Tuesday, January 3, 2012

How to study organic chemistry for an A-Plus

Organic chemistry is the most interesting and easier to learn the subject matter of all sciences. However many students find intimidating. Organic chemistry fear permeates a lack of understanding of the subject which itself derives from an erroneous approach to learning object.

There are two stages of organic chemistry - based reasoning and the practice of learning. My twenty years of experience teaching chemistry, I have come across a large number of students, and they did very well themselves by simply following the steps above. Basic learning chemistry considerations are as follows:

Organic chemistry is not hard: Organic chemistry is not difficult at all. This is one of the most systematic sciences. This subject more practical needs and if you're comfortable in writing and identification of compounds from different structural ratings, I'm sure will provide you a good start. Don't forget that 40 per cent of organic chemistry is done if you're well in identification of structural formulas or write structures.

Don't Just read - practice it: You can delete your concepts only by questions. You must make a minimum of 10-20 different questions for each concept to master. For example, Carbocations concept governs product training preferred in many a reactions. Each book has a detailed explanation of the meaning and stability of carbocations. Understand the concept of the order of stability is very simple and straightforward, however the problem occurs when you start to apply the concepts of the problem. Identification of the different types of carbocations becomes easy until you have practiced various questions based on it.

Collapse your chapter in small matters: For example, write Lewis Structures, resonance, tautomerism etc. You must not cram the given language in the book, read many topics to understand. For example in headings of resonance, there are many: s and: s. not pas cram them because will you achieve anything by. What is important is to put into practice the rules by making questions based on the notes. As you rest assured that no Professor only asks you to list of rules for the review - but rules much a teachers favorite issues Forms applications. They may also appear in the true / false statements, but they may be tempted very well if the topic was understood by you.

Understanding organic reactions master their: Organic reactions can only be mastered by understanding. Each atom has a reaction with good reason. It is important to understand the reason why each reaction. For example, if Markovnikov rule, you can try to memorize this - you can also learn by heart. However you will find very difficult to keep them long term. However if you have learned with her reasoning for the stability of the carbocation, believe me you will hold it in your memory long long.

M aster reactions by practising with different examples: my advice is to practice at least ten examples of each reaction to master. For example consider the reaction for reduction of alkene with a metallic catalyst. If you practice different 10 examples of the response, the response will be get etched in your memory as an addition of hydrogen on the double bond cis. Anytime later when you come across this reaction, believe me, the reaction will snap back you immediately.

Exceptions or clumsy behaviour: In organic chemistry, as in every part of life, there are a few exceptions or clumsy behaviour of certain molecules. -Thank God there are only a handful of them. You will find all the books an explanation for their exceptional behavior. Simply include that the reasoning behind it, the exception appears more an exception for you.

Ultimately, I can summarize my experience teaching in organic chemistry in a single sentence:


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Sunday, January 1, 2012

The Energy Conservation Act

Energy conservation law States that energy can neither be created nor destroyed. Energy can only be conserved. Consider throwing a upwards. What are the forces acting on the ball? One is the weight of the ball can be pulled down. The next is the force behind rising on the balloon provided when the ball leaves the hand.

How far the ball travels upward? Answering this question can be easily to the law of conservation of energy. When the ball is thrown energy there is due to its speed and kinetic energy. His movement upwards the ball has potential energy and kinetic energy. Speed or velocity of the ball will be maximum when it leaves the hand and will be zero when it reaches the maximum height. Anytime speed reduce slowly (this is called as a delay versus acceleration where the veolocity increases continuously). The maximum height traveled by ball then can be determined by equating the potential energy of the destination (here kinetic energy is equal to 0, the ball is at rest), kinetic energy at the beginning of the query (here the potential energy is equal to 0).

Without the force of gravity, we would not drawn down for our planet and therefore we will be always floating upwards. The LCE can be used to make substantial diversions even the object lifted upwards does not in the vertical direction (or goes into a parabolic trajectory).

