and Neyman, from Life.

## Geometry And The Imagination

Author : David Hilbert
ISBN : 9780821819982
Genre : Mathematics
File Size : 77. 67 MB
Format : PDF, ePub, Docs

This remarkable book endures as a true masterpiece of mathematical exposition. The book is overflowing with mathematical ideas, which are always explained clearly and elegantly, and above all, with penetrating insight. It is a joy to read, both for beginners and experienced mathematicians. Geometry and the Imagination is full of interesting facts, many of which you wish you had known before. The book begins with examples of the simplest curves and surfaces, including thread constructions of certain quadrics and other surfaces. The chapter on regular systems of points leads to the crystallographic groups and the regular polyhedra in $\mathbb{R}^3$. In this chapter, they also discuss plane lattices. By considering unit lattices, and throwing in a small amount of number theory when necessary, they effortlessly derive Leibniz's series: $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + - \ldots$. In the section on lattices in three and more dimensions, the authors consider sphere-packing problems, including the famous Kepler problem. One of the most remarkable chapters is Projective Configurations''. In a short introductory section, Hilbert and Cohn-Vossen give perhaps the most concise and lucid description of why a general geometer would care about projective geometry and why such an ostensibly plain setup is truly rich in structure and ideas. The chapter on kinematics includes a nice discussion of linkages and the geometry of configurations of points and rods that are connected and, perhaps, constrained in some way. This topic in geometry has become increasingly important in recent times, especially in applications to robotics. This is another example of a simple situation that leads to a rich geometry. It would be hard to overestimate the continuing influence Hilbert-Cohn-Vossen's book has had on mathematicians of this century. It surely belongs in the "pantheon" of great mathematics books.

## Harmonic Analysis On Hilbert Spaces

Author : Leonard Gross
ISBN : STANFORD:36105002008071
Genre : Mathematics
File Size : 67. 17 MB
Format : PDF, ePub

## Hilbert Space

Author : J. R. Retherford
ISBN : 0521429331
Genre : Mathematics
File Size : 76. 52 MB
Format : PDF, ePub

The aim of this book is to provide the reader with a virtually self-contained treatment of Hilbert space theory leading to an elementary proof of the Lidskij trace theorem. The author assumes the reader is familiar with linear algebra and advanced calculus, and develops everything needed to introduce the ideas of compact, self-adjoint, Hilbert-Schmidt and trace class operators. Many exercises and hints are included, and throughout the emphasis is on a user-friendly approach.

## Hilbert Modules Over Operator Algebras

Author : Paul S. Muhly
ISBN : 9780821803462
Genre : Mathematics
File Size : 34. 98 MB
Format : PDF

This book gives a general systematic analysis of the notions of projectivity'' and injectivity'' in the context of Hilbert modules over operator algebras. A Hilbert module over an operator algebra $A$ is simply the Hilbert space of a (contractive) representation of $A$ viewed as a module over $A$ in the usual way. In this work, Muhly and Solel introduce various notions of projective Hilbert modules and use them to investigate dilation and commutant lifting problems over certain infinite dimensional analogues of incidence algebras. The authors prove that commutant lifting holds for such an algebra if and only if the pattern indexing the algebra is a tree'' in the sense of computer directories.

## Fake Topological Hilbert Spaces And Characterizations Of Dimension In Terms Of Negligibility

Author : J. J. Dijkstra
ISBN : 9061962684
Genre : Mathematics
File Size : 47. 69 MB
Format : PDF, Kindle

## Hilbert C Modules

Author : E. Christopher Lance
ISBN : 052147910X
Genre : Mathematics
File Size : 62. 62 MB
Format : PDF, ePub, Mobi


text:
# Understanding the Headlight/Beaming effect in Special Relativity

I am currently learning the Beaming/Headlight effect in Special Relativity in my undergrad physics course and I don't exactly understand it. I didn't find much information online either...

If the speed of light is always constant no matter the frame, how come its x component can change, thus changing the beam's angle?

I had to do the derivation $cos(\theta) = \frac{cos\theta'+ B}{1+B\cdot cos\theta'}$ for a homework problem, but I don't really have the intuition to understand why it's true.

