text: 3 linear transformations which are surjective but not injective, iii. December 14, 2020 by Sigma. See the answer. Functions . How can this be shown? To be surjective but not injective ℕ → ℕ you need a function f: x ∈ ℕ → y ∈ ℕ : ∀ y ∃ x but ∄ x : ∀ x ∃ y. i.e. Lv 5. Previous question Next question Transcribed Image Text from this Question. Diana Maria Thomas. There can be many functions like this. Add to My Favourites. Hope this will be helpful. Then is neither injective nor surjective, is surjective but not injective, is injective but not surjective, and is bijective. 2 0. Strand: 5. Can you have a purely surjective mapping where the cardinality of the codomain is the same as that of the range? Then, at last we get our required function as f : Z → Z given by. Clearly, f is a bijection since it is both injective as well as surjective. injective. In other words, we’ve seen that we can have functions that are injective and not surjective (if there are more girls than boys), and we can have functions that are surjective but not injective (if there are more boys than girls, then we had to send more than one boy to at least one of the girls). MEDIUM. The only possibility then is that the size of A must in fact be exactly equal to the size of B. [End of Exercise] Theorem 4.43. This problem has been solved! It's not injective and so there would be no logical way to define the inverse; should $\sin^{-1}(0) ... \rightarrow \mathbb{R}$ then it is injective but not surjective. So f(1) = f(2) = 1, f(3) = f(4) = 2, f(5) = f(6) = 3, etc. United States Military Academy West Point. Definition of Function; Injective; Surjective; Bijective; Inverse; Learn More; Definition of Function. Switch; Flag; Bookmark; Check whether the relation R in R defined by R = {(a,b) : a ≤ b 3} is refleive, symmetric or transitive. Rate this resource. generalebriety Badges: 16. n!. MHF Helper. Functions. We say that Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. surjective) maps defined above are exactly the monomorphisms (resp. This relation is a function. Injective but not surjective. Answer #1 | 24/08 2015 00:38 f from integers to whole numbers, f(n) = n^2 Positive: 68.75 %. A member of “A” only points one member of “B”. Please Subscribe here, thank you!!! Number of one-one onto function (bijection): If A and B are finite sets and f : A B is a bijection, then A and B have the same number of elements. It is injective (any pair of distinct elements of the … A map is an isomorphism if and only if it is both injective and surjective. (one-to-many is not allowed. Injective, but not surjective; there is no n for which f(n) = 3=4, for example. Finally, a bijective function is one that is both injective and surjective. Injective, Surjective & Bijective. #18 Report 8 years ago #18 Shame I can't rep that post by nuodai. Well, no, because I have f of 5 and f of 4 both mapped to d. So this is what breaks its one-to-one-ness or its injectiveness. SC Mathematics. Powerpoint presentation of three different types of functions: Injective, Surjective and Bijective with examples. If the restriction of g on B is not injective, the g is obviously also not injective on D_g. injective but not surjective (b.) How does light 'choose' between wave and particle behaviour? However the image is $[-1,1]$ and therefore it is surjective on it's image. f is not onto i.e. all of ℕ is reachable from ℕ under f, but not all of ℕ can reach ℕ under f. I think that might be a contradiction. The injective (resp. One example is $y = e^{x}$ Let us see how this is injective and not surjective. 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A subset of C, f: Z → Z which is bijective A- > is. B be non-empty sets and f: Z → Z given by one element Y!


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# Assemble one battery from two batteries

I have two 18V 1.5Ah batteries for power tool, which consists of five 18650 elements in serial (5s1p configuration). 18650 elements are of the same model, age and condition. I want to make one battery from these two (so it become 5s2p), but my concerns are about charging the new battery. Can I use the same charging controller? Or I need to buy a new one? I want to use the same charger for the battery.

As long as you connect them correctly, this should be no problem.

You need to connect the point between cell 1 and cell 2 of battery pack A to the point between cell 1 and 2 of battery pack B. Connect the point between cell 2 and 3 of A to the point between cell 2 and 3 of B.

Pack A:
[cell1] | [cell2] | [cell3] | [cell4] | [cell5]
Pack B:
[cell1] | [cell2] | [cell3] | [cell4] | [cell5]

^         ^         ^         ^
this       this      this      this

And so on.

What you'll end up with is a 5s2p battery that can be balance charged in the same way as a 5s1p battery.

Pack AB:
[cell1] | [cell2] | [cell3] | [cell4] | [cell5]
[cell1] | [cell2] | [cell3] | [cell4] | [cell5]

And of course, connect the very ends of pack A to the corresponding ends of pack B as well.

Pack AB:
| [cell1] | [cell2] | [cell3] | [cell4] | [cell5] |
| [cell1] | [cell2] | [cell3] | [cell4] | [cell5] |

Before you make a connection, you should ensure that both connection points is at the same voltage potential (in other words, before you hook up the point between cell 1 and cell 2 to each other, check that cell 1 in battery pack A is at the same voltage as cell 1 in battery pack B).

You don't want huge amounts of current to flow to even out the charges as soon as you hook the cells up to each other, because that can become hot and/or damage the cells.

At this point, the charger should not know or care whether it's a 5s1p battery or a 5s2p battery, other than that the capacity is roughly doubled.

• The charger could possibly care, e.g. it could hit a safety timeout on a larger capacity pack, and any fuel gauge will probably be confused, and balancing may suffer, etc (depending on which parts of the BMS are on the pack vs. in the charger) – Bill Dubuque Jun 11 '16 at 14:31

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# Flat Plate External Free Convection Problem

1. Aug 3, 2014

### Libertiene

I am trying to calculate heat transfer from an inclined flat plate under free convection conditions. I am referring to Fundamentals of heat and mass transfer (Incropera / DeWitt). The equations and a spreadsheet which can be used to find a solution can be found at:

http://www.brighthubengineering.com...transfer-coefficient-estimation-calculations/

"Natural Convection Heat Transfer from an Inclined Plate"

These calculations involve calculating the Rayleigh number and then using a correlation to find the Nussult number and thus the heat transfer coefficient.

The problem is that i don't understand what happens when the plate is cooled (at lower temperature than the surrounding air). The plate i am trying to calculate this for is an evaporator. The because the temperature between the plate and air is negative this makes the Rayleigh number negative and i don't think this is correct. The correlation is then erroneous because you need to take Ra^(1/4), which clearly is incorrect for a negative number.

If you make the surface temperature of the plate lower than the air in the example spreadsheet reference above it goes into error so the spreadsheet has not addressed the problem.

I am presuming that the Rayleigh number is negative because the air is being cooled rather than heated, or related to that cooling is occurring.

I am thinking I should just times the Rayleigh number by -1 to make it positive and calculate the heat transfer co-efficient from this but i have no idea if this is valid.

The book i am referring to has a diagram of an inclined plate being cooled in the same chapter, so it would be strange if this cannot be made to work.

My values are as follows: Air temp = -0.4 deg C, plate temp = -50 deg C, Tf = -25.2 deg C, g = 9.81cos60 = 4.905, β=1/(-25.2+273) = 0.00404, L^3 = 4.913, v = 1.132 * 10^-5 (m^2/s), α = 15.9*10^-6 (m^2/s), pr = 0.72.

Any ideas?

2. Aug 3, 2014

### jlefevre76

If you want the long and short of it, read section 9.6.2 of Incropera and Dewitt (great textbook by the way). If you read this section, they basically say you handle either the top side of a cool surface (relative to surroundings) or the bottom side of a hot surface (relative to surroundings) using equation 9.26 and assuming gravity is equal to $gcos(\theta)$. Bear in mind, that equation doesn't distinguish whether the vertical surface is warm or cold, so h should be the same either way for a given temperature difference and thermal properties (which in turn affect the Raleigh number, which will be positive as you use the magnitude of the temperature difference). There are no equations in the book to handle the lower side of a cool surface or the upper side of a hot surface for an inclined surface, at least not in the edition that I'm using (6th).

I hope that helps somewhat. Best of luck with that problem you're working on.

3. Aug 4, 2014

### Libertiene

Hi Jlefevre, Yes if i use the magnitude of the temperature difference i get the same answer but the Raleigh number is positive. I have got a heat transfer coeff however this seems a bit low. That's good enough for now i think.

4. Aug 4, 2014

### jlefevre76

Yeah, it kind of depends on what you're looking at in terms of application. Is this for a project or research? Or is this for a homework problem? If it's for homework, then this is as good as your textbook gives you in terms of options. If it's for something else, I can discuss some possibilities for approximating something closer to what your setup is (if you can describe it in detail).

5. Aug 5, 2014

### Libertiene

Yes its for a research project. We have a flat plate evaporator on a roof, it is a micro-channel style heat exchanger. This is then used as part of a heat pump for hot water generation. The heat transfer from the evaporator is affected both by radiation from the sun and convective heat transfer from the air flowing across the panel.

We are measuring wind speed, irradance & panel temperature. But the difficulty is in estimating the different convective and radiation heat transfer elements or the proportion of which with respect to external conditions. I need to be-able to estimate these both accurately so that i can generate a simulation model for a given set of climate data.

I'm interested in free convective conditions because often the wind speed is very low or 0 and i believe that this is driving the panel heat transfer. Other research suggests the convective heat transfer is a high proportion of the total circa 70%.

Currently the model is grossly underestimating the reality of the situation (heat transfer).

6. Aug 5, 2014

### jlefevre76

You can use the Duffie and Beckman equations to estimate a convection coefficient, that would be the easiest. One such example is h=max(5, 8.6V^0.6/L^0.4) where V is wind speed in m/s, and L is a characteristic length, the cube root of the volume that would enclose your system.

Otherwise, it gets more complicated than I have time to address right this second (I'm responding from a tablet). I have an old method I was using to do something similar, I'll look it up and see if I can find it. Is it safe to assume there is convection off the top and bottom of the collector, or just the top?

7. Aug 5, 2014

### jlefevre76

Okay, back on the laptop here and I can address this using a little bit more detail.

