let no, na,nw be the refractive indices of oil, air and waterrespec.pdf

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let no, na,nw be the refractive indices of oil, air and waterrespectively then from snells law Solution let no, na,nw be the refractive indices of oil, air and waterrespectively then from snells law.

$let no, na,nw be the refractive indices of oil, air and waterrespectively then from snells law Solution let no, na,nw be the refractive indices of oil, air and waterrespectively then from snells law$

1) (i)Finance (H) (ii) Corporation (J) (iii) Treasurer (G) (iv) Limited Liability (I) (v) Business Ethics (C) (vi) Limited Partner (D) (vii) Double taxation of dividends (A) (viii) Shareholder wealth maximisation (B) (ix) Stakeholder (E) (x) Value (F) 2) The CFO in a company is responsible for Accounting, investments and Research and development departments. Solution 1) (i)Finance (H) (ii) Corporation (J) (iii) Treasurer (G) (iv) Limited Liability (I) (v) Business Ethics (C) (vi) Limited Partner (D) (vii) Double taxation of dividends (A) (viii) Shareholder wealth maximisation (B) (ix) Stakeholder (E) (x) Value (F) 2) The CFO in a company is responsible for Accounting, investments and Research and development departments..

sol Quantum numbers describe an electrons shell.pdf

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sol Quantum numbers describe an electron\'s shell number, sublevel, orbital, and spin. Each electron in an atom has a unique set of quantum numbers that cannot be duplicated by any other electron in the atom. The first quantum number, n, tells you which shell the electron is in. n is always a positive whole number, so it could be 1, 2, 3, 4, etc . If the electron is in the first shell (the shell closest to the nucleus), then n = 1 for that electron. n increases for electrons in higher shells. The second quantum number, L, tells you what sublevel the electron is in. Recall that there are four common sublevels in atoms. The value of L is also an integer. It can be anywhere from 0 to n - 1. So in n = 1, L can only be 0. If n = 2, L can be 0 or 1. If n = 3, L can be 0, 1, or 2. The value of L corresponds to a particular sublevel. For example, if an electron is in an s-type sublevel, then its L value is 0. If it\'s in a p-type sublevel, its L value is 1. d-sublevel electrons have L = 2 and f-sublevel electrons have L = 3. The third quantum number, m, tells you which orbital an electron is in. Each of the sublevels is further divided into orbitals, which are regions of probability where electrons exist. The value of m can be anywhere from -L to +L. If L = 0 (s-type sublevel), there is only one possible value of m: 0. . If L = 1 (p- type sublevel), then there are three possible values of m: -1, 0, and +1. If L = 2 (d-type sublevel), there are five possible values of m: -2, -1, 0, +1, and +2. If L = 3 (f-type sublevel), there are seven possible values of m: -3, -2, -1, 0, +1, +2, and +3. The last quantum number is the spin quantum number, s. Each of the orbitals described before can hold up to two electrons, but they MUST have opposite spins. The spins are defined as +1/2 and -1/2. By convention, the first electron in an orbital is assigned a spin of +1/2, and the second electron (if there is one) gets a spin of -1/2. Solution sol Quantum numbers describe an electron\'s shell number, sublevel, orbital, and spin. Each electron in an atom has a unique set of quantum numbers that cannot be duplicated by any other electron in the atom. The first quantum number, n, tells you which shell the electron is in. n is always a positive whole number, so it could be 1, 2, 3, 4, etc . If the electron is in the first shell (the shell closest to the nucleus), then n = 1 for that electron. n increases for electrons in higher shells. The second quantum number, L, tells you what sublevel the electron is in. Recall that there are four common sublevels in atoms. The value of L is also an integer. It can be anywhere from 0 to n - 1. So in n = 1, L can only be 0. If n = 2, L can be 0 or 1. If n = 3, L can be 0, 1, or 2. The value of L corresponds to a particular sublevel. For example, if an electron is in an s-type sublevel, then its L value is 0. If it\'s in a p-type sublevel, its L value is 1. d-sublevel electrons have L = 2 and f-sublevel electrons have L = 3. The third .