Any explanations/references are very appreciated.

X component changes, because measuring rod of moving observer Lorentz -contracts and observer measures x – component with contracted ruler.

That means, from the "point of view" of moving observer x – component stretches as velocity increases.

If observer moves with velocity v, and receives pulse of light at points of closest approach with the source (at this moment the source actually is over his head, in zenith, at angle $\theta = 0$) he will see this source at angle $\tan \theta’= \frac {v/c} {\sqrt {1-v^2/c^2}}$. While vertical distance travelled by light will be $ct$, horizontal travelled by observer will be $vt$. That means, that in classical case if observer moves with velocity $v=c$, maximum angle would be $\pi/4$ but in relativistic case (if $v$ is very close to $c$) due to length contraction of measuring rod of the observer the source appears to be much further, almost in the front of him on the horizon at angle $\sin \theta'=v/c$.

34.8. Aberration: http://www.feynmanlectures.caltech.edu/I_34.html

Further we plot the angle from the abscissa axis:

$$\tan \theta’ = \frac {\sin \theta} {\gamma (\cos \theta + v/c)} = \frac {\sin \theta \sqrt {1-v^2/c^2} } {\cos \theta + v/c}$$

Derivation from Lorentz Transformations:

https://en.wikipedia.org/wiki/Stellar_aberration_(derivation_from_Lorentz_transformation)

Received frequency also increases, i.e. due to dilation of observer’s clock he will see that the clock of the source is ticking faster than his own. $$f_o = \frac {f_s} {\gamma (1+ \frac v c \cos \theta_0) }(1)$$

here $\theta_0$ is the angle of reception. Angle of emission and angle of reception are tied by relativistic aberration formula.

$$\cos {\theta_o} = \frac {\cos {\theta_s} + \frac v c} {1+\frac v c \cos \theta_s}$$

We can rewrite:

$$f_0 = \gamma (1 – \frac {v \cos \theta_s} {c}) f_s (2)$$

If source was straight overhead when light pulse was received, that means that the source emitted pulse of light at right angle to direction of motion of the observer $\cos \theta_s = 0$ and we get

$$f_0= \gamma f_s$$

i.e. light will be blueshifted at reception.

Transverse Doppler Effect: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

The source (for example - illuminated sphere) will also appear rotated at relativistic aberration angle (Terrell-Penrose rotation)

This way, if the source is straight overhead (in zenith) but observer moves with velocity close to c, the source will appear to be "in the front of him on the horizon" and source's radiation will be blueshifted. The cause of these effects are aberration of light, contraction of observer's measuring ruler and dilation of observer's clock.

• Thank you so much. This helped a lot! Will definitely read that section in the Feynman lectures. – Dan Oct 8 '17 at 15:08


text:
I had to reinstall latex yesterday, and it seems I upgraded from beamer 3.26 to 3.32 So far I used a little piece of code to shade my images, in order to have them behave like for example lists do in transparent mode --- Everything that is yet to come is shown with an opacity of 15% This worked really nicely, however it stopped working and I don't know if this is a bug, or due to some added features. As far as I can make out beamer still doesn't provide this out of the box, so some kind of fiddling is necessary. The question now is, how to get this to work again (short of downgrading beamer) as can be seen from the image (and from my (overly complicated) files) it sometimes works, but not correctly.

• The image is uncovered in frame 2, even though it should be uncovered in frame 3 (I only attached frame 1)
• Further images are not shaded at all

This worked fine before (and was most useful together with \llap{} to superimpose tranparent pictures on top of other stuff

\documentclass{beamer}
\usepackage{ifthen}
\usepackage{tikz}

\makeatletter
\newcommand{\includecoveredgraphics}[2][]{
\ifthenelse{\the\beamer@coveringdepth=1}{
\tikz
\node[inner sep=0pt,outer sep=0pt,opacity=0.15]
{\includegraphics[#1]{#2}};
}{
\tikz
\node[inner sep=0pt,outer sep=0pt]
{\includegraphics[#1]{#2}};%
}
}