So, one way to APPROXIMATE (this is not physically realistic, but it won't make you end up too far off), is to use a weighting function that interpolates between the results you get for cases you know (100% vertical or 100% horizontal). This becomes increasing complicated because the upper and lower surface behave differently in terms of free convection, so you actually have to calculate 3 Nusselt numbers (with three different Raleigh numbers because they use different characteristic lengths I think). You then use your three Nusselt numbers to get your three convection coefficients, and then can use the averaging shown in the figure.

Again, this is likely non-physical, but you're going to have issues anyway as far as I can tell, because wind direction is going to change the boundary layer (theoretically) and therefore change your convection coefficients. Incropera and Dewitt give instructions for combining free and forced convection, but in your case they'll likely be non-physical anyway because the way the wind flows across the surface will change depending on your collector's orientation.

Theta in the figure is the inclination angle from horizontal, in case you needed to know (0 = horizontal, 90 = vertical).

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8. Sep 7, 2014

### Libertiene

I have actually had a chance to work through this now. I think I see what you mean. If you like you could perhaps have a look at the attachment and see if you think I have interpreted this correctly (cooled plate).

This is way more complex than I expected and its easy to get this sort of thing wrong.

I think for the purposes of my simulation this is more than enough. I think some of the detail given in the Duffie and Beckman book is also very useful and is far more relevant than what I have found in the general heat transfer books. Even if the attached calculation is inaccurate and i don't use it, I am fairly happy that I can justify with reference to the Duffie and Beckman book.

Thanks a lot for the help here this may have saved me untold hours.

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# Showing prime and maximal ideal

In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$.

Is $I$ a Prime ideal of $\mathbb{Z}\left [ x \right ]$?

Let $f_{1}\left ( x \right ),f_{2}\left ( x \right ) \in \mathbb{Z}\left [ x \right ].$

Let $f_{1}\left ( x \right )f_{2}\left ( x \right ) \in I$.

Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I.

Now, $f_{1}\left ( 0 \right )f_{2}\left ( 0 \right )$ is an even integer in I. Hence, either $f_{1}\left ( 0 \right )$ has an even constant integer or $f_{2}\left ( 0 \right )$ has an even constant integer. Thus, I is a prime ideal by definition.

Is this correct?

I would now like to show if I is a maximal ideal.

Again, clearly, I is a proper ideal by the above argument. Let B be an ideal of $\mathbb{Z}\left [ x \right ]$ and I $\subseteq B\subseteq \mathbb{Z}\left [ x \right ]$. How should B be defined so that I would conclude either $B=I$ or $B=\mathbb{Z}[x]$?

• Hint: $I = (2, x)$. What is $\mathbb{Z}[x]/(2,x)$ isomorphic to? – André 3000 Jul 14 '17 at 6:05
• It is isomorphic to the image of Z[x].@Quasicoherent – Mathematicing Jul 14 '17 at 6:10
• @Mathematicing Isn't it isomorphic to $\mathbb{Z}/2\mathbb{Z}$? – RideTheWavelet Jul 14 '17 at 7:07
• "Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I." If I were writing this, I would use a specific example, such as $f(x) = 1$, to demonstrate. There are other minor details like this, but I think your proof is correct. – Arthur Jul 14 '17 at 7:29
• @Quasicoherent are you referring to my question asking for hep to prove that I is a maximal ideal? – Mathematicing Jul 14 '17 at 7:36

Clearly if $f(0)$ is not even, then $f(0)-1$ is even. This shows: If $f \notin I$, then $f-1 \in I$, hence $1 \in (I,f)$. Thus $I$ is maximal by the very definition.
This argument also immediately yields $\mathbb Z[X]/I = \{0,1\} \cong \mathbb F_2$.
For maximality, consider the two ring homomorphisms \begin{align} \alpha&\colon \mathbb{Z}[x]\to\mathbb{Z} & p(x)&\mapsto p(0) \\[6px] \beta&\colon \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z} & m&\mapsto m+2\mathbb{Z} \end{align} and look at the kernel of $\beta\circ\alpha$.


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# Tag Info

Yes, $E$ will be always be secure. This follows from a standard type of proof called a hybrid argument. Giving the full details would be tedious, so here is a sketch in case you are familiar with hybrid arguments: We define games $H_0,H_1,H_2$. We let $H_0$ be the IND-CPA game, but with the game's secret bit hardcoded to $0$. So the game always outputs ...


text:
Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

Chapter 7 - Exponents and Exponential Functions - 7-4 Division Properties of Exponents - Practice and Problem-Solving Exercises - Page 445: 94

Answer B

Work Step by Step

We are given $\frac{(2x)^5}{x^3}$ $=\frac{32x^5}{x^3}$ $=32x^{5-3}$ $=32x^2$ Answer B is correct.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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### Curve Fitter 2

Can you construct a cubic equation with a certain distance between its turning points?

# Curve Fitter

##### Stage: 5 Challenge Level:

The following function fits exactly, but has the wrong asymptotic behaviour. $$y= 0.75x^4-2.5 x^3+0.75x^2+3x$$

The following cubic nearly fits. $$2 y = x^2(2x-9)+11x$$

This following quintic exactly fits. $$4y = 4x^5-21x^4+42x^3-45x^2+28x$$

The differential set to zero gives $$20x^4-84x^3+126x^2-90x+28 = (x-1)(x-2)(20x^2-24x+14)=0\;.$$

This is by choosing the coefficient of $x^5$ to be $1$.

This curve only has 2 real turning points.

You can see these graphs on the attached xls.


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# BME100 f2017:Group4 W1030 L2

BME 100 Fall 2017 Home
People
Lab Write-Up 1 | Lab Write-Up 2 | Lab Write-Up 3
Lab Write-Up 4 | Lab Write-Up 5 | Lab Write-Up 6
Course Logistics For Instructors
Photos
Wiki Editing Help

# OUR TEAM

 Name: Quentin Ellis Name: Nazima Ansari Name: Jacob Schuler Name: Alexis Chavez Name: Ryan Dougherty

# LAB 2 WRITE-UP

## Device Image and Description

Image 1 is the top left. Image 2 is the top right. Image 3 is center left. Image 4 is center right. Image 5 is bottom left.

IMAGES 2, 3, 4, WERE FILES FROM THINGIVERSE FROM A PROJECT BY INTEL WHICH HAS NOW BEEN CLOSED. I DID NOT MAKE THESE AND HAVE PERMISSION FROM THE PROJECT MANAGER TO USE THESE IN OUR DESIGN.

In image 2, 3, and 4, 3 parts are seen that form an entire finger. These are to be assembled for a total of 5 fingers.

In image one and five the two large cylinders seen are the outside components of the arm. As the arm is compressed the cylinder with the piston is pushed around the larger cylinder, and the piston deeper into it. The two sets of teeth seen in image one are placed inside the larger cylinder. The 2 very small pieces that look like a cylinder attached to a triangle (one is larger than the other) fit together with a spring in between and are assembled in a hole 3 inches above the base of the piston. The block with 5 holes cut is attached to the top of the smaller cylinder (functions as the palm). A finger is placed in each hole and the remaining two small cylinders (pins) are slid through a circular openings in the side to keep that fingers in place. This entire assembly amounts to a forearm with function fingers for grasping objects. This address the health issue because of the limited technology necessary and basic parts it ensures costs will be cheaper than other arms on the market. Because of the mechanism, it gives more functionality than other arms on the market.

## Technical and Clinical Feasibility

Technical Feasibility-3
The design is completely feasible, even for freshman students with no experience. This device is currently being 3D printed and assembled. This prosthetic arm operates off of applying force to any given object. As the force is applied, the forearm compresses pushing a piston against a spring. 9 teeth line either side of the large cylinder for 3 inches (total compression length) and two separate teeth, in opposite orientation, are connected by a spring through the center of the piston 3 inches from the base of it. As the piston is pushed through, these teeth on the piston will be pushed against a slope that is the inverse to its own. This compresses the spring in the piston and pushes the teeth in the piston together. After traveling a length of .33 inches, these inner teeth will extend back out once more to be firmly lodged against the underside of the previous tooth. This process can occur for a total 2.97 inches (9 connected outer teeth). In order for he forearm to expand, the forearm is turned 90 degrees counter-clockwise. The spring at the bottom of the large cylinder and piston will then provide upward force and thrust the upper portion back to its original position. There is a string connected to the end of the piston that runs up through the forearm and into the center of the palm. From the center of the palm, the string separates into 5 separate strings that attach to each portion of the finger. Elastic bands on the back side of the fingers insure that the fingers always stay in an extend position when to in use. As the piston is pushed the the string is subsequently pulled, which in turn slowly closes the fingers in a grabbing like motion. To let go off an object the forearm is rotated 90 degrees, which allows the forearm to re-expand, the tension on the string to be alleviated, and the elastic bands to pull back the fingers to their initial position. The forearm can then be turned 90 degrees clockwise to reset the mechanism. It relies on 2 springs of different spring forces, and string and elastic tension. These are basic physics concepts and do not present a challenge. Errors that can occur with the device are simply pieces breaking or becoming worn out. The elastic can lose its elasticity and the spring can slowly lose its ability to compress.

Clinical Feasibility-3
Because of the basic concepts used within the design, we expect it to succeed in clinical trials. Because this device doesn't enter the body and attaches to a sleeve molded to the patient's amputation the only risk is the chance of infection at the amputation site. This is a common risk associated with all prosthetics. Lower arm non-invasive prosthetics have been brought to market before and are considered class 1 devices. This means that they do no have to be submitted to the FDA and require no clinical trials (Center for Devices and Radiological Health).