There should be a table in your book with these L.pdf

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There should be a table in your book with these Look at the table and see ultraviolet is higher frequency (shorter wavelength) than infrared ultraviolet is shorter wavelength than microwaves visible is shorter wavelength than radiowaves visible is shorter wavelength than infrared Solution There should be a table in your book with these Look at the table and see ultraviolet is higher frequency (shorter wavelength) than infrared ultraviolet is shorter wavelength than microwaves visible is shorter wavelength than radiowaves visible is shorter wavelength than infrared.

single displacement occurs when an element takes .pdf

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single displacement occurs when an element takes the place of an element in a compound. Metals displace metals, nonmetals displace nonmetals. #1 no displacement; both reacants combined (synthesis) #2 the element, hydrogen, replaced the tungsten (displacment) #3 the cadmium and sulfur both combined with the oxygen (oxidation-reduction) #4 the element chromium displaced the hydrogen in the compound (displacement Solution single displacement occurs when an element takes the place of an element in a compound. Metals displace metals, nonmetals displace nonmetals. #1 no displacement; both reacants combined (synthesis) #2 the element, hydrogen, replaced the tungsten (displacment) #3 the cadmium and sulfur both combined with the oxygen (oxidation-reduction) #4 the element chromium displaced the hydrogen in the compound (displacement.

pH Electrode Maintenance The entire glass membra.pdf

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pH Electrode Maintenance The entire glass membrane must always be clean. Rinsing the membrane with distilled water will often suffice for aqueous solutions. Rinsing the electrode with a mild detergent solution once a week will be beneficial. An alkaline hypochlorite solution can be used to clean electrode membranes subjected to solutions containing fat or proteins. Between measurements, store the glass electrode in a pH buffer with pH < 7. High temperature measurements, compounded by constant use in strong alkaline solutions or weak solutions of hydrofluoric acid will drastically reduce the lifetime of the electrode, since the glass membrane will slowly dissolve. Dry storage is recommended if the electrode will not be used for two weeks or more. Before use, the electrode should be soaked well. Trapped air bubbles around the inner reference electrode will produce an unstable reading. Swing the electrode in an arc or tap it gently to remove the bubbles. The electrode may have to be heated gently to approximately 60° C in a water bath if the air bubbles are trapped by KCl crystals. In order to establish a stable, hydrated glass membrane, new or dry-stored electrodes should be soaked overnight in 0.1 M HCl. After overnight soaking, rinsing, soaking in a buffer of pH = 4, and again rinsing, the electrode should be ready for use. If a shorter soaking time is necessary, the electrode should be calibrated frequently to adjust for drifting potentials. A sluggish response for a glass electrode, even after proper maintenance has been performed, may dictate the need for a slight etching of the outer glass layer of the membrane. The following treatment is only recommended after all other measures have been used to improve response and have failed. Soaked the glass membrane portion of the glass electrode in a 20% ammonium bifluoride solution for one minute, followed by 15 seconds in 6 M hydrochloric acid. ( Since hydrofluoric acid can be formed during this procedure, be careful if choosing this method of electrode rejuvenation. ) Rinse the electrode well and soak for 24 hours in a pH buffer with pH < 7. The proper functioning of the glass electrode depends on the hydration of the glass layer that takes place on the surface of the pH sensitive glass membrane during soaking and measurement in aqueous solutions. As long as the electrode is frequently rehydrated, accurate measurements in non-aqueous or partly aqueous solutions are also possible. This can be accomplished by soaking in a slightly acidic buffer. In non-aqueous solvents completely immiscible with water and before soaking, the electrode should first be rinsed with a solvent which is miscible with both water and the solvent before rinsing with water. The electrode cable and the electrode plug must be kept clean and dry if reliable measurements are to be obtained because of the very small electrode currents which pass through the glass electrode. A number of factors dictate the useful lifetime of the.