Imagine a marble is just rolling down a slope and is made to move in a circular ring, energy is conserved in this case as well. Can use the energy conservation act to determine the maximum time that the marble will be inside the circular ring.

How are we able to walk? We walk by the force of friction between foot and land surface. If there was no friction, we would be unable to stop walking. A ball which is rolling on stops floor at any time due to friction between the surface of land and the ball.

Friday, December 30, 2011

Teaching dynamic rotation

Then, I describe how I teach problem solutions for dynamic rotation in terms of basic principles. As always, the statement of principles will be the first line of the solution to the problem. I chose two problems to illustrate my approach. They will require the application of the second law of Newton, angular velocity versus the angular position and the equation of motion of rotation.

In previous articles, I described the unusual scoring, I used. It is necessary due to the limitations of the text editor. Notation is dispersed on four different articles, which makes accessing very annoying. To overcome this difficulty, I summarize this notation here:

1. Most of the kinematic variables are represented by letters. For example, v is represented by the speed and acceleration by a.

2 Indices are designated using lowercase letters. For example, initial speed V-naught"is Vo.

3 Powers are represented by *. For example, "time squared" is T * 2.

4. An amount is represented by the sum (the conditions to be summarized).

5 Ft is represented by FT.

6 Static and kinetic friction coefficients are represented by the United Kingdom and United States, respectively.

7 Angles are represented with a script italic examples, I represent "Theta" by th and "phi" by phi.

8. Rotational quantities are represented with a script in italics. Specifically, the angular position is represented by Q, angular speed W, angular acceleration by A net external forces by text, torque moment of inertia by I.

Examine problems and their solutions.

Problem. Vertical grinder (I = 15 kgm * 2) has a RADIUS R = 80 cm. The wheel turns 80 rev/min when a friction brake is pushed against the rim. The wheel and then stops after five additional revolutions. The coefficient of kinetic friction between wheel and brake pad is Uk = 0.70. What is the normal force n applied to the brake?

Analysis. We have the angular velocity at the beginning and the end of five revolutions. Therefore, if we relate speed angular angular position, we should be able to determine the angular acceleration. Then, we can use the rotation of the motion equation to determine the force of friction and the normal force.

The wheel is supported to the axis of rotation (Group B) bearings, thrust by brake (normal force n and friction force f = UkN), and the acts of its weight to its centre of mass (which is also in the axis of rotation). Only the friction force exercises a couple around the central axis of rotation of the wheel.

Firstly, we must obtain 80.0 rev/min and 5 revolutions in appropriate units:

............................80 rev/min = (80 rev/min) (are rad/rev) (1 min / 60 s) = 8.38 rad/s.

... .rev 5 = (rev 5) (are rad/rev) = 31.4 rad.

Now, we can apply the basic principles:

....Angular velocity Versus the angular Position...Rotation equation of motion

..........W**2 = Wo**2 + 2A(Q - Qo)...........................................Sum (Text) = IA

..........0**2 = Wo**2 + 2A(Q - Qo).............................................fR + B(0) + N(0) = IA

..........0 * 2 = (8.38 rad/s) * 2 + 2 A (31.4 rad)... UkNR = IA

therefore...A = - 1.12 rad/s * 2.

With this a Uk = 0.70 with I = 15 kgm value inserted in the rotation of the motion equation, we have

N = - IA / UkR (kgm 15 * 2) = (-1.12 rad/s**2)/(0.70)(0.80 m) = 30 n.

Problem. A pulley radius r and moment of inertia is free to rotate without friction around a horizontal axis through the Centre of the pulley. A string is wrapped around the pulley and the free end of the chain is linked to a mass M. cylinder when the bottle was released, falls and turns the pulley. What are the cylinder acceleration and angular acceleration of the pulley? What is the chain tension?

Analysis. We can apply Newton's second law to the bottle and the equation of rotation of the movement to the pulley. The angular acceleration of the pulley and linear acceleration of the cylinder are bound by a = AR.