Sources: Center for Devices and Radiological Health. (n.d.). Implants and Prosthetics - Regulatory History. Retrieved September 20, 2017, from https://www.fda.gov/MedicalDevices/ProductsandMedicalProcedures/ImplantsandProsthetics/ucm452270.htm

## Market Analysis

Value Creation
Our product seeks to provide not only a cheaper alternative to most prosthesis, but an alternative that is both durable and multi-functional. The hand is made of a sturdy machined metal (TBD), and the system of its hand movements is purely mechanical. The system in itself is composed of a ratchet system, with tensions springs as a way of maintaining grip on an object that has been pressed into the palm of the device. In terms of price, the device would cost a total of $5000 dollars to the consumer, far cheaper then the average$10,000 dollars needed for a hook based prosthesis. (How Much Does a Prosthetic Arm Cost)

Manufacturing Cost
The cost of materials for the device is relatively low, but machining, manufacturing, and delivery to a foreign country drive that price up to $1000 dollars. 1.$150 for Parts 2. $300 for labor 3.$500 for Shipping 4. $50 for tariff/taxes Whether machining is done in house or at a manufacturing plant is still up for debate, but the price still stays the same. Sales Price The actual sales price of the item will be 5000, according to the idea that that would actually make us profit and allow for growth into different sectors. We were also told to do this a general guideline by the professors of BME 100 lab. Market Size We calculated the market size by taking the percentage of China that makes up the world population, or 18.3%, and multiplying that by the estimated number of arm amputees in the world, 3,000,000. This roughly equally 555,685. Now, assuming 5% market penetration with$5000 dollars per unit sold, to an approximate population of 555,685 we approach the number of \$138,921,250 total market. This would provide us the score of a 1 on market value.

Sources: How Much Does a Prosthetic Arm Cost? - CostHelper.com. (n.d.). Retrieved September 20, 2017, from http://health.costhelper.com/prosthetic-arms.html

## Fundability Discussion

Scores: Customer validation: 1

Market size: 1

Competition: 2

IP postion: 2

Technical Feasibility: 3

Clinical Feasibility: 3

Regulatory Pathway: 3

Our total score is 108. The Reimbursement category has not yet been filled out. As of currently our device will not be funded through the criteria provided by the investor. If market size were larger and this product were given the expected score of 2, that would increase the over all score to 432, which is easily fundable. However, we believe this device to be unique and solve the issues of durability and cost on the prosthetic market, and thus make it fundable.


text:
## anonymous 5 years ago I need to solve the following problem, 3x over x-2 = 3x+5 over x-1, can anyone help me?

1. anonymous

In this case, you can cross multiply (or multiply both sides by (x-2)(x-1)). When you do that, you get 3x^2 - 3x = 3x^2 -x -10 When you combine like terms, you get 10=2x, so x=5

2. anonymous

Guess you mean: $\frac{3x}{x-2}=\frac{3x+5}{x-1}$ so you need to say, it's just solvable if x !=2 and x!=1 By cross multiplying u get: $3x(x-1) = (3x+5)(x-2)$ $3x^2 -3x = 3x^2 +5x -6x -10$ $-3x = -1x -10$ $-2x = -10$ $x = 5$

3. anonymous

Thank you.


text:
# Developer Guide

Application packages can contain Java components to be run in container clusters. The most common component types are:

• Searchers, which can modify or build the query, modify the result, implement workflows issuing multiple queries etc.
• Document processors that can modify incoming write operations.
• Handlers that can implement custom web service API's.
• Renderers that are used to define custom result formats.

Components are constructed by dependency injection and are reloaded safely on deployment without restarts. See the container documentation for more details.

See the sample applications in getting started, to find examples of applications containing Java components. Also see troubleshooting.

## Developing Components

The development cycle consists of creating the component, deploying the application package to Vespa, writing tests, and iterating. These steps refer to files in album-recommendation-java:

Build All the Vespa sample applications use the bundle plugin to build the components. A key Vespa feature is code and configuration consistency, deployed using an application package. This ensures that code and configuration is in sync, and loaded atomically when deployed. This is done by generating config classes from config definition files. In Vespa and application code, configuration is hence accessed through generated config classes - read more in the Cloud Configuration System. The Maven target generate-sources (invoked by mvn install) uses  metal-names.def to generate target/generated-sources/vespa-configgen-plugin/com/mydomain/example/MetalNamesConfig.java. After generating config classes, they will resolve in tools like IntelliJ IDEA. Examples unit tests are found in MetalSearcherTest.java. testAddedOrTerm1 and testAddedOrTerm2 illustrates two ways of doing the same test: The first setting up the minimal search chain for YQL programmatically The second uses  com.yahoo.application.Application, which sets up the application package and simplifies testing Read more in unit testing. To connect a debugger, use Java Debug Wire Protocol (JDWP) - instructions.

## IntelliJ plugin

If you use IntelliJ, you can install the Vespa IntelliJ plugin to simplify working with schema files.


text:
## Illustration regarding payment of surcharge in challan form VAT-2B.

In view of calculation of amount for Challan-2, 2A and 2B is as follows :-

Suppose sale is Rs. 1 lakh

Tax @ 5% = 5000

additional tax @ 10% of tax = Rs. 500

Total Tax is Rs. 5500

Suppose a client comes to you and tell you that his total Tax + Surcharge is = Rs. 5500, then you can calculate VAT and Surcharge as follows :-

5500 x 10/11 = 5000 – VAT

5500/11 = 500 – Surcharge

Now you have to pay Rs. 5500 in which is 5000 is VAT and 500 is Surcharge. This can be paid as follows :-
On Challan-2 = Rs. 4900 (you can deposit in cash or through cheque/draft in favour of Assistant Excise and Taxation Commissioner)

On Challan 2A = Rs. 500 (you can deposit in cash or through cheque/draft in favour of Punjab Municipal Fund)

On Challan 2B = Rs. 100 (you can deposit in cash or through cheque/draft in favour of Punjab Municipal Infrastructure Development Fund)

These figures represent percentage wise as follows :-

On Challan 2 = 98% of VAT (excluding surcharge) (i.e. Rs. 4900)
On Challan 2A = 10% of VAT (excluding surcharge) (i.e. Rs. 500)
On Challan 2B = 2% of VAT (excluding surcharge) (i.e. Rs. 100)

[Note : Practically the above solution is the only solution. Although if we go by the exact wording of the law, - tax, Municipal Fund and Municipal Infrastructure Development Fund cannot be calculated properly.]

See here relevant notification on form VAT 2B


text:
# A simple question of a property of norms

For a normed space $X$, when does $$\lim_{n \to \infty} \| x_n \|_X = \| x \|_X$$ imply $$\lim_{n \to \infty} \| x_n - x \|_X = 0 \;\;?$$

-
First of all, this would mean that if $||x||=||y||$ then $||x-y||=0$, and hence $x=y$. So your norm would have to be $1-1$ –  Thomas Andrews Nov 13 '12 at 14:34

Following on Thomas Andrews' comment, suppose $X$ has the property that $\lim_n\|x_n\|=\|x\|$ implies that $\lim_n\|x_n-x\|=0$. By considering constant sequences, this shows that $\|x\|=\|y\|$ implies $x=y$. Then $X$ is one-dimensional: indeed fix $x_0\in X$ with $x_0\ne0$. Write $c=\|x_0\|$. For any other nonzero $x\in X$, $\|x\|=\|\,\|x\|/c\,x_0\|$, so $x=\|x\|/c\,x_0$.

Now using Tao's example, we consider the sequence $\{(-1)^nx_0\}$. As $\|(-1)^nx_0\|=\|x_0\|$, we get $\lim_n\|(-1)^nx_0-x_0\|=0$. The only way this can converge is when $x_0=0$.

So the only possible space with this property is the trivial space $\{0\}$.

-

maybe this is wrong counterexample: $x_n=-1,1,-1,1,-1,1----$, then we take the norm the absolute value. $X=R$.......

-
the converse of the result is right duo to the simple inequality –  Tao Nov 13 '12 at 14:23

Just take $X= \mathbb{R}$ with the Euclidean norm. Then $x_n= -1$ and $x=1$ satisfies the first but not the second requirement.

-


text:
Variadic arguments are nothing new. They power all of our Printf statements. These are just syntactic sugar for arrays. For example, here is a variadic function that prints all of the strings passed to it:

So, if I call printStrings("one", "two", "ten") or printStrings("one", "twelve"), it will print everything out!

In Golang, interface{} is a valid type. Everything conforms to the empty interface, so it can represent any object. So, this is true:

And this is, too…

So, what happens if I attempt to wrap a variadic function with another variadic function?

If they are both of type interface{}, it works (with a caveat):

Output:

A direct call to a function with variadic arguments:
Data= [one two]
Direct Type= []interface {}

An indirect call to a function with variadic arguments:
Middleman Data= [one two]
Data= [[one two]]
Direct Type= []interface {}
Indirect Type= []interface {}

Specifically, the second call is nesting an array of interface{}s:

However, both show the type []interface{}, because a slice of []interface{} is still an interface{}. Weird, right?

In contrast, if I change it to specifically use strings, I get a nice error:

Specifically, a slice of strings is not the same as passing arguments to the variadic function:

prog.go:18: cannot use ok (type []string) as type string in argument to direct

Of course, I could just use slices. This is not very different, it is just not very appealing:

indirect([]string{"one", "two"})

However, this approach is much more extensible. I can wrap it as much as I want without added complexity.

## Questioning Design

However, this really made me question Go. It makes sense that the interface{} one works, but it can suppress important errors that may indicate a problem down the road. Even using reflection, I cannot tell the immediate difference between []interface{} and one nested within another. They are both of the interface{} type.

This has led me to be incredibly cautious when it comes to the interface{} type. This concept is far from strictly typed, and can introduce an array of problems into a program. It seems to be used for one thing: a compromise. It removes the need for generics, but, if it is used in excess, it undermines the beauty of Go’s type system.