O-O+ +e- is the correct answer as first ionizati.pdf

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O and H always make covalent bond. You can tell .pdf

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O and H always make covalent bond. You can tell because they are both non metals. Two nonmetals always make a covalent bond.(If you see two non metals together, you know that they can make a polar covalent or non polar covalent bond, but never ionic). Only when metal combine with nonmetal, do they make a ionic bond. One other thing, Oxygen is very electronegative atom (this means that it likes electrons a lot and attracts them toward itself more strongly them most other atoms) therefore it tends to form polar covalent bonds. Polar covalent bond is a kind of covalent bond where electrons are sheared unevenly. So the most complete answer to your question is: oxygen and hydrogen make a polar covalent bond. Solution O and H always make covalent bond. You can tell because they are both non metals. Two nonmetals always make a covalent bond.(If you see two non metals together, you know that they can make a polar covalent or non polar covalent bond, but never ionic). Only when metal combine with nonmetal, do they make a ionic bond. One other thing, Oxygen is very electronegative atom (this means that it likes electrons a lot and attracts them toward itself more strongly them most other atoms) therefore it tends to form polar covalent bonds. Polar covalent bond is a kind of covalent bond where electrons are sheared unevenly. So the most complete answer to your question is: oxygen and hydrogen make a polar covalent bond..

NiCO3 (s) + H2SO4 (aq) NiSO4 (s) + CO2 (g) + H2.pdf

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When I was getting ready at home, I was excited and anxious, but as I was driving down the road, I began to get nervous. I had to take the interstate, and my first thought was traffic. I was scared that there was going to be a horrific accident on the busy interstate and that I would be late on my first day. I was very relieved when I saw that the interstate was not as busy as I thought it was going to be, and the drive down was a breeze. All of my favorite songs came on the radio, and I began to grown excited and anxious again. When I drove up on the Macon State campus, that terrible feeling came back again. I saw all of the cars and thought to myself frantically, “Oh no!, what if I can’t find a parking spot and be late to class after all.” Luckly, I found an excellent parking spot that wasn\'t too far away from where I needed to be. My main fears were whether I would be able to find any topics of discussion while talking to the new people I would meet; whether they would be receptive of me and would I be able to make friends here. Would I be able to share my thoughts with anyone; will the teachers like me and whether I would be able to break the ice with my classmates. There were a lot of all kinds of people there. Young teenagers, and older men and women. When I arrived at my 9:30 class, I was very glad to see two of my friends from high school. My professor was very nice and cordial. She introduced herself and explained very clearly what the class was about and what was expected. The room we were in was very cold, and as my professor continued her precise conversation, I began to get very sleepy. During the course of the rest of my day, I saw some more of my friends from high school. My next two professors were also nice. I began to get kind of worried when my History professor told us that reading is a major part of this class. I guess I\'ll have to get used to reading more because I want to pass. In conclusion, my first day of college turned out to be okay. I went from being excited and anxious, to nervous, and then to calm when I realized that everything was good. My first day was a great experience, and I hope that I will continue to enjoy it and be successful. Solution When I was getting ready at home, I was excited and anxious, but as I was driving down the road, I began to get nervous. I had to take the interstate, and my first thought was traffic. I was scared that there was going to be a horrific accident on the busy interstate and that I would be late on my first day. I was very relieved when I saw that the interstate was not as busy as I thought it was going to be, and the drive down was a breeze. All of my favorite songs came on the radio, and I began to grown excited and anxious again. When I drove up on the Macon State campus, that terrible feeling came back again. I saw all of the cars and thought to myself frantically, “Oh no!, what if I can’t find a parking spot and be late to class after all.” Luckly, I found an excellen.