The cylinder is affecting only the string (strain Tupward). the weight of the cylinder is MG. The pulley is affecting string (strain Tdownward) and support its central axis (B). Pulley weight acts to the axis of rotation of the pulley. Only the voltage (point R arm) exerts a couple around the central axis of the pulley.

Using free body for two-body diagrams, we apply the second law of Newton (with the positive direction downwards) to the bottle and the equation of rotation movement pulley (with positive clockwise):

...............................The second law of Newton...Rotation equation of motion

..................................SUM(Fy) = MAy....................................Sum (Text) = IA

...................................MG - T = McAy....................................TR = IA

After combining A = AR, and these two equations find us, after a bit of arithmetic, which

..................A = MGR * 2 / (M. * 2 + I)...T = GIG / (M. * 2 + I)...A = MGR / (M. * 2 + I)

Again, we see physical problems solved by consulting basic principles. In short, this is my message. Teach students to resolve Physics problems starting with the basics.

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed fundamental principles consulting


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Wednesday, December 28, 2011

Introduction to nanotechnology

Nanoscience is the study of phenomena and manipulation of matter at the nanoscale (0.1-100 nm), while nanotechnology is design, characterization, production and application of structures, devices and systems by control form size and nanoscale. The term commonly used nanotechnology to refer to both disciplines.

A nanometer is a unit of length equal to one millionth of a meter (1 x 10-9 m). At this scale, physical, chemical or biological properties of materials, objects, systems, etc., fundamentally different properties of the same size micro / macroscopic so that research and development of nanotechnology aims to understand and to create improved materials, devices and systems that exploit these new properties. In this regard, nanotechnology promises a better understanding of the nature and life itself, where the size and shape are important.

In turn, physics, chemistry, materials science, simulation and Engineering Computing, converge to the same principles theoretical and technical driver enabling the technological advance exceptional synergy and interdisciplinary initiatives taken by different sectors and countries.

In General, experts worldwide agree that nanotechnology has the potential to develop tools for making and influence procedures and even doctors unprecedented sur la société et des relations internationales. Therefore, it is considered a mega trend and a disruptive technology. For example, nanotechnology promises to increase the efficiency of the industry and develop traditional radical news from emerging technology applications. Internet, until recently called the revolution of the next generation unlike this new phenomenon blades.

There are many areas in which nanotechnology have potential applications. Progress will affect a wide range of industries such as cosmetics, pharmaceutical products, appliances, medical personnel, building, communications, security and defence, automotive and aerospace industries, among others. The environment will also benefit, while energy production will be less expensive and cleaner and more environmentally friendly materials were used.

The market is now available for such applications. For example, nanostructured materials are already used in products like tennis balls, golf or bowling (a way to reduce the number of tours are the same), to tire manufacturing high performance in the manufacture of fabrics with properties anti blemish or wrinkles, cosmetics, medicines and new therapeutic treatments, nanostructured membrane filtration water and sanitation environment; improving production processes through the introduction of more sustainable materials and efficient (industrial and agro-industrial) in the design of new materials for use in electronics, the aeronautics industry and almost all of the transport.

Now with nanoscience and nanotechnology are still at an early stage of research and development, where research and most of the investments are directed towards understanding of phenomena in nanoscale, processes and the creation of new materials or nano-structures. Technology trends towards the year 2020 in the world pointing to the transition from nanomaterials Nanosystems, thanks to the construction of systems at the nanoscale requires the combined use of the statutes of the nanoscale, biological principles, information technology and systems integration.

Nanotechnology-oriented innovations will meet a large number of current problems and needs of society and provide an enormous challenge for the future industrial and economic activities which often and is considered to be the engine of the next industrial revolution. Among the wide range of possibilities and after execution of the dispersion of initial positions it is illegal to questions about the areas where she focuses on our knowledge of our Paris and challenges assumed in order risk, development and implementation expectations.

Speakers in detail the origins of nanotechnology since its origin, classification, materials and their applications in various industries all with the intent to understand fully the nanotechnology research area and, thus, to calculate its vital importance in society.