What are your thoughts? Let me know.


text:
ISSN 0439-755X
CN 11-1911/B

中国科学院心理研究所

• 论文 •

### 1~6秒时距认知分段性特征

1. (1重庆师范大学教育科学学院, 重庆沙坪坝 401331) (2西南大学心理学部, 重庆北碚 400715)
• 收稿日期:2015-08-14 出版日期:2016-09-25 发布日期:2016-09-25
• 通讯作者: 黄希庭, E-mail: xthuang@swu.edu.cn
• 基金资助:

教育部人文社科青年基金(12YJC190035); 第53批中国博士后面上项目(2013M530759); 重庆市教委科技项目(KJ120618, KJ130365); 重庆师范大学校级国家自然科学基金预研项目(14XYY013); 西南大学211工程项目(NSKD11001)资助。

### The characteristic of 1~6 s duration cognition segmentation

YIN Huazhan1; LI Dan1; CHEN Yingyu1; Huang Xiting2

1. (1 School of Education Science, Chongqing Normal University, Chongqing 401331, China) (2 School of Psychology, Southwest University, Chongqing 400715, China)
• Received:2015-08-14 Online:2016-09-25 Published:2016-09-25
• Contact: Huang Xiting, E-mail: xthuang@swu.edu.cn

Abstract:

There are two hypotheses concerning the temporal cognition segmentation: One is segmentation hypothesis and the other is non-segmentation hypothesis. The former one holds that the processing mechanism and representation of different length of time are different. The latter one holds that the processing mechanism and representation of different length of time are identical. However, previous studies have indicated that the two kinds of hypotheses are supported by evidence, respectively. In this research, we explored that the temporal cognition segmentation of 1~6 s through two relatively novel ways. In Experiment 1, A total of 44 participants were engaged in a 2 (memory group: High WMC vs. Low WMC) ´ 2 (Modality: Visual vs. Auditory) ´ 6 (sample duration: 1 s, 2 s, 3 s, 4 s, 5 s vs. 6 s) mixed design experiment, to study the temporal cognition segmentation of 1~6 s, by completing time reproduction task. Three-way ANOVAs were performed using standard duration and modality as within subject factors, memory group as a between-subject factor and mean reproduced interval, ratio (of reproduced interval and sample duration) and coefficient of variation as dependent variables. In Experiment 2, 48 participants were engaged in a mixed design experiment. The procedure was identical to that of experiment 1, except that time reproduction task were instead of time production task. The results showed that 1) in experiment 1, significant main effects of sample duration, modality, and memory group, were found, for the mean reproduced interval, the ratio score, and the coefficient of variation. Specifically, high WMC compared with low WMC, long duration with short duration and auditory with visual, the mean reproduced interval was longer, the ratio score higher and the coefficient of variation smaller. Furthermore, there was a significant interaction between duration and modality, and between duration and memory group for the mean reproduced interval, the ratio score, and the coefficient of variation. For 1 s and 2 s, for the mean reproduced interval, the ratio score, and the coefficient of variation, there was no difference between the high WMC group and the low WMC group, or visual group and auditory group. For 3 s, 4 s, 5 s and 6 s, the mean reproduced interval of the high WMC group was significantly longer, the ratio score was significant higher, and the coefficient of variation was significant smaller than those of the low WMC group. For 3 s, 4 s, 5 s and 6 s, the mean reproduced interval of the auditory group was significantly longer, the ratio score was significant higher, and the coefficient of variation was significant smaller than those of the visual group. 2) in experiment 2, significant main effects of sample duration, modality, and memory group, were found, for the mean produced interval, the ratio score, and the coefficient of variation. Specifically, high WMC compared with low WMC, long duration with short duration and auditory with visual, the mean produced interval was shorter, the ratio score lower and the coefficient of variation smaller. Furthermore, there was a significant interaction between duration and modality, and between duration and memory group for the mean produced interval, the ratio score, and the coefficient of variation. For 1 s and 2 s, for the mean produced interval, the ratio score, and the coefficient of variation, there was no difference between the high WMC group and the low WMC group, or visual group and auditory group. For 3 s, 4 s, 5 s and 6 s, the mean produced interval of the high WMC group was significantly shorter, the ratio score was significant lower, and the coefficient of variation was significant smaller than those of the low WMC group. For 3 s, 4 s, 5 s and 6 s, the mean produced interval of the auditory group was significantly shorter, the ratio score was significant lower, and the coefficient of variation was significant smaller than those of the visual group. The results demonstrated that the temporal cognition of 1 to 6 seconds is segmented, and the critical point may be 2~3 s, which support segmentation hypothesis. The segmentation of temporal cognition can also be explained by the time working memory hypothesis, which is the form of temporal cognition segmentation model explaining the cognitive mechanism of short-duration processing.


text:
# How to calculate theta from tantheta

1. Nov 23, 2007

### chemboy

tan theta = 0.107
Therefore theta = 6.14 degrees

How do I come to this conclusion using a calculator?

Thanks

2. Nov 23, 2007

### pinkyjoshi65

on you calculator, there must be a button saying tan$$^{1}$$. After pushing that, enter the tanthetha value, and you will get the angle.

3. Nov 23, 2007

### pinkyjoshi65

opps the button is tan$$^{-1}$$

4. Nov 23, 2007

### chemboy

I thougt that was it, but my calculator returns .016594434 when tan$$^{-1}$$(.107) is entered

5. Nov 23, 2007


text:
# Use of three way Anova or Ancova in R?

I doing a three way Anova (or Ancova - I'm not sure, hence my question) in R. I am testing how temperature, the development stage and the size of a carcass affect the development rate of maggots.

My response variable is Duration (a measurement of hours) and my factors are Size (2 levels = small and large), and Stage (7 levels = eggs, 1st instar, 2nd instar, 3rd instar, postfeed, pupa and total) and Temperature (4 levels = 15, 20, 25, 30).

I intend to examine how Duration varies with Temperature and Size and Stage.

I imported my data set (AnovaTWD). One function I have tried is:

model1 <-(aov(Duration~Stage*Temperature*Size, AnovaTWD))
summary(model1)

This gave me:

                        Df  Sum Sq Mean Sq F value   Pr(>F)
Stage                    6 7206782 1201130 149.059  < 2e-16 ***
Temperature              1 1924926 1924926 238.881  < 2e-16 ***
Size                     1   78491   78491   9.741  0.00205 **
Stage:Temperature        6 2500293  416716  51.714  < 2e-16 ***
Stage:Size               6  120539   20090   2.493  0.02367 *
Temperature:Size         1  140090  140090  17.385 4.43e-05 ***
Stage:Temperature:Size   6  184679   30780   3.820  0.00122 **
Residuals              214 1724431    8058
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Am I right in thinking that R is recognising Temperature as a continuous variable here?

Another function I have tried is:

AnovaTWD$Temperature <- factor(AnovaTWD$Temperature)
model2 <-(aov(Duration~Stage*Temperature*Size, AnovaTWD))
summary(model2)

This gave me:

                        Df  Sum Sq Mean Sq F value   Pr(>F)
Stage                    6 7206782 1201130 527.084  < 2e-16 ***
Temperature              3 2399926  799975 351.048  < 2e-16 ***
Size                     1  116577  116577  51.157 1.90e-11 ***
Stage:Temperature       18 3068185  170455  74.800  < 2e-16 ***
Stage:Size               6  157139   26190  11.493 6.34e-11 ***
Temperature:Size         3  214981   71660  31.446  < 2e-16 ***
Stage:Temperature:Size  18  292783   16266   7.138 1.10e-13 ***
Residuals              186  423861    2279
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

I know that R is treating Temperature as a fixed factor now. However, I have been advised to treat Temperature as a covariate. I read "ANCOVA is easily reached using the aov() function using the syntax + variable name to indicate that the predictor variable is a covariate."

So I tried:

 summary(aov(Duration~Size*Stage+Temperature, AnovaTWD))

This gave me:

Df  Sum Sq Mean Sq F value Pr(>F)
Size          1   87904   87904   4.928 0.0274 *
Stage         6 7223925 1203988  67.497 <2e-16 ***
Temperature   3 2411455  803818  45.063 <2e-16 ***
Size:Stage    6  143505   23917   1.341 0.2400
Residuals   225 4013443   17838
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Is this now ANCOVA and treating Temperature as a covariate? Or is it not testing an interaction with the other factors?

I could simply treat Temperature as a fixed variable as my experiment design used the 4 constant temperatures (15, 20, 25, 30). However, I know this approach is not giving all the information possible, hence I am looking into ANCOVA as it is usually used when there are both continuous and categorical predictors.

The advice you were given - use '+ variable name to indicate that the predictor variable is a covariate - is not correct. + means that you are including the variable as an additive effect. In your examples, it means you are saying you do not want to interact Temperature with the other variables. The change between continous versus categorical treatment of the variable is determined by whether it is a factor or not.

In your first model Temperature is treating temperature as a continuous variable and specified interactions between all the predictors.

In your second model Temperature is a factor, and you have specified interactions between all the predictors.

In the third model, Temperature is still a factor, but you have specified no interactions between Temperature and the other variables.

ANCOVA just means that one of the variables is not a factor.

• Thank you for your answer, it was very clear. Just to add to this, does this mean that Temperature is a 'covariate' in the first model as it's being treated as continuous? Thank you Feb 16 '17 at 10:50
• Yes, that's right.
– Tim
Feb 16 '17 at 22:21


text:
# Find a Profile-most Probable k-mer in a String solved by 705

July 29, 2015, 12:29 a.m. by Rosalind Team

Given a profile matrix Profile, we can evaluate the probability of every k-mer in a string Text and find a Profile-most probable k-mer in Text, i.e., a k-mer that was most likely to have been generated by Profile among all k-mers in Text. For example, ACGGGGATTACC is the Profile-most probable 12-mer in GGTACGGGGATTACCT. Indeed, every other 12-mer in this string has probability 0.

In general, if there are multiple Profile-most probable k-mers in Text, then we select the first such k-mer occurring in Text.

## Profile-most Probable k-mer Problem

Find a Profile-most probable k-mer in a string.

Given: A string Text, an integer k, and a 4 × k matrix Profile.