Viruses are those biological entities which are intermediate between.pdf

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Viruses are those biological entities which are intermediate between living and non-living. Outside the host, they are non-living. Even in non-living state they can survive for millions of years. So, there is no strategy in the world which will prove fatal to virus\' future in the world. Second, viruses have a protein capsid and genetic material. There is no machinery like mitosis, proofreading etc. in their body which will maintain their capsid and genetic material as it is. In fact, their genetic material is highly subject to mutation depending on the host they infect. So, no matter what immune mechanisms human evolve, there is no immune mechanism which will combat these unknown parasites. Humans can only prevent the spread of some viruses, most of which are those that had previously attacked humans; but for new viruses, humans will not be able to develop immune system. So, complete protection is impossible. Third, disease does not confer any kind of protection or survival to the virus. In fact the virus has to leave this host and infect a new healthy host. So, it is true that viruses cause disease accidently, but this does not mean that viruses\' future is at stake because of this. Viruses will survive till the date biological world will survive. Solution Viruses are those biological entities which are intermediate between living and non-living. Outside the host, they are non-living. Even in non-living state they can survive for millions of years. So, there is no strategy in the world which will prove fatal to virus\' future in the world. Second, viruses have a protein capsid and genetic material. There is no machinery like mitosis, proofreading etc. in their body which will maintain their capsid and genetic material as it is. In fact, their genetic material is highly subject to mutation depending on the host they infect. So, no matter what immune mechanisms human evolve, there is no immune mechanism which will combat these unknown parasites. Humans can only prevent the spread of some viruses, most of which are those that had previously attacked humans; but for new viruses, humans will not be able to develop immune system. So, complete protection is impossible. Third, disease does not confer any kind of protection or survival to the virus. In fact the virus has to leave this host and infect a new healthy host. So, it is true that viruses cause disease accidently, but this does not mean that viruses\' future is at stake because of this. Viruses will survive till the date biological world will survive..

The most probable answer if this question seems to be option B. The .pdf

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The most probable answer if this question seems to be option B. The condition is also known as ciliary dyskinesia in which movement in cilia does not occur resulting in inhibiting with the process of sweeping of mucus and growing particles out of the lungs. Solution The most probable answer if this question seems to be option B. The condition is also known as ciliary dyskinesia in which movement in cilia does not occur resulting in inhibiting with the process of sweeping of mucus and growing particles out of the lungs..

Exactly how you have drawn it, this compounds I.pdf

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The contamination could be due to the presence of yeast in the cultu.pdf

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The contamination could be due to the presence of yeast in the culture. Cultures contaminated with yeasts become turbid with no pH change. Under microscope, yeast appear as small spherical particles. It could not be bacterial contamination because bacterial contanmination is accompanied with sudden drop in the pH of the culture medium. It could not be virus or mycoplasma contamination because the media does not change turbid in the case of mycoplasma contamination. They are very difficult to detect and they do not be appear as small round dots under microscope. It could not be mold contamination because the mycelia usually appear as thin, wisp-like filaments under microscope. Solution The contamination could be due to the presence of yeast in the culture. Cultures contaminated with yeasts become turbid with no pH change. Under microscope, yeast appear as small spherical particles. It could not be bacterial contamination because bacterial contanmination is accompanied with sudden drop in the pH of the culture medium. It could not be virus or mycoplasma contamination because the media does not change turbid in the case of mycoplasma contamination. They are very difficult to detect and they do not be appear as small round dots under microscope. It could not be mold contamination because the mycelia usually appear as thin, wisp-like filaments under microscope..