Monday, December 26, 2011

Teach Newton's second law

In three previous articles, I described how I deal with all problems of kinematics of the same three basic equations. I also said that this approach encourages students to discuss the options of physical problem in terms of fundamental principles. The purpose of this paper is to show how I continue to focus on basic-principle approach when the laws of Newton. Future articles will demonstrate how I turn to other topics such as dynamic rotation, mechanical energy conservation and static equilibrium.

Because my options are limited by the Equation Editor, I use the rather unusual notation. Much of it is summarized in an article entitled "teaching kinematics". Additional notation introduced for the first time in this article is: 1. an amount is represented by sum (the conditions to be summarized). 2 Pi is represented by FT. 3 Static and kinetic friction coefficients are represented by the United Kingdom and United States, respectively. A typical problem solution cinematic starts with the statement of one or more of the three fundamental equations like this:

...Position to time...Speed time.

...X = Xo + VoT + (AT**2)/2................................V = Vo + AT)

...Equation specific specific libellésÉquation

Many students who accept my message comment how easily they can solve most of their duties kinematic problems. More important still, many of these students start actually solve the problems of Physics in terms of fundamental principles. Let me show you two examples.

Problem. A block of mass Mj is pushed rightward by a horizontal force P. A small block of mass Mi rests on it. Coefficients of static and kinetic friction on all surfaces are the United Kingdom and United States, respectively. If the block at the top of the page does slips on the bottom one, what is the acceleration of the blocks? (b) if the block at the top of the page does not bottom one slip out, what is the maximum p can be?

Analysis. We will use a Galilean repository at rest to Earth with horizontal and vertical x axis y axis. If the upper block is not dragging, two blocks have the same Aix = Ajx = acceleration. The upper block is receive (i) just lower block (normal force neither pointing upwards and friction force fi right-pointing). (ii) the weight of the upper block is MiG. The lower block is touch: (i) block higher (normal force nor pointing downwards and frictional force fi pointing Leftwards - what reaction forces to those of the lower block on the upper block); (ii) the floor (normal force Nj and strength friction fj left-pointing). (iii) the weight of this block is MjG. (iv) there is also a P, force pushing the lower block on the right. Newton second

................................................Newton's second law

...............................Block i..........................................................Block j

.Sum (Fix) = MiAix....Sum (FIY) = MiAiy...Sum (Fjx) = MjAjx...Sum (Fjy) = MjAjy

.....fi = MiA..................NI - MiG = 0... P - fi - fj = MA...New Jersey - Ni - MjG = 0

the first equation gives neither = MiG. When put us this into the equation y second and then solve for Nj, we find Nj = (Mi + Mj) g. Given that the lower block is rolling around the floor, fj = UkNj = Uk (Mi + Mj) g. After placing it with the x-fi first equation in the second equation x, we

...P-MiA - Uk (Mi + Mj) G = MDY


...A = (P - Uk (Mi + Mj) G) /(Mi_+_Mj).

(b) the top block slides when it static friction force reaches fimax = UsNi = UsMiG. With this inserted in the first equation-x, we have Amax = fimax/Mi = UsMiG/Mi = UsG. Finally, after placing the result in the equation for acceleration found in (a) and problems for P, we find

...Pmax = (US + Uk)(Mi_+_Mj) g.

Problem. Coin is sitting on a rotating horizontal platform distance R = 20 cm from the Centre of the platform. If the platform is a T = 1.1 s rotation period, which is the minimum coefficient of friction required to maintain the piece in place?

Analysis. We study the motion of the room. We'll use radial (outward) and the axes of vertical reference (upwards). Since there is no tangential acceleration, so can be ignored. Room is affected just the platform (vertical upward normal inward force and n radial static force of friction f). The weight of the piece is MG... With the second law of Newton in radial and vertical directions, we have

.................................Newton's second law

...SUM(Fr) = MAr......................................................Sum (Fz) = MAz

...-f = M(-V**2)/R.....................................................N MG = 0

Since we want we have minimum coefficient of friction, f = UsminN, which, when combined with radial and vertical, equations yields UsMG = (MV * 2) / r. Thus the minimum coefficient of friction is

...Usmin = (V * 2) / RG.