Return: A Profile-most probable k-mer in Text. (If multiple answers exist, you may return any one.)

## Sample Dataset

ACCTGTTTATTGCCTAAGTTCCGAACAAACCCAATATAGCCCGAGGGCCT
5
0.2 0.2 0.3 0.2 0.3
0.4 0.3 0.1 0.5 0.1
0.3 0.3 0.5 0.2 0.4
0.1 0.2 0.1 0.1 0.2

## Sample Output

CCGAG


text:
How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colorless pungent smelling gas HCl , but if the mixture contains bromide then we get red vapour of bromine. Why?

Asked by Pragya Singh | 1 year ago |  54

##### Solution :-

(a) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.

(i) In a neutral medium, $$OH^{ – }$$ ions are produced in the reaction. Due to that, the cost of adding an acid or a base can be reduced.

(ii) $$KMnO_{ 4 }$$ and alcohol are homogeneous to each other as they are polar. Alcohol and toluene are homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium compared to heterogeneous medium. Therefore, in alcohol,

$$KMnO_{ 4 }$$ and toluene can react at a faster rate.

(b) When concentrated $$H_{ 2 }SO_{ 4 }$$ is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces $$H_{ 2 }SO_{ 4 }$$ to $$SO_{ 2 }$$ with the evolution of bromine’s red vapour.

$$2 \; NaBr \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \;$$

$$2 \; HBr \; + \; H_{ 2 }SO_{ 4 } \; \rightarrow \; Br_{ 2 } \; + \; SO_{ 2 } \; + \; 2 \; H_{ 2 }$$

When concentrated $$H_{ 2 }SO_{ 4 }$$ I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce $$H_{ 2 }SO_{ 4 }$$ to $$SO_{ 2 }$$.

$$2 \; NaCl \; + \; 2 \; H_{ 2 }SO_{ 4 } \; \rightarrow \; 2 \; NaHSO_{ 4 } \; + \; 2 \;$$

Answered by Abhisek | 1 year ago

### Related Questions

#### Fluorine reacts with ice and results in the change:

Fluorine reacts with ice and results in the change:

$$H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) }$$

Justify that this reaction is a redox reaction

#### Depict the galvanic cell in which the reaction is

Depict the galvanic cell in which the reaction is:

$$Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )}$$

Further show:

(i) which of the electrode is negatively charged?

(ii) the carriers of the current in the cell.

(iii) individual reaction at each electrode.

#### Given the standard electrode potentials,

Given the standard electrode potentials,

$$\dfrac{ K^{+}}{K}$$= –2.93V

$$\dfrac{ Ag^{+}}{Ag}$$ = 0.80V

$$\dfrac{Hg^{2+}}{Hg}$$ = 0.79V

$$\dfrac{ Mg^{2+}}{Mg}$$ = –2.37V

$$\dfrac{Cr^{3+}}{Cr}$$ = –0.74V

Arrange these metals in their increasing order of reducing power.

#### Arrange the given metals in the order in which they displace each other from the solution of their salts.

Arrange the given metals in the order in which they displace each other from the solution of their salts.

Al, Fe, Cu, Zn, Mg

#### An aqueous solution of AgNO3 with silver electrodes

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $$AgNO_{3}$$ with silver electrodes

(ii) An aqueous solution $$AgNO_{3}$$ with platinum electrodes

(iii) A dilute solution of $$H_{2}SO_{4}$$ with platinum electrodes

(iv) An aqueous solution of $$CuCl_{2}$$with platinum electrodes.


text:
Alan Braslau braslau.list at comcast.net
Fri Aug 2 20:29:29 CEST 2019

Hello,

\setupheadertexts [{left odd}] [{right odd}] [{left even}] [{right even}]
sets the left and right headers for even and odd pages
(perhaps I have this backwards?)

(1) It would be useful to have
but this does not work.
In this case, we probably also need a *six* argument version (even/odd).

(2) It would also be useful that
NOT wipe-out left and right if they are set
(but I could live with this perhaps historical behavior if we were to have (1).)

(3) It is a bit messy that one must use
*two calls* in order to have no centered header.
(Again, I could live with
if we were to have (1).)

Similarly for footertexts, of course.
What is the reasoning, presently?

ALTERNATIVELY (or additionally), we could have


text:
### Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

### Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
What are Zeolites?

Zeolites are aluminosilicates, i.e., three dimensional network silicates in which some silicon atoms are replaced by aluminium atoms. They are shape selective catalysts, i.e., only those molecules having a size, small enough to pass their pores are adsorbed by them.

Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?

The Plantium and palladium are used for electrolysis of aqueous solution due to the chemically inert in nature and good adsorptive capacity for hydrogen.
Because Gases like hydrogen are absorbed on the surface of platinum and palladium (adsorption) and then passes into the bulk of the adsorbent (platinum and palladium).

Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?

Ester + water ---> Acid + Alcohal
The acid produced in the reaction acts as a catalysis and makes the reaction faster.

Why are powdered substances more effective adsorbents than their crystalline forms?

Powderded substance are more effective adsorbent than their crystalline forms because when a substance is powdered its surface area increase and physisorption is directly proportional to the surface area of adsorbent,

Why does physisorption decrease with the increase of temperature?


text:
# Showing holomorphic functions are preserved under pullback by a holomorphic map

Let $f: X\rightarrow Y$ be a holomorphic mapping of complex manifolds and assume for simplicity that $dim(X)=dim(Y)=1$. I want to show that it preserves holomorphic functions under pullback. We define a function $\xi: V \rightarrow \mathbb{C}$, where $V$ is open in $Y$, to be holomorphic, if for any chart $\psi: V' \rightarrow \mathbb{C}$, the map $\xi \circ \psi^{-1} : \psi(V \cap V') \rightarrow \mathbb{C}$ is holomorphic.

Let $\xi: V \rightarrow \mathbb{C}$ be holomorphic. The pullback of $\xi$ under $f$ is $\xi \circ f : f^{-1}(V) \rightarrow \mathbb{C}$. I want to show that for any chart $\phi:U \rightarrow \mathbb{C}$ of $X$ the function $\xi \circ f \circ \phi^{-1} : \phi(U \cap f^{-1}(V)) \rightarrow \mathbb{C}$ is holomorphic.

Edited:

Let $p \in V$. Since $f$ is holomorphic, there exist chart $(V_1,\psi_1)$ of $Y$ and chart $(U_1,\phi_1)$ of $X$ such that $p \in V_1$ and $\psi_1 \circ f \circ \phi_1^{-1}$ is holomorphic.

Then $\xi \circ f$ can locally be written as $\xi \circ f=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1$,

i.e.

$\xi \circ f|_{f^{-1}(V \cap V_1)\cap U_1}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1|_{f^{-1}(V \cap V_1)\cap U_1}$. Thus $\xi \circ f \circ \phi^{-1}|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ [\phi_1 \circ \phi^{-1}]|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}$.

The latter is holomorphic as composition of holomorphic maps. Hence $\xi \circ f$ is holomorphic.

Questions:

Is the above argument correct?

How can i make it more rigorous?

Thanks :-)

-
"the map $\xi \circ \psi_a^{-1} : \psi_a(V_a) \rightarrow \mathbb{C}$ is holomorphic" This does not make sense because $\xi$ may not be defined on the whole $V_a$. – Makoto Kato Sep 18 '12 at 19:52
Yea...i am still working on how to fix this. – Manos Sep 18 '12 at 20:17
@MakotoKato: I refined the argument, with what now seems something more accurate... – Manos Sep 18 '12 at 20:36
"if for any chart $\psi: V \rightarrow \mathbb{C}$," There may not be such a chart. – Makoto Kato Sep 18 '12 at 21:04
There may not be such a chart. Take for example $V = Y$. – Makoto Kato Sep 19 '12 at 13:39

Let $p \in f^{-1}(V)$. There exist chart $(U,\phi)$ of $X$ and chart $(W,\psi)$ of $Y$ such that $p \in U$ and $f(p) \in W$. Since $f$ is holomorphic, $\psi \circ f \circ \phi^{-1}$ is holomorphic on a neighborhood of $\phi(p)$.
Since $\xi \circ f\circ \phi^{-1} = (\xi \circ\psi^{-1})\circ (\psi\circ f\circ \phi^{-1})$, $\xi \circ f$ is holomorphic on a neighborhood of $p$. Hence $\xi \circ f$ is holomorphic on $f^{-1}(V)$.


text:
+0

# how do you find the tangent

0
291
1

how do you find the tangent of a right triangle

Guest May 8, 2015

#1
+79
+5

The tangent is the ratio between the opposite side and the adjacent side of a right triangle.

$$\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{x}}\right)} = {\frac{{\mathtt{opposite}}}{{\mathtt{adjacent}}}}$$

Squaddy  May 8, 2015
#1
+79
+5

The tangent is the ratio between the opposite side and the adjacent side of a right triangle.

$$\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{x}}\right)} = {\frac{{\mathtt{opposite}}}{{\mathtt{adjacent}}}}$$

Squaddy  May 8, 2015


text:
Passer au contenu

Je donne

# Godin, Jonathan

Faculty of Arts and Science - Department of Mathematics and Statistics

André-Aisenstadt Office 4225

### Courses

• MAT2450 H - Mathématiques et technologie
• MAT1410 H - Calcul 2
• MAT2100 H - Analyse 3

### Student supervision Expand allCollapse all

Classification analytique des points fixes paraboliques de germes antiholomorphes et de leurs déploiements Theses and supervised dissertations / 2020-12
Godin, Jonathan
Abstract
We are interested in the dynamics in a neighbourhood of a fixed point of an antiholomorphic function of one variable. First, we want to describe and understand the space of orbits in a neighbourhood of a multiple fixed point, called a parabolic point, and to explore the geometric properties preserved by changes of coordinate. In particular, we solve the problem of analytical classification of parabolic fixed points. To solve this problem, we define a complete modulus of classification that allows to determine whether two germs of antiholomorphic diffeomorphisms are analytically conjugate in a neighbourhood of their parabolic fixed point. We also consider the applications of the modulus to different problems: i) extraction of an n-th antiholomorphic root, ii) existence of an invariant real analytical curve under the dynamics of a parabolic antiholomorphic germ, and iii) centraliser of a parabolic antiholomorphic germ. In the second part, we study generic unfoldings of a double fixed point, i.e. a parabolic point of codimension 1. The questions are similar in nature, namely to understand the space of orbits and the geometric properties of unfoldings. In order to classify generic unfoldings, the modulus of classification of the parabolic point is unfolded, thus providing the necessary and sufficient conditions to determine when two generic unfoldings are equivalent.


text:
# What keeps quarks separate (strong force pulls, but what repels to equal out)

We know that the strong force keeps quarks together, that is mediated by gluons (and their charge is called color charge). We know that the residual strong force keeps neutrons and protons together in the nucleus (called the nuclear force), and that is mediated by Pions (quark and anti-quark). We know that electric charge can repel (same charge) or pull (opposite charge). But I do not see anywhere if color charge can repel, i only see that it can pull. We know that protons and neutrons are stable together in a nucleus, because two forces equal out (nuclear force pulls and electric charge repels).