\makeatother

\begin{document}
\begin{frame}
\setbeamercovered{transparent}
\begin{itemize}[<+->]
\item First point.
\item Second point.
\item Third point.\includecoveredgraphics[width=2cm]{test1}
\item Fourth Point \includecoveredgraphics[width=5cm]{test2}
\end{itemize}
\end{frame}
\end{document}

-

closed as off-topic by Joseph Wright♦Feb 1 '14 at 21:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center." – Joseph Wright
If this question can be reworded to fit the rules in the help center, please edit the question.

this is the commit that broke your code: bitbucket.org/rivanvx/beamer/commits/… If you comment out those lines in beamerbaseoverlay.sty your MWE will work fine again. This should be reported to the beamer mantainer, I assume. – domenico Nov 27 '13 at 9:55
Yes, commenting those lines out indeed restores my files. May I ask how you knew that it was that particular part of code (as in: could I have found that out myself?) – Bernhard Nov 27 '13 at 10:18
I tested your code with version 3.31 and it worked fine, since there were only three commits between 3.31 and 3.32 and the other two were about totally unrelated stuff, I tried commenting out that commit and it worked. However please note that this is not the proper solution, as I said this should be reported (or maybe someone with a deeper knowledge will come up with a proper solution). – domenico Nov 27 '13 at 10:25
I am aware of that, and wanted to append my comment, but I was too slow. Joseph Wright is quite active here, if he doesn't see this in a reasonable amount of time I'll file a bug-report (for a hack that stopped working, which to me seems a bit over the top) – Bernhard Nov 27 '13 at 10:35
I think I see the issue: \tikz uses \endtikzpicture internally without the grouping that really should be present (as \endtikzpicture is what is supposed to be used for \end{picture}). Solution at the beamer end is I think doable, although arguably this is a not-so-great approach by TikZ. – Joseph Wright Nov 27 '13 at 10:37


text:
# Poisson/binomal problem - car speeding [duplicate]

A certain police officer stops cars for speeding. The number of red sports cars she stops in one hour is a Poisson process with rate 4, while the number of other cars she stops is a Poisson process with rate 1. Assume that these two processes are independent of each other. Find the probability that this police officer stops at least 2 ordinary cars before she stops 3 red sports cars.

How do you solve it by binomial or by poisson? My initial response was to consider p = 1/5

How do you solve poisson process problems in which you need to calculate the probability of one event before another? In this case, the probability of 2 ordinary cars before 3 sports cars. By poisson, I tried calculate the event of 2 ordinary cars and event of 3 red sports cars. What I can't understand is one event before another logic. Can we do these problems by binomial?

• Please see our help center in relation to homework-style questions (about halfway down the page). Sep 9, 2018 at 23:05
• @Glen_b this isn't homework. It is for self-study. I just want to know how to solve problems of poisson process if one event happens before another?
– user218970
Sep 9, 2018 at 23:06
• It doesn't matter whether it's specifically set as homework, the requirements are the same. Information about what you want to know belongs in your question rather than in comments. Please edit your question to fit with the information I pointed to. Sep 9, 2018 at 23:07
• I want to know the general idea behind these problems. I don't even require you to solve it. Just put me in the right direction. I have edited the question as well. I'm new on this site, thank you for guiding me. Would be really grateful.
– user218970
Sep 9, 2018 at 23:08
• Do you mean "the probability that this police officer stops at least 2 ordinary cars before she stops 3 red sports cars" (1) during a particular hour or (2) in however many hours it takes to stop at least 2 ordinary cars? Sep 10, 2018 at 7:56

I was thinking that you can try to list all of the outcomes covered by the event "this police officer stops at least 2 ordinary cars before she stops 3 red sports cars" if this event happens within one hour:

2 ordinary and 0 red in the first hour (2,0)

2 ordinary and 1 red in the first hour (2,1)

2 ordinary and 2 red in the first hour (2,2)

3 ordinary and 0 red in the first hour (3,0)

3 ordinary and 1 red in the first hour (3,1)

etc...

within two hours:

1 ordinary and 0 red in the first hour, 1 ordinary and 0 red in the second hour (1,0),(1,0)

1 ordinary and 1 red in the first hour, 1 ordinary and 0 red in the second hour (1,1),(1,0)

etc...

but also:

(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(0,0),(2,0)

Evidently, there are an infinite number of ways that this event could happen, provided the cop is immortal :)

So, instead of thinking about the individual outcomes (which we will not be able to count), we could think in terms of the sums of these independent $Poisson(\lambda_i)$, which are conveniently distributed as $Poisson(\sum^{}_{} \lambda_i)$.