Systems Development Life Cycle(SDLC) is the step by step process whi.pdf

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Systems Development Life Cycle(SDLC) is the step by step process which we are following to complete software project that includes development and testing. There are 6 different phases available in SDLC. 1.Requirements 2.Analysis 3.System Design 4.System Implementation / Coding 5.System Testing and Integration 6.Release or System maintenance 1.Requirements: This is the first phase in SDLC,once the project has been confirmed between client and company will provide directly requirements to the company BA team. Defining the problems,adjectives such as resources and personal costs. Studying the ability of providing alternative solutions after meeting with clients,suppliers,c consultants and employees. After analyzing this you have a three choices i.e. develop a new system,improve the current system/project or leave a system. 2.System Analysis: End users requirements should be determined and documented,what their expectations are there for the system,and how it will perform. It was very important to maintain strong communication level with the clients to make sure you have a clear vision of a product. 3.System Design: In this phase defines the elements of a system,the components,the security levels and the modules,architecture and different interfaces. In this design phase high level and low level designers will design system process. 4.System Implementation / Coding : In this phase development team will involving to write the actual coding functional modules. In this phase system is ready to be deployed and installed customer premises. Training will be given end users depends upon the project 5.System testing and Integration testing During this phase testing team involved to ensures customer satisfaction and it will no required coding knowledge. Testing team performed real users it was systematic process in testing and done the integration testing and automation testing if required. 6.Release and system maintenance: During this phase technical team will be involving delay the application into production environment. Technical team and testing team will be involving to provide support to client while using application production. In the above question is that critical activity may include the coding of the application as per client requirement is very difficult and gathering the requirements from the client also very effecient task in SDLC process. My position include coding of a project and in that project client need the minimum 3 months of time for deploying the application into the production then our manager and team lead will mainly focus on the developer in the project. If we develop project then the testing team will find the errors and then that errors we will again rectify and where that bugs are impacting we will find. B. In project we are following the agile methodology.this will includes the following Customer satisfaction by rapid delivery of useful software Welcome changing requirements,e Ben late in development Working software is .

d. spontaneous at all temperatures .pdf

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MineSweeper.java public class MS { public static void main(Strin.pdf

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MineSweeper.java public class MS { public static void main(String[] args) { Gamegm= new Game(); } } --> Game.java public class Game { private Bd bd; boolean fnsh = false; boolean w = false; int tn=0; public jg(){ bd = new Bd(); Py(bd); } public void Py(Bd bd){ do{ tn++; System.out.println(\"Turn \"+tn); bd.show(); fnsh = bd.setPosition(); if(!fnsh){ bd.openNeighbors(); fnsh = bd.w(); } }while(!fnsh); if(bd.w()){ System.out.println(\"Congratulations, you let the 10 fields with the mns in \"+tn+\" tns\"); bd.showMines(); } else { System.out.println(\"Mine! You lost!\"); bd.showMines(); } } } --> Bd.java import java.util.Rnm; import java.util.Scanner; public class Bd { private int[][] mns; private char[][] bdgame; private int Line, Clm; Rnm rnm = new Rnm(); Scanner input = new Scanner(System.in); public Bd (){ mns = new int[10][10]; bdgame = new char[10][10]; stMs(); rnmMines(); fillTips(); startBd(); } public boolean w(){ int ct=0; for(int line = 1 ; line < 9 ; line++) for(int clm = 1 ; clm < 9 ; clm++) if(bdgame[line][clm]==\'_\') ct++; if(ct == 10) return true; else return false; } public void oN(){ for(int i=-1 ; i<2 ; i++) for(int j=-1 ; j<2 ; j++) if( (mns[Ln+i][Clm+j] != -1) && (Ln != 0 && Ln != 9 && Clm != 0 && Clm != 9) ) bdgame[Ln+i][Clm+j]=Character.forDigit(mns[Ln+i][Clm+j], 10); } public int getPosition(int Ln, int Clm){ retn mns[Ln][Clm]; } public boolean setPosition(){ do{ System.out.print(\"\ Ln: \"); Ln = input.nextInt(); System.out.print(\"Clm: \"); Clm = input.nextInt(); if( (bdgame[Ln][Clm] != \'_\') && ((Ln < 9 && Ln > 0) && (Clm < 9 && Clm > 0))) System.out.println(\"Field already shown\"); if( Ln < 1 || Ln > 8 || Clm < 1 || Clm > 8) System.out.println(\"Choose a number between 1 and 8\"); }while((Ln < 1 || Ln > 8 || Clm < 1 || Clm > 8) || (bdgame[Ln][Clm] != \'_\') ); if(getPosition(Ln, Clm)== -1) retn true; else retn false; } public void show(){ System.out.println(\"\ Lns\"); for(int Ln = 8 ; Ln > 0 ; Ln--){ System.out.print(\" \"+Ln + \" \"); for(int Clm = 1 ; Clm < 9 ; Clm++){ System.out.print(\" \"+ bdgame[Ln][Clm]); } System.out.println(); } System.out.println(\"\ 1 2 3 4 5 6 7 8\"); System.out.println(\" Clms\"); } public void fillTips(){ for(int ln=1 ; ln < 9 ; ln++) for(int clm=1 ; clm < 9 ; clm++){ for(int i=-1 ; i<=1 ; i++) for(int j=-1 ; j<=1 ; j++) if(mns[ln][clm] != -1) if(mns[ln+i][clm+j] == -1) mns[ln][clm]++; } } public void showMines(){ for(int i=1 ; i < 9; i++) for(int j=1 ; j < 9 ; j++) if(mns[i][j] == -1) bdgame[i][j]=\'*\'; show(); } public void startBd(){ for(int i=1 ; i Solution MineSweeper.java public class MS { public static void main(String[] args) { Gamegm= new Game(); } } --> Game.java public class Game { private Bd bd; boolean fnsh = false; boolean w = false; int tn=0; public jg(){ bd = new Bd(); Py(bd); } public void Py(Bd bd){ do{ tn++; System.out.println(\"Turn \"+tn); bd.show(); fnsh = bd.setPosition(); if(!fnsh){ bd.openNeighbors(); fnsh = bd.w(); } }while(!fnsh); if(bd.w()){ System.out.pr.