Questions:

1. Since strong force (mediated by gluons) pulls quarks together, what keeps quarks separate from each other, meaning why are quarks not coming closer together and crush into each other? I only see the strong force pulling, but what is the other force, that repels here and equals out?

2. I understand that in case of two Protons, two forces equal out, electric force repels, and nuclear force pulls. That is why two protons are stable in a nucleus and are not flying away and are not crushing into each other either. In case of a Neutron, there is no electric force to repel, but there is still nuclear force to pull, so a neutron is pulled together to another neutron or a proton, but what keeps the neutron from crushing into another neutron or proton?

Since strong force (mediated by gluons) pulls quarks together, what keeps quarks separate from each other, meaning why are quarks not coming closer together and crush into each other? I only see the strong force pulling, but what is the other force, that repels here and equals out?

To start with quarks , in contrast to protons and neutrons, are not composite, they are elementary particles in the standard model of particle physics which describes the data up to now .

Here is an illustration that describes what is happening within the composite proton:

Quarks and antiquarks and gluons dance around and annihilate and pair produce in a non stop manner, so they do "overlap" in the feynman diagrams of the individual interactions, and annihilate. The three valence quarks are lost in the soup, and in any case it is a matter of conservation of quantum numbers, there should be an excess of one down and two up for the proton.

So it is not a matter of repelling, it is just that overall the quarkness up and down should add up to the valence quarks of a proton, and the same holds for the neutron two down one up excess in the soup .

I understand that in case of two Protons, two forces equal out, electric force repels, and nuclear force pulls. That is why two protons are stable in a nucleus and are not flying away and are not crushing into each other either. In case of a Neutron, there is no electric force to repel, but there is still nuclear force to pull, so a neutron is pulled together to another neutron or a proton, but what keeps the neutron from crushing into another neutron or proton?

A neutron, as well as a proton, is a bound state of QCD. As bound as a hydrogen atom. For the same reason that if you hit two hydrogen atoms on each other at low energies they remain hydrogen atoms, hitting two neutrons at low energy on each other they remain neutrons, a specific(complicated) bound state of quarks. At high energy they will create a lot of quark antiquark pairs, the same as the results seen in the LHC proton proton scatters, though baryon number conservation holds in all elementary particle interactions.

In conclusion it is not about pushing and repelling but about conserved quantum numbers and/or bound states.

In lattice QCD they assume a potential and there they can approximately solve to find masses for pions and kaons, within the limits of the model.

• There are actually two more features in the soup that might address the OP's question. First, as two quarks get far too close, asymptotic freedom makes their gluon interaction essentially insignificant -- their are freed from each other. And if they are the same kind they exclude each other by the Pauli principle; if not some type of antisymmetrization may also be provided by a generalized version of that principle. It may be his vision of "crushing" quarks that's obscure... – Cosmas Zachos Mar 27 '18 at 14:12
• Dear Anna v, thank you so much, now i understand that there is a sea of quarks and that they pop out of vacuum and annihilate all the time so there is no need for repulsion so I understand your answer for question #1. But what I do not understand is, the answer to #2. In the case of protons it is obvious, because they repel each other electomagnetically, and stick together because of the nuclear force. But in the case of neutrons, there is no electomagnetic repulsion. Just the residual force pulling them. So why are two neutrons not pulling each other too close and becoming one sea of quarks? – Árpád Szendrei Mar 27 '18 at 16:38
• So what keeps two neutrons separate? Why is the two seas of quarks not coming together into one sea? What is the separating line between two neutrons? Why is a neutron strictly a certain amount of quarks making a certain size of sea, that is confined by what? So why cannot two neutrons, become a bigger sea of quarks? There must be something separating them or something confining the separate neutrons and a strict rule that prohibits neutrons from having more then a certain number of quarks? – Árpád Szendrei Mar 27 '18 at 16:43
• I thought I explained that. Because the neutron is a bound state, and bound states need energy to free the contents (quarks) from their potential trap , and the energy within nuclei is not enough for that. – anna v Mar 27 '18 at 16:43
• Oh OK, so you are saying that this is the strict rule that I was asking about, so this bound state is what determines how many quarks can be in the neutron? I understand that the bound state traps the quarks into a neutron. Does this bound state have any explanation? Is this like a rule that says a certain amount of quarks can be together, not less and not more, and that is a neutron? Anything else is not stable? Or is this bound state explained with a boundary layer around the neutron? – Árpád Szendrei Mar 27 '18 at 16:50
1. The strong force does pull quarks together, but it also gets weaker as the quarks get closer (i.e. it acts sort of like a spring), in a phenomenon known as "asymptotic freedom." In this way, the strong force is very different than electromagnetism, where the force gets stronger if the charges are closer together. As such, there's no reason to expect that quarks which are placed close together will immediately annihilate, as there's just not a lot of force on them in the first place.

2. The force that holds nucleons together is described by the nucleon-nucleon (NN) potential, which looks like this (horizontal axis is $r$, vertical axis is $V$):

The NN potential is a residual interaction resulting from very long-range reactions between quarks in adjacent nucleons. Since the reaction is long-range, color charge (and therefore gluons) cannot be exchanged. As such, the mediators for this force are color-neutral, and consist of the lighter mesons (like $\pi$, $\rho$, and $\sigma$). Because of this, the nature of this residual interaction is completely different from the quark-level strong interaction. In particular, note the strong repulsion that happens at distances less than 1 fm (i.e. the diameter of a nucleon). This repulsion, mediated by vector-meson ($\rho$) exchange, is what keeps protons and neutrons apart.

• Dear probably_someone, you are saying that the residual strong force is a repulsion? So the pion that is mediated between two neutrons is making them repulse? So then what keeps neutrons stick together? I thought it is the nuclear force (residual strong force) that keeps neutrons together? Is it then repulsive or attractive? – Árpád Szendrei Mar 27 '18 at 16:30
• No, I'm saying that the residual strong force obeys the potential I displayed. It's attractive at long distances and repulsive at short distances. – probably_someone Mar 27 '18 at 16:31
• OK so one force acts like two forces, depending on the distance? It mediates the same particles, pions, and the same pions mediate repulsion at short distances and mediate attraction at long distances, and that equals out in a middle distance? – Árpád Szendrei Mar 27 '18 at 16:53
• I'm saying that I'm not sure that you can picture individual mesons as contributing only attraction or repulsion (as opposed to some combination of both). Look at the $\sigma$-exchange region, for instance. – probably_someone Mar 27 '18 at 17:16
• I'm saying that the notion that an individual force must be either purely attractive or purely repulsive is incorrect. Even in classical mechanics, the force on a mass attached to a wall by a spring is attractive if the mass is further from the wall than the spring's equilibrium length, and is repulsive if the mass is closer to the wall than the equilibrium length. – probably_someone Mar 27 '18 at 17:21


text:
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text:
# 9.5.4 How Many Minutes for Chatting?

## Activity: How Many Minutes for Chatting?

Recall the formula for the monthly cost in $of a mobile phone that we used earlier: The owner wishes to stick to a monthly budget of$45. Starting with the “total number of minutes,” draw a doing diagram to show the operations to find the “monthly cost.” Put the monthly cost equal to $45 and then draw an undoing diagram to work out how many minutes of phone calls can be made if the monthly cost is$45.

Here is the doing diagram.

Undoing each step in turn gives the following undoing diagram.

Applying these operations in turn gives:

• Subtracting 25 gives: .

• Dividing by 0.2 gives: .


text:
장소: 129동 310호 / zoom: 946 805 8300

We prove that groups acting properly on 2-dimensional CAT(0) complexes with a bound on the orders of cell stabilisers satisfy the Tits Alternative, that is, their finitely generated subgroups are either virtually free abelian or contain a (non-abelian) free subgroup. This is joint work with Piotr Przytycki (McGill).


text:
Important Questions of Organic Chemistry Class 11-With Slutions

Class 9-10, JEE & NEET

Q. Give the IUPAC names of the following compounds :

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Sol. 2-bromo- 4-methyl pentan- 3-one

Q. Give the IUPAC name of the alkane having the lowest molecular mass that contain a quaternary carbon.

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Sol.

Q. Write the structural formula of (i) sec-butyl and (ii) isobutyl groups.

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Sol.

Q. Lassaigne’s test is not shown by diazonium salts, though they contain nitrogen. Why ?

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Sol. Diazonium salts $\left(C_{6} H_{5} N_{2}^{+} X^{-}\right)$ readily lose $N_{2}$ on heating before reacting with fused sodium metal. Therefore, these do not give positive Lassaigne’s test for nitrogen.

Q. Suggest a suitable technique of separating naphthalene from kerosene present in a mixture.

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Sol. By differential extraction.