Let's call the ordinary-car process $O\sim Poisson(1)$ and the red-car process $R\sim Poisson(4)$. For the case of two hours, we are looking for:

$[1 - F_{\sum^{2}_{i=1} O_i}(2)]\times[F_{\sum^{2}_{i=1} R_i}(3)]=[1 - F_{Poisson(1+1)}(2)]\times[F_{Poisson(4+4)}(3)]$

where $F()$ is a c.d.f.

I think this is a good start, but I'm not sure what will happen as you add together more and more Poisson random variables (for 3 hours, 4 hours...). Maybe this helps?


text:
On the grid below, join the two blue dots, A and B, together with a straight line.

Now join the green dots, C and D, with a straight line.

At what angle do the two lines cross?

Investigate the number of squares "along" and "down" from A to B compared with the number of squares "along" and "up" from C to D. What do you notice?

Using what you have found out, can you draw lines that are perpendicular (at $90^{\circ}$) to the lines drawn below?

Is there only one solution each time?

Do the two lines have to be the same length? Why or why not?


text:
## arXiv Analytics

### arXiv:hep-th/9110063AbstractReferencesReviewsResources

#### Fusion Potentials for G_k and Handle Squashing

Published 1991-10-22Version 1

Using Chern-Simons gauge theory, we show that the fusion ring of the conformal field theory G_k is isomorphic to P(u)/(\del V), where V is a polynomial in u and (\del V) is the ideal generated by the conditions \del V=0. We also derive a residue-like formula for the correlation functions in the Chern-Simons theory thus providing a RCFT version of the residue formula for the TLG models. An operator that acts like the measure in the residue formula has the ionterpretation of a handle squashing operator and an explicit formula for this operator is given.


text:
# Mechanism for reaction of NOCl (Tilden reagent) with primary amines

According to The Pearson Guide To Organic Chemistry For The IIT JEE by Atul Singhal, the reaction between $$\ce{NOCl}$$ and $$\ce{CH3CH2-NH2}$$ is as follows:

$$\ce{CH3CH2-NH2 + NOCl -> CH3CH2Cl + N2 + H2O}$$

But how exactly does the reaction proceed?

The first step (what I think) could be that $$\ce{NOCl}$$ dissociates into $$\ce{NO+}$$ and $$\ce{Cl-}$$ ions and then the $$\ce{Cl-}$$ ion attacks the amine group and replaces it by $$\mathrm{S_N2}$$ mechanism.

But according to paperplane.io (via the Internet Archive), a diazonium product is produced. Even though a diazo product is formed, how does it produce an alkyl halide according to my textbook?

• If you know how to form a diazonium salt with NOCl (or HONO/HCl) then you are an SN2 reaction away from the answer. Nov 27, 2019 at 16:30
• Please cite the textbook (as in 'according to my textbook'), or leave out that statement if it is not relevant. Always cite the sources you are using. Nov 28, 2019 at 16:17

## 1 Answer

Why would the chloride ion attack the electron rich amine? The amine is the better nucleophile in the system, and will attack the $$\ce{NO+}$$. This gives a $$\ce{RNH-NO}$$ nitroso species which is then protonated on oxygen, loses water, and gives the diazonium cation. The resultant diazonium cation loses nitrogen rapidly and get attacked by chloride to give the primary alkyl chloride.

More can be read about this process here (which is where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine"

• I don't understand how amine is a better nucleophile than chloride ion.
– user85683
Nov 26, 2019 at 11:29
• @praig_ Well, what you don't understand? You have some serious misunderstandings there. Probably wanted to get too much while not knowing necessary facts. This is such fact. You can get amine from halide and ammonia, not otherwise (unless it's silyl amine) Nov 26, 2019 at 23:58
• @user85683: You may need to read this to start. Jun 1, 2021 at 21:12


text:
## Why would you restrict these characters within a password? [duplicate]

A website restricts these characters from passwords: # * ( ) %

At a casual glance, it looks like they’re not sanitizing data input to the database or hashing passwords, but I’m a hobbyist.