It is truebecause The SMTP server cannot find the destination emai.pdf

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Homo neanderthalensis are evolved and lived ina) Europe andor sou.pdf

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B. note aldehydes hydrolysis gives diol products.pdf

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Hi,I have updated your code. It is working fine now. Highllighted .pdf

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Hi, I have updated your code. It is working fine now. Highllighted all code changes below. Shapes.java import java.util.Scanner; public class Shapes { private int radius; public Shapes(int radius){ this.radius = radius; } public void setradius( int radius ) { } public int getradius() { return radius; } public void report() { System.out.printf( \"Circle perimeter, area / Sphere volume:\"); System.out.println(); System.out.printf( \" perimeter : \" + 2. * Math.PI * getradius()); System.out.println(); System.out.printf( \" area : \" + Math.PI * getradius() * 1.); System.out.println(); System.out.printf( \" volume : \" + (4 * Math.PI * Math.pow(getradius(),3)/3)); System.out.println(); System.out.printf( \"Square perimeter, area:\"); System.out.println(); System.out.printf( \" perimeter : \" + 2 *Math.E * 1.); System.out.println(); System.out.printf( \" area : \" + Math.E * 1.* 1); System.out.println(); } public static void main( String[] args ) { //System.out.printf( \"Radius (as integer) of bounding sphere?\ %s!\ \", getradius() ); Scanner input = new Scanner( System.in ); System.out.println(\"Enter the radius:\"); int radius = input.nextInt(); Shapes s = new Shapes(radius); s.report(); } } Output: import java.util.Scanner; public class Shapes { private int radius; public Shapes(int radius){ this.radius = radius; } public void setradius( int radius ) { } public int getradius() { return radius; } public void report() { System.out.printf( \"Circle perimeter, area / Sphere volume:\"); System.out.println(); System.out.printf( \" perimeter : \" + 2. * Math.PI * getradius()); System.out.println(); System.out.printf( \" area : \" + Math.PI * getradius() * 1.); System.out.println(); System.out.printf( \" volume : \" + (4 * Math.PI * Math.pow(getradius(),3)/3)); System.out.println(); System.out.printf( \"Square perimeter, area:\"); System.out.println(); System.out.printf( \" perimeter : \" + 2 *Math.E * 1.); System.out.println(); System.out.printf( \" area : \" + Math.E * 1.* 1); System.out.println(); } public static void main( String[] args ) { //System.out.printf( \"Radius (as integer) of bounding sphere?\ %s!\ \", getradius() ); Scanner input = new Scanner( System.in ); System.out.println(\"Enter the radius:\"); int radius = input.nextInt(); Shapes s = new Shapes(radius); s.report(); } } Solution Hi, I have updated your code. It is working fine now. Highllighted all code changes below. Shapes.java import java.util.Scanner; public class Shapes { private int radius; public Shapes(int radius){ this.radius = radius; } public void setradius( int radius ) { } public int getradius() { return radius; } public void report() { System.out.printf( \"Circle perimeter, area / Sphere volume:\"); System.out.println(); System.out.printf( \" perimeter : \" + 2. * Math.PI * getradius()); System.out.println(); System.out.printf( \" area : \" + Math.PI * getradius() * 1.); System.out.println(); System.out.printf( \" volume : \" + (4 * Math.PI * Math.pow(getradius(),3)/3).