Q. How will you separate a mixture of o-nitro-phenol and p-nitrophenol ?

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Sol. A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation. o-nitrophenol being less volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Q. Will $C C l_{4}$ give white precipitate of $A g C l$ on heating with silver nitrate? Give reason for your answer.

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Sol. The precipitate of $A g C l$ will not be formed because $C C l_{4}$ is covalent compound and does not ionise to give $C l^{-}$ ions to react with $A g N O_{3}$. $C C l_{4}+A g N O_{3} \longrightarrow$ No reaction

Q. Which is expected to be more stable, $\mathrm{O}_{2} N \mathrm{CH}_{2} \mathrm{CHO}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ and why?

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Sol. $\mathrm{O}_{2} \mathrm{N} \mathrm{CH}_{2} \mathrm{CHO}$ is expected to be more stable because each atom has complete its octet and has no charge.

Q. How many $\sigma$ and $\pi$ bonds are present in each of the following molecules? (i) $\quad \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$ (ii) $\quad \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}$

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Sol. (i) $\quad \sigma_{C-C}=4, \sigma_{C-H}=6, \pi_{C=C}=3$ (ii) $\quad \sigma_{\mathrm{C}-\mathrm{C}}=4, \sigma_{\mathrm{C}-\mathrm{H}}=6, \pi_{\mathrm{C}=\mathrm{C}}=3$

Q. Which of the following pairs of structures do not constitute resonance structures :

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Sol. Pair of structures shown in (i) and (iv) do not constitute resonance structures.

Q. Classify the following molecules/ions as nucleophiles or electrophiles: $H \ddot{S}^{-}, B F_{3}, C H_{3} C H_{2} O^{-}$ $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}, \mathrm{Cl}^{+}, \mathrm{CH}_{3} \mathrm{C}=\mathrm{O}, \mathrm{H}_{2} \mathrm{N}:, \mathrm{NO}_{2}^{+-}$

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Sol. Electrophiles : $B F_{3} . C l^{+}, C H_{3}^{+} C=O, N O_{2}^{+}$ Nucleophiles : $H S, C H_{3} C H_{2} O^{-},\left(C H_{3}\right)_{3} \ddot{N}, \dot{H}_{2} \bar{N}$:

Q. In $s p^{3}, s p^{2}$ and sp hybrid orbitals which hybrid orbitals show the electronegativity?

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Sol. $s p^{3}$ hybrid orbitals.

Q. What type of hybridization is shown by second carbon atom.

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Sol. sp hybridization.

Q. Explain why $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$ is more stable than $C H_{3}^{+} C H_{2}$ and $^{+} C H_{3}$ is the least stable cation.

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Sol. the nodal plane of the vacant 2p orbitals and hence cannot overlap with it. Thus lacks hyperconjugative stability.

Q. State bond length and strength of $s p, s p^{2}$ and $s p^{3}$ hybrid orbitals.

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Sol. The sp hybrid orbital contains more $s$ character and hence it is closer to its nucleus and forms shorter and stronger bonds than the $s p^{3}$ hybrid orbital. The $s p^{2}$ hybrid orbital is intermediate between $s p$ and hybrid orbitals, hence the length and enthalpy of the bonds it forms is also intermediate between them.

Q. On the basis of type of hybridization, predict the shape of the following molecules : (i) $\quad H_{2} C=O \quad$ (ii) $C H_{3} F \quad$ (iii) $\mathrm{HC} \equiv \mathrm{N}$

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Sol. (i) $s p^{2}$ hybridised carbon and therefore, shape is trigonal planar. (ii) $s p^{3}$ hybridised carbon and shape is tetrahedral. (iii) $s p$ hybridised carbon and shape is linear.

Short Answer (2 or 3 Marks)
Q. Write IUPAC nomenclature of the following :

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Sol. (i) 2-ethyl 3-methyl pentan-1-ol (ii) 1-chloro propan-2-one (iii) Hexa 1, 3-diene-5-yne (iv) 2, 4, 6 – tri bromo phenol

Q. Give IUPAC name for the following :

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Sol. (i) 2 methyl propan-1-ol (ii) 4-ethyl-2-methyl aniline

Q. Give IUPAC name of the following organic molecules :

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Sol. (i) 5-methoxy-2-nitrophenol (ii) 4-ethyl-2-methyl anisol (iii) 2-chloro-4-methoxy benzoic acid (iv) 2-ethoxy-4-methyl aniline

Q. Write the IUPAC names of :

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Sol. (i) Trans-2-hexene (ii) Cis-2-hexene (iii) (E)-3-methyl-2-pentene (iv) (Z)-3-methyl-2-pentene (v) Trans-1, 4-dimethyl cyclo hexane.

Q. Give the IUPAC name of the following compounds :

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Sol.

Q. Which of the following represents the correct IUPAC name for the compounds concerned : (i) 2, 2–dimethylpentane or 2–dimethylpentane. (ii) 2, 3–dimethyl pentane or 3, 4–dimethyl pentane. (iii) 2, 4, 7–trimethyloctane or 2, 5, 7–trimethyloctane, (iv) 2 –chloro–4–methylpentane or 4–chloro 2–methylpentane, (v) But–3–yn-1-ol or But–4–ol–1–yne. [NCERT]

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Sol. (i) 2, 2-dimethyl pentane (ii) 2, 3-dimethyl pentane (iii) 2, 4, 7-trimethyl octane (iv) 2-chloro-4-methyl pentane (v) But-3-yne-1-ol

Q. Write IUPAC name of the following :

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Sol. (i) 1-ethoxy propan-2-ol (ii) Ethyl benzoate (iii) 2-phenyl propanal (iv) 6-chloro-3-methyl-hexan-2-one

Q. Write bondline formulae for : tert–Butylcyclopentane, Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one, Cyclohexanone.

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Sol.

Q. Draw formulae for the first five members of each homologous series beginning with the following compounds. (i) $\mathrm{HCOOH},$ (ii) $\mathrm{CH}_{3} \mathrm{COCH}_{3},\left(\text { iii) } \mathrm{CH}_{2}=\mathrm{CH}_{2}\right.$

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Sol.

Q. Identify the functional groups in the following compounds:

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Sol. (iii) N-Substituted amide and acid chloride

Q. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate products as free radical, carbocation and carbanion. [NCERT]

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Sol. (i) Homolysis, Free radicals are formed.

Q. Explain the order of stability of primary, secondary and tertiary carbonium ions.

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Sol. $3^{\circ}>2^{\circ}>1^{\circ}$ is order of stability of carbocation. 3 ” carbocation are most stable due to maximum positive inductive effect of three alkyl groups. $2^{\circ}$ carbocation is morestable than $1^{\circ}$ carbocation due togreater inductive effect of two alkyl groups as compared to $1^{\circ} C$ carbocation in which there is one alkyl group.

Q. 0.378 g of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapours. Calculate the percentage of C and H. [NCERT]

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Sol. Mass of organic compound $=0.378 \mathrm{g}$ Mass of $\mathrm{CO}_{2}$ formed $=0.264 \mathrm{g}$ Mass of $\mathrm{H}_{2} \mathrm{O}$ formed $=0.162 \mathrm{g}$ (i) Percentage of carbon $44 \mathrm{g}$ of contains carbon $0.264 \mathrm{g}$ of contains carbon $=12 \mathrm{g}$ $=\frac{12}{44} \times 0.264=0.072 \mathrm{g}$ Percentage of carbon $=\frac{0.072}{0.378} \times 100=19.04 \%$ $18 g$ of contains hydrogen $=2 g$ $0.162 g$ of contains hydrogen $=\frac{2}{18} \times 0.162=0.018 \mathrm{g}$ Percentage of hydrogen $=\frac{0.018}{0.378} \times 100=4.76 \%$ (ii) Percentage of hydrogen

Q. During nitrogen estimation of an organic compound by Kjeldahl’s method, the ammonia evolved by $0.5 g$ of the compound neutralised $10 \mathrm{mL}$ of $1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .$ Calculate the percentage of nitrogen in the compound.

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Sol. $1 \mathrm{M}$ of $10 \mathrm{mL} \mathrm{H}_{2} \mathrm{SO}_{4} \equiv 1 \mathrm{M}$ of $20 \mathrm{mL}$ of $\mathrm{NH}_{3}$ $1000 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contains $=14 \mathrm{g}$ nitrogen $20 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contain $=\frac{14 \times 20}{1000} \mathrm{g}$ nitrogen $\therefore$ Percentage of nitrogen $=\frac{14 \times 20}{1000 \times 0.5} \times 100=56.0 \%$

Q. (i) In sulphur estimation, 0.157 g of organic compound gave $0.4813 \mathrm{g}$ of $\mathrm{BaSO}_{4} .$ What is the percentage of sulphur in organic compound? (ii) $0.092 g$ of organic compound on heating in carius tube and subsequence ignition gave $0.111 g$ of $M g_{2} P_{2} O_{7}$. Calculate the percentage of phosphorus in organic compound.