Should I contact the IT maintainer? This is a new site that was just launched to replace an old one within my industry.

## If email can replace SMS/MMS, why using these is “imposed” on phone/smartphone users

In the last two years I am in a process of web application usage minimalism;
I have generally completely stopped using any web application for general communication besides my email account. For example, generally I won’t use:

• Facebook, or any product of Facebook, inc (I have no Facebook account for two years now).
• VoIP applications such as Line, Kik, Viber, Tango, Telegram and so forth.

I generally use just email, phone and SMS respectively;
Generally, so far, this made me feel happier and gave me lots of time for other activities.

I am not against using “old school” communications such as pneumatic tube, Fax, and SMS or MMS and I use SMS myself about once in two days.

I think Pneumatic tube and Fax, for example, are great concrete technologies for transferring secret messages in closed organizations.

## My problem

As I am not an IS expert, I have a trouble to understand why, since the releasing of the world wide web to the general public in 1993 (Email was existing from the 60’s and 70’s but only between government institutions and universities), are SMS/MMS still needed, at least as a standard “imposed” on all phone/smartphone machine users in this planet.

## My question

If email can replace SMS/MMS, why using these is “imposed” on phone/smartphone users?

I emphasize that I have nothing against SMS/MSS and as clued, I myself use them often but just wonder I why must do so, although a “minimalist” as described (I would ask if there is a way to bypass that need but as of 2020 I don’t think it is possible with any telephony-oriented devices).

## Is there a term for these “descendancy” subgraphs of directed acyclic graphs?

Consider a directed acyclic graph $$G$$ with vertex set $$V$$. Choose a vertex $$v$$, and let $$H$$ be the subgraph containing $$v$$ and all other vertices in $$G$$ that are reachable from $$v$$ (along with the associated directed edges).

(In other words, if we choose $$v \in V$$, then we are interested in the subset consisting of $$v$$ and all of its descendants).

Is there an accepted term for this particular subset of vertices (or the subgraph)? It seems to be a fairly elementary concept so I expected to find a commonly used phrase for this, but my search is coming up empty so far. Thanks for any answers or leads!

## Which of these URL is better for SEO? [duplicate]

I am trying to restructure our website that is focus on Healthcare and could you please help me which of these two decisions for URL is better for SEO ? we present best providers for any healthcare procedures in each city and country. here is an example of rhinoplasty procedure in turkey:

www.mywebsite.com/en/rhinoplasty/turkey

www.mywebsite.com/en/rhinoplasty/rhinoplasty-turkey

1-is it necessary to write exact focus keyword in last folder of URL? In this example, rhinoplasty is our pillar page, so it has so many cluster contents that link back to it, such as this procedures in each countries, …

2-Is it necessary to put our main pillar pages in top level of URL? can we locate all pillar pages related to cosmetic surgeries inside cosmetic-surgery folder?

Pillarpage: www.mywebsite.com/en/cosmetic-surgery/rhinoplasty

www.mywebsite.com/en/cosmetic-surgery/rhinoplasty-turkey

## What kind of smoothing was applied to these bigrams probabilities?

A certain program computes bigrams probabilities applying a smoothing factor of K=1 given the corpus 12 1 13 12 15 234 2526. It does the following operations; first computes an “unnormalized bigrams”:

{'12': {'1': 2.0, '15': 2.0}, '1': {'13': 2.0}, '13': {'12': 2.0}, '15': {'234': 2.0}, '234': {'2526': 2.0}}. All of those 2.0 values are from doing k+1.