Domain = [ -, +]SolutionDomain = [ -, +].pdf

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Debit CreditGovermental Activities Funds XExpenditure XCap.pdf

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Capacitive Proximity Sensors should be used in detecting plastic sub.pdf

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Capacitive Proximity Sensors should be used in detecting plastic subtance such as plastic cup. Capacitive proximity sensors are similar to inductive proximity sensors. The main difference between the two types is that capacitive proximity sensors produce an electrostatic field instead of an electromagnetic field. Capacitive proximity switches will sense metal as well as nonmetallic materials such as paper, glass, liquids, and cloth. The sensing surface of a capacitive sensor is formed by two concentrically shaped metal electrodes of an unwound capacitor. When an object nears the sensing surface it enters the electrostatic field of the electrodes and changes the capacitance in an oscillator circuit. As a result, the oscillator begins oscillating. The trigger circuit reads the oscillator’s amplitude and when it reaches a specific level the output state of the sensor changes. As the target moves away from the sensor the oscillator’s amplitude decreases, switching the sensor output back to its original state Standard targets are specified for each capacitive sensor. The Dielectric Constant standard target is usually defined as metal and/or water. Capacitive sensors depend on the dielectric constant of the target. The larger the dielectric number of a material the easier it is to detect. Solution Capacitive Proximity Sensors should be used in detecting plastic subtance such as plastic cup. Capacitive proximity sensors are similar to inductive proximity sensors. The main difference between the two types is that capacitive proximity sensors produce an electrostatic field instead of an electromagnetic field. Capacitive proximity switches will sense metal as well as nonmetallic materials such as paper, glass, liquids, and cloth. The sensing surface of a capacitive sensor is formed by two concentrically shaped metal electrodes of an unwound capacitor. When an object nears the sensing surface it enters the electrostatic field of the electrodes and changes the capacitance in an oscillator circuit. As a result, the oscillator begins oscillating. The trigger circuit reads the oscillator’s amplitude and when it reaches a specific level the output state of the sensor changes. As the target moves away from the sensor the oscillator’s amplitude decreases, switching the sensor output back to its original state Standard targets are specified for each capacitive sensor. The Dielectric Constant standard target is usually defined as metal and/or water. Capacitive sensors depend on the dielectric constant of the target. The larger the dielectric number of a material the easier it is to detect..

C is correct. Packet is the PDU at the Network layer.A, B, D, and .pdf

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C is correct. Packet is the PDU at the Network layer. A, B, D, and E are incorrect. A is incorrect because the data is the PDU for the top three layers. B is incorrect because the Layer 4 PDU is called a segment. D is incorrect because the frame is the PDU for the Data Link layer. E is incorrect because the bit is the PDU for the Physical layer. Solution C is correct. Packet is the PDU at the Network layer. A, B, D, and E are incorrect. A is incorrect because the data is the PDU for the top three layers. B is incorrect because the Layer 4 PDU is called a segment. D is incorrect because the frame is the PDU for the Data Link layer. E is incorrect because the bit is the PDU for the Physical layer..