Speed is related to the period by V = 2 (pi), R/t. insert this and given in equation Usmin values we find that

...Usmin = 0.67.

Two problems I just use examples, major emphasis has been on a basic principle, in this case Newton's second law. If you look at how I teach students to approach constant-acceleration kinematics, you'll see that my approach here is an obvious extension to what I was doing it. I try very hard to keep students from ever "formula hunting". You can find many more examples in the physics Web site help. In the next article, I demonstrate how I approach the problem of work-energy theorem solutions.

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.

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Saturday, December 24, 2011

Dissolve Fat - 5 Secrets to lose fat forever

Depending on the degree of seriousness, you're dissolve your fat and be pleased that once again, when it comes to cellular nutrition, it is a miracle of modern science. This is why many doctors recommend cellular NUTRITION, because it is a proven formula that works for the fight against obesity and its complications have become a wold wide epidemic.

More than two-thirds of adult United States are overweight or obese.People who are obese, tend to pay 42% higher health care costs.From 1960 to 2006, obesity in the United States rose from 13.4 to 35.1% among adults 20 years at 74. Obesity in the United States is 32.2% of adult men and 35% in the United States women.16% adults, children are obese and 11% are extremely obese.

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Thursday, December 22, 2011

Master of Science for an emerging Graduate Professional degree

With growth in the health care industry, emerges from a relatively new diploma. He is known as the Science of a professional or MMP - masters degree and it is designed so that students can go beyond making advance their understanding of science and mathematics. The Professional Science master's degree provides students business skills they might need in the workplace working.

This combination of skills is a key to the success of many businesses in biotechnology in particular, according to what one researcher Senior Council of graduate schools project residence and professional master's programs Director says BioWorld today. However, the majority of medical products companies is established by scientists who might consider research honored profession and business as the "Devil's work", underlined article October BioWorld today. Up to now, more than 100 colleges and universities PSM degrees make available proposed an announcement by the Council of graduate schools.

The Council of graduate schools strives to promote the PSM degree so that it becomes as commonplace as other degrees. In an article in the New York Times in December, the Dean of the University in Upstate New York called a "21st century degree." He compared this degree with a hybrid, expands in how MBA evolved over 100 years ago.

Professional Science Master degrees extend beyond areas generally associated with mathematics and science. Certainly, there are degrees PSM in areas such as biotechnology, medical related sciences, physics and geological and mathematical and statistical sciences. There are also PSM degrees in forensic science, national defence, computational sciences as the analytical science and imaging and sciences de l'environnement such as coastal and watershed science policy, science and climate solutions.

MMP levels are created in collaboration with employers. Some of the skills taught in courses or other offers of vocational training that accompanies these programs are communications, business and regulatory affairs, noted the announcement of the Council of graduate schools.

The Council of graduate schools plans to establish a means to gather information for registration in the mastery of Science professionals as well as diplomas awarded degrees. The Council also plans to gather information about the students who participate in these programs, where they work and how their work and their wages are progressing in the five years between the time where they get their professional Science master's degree.

A not-for-profit foundation known as the Alfred P. Sloan Foundation provides the Council more than $490,000 for his efforts. The Board plans to review the Web site, sciencemasters.com, of the plans of awareness for new master Science professionals. Currently, the New York Times article noted, the majority of students participating in professional Science masters degree programs are Americans, while a solid number of foreign students is in fact the same.

PSM degrees have come a long way since their introduction of the 1990s. This diploma more recent is taking off in popularity. Those who are interested in potentially earn a master's degree in this area can explore their options with indications of connection of education; they can be paired with schools and programs that meet their interests. Michael Teitelbaum is responsible for professional Foundation Sloan Science master's program. He has been quoted in the announcement of the Council of graduate schools by saying that "The rapid growth of the program," Teitelbaum said, "speaks volumes for efforts (the Council of graduate schools) and its members have already made and we look forward to success."

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