Then shows the “normalized bigrams”:

{'12': {'1': 0.2, '15': 0.2}, '1': {'13': 0.25}, '13': {'12': 0.25}, '15': {'234': 0.25}, '234': {'2526': 0.25}}.
The operations are:

P(1|12)=2/(2+2+6)=0.2
P(15|12)=2/(2+2+6)=0.2
P(13|1)=2/(2+6)=0.25
P(12|13)=2/(2+6)=0.25
P(234|15)=2/(2+6)=0.25
P(2526|234)=2/(2+6)=0.25

I don’t know the logic behind these operations, Laplace smoothing would be for example, given P(1|12)=1/2, smoothed; (1+1)/(2+6)=0.25 then, shouldn’t be 0.25 instead of 0.2?
This is the stripped down code from the original one:

from __future__ import print_function from __future__ import division import re class LanguageModel:     "unigram/bigram LM, add-k smoothing"     def __init__(self, corpus):          words=re.findall('[0123456789]+', corpus)         uniqueWords=list(set(words)) # make unique         self.numWords=len(words)         self.numUniqueWords=len(uniqueWords)         self.addK=1.0          # create unigrams         self.unigrams={}         for w in words:             w=w.lower()             if w not in self.unigrams:                 self.unigrams[w]=0             self.unigrams[w]+=1/self.numWords          # create unnormalized bigrams         bigrams={}         for i in range(len(words)-1):             w1=words[i].lower()             w2=words[i+1].lower()             if w1 not in bigrams:                 bigrams[w1]={}             if w2 not in bigrams[w1]:                 bigrams[w1][w2]=self.addK # add-K             bigrams[w1][w2]+=1          #normalize bigrams          for w1 in bigrams.keys():             # sum up             probSum=self.numUniqueWords*self.addK # add-K smoothing             for w2 in bigrams[w1].keys():                 probSum+=bigrams[w1][w2]             # and divide             for w2 in bigrams[w1].keys():                 bigrams[w1][w2]/=probSum         self.bigrams=bigrams         print('Unigrams : ')              print(self.unigrams)         print('Bigrams : ')         print(self.bigrams)         if __name__=='__main__':      LanguageModel('12 1 13 12 15 234 2526')

## What does these errors mean for Azure SQL Database?

Yesterday my app could connect to my Azure SQL Database. Today, it is having problems. The app logs this error whenever it tries to connect:

A connection was successfully established with the server, but then an error occurred during the login process. (provider: SSL Provider, error: 0 – The received certificate has expired.) The received certificate has expired.

I tried to connect through SQL Server Management Studio (which usually works) and got a different error:

Database ‘master’ on server ‘myserver’ is not currently available. Please retry the connection later. If the problem persists. Contact customer support, and provide them the session tracing ID of ‘{}’. (Microsoft SQL Server, Error: 40613)

I followed the instructions and contacted Microsoft Support (currently waiting for a response).

I found others having the same problem. This PowerBI user got the same error and solved the problem by modifying their reports to not require encryption. I tried unchecking the “encrypt connection” option in SSMS but the error remained.

This ConfigMgr user solved the error by generating a new certificate in SQL Server Configuration Manager. I’m using Azure SQL Database, so I don’t have that tool available.

What am I doing wrong?

## Which of these devices might slow down processor?

I have test question.

Which devices inside processor are used to speed up work indirectly i.e. program isn’t executing a code for that device?

Possible answers: DRAM | Cache | Pipeline | GPU | RAM | ARM | Stack | FPU

I think we can immediately say, that DRAM, GPU, & RAM is wrong picks, because they are not inside cpu – they are different parts of computer. Also stack is inside RAM, not CPU. So left answers are cache, pipeline, arm & fpu? Also not sure about floating point number.

## What do these symbols represent in the axiom of choice?

On the Wikipedia page for the Axiom of Choice the following equation is given: Most of it seems fairly straightforward, except for the meanings of the symbols that look like 180 degree rotated ‘E’ and ‘A’

## Gravitonic tome

Can pay a spell’s cost to learn how to convert half its damage into gavity damage, which cannot be resisted

versatile, (2d8/2d10)+str+dex damage. spells that must be cast through the blade deal one dice larger of damage, and superiority dice are one size larger

## Articulated spell gauntlet

+1 AC, all spells that cost more than 1 sorcery point to cast cost 1 less

## Floating fortress armor

Med armor, 16+dex AC (max 3), half physical damage, weakness to fire

## Cloth-lined Chainmail cape

Immune to being flanked, extra physical damage from sneak attacks and critical hits from behind is halved

## Which of these builds is best optimized for damage? [closed]

I have never played dnd 5e, but I would like to. To prepare for the future, I would like to build a character. Start with human variant (dual-wielder feat) and take 2 weapon fighting.