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Sol. (i) Mass of $B a S O_{4}=0.4813 \mathrm{g}$ Mass of organic compound $=0.157 \mathrm{g}$ $\% \mathrm{S}=\frac{32 \times W_{B a S O_{4}} \times 100}{233 \times W_{\mathrm{Substance}}}=\frac{32 \times 0.4813 \times 100}{233 \times 0.157}=42.10$ (ii) Mass of organic compound $=0.092 \mathrm{g}$ Mass of $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}=0.111 \mathrm{g}$ $\mathrm{g}_{\mathrm{bof}} \mathrm{P}=\frac{62 \times \mathrm{W}_{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}}}{222 \times \mathrm{W}_{\mathrm{Substance}}} \times 10 \mathrm{C}=\frac{62 \times 0.111 \times 100}{222 \times 0.092}=33690024 Q. (i) A mixture contains benzoic acid and nitrobenzene. How can this mixture be separated into its constituents by the technique of extraction using an appropriate chemical reagent? (ii) Write the formula of iron (III) hexa cyanoferrate (II). [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) The mixture is shaken with a dilute solution of$\mathrm{NaHCO}_{3}$and extracted with ether or chloroform when nitrobenzene goes into the organic layer. Distillation of the solvent gives nitrobenzene. The filtrate is acidified with dil. HCl when benzoic acid gets precipitated. The solution is cooled and benzoic acid is obtained by filtration (ii)$\quad F e_{4}\left[F e(C N)_{6}\right]$Q. Give hybridization state of each carbon in following compounds :$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}, \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{CH}_{2}=\mathrm{CHCN},\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}, \mathrm{CH}_{3} \mathrm{CH}_{2}^{+}, \mathrm{CH}_{3} \mathrm{CH}_{2}$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. (i) Giving justification categorise the following species as nucleophile or electrophile :$\begin{array}{ll}{\text { (a) } B F_{3}} & {\text { (b) } C_{2} H_{5} O^{-}}\end{array}$(ii) Write the IUPAC names of the following compounds : Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i)$\quad$(a)$B F_{3}$is electrophile because it is electron deficient, i.e. octet of boron is not complete. (b)$C_{2} H_{5} O^{-}$is nucleophile because it is negatively charged. Q. Indicate the$\sigma$and$\pi$-bonds in the following molecules: (i)$\quad C_{6} H_{6}$(ii)$\quad C_{6} H_{12}$(iii)$C H_{2} C l_{2}$(iv)$\quad \mathrm{CH}_{3} \mathrm{NO}_{2}$(v)$\quad H C O N H C H_{3}$[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i)$\quad C_{6} H_{6}$is benzene and structural formulae is: Q. Draw the complete structures of bromomethane, bromoethane. 2–bromopropane and tert–butylbromide. Arrange them in order of decreasing steric hindrance. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. Draw the resonating structures for the following compounds. Show the electron shift using curved arrow notation. (i)$\quad C_{6} H_{5} O H$(ii)$\quad C_{6} H_{5} N O_{2}$(iii)$C_{6} H_{5} C^{\oplus} H_{2}$(iv)$\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$(v)$\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}^{\oplus}$(vi)$\quad C_{6} H_{5} \mathrm{CHO}$(vii)$C H_{2}=C H O C H_{3}$[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Q. Explain why alkyl groups act as electron donors when attached to a$\pi-$system. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Alkyl group has$s p^{3}$hybridization whereas$\pi$-bond atom is$s p^{2}$hybridized which is more electronegative therefore alkyl group acts as electron donors. Q. Classify the reagents shown in bold in the following equation as nucleophiles or electrophiles. Use curved-arrow notation to show the electron movement. (i)$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HO}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}$(ii)$\mathrm{CH}_{3} \mathrm{COCH}_{3}+^{-} \mathrm{NC} \longrightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{CN}) \mathrm{OHCH}_{3}$(üii)$\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{C}^{+} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}$[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i)$\quad O H^{-}$is nucleophile Q. Classify the following reactions in one of the reaction type: (i)$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HS}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{Br}^{-}$(ii)$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HCl} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{ClC}-\mathrm{CH}_{3}$(iii)$\left.\mathrm{(CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBr}_{2} \mathrm{CH}_{3}$(iv)$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HO} \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}$[NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) Nucleophilic Substitution reaction (ii) Addition reaction (iii) Nucleophilic Substitution reaction (iv) Elimination reaction. Q. Explain with the help of examples that Geometrical isomerism is different from conformation. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Conformation are obtained by rotation around$\sigma$-bond, e.g., eclipsed and staggered conformation of ethane. Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers. Long Answer (5 Mark) Q. (i) Given the IUPAC name for the amine. (ii) Discuss the hybridization of carbon atoms in allene$\left(C_{3} H_{4}\right)$and show the$\pi$– orbital overlaps. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i)$3, N, N$-Trimethylpentan-$-3$-amine. (ii)$\quad$The structure of allene$\left(C_{3} H_{4}\right)$is The carbon atoms 1 and 3 are$s p^{2}$-hybridized since each one of them is joined by a double bond. In contrast, carbon atom 2 is -hybridized since it has two double bonds. Thus, the two$\pi-$bonds in allene like in acetylene are perpendicular to each other as shown below : Whereas$H_{c}$and$H_{d}$lie in the plane of the paper while$H_{a}$and$H_{b}$lie in a plane perpendicular to the plane of the paper. Q. Give IUPAC names of the following compounds : Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. In all the questions given above the ring phenyl group. In case the phenyl group is further substituted, the carbon atoms of the ring are separately numbered starting from the carbon atom of the ring directly attached to the parent chain in such a way that substituent on ring gets lowest possible position. According the IUPAC name are : (i) 2-(4-chlorophenyl) propanoic acid (iv) If the compound contains more than one similar complex radicals/substituents, then the numerical prefixes such as di, tri, tetra, etc. are replaced by bis,tris, tetrakis, etc. respectively. Hence, the IUPAC name is : 1, 1, 1-Trichloro-2, 2-bis-(chlorophenyl) ethane. Q. (i) A mixture contains two components A and B. The solubilities of A and B in water near its boiling point are 10 grams per 100 ml and 2 g per 100 ml respectively. How will you separate A and B from this mixture? (ii) Explain why an organic liquid vapourizes at a temperature below its boiling point in steam distillation? (iii) Which technique can be used to separate naphthalene from kerosene oil present in its mixture. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) A and B from the mixture can be separated by using Fractional crystallization. When the saturated hot solution of this mixture is allowed to cool, the less soluble component B crystallizes out first leaving the more soluble component A in the mother liquor. (ii) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. During steam distillation, the mixture boils when sum of the vapour pressure of water and organic liquid becomes equal to the atmospheric pressure. Since the vapour pressure of water is appreciably higher than that of organic liquid, therefore, the organic liquid will vapourize at a temperature much lower than its normal boiling point. (iii) Since naphthalene and kerosene have large difference in their boiling points. Hence they can be separated by simple distillation. Q. (i) The$R_{f}$value of A and B in a mixture determined by TLC in a solvent mixture are 0.65 and 0.42 respectively. If the mixture is separated by column, chromatography using the same solvent mixture as a mobile phase, which of the two components, A or B, will elute first ? Explain. (ii) A mixture contains 71 percent calcium sulphate and 29 percent camphor. Name a suitable technique of separation of the components of this mixture. (iii) Without using column chromatography, how will you separate a mixture of camphor and benzoic acid? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) Since the$R_{f}$value of A is 0.65, therefore, it is less strongly adsorbed as compared to compound B with value of 0.42. Therefore, on extraction of the column, A will elute first. (ii) Sublimation process can be used for the separation of camphor and calcium sulphate. Camphor sublimes where as is non-volatile. (iii) Sublimation cannot be used since both camphor and benzoic acid sublime on heating. Therefore, a chemical method using solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dil. HCl to get benzoic acid. Q. What are reactive intermediates ? How are they generated by bond fission? [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Generally, the organic reactions complete through the involvement of certain chemical species which are short lived ($10^{-6}$seconds to a few seconds) and highly reactive. The short lived and highly reactive species are called reactive intermediates such as carbocations, carboanions, free radicals, carbenes and nitrenes. Heterolytic fission of covalent bond results in carbocations and carboanions. Free radicals are formed by homolytic fission. Carbenes are neutral carbon species in which carbon atom is bonded to two monovalent atoms or groups and also contains one non-bonding pair of electrons. They are produced in photolysis (irradiation with UV light or thermolysis or pyrolysis (action of heat) on diazoalkanes or ketones.) Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers. Q. What are the different types of structural isomerism ? Describe them with suitable examples. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. Structural isomerism is of following types : (i) Chain isomerism : “When two or more compounds have same molecular formula but different carbon skeleton are said to have chain isomerism.” For example,$\mathrm{C}_{5} \mathrm{H}_{12}$shows the following three isomers: (ii) Position Isomerism : ‘When two or more compounds have same molecular formula but different position of substituent atoms or groups on the carbon skeleton, are said to have position isomerism.’ For example,$C_{3} H_{8} O$shows the following two position isomers of alcohol : (iii) Functional group isomerism : ‘When two or more compounds have same molecular formula but different functional groups, are said to have functional group isomerism.’ For example,$C_{4} H_{10} O\$ represents the following two compounds.

Q. Why do alkenes and cycloalkanes show cis-trans isomerism? Or What is the origin of geometrical isomerism in alkenes?

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Sol. Presence of carbon-carbon double bond, in a molecule is the cause of geometrical isomerism. This is because when a carbon-carbon double bond is present, the molecule cannot rotate freely about the double bond. It is called hindered or restricted rotation of the molecule about the double bond. During the formation of double bond, if two similar atoms or groups are present on the same side of the plane of the double bond it is called a cis isomer and if present on the opposite sides, it is called a trans isomer. Because of restriction in free rotation one form cannot change to the other. Therefore presence of C = C bond is the origin of geometrical isomerism . This restriction in rotation may be present in rigid of cyclic molecules. Therefore cyclic compounds also shows geometrical isomerism.

• May 9, 2021 at 11:41 am

Super so much useful all

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• March 12, 2021 at 11:05 pm

very easy

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• March 1, 2021 at 6:11 am

Good questions 👍💯

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• February 27, 2021 at 5:25 pm

Poor explanation in some questions

3
• February 27, 2021 at 5:24 pm

Good questions
But some answers are incorrect, ithink
So plz. Look at this
There should be some MCQ s
As per cbse
Thanks….

1
• February 12, 2021 at 1:20 pm

Hume sirf 11th ka nomenclature chahiya only question

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• February 9, 2021 at 11:15 pm

Thanks

2
• January 31, 2021 at 8:07 pm

Thanks

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• November 25, 2020 at 6:13 pm

Very good questions…

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• May 26, 2020 at 7:40 pm

2_bromo 2-methyle 3pentenon

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