Which of these builds is best optimized for damage?

rouge 1/ fighter 1/ arcane trickster 3/rune master 5/ arcane trickster 15

rouge 1/ fighter 1/ arcane trickster 3/battle master 5/ arcane trickster 15

rouge 1/ ranger 2/ arcane trickster 3/ gloom stalker 5/ arcane trickster 15

rouge 1/ ranger 2/ arcane trickster 3/ beast master 5/ arcane trickster 13

No home rules.


text:
# User:EsalazioeHg/Sandbox/Thrust-to-weight ratio/it

La TWR è il rapporto tra FT e FG. F è rivolta verso l'alto se TWR > 1, verso il basso se TWR < 1 e nulla se TWR = 1

Il Rapporto spinta/peso (TWR, Thrust-to-Weight Ratio) è il rapporto che definisce la potenza del motore di una navicella in relazione al suo peso. Se una navicella deve raggiungere un'orbita stabile od atterrare in sicurezza sul corpo celeste attuale senza utilizzare paracadute, allora i suoi motori dovranno produrre una spinta maggiore del suo attuale peso per contrastare la gravità. In termini di rapporto, una navicella in grado di produrre una spinta maggiore del proprio peso avrà una TWR maggiore di 1. Il peso dipende dalla massa di questa e dall'accelerazione gravitazionale locale che solitamente è la gravità superficiale del corpo celeste nella cui sfera di influenza si trova l'astronave. In un'orbita stabile il rapporto spinta/peso non è importante, ma il suo valore può esser usato per stimare l'accelerazione massima possibile.

Se il rapporto è minore di 1 e la navicella è sulla superficie, allora questa non sarà in grado di decollare dal terreno senza l'ausilio di portanze aerodinamiche (i.e. profili alari). Viceversa, se questa sarà in caduta libera verso la superficie, i motori non potranno generare una spinta sufficiente ad ottenere un atterraggio morbido.

## Formula

${\displaystyle {\text{TWR}}={\frac {F_{T}}{m\cdot g}}>1}$
Dove: {{{where}}}

Quando la TWR e l'accelerazione gravitazionale di un corpo celeste (A) sono note, è possibile calcolare la TWR per la gravità superficiale di un altro corpo celeste (B). Nello specifico se la TWR conosciuta è relativa a Kerbin, è possibile utilizzare il valore della gravità superficiale espressa in g-force agente sul secondo corpo.

${\displaystyle {\text{TWR}}_{A}\cdot {\frac {g_{B}}{g_{A}}}={\text{TWR}}_{B}}$
${\displaystyle {\text{TWR}}_{\text{Kerbin}}\cdot g_{B}={\text{TWR}}_{B}}$, l'accelerazione gravitazionale ${\displaystyle g_{B}}$ è data in multipli di ${\displaystyle g_{\text{Kerbin}}}$ (g-force).

When the TWR and surface gravity for a celestial body (A) is known, it is possible to calculate the TWR for the surface gravity of another celestial body (B). Especially if the known TWR is for Kerbin, it is possible to use the surface gravity given in g-force acting on the second body.

${\displaystyle {\text{TWR}}_{A}\cdot {\frac {g_{B}}{g_{A}}}={\text{TWR}}_{B}}$
${\displaystyle {\text{TWR}}_{\text{Kerbin}}\cdot g_{B}={\text{TWR}}_{B}}$, the gravitational acceleration ${\displaystyle g_{B}}$ is given in multiples of ${\displaystyle g_{\text{Kerbin}}}$ (g-force).

To estimate the maximum acceleration (${\displaystyle a}$) at launching vertically only from knowing the TWR and gravitational acceleration the following formula can be used:

${\displaystyle a={\frac {F_{T}-F_{G}}{m}}={\frac {F_{T}-mg}{m}}={\frac {F_{T}}{m}}-g=g({\text{TWR}}-1)}$
Dove:
• ${\displaystyle {\text{TWR}}}$ the thrust-to-weight ratio for the given ${\displaystyle g}$
• The rest are the same from the original formula

## Physical background

To lift off, the engines need to supply enough force in the opposite direction of the gravitational pull to counteract it. Usually the total thrust of all engines in the current stage running at full throttle is used in the calculation to find the largest possible ratio. The gravitational pull is the weight of the craft which can be calculated by multiplying the mass with the current gravitation. To make the formula easier, the surface gravity of the celestial body in question is used.

{\displaystyle {\begin{aligned}\sum \limits _{i}F_{T_{\text{engine i}}}=F_{T}&>F_{G}=m\cdot g\\{\frac {F_{T}}{m\cdot g}}&>1\end{aligned}}}

This value isn't constant over a flight because of five reasons:

1. As the engines consume resources, the craft becomes lighter over time, raising the ratio
2. The gravitational pull is lower the farther from a body, so the ratio increases with altitude
3. On certain engines the thrust can be throttled, so lowering the thrust during flight leads to a lower ratio than one calculated for full throttle
4. As previous stages are removed from a multistage craft, it becomes lighter as parts are removed and thrust changes as previous engines are removed and any subsequent engines start operating
5. Both thrust and weight can increase from docking
The engine is tilted by ${\displaystyle \alpha =30^{\circ }}$, reducing the TWR

As soon as a craft starts with the gravity turn only a portion of the craft's thrust is applied to counteract gravity, reducing the TWR. To calculate how much thrust is used to counteract gravity the pitch of the engine can be included:

${\displaystyle F_{\mathit {eff}}=F_{T}\cdot \cos(\alpha )}$
Dove:
• ${\displaystyle F_{\mathit {eff}}}$ is the effective thrust to counteract gravity
• ${\displaystyle F_{T}}$ is the engine's thrust
• ${\displaystyle \alpha }$ is the pitch of the engine (0° straight downward, 90° straight sideways)

This can also be used to calculate the thrust for engines that are placed angled on the craft. Technically it is like they are already pitched. Usually the engines on the other side are angled too, to thrust only upwards reducing the efficiency of the engines, because some thrust is cancelled out by them.

## Examples

The Kerbal X with a mass of 130.94 t, 6 LV-T45 Liquid Fuel Engines and 1 Rockomax "Mainsail" Liquid Engine on the launch pad of the Kerbal Space Center has a TWR of:

${\displaystyle {\text{TWR}}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{130.94{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=2.102}$

A TWR of 2.102 is above 1 and means liftoff!

The second stage of a Kerbal X with a mass of 16.12 t and the Rockomax "Poodle" Liquid Engine with 220 kN thrust can lift off only with full throttle from Kerbin but it lifts off quite well from the Mun:

${\displaystyle {\text{TWR}}_{\text{Kerbin}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Kerbin}}}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=1.391}$
${\displaystyle {\text{TWR}}_{\text{Mun}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Mun}}}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot 1.63{\frac {\text{m}}{{\text{s}}^{2}}}}}=8.373}$

If the engine has been worked with the thrust of LV-909 Liquid Fuel Engine which produces only 50 kN thrust the stage itself wouldn't be able to lift off Kerbin, but still could lift up from Mun.

${\displaystyle {\text{TWR}}_{\text{Kerbin}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Kerbin}}}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=0.316}$
${\displaystyle {\text{TWR}}_{\text{Mun}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Mun}}}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot 1.63{\frac {\text{m}}{{\text{s}}^{2}}}}}=1.903}$

## Practical illustration

Test craft massing 20 tonnes

The test craft shown here has a mass of 20 metric tons (20,000 kg); it is powered by a stock LV-T45 engine rated at 200 kN of thrust. As you can see, this yields a TWR at Kerbin surface just sufficient to lift off the pad.

Because the gravitational acceleration on Kerbin's surface is roughly 10 m/s², 10 kN per ton or 100 kg per unit of thrust result in a thrust-to-weight ratio of about 1. This represents the minimum for launch; a TWR in the range 1.5 to 2.5 is better.


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