Brief background of Qwest (SEC filings)After its initial public of.pdf

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Brief background of Qwest (SEC filings) After its initial public offering in 1997, Qwest touted itself as a progressive, new-generation technology company with enormous growth potential. Beginning in 1999, in fact, Qwest’s CEO consistently predicted publicly that Qwest would achieve double-digit revenue and earnings growth. By mid-1999, it became clear to Qwest senior management that the market for telecommunications services was declining and that revenue from those services would not sustain Qwest’s projected revenue and earnings growth. To “fill the gap” between its actual and projected revenue, Qwest, at the direction of its senior management, began selling indefeasible rights of use (“IRUs”). An IRU is an irrevocable right to use a specific fiber strand or specific amount of fiber capacity for a specified time period. Thus, to meet revenue expectations that it created, Qwest sold what the company had previously identified in Commission filings and press releases as its “principal asset.” When the demand for IRUs declined, Qwest engaged in IRU “swaps” whereby Qwest bought IRUs from other companies in exchange for agreements from those companies to buy IRUs from Qwest. As another “gap filler,” Qwest sold capital equipment. Both IRU and equipment sales were referred to internally as “one hit wonders.” Indeed, the investment community generally discounted such non-recurring revenue sources when valuing telecommunications companies because non-recurring revenue sources were not sustainable. Qwest’s use of one-time transactions to fill the gap between actual and projected revenue became so common that many Qwest employees likened the practice to an “addiction” and the non- recurring IRU and equipment sale transactions as Qwest’s “heroin.” In Commission filings and other public statements, Qwest fraudulently characterized non- recurring revenue from IRU and equipment transactions as recurring \"data and Internet service revenues,\" thereby masking its declining financial condition and artificially inflating its stock price. Fraudulent Accounting for IRU and Equipment Sale Transactions In addition to fraudulently characterizing non-recurring revenue as recurring revenue, Qwest ignored generally accepted accounting principles (“GAAP”) by recognizing upfront revenue from IRU transactions and equipment sales. Qwest, in fact, employed fraudulent devices such as backdated contracts and secret side agreements to conceal the fact that its IRU and equipment transactions did not meet GAAP’s requirements for upfront revenue recognition. Under GAAP, Qwest should either have not recognized any revenue on these transactions or recognized revenue ratably over the lives of the contracts. Other Fraudulent Conduct Qwest engaged in a variety of other fraudulent conduct. In particular: Paragraph 67 of PCAOB Auditing Standard No. 12: Paragraph 67 of PCAOB Auditing Standard No. 12 relates to Identifying and Assessing Risks of material misstatements. Relevant text is.

4096Solution4096.pdf

aniyathikitchen

When a metal cation combines with a Lewis base, a.pdf

aniyathikitchen

When a metal cation combines with a Lewis base, a complex ion forms. We can define a complex ion as an ion containing a central metal cation bonded (covalently) to one or more molecules or ions called ligands. Transition metal ions have a particular tendency to form complex ions because they have more than one oxidation state. This property allows them to act effectively as Lewis acids in reactions with many molecules or ions that serve as electron donors or as Lewis bases. The formation of a complex ion tends to increase the solubility of a salt because the metal cations are removed from the solubility equilibria to form the complex ion. Copper ammonium complex-> Cu2+(aq) + 4 NH3(aq) <==> Cu(NH3)42+(aq) Nickel- ammonium complex ->Ni(NH3)62+ Solution When a metal cation combines with a Lewis base, a complex ion forms. We can define a complex ion as an ion containing a central metal cation bonded (covalently) to one or more molecules or ions called ligands. Transition metal ions have a particular tendency to form complex ions because they have more than one oxidation state. This property allows them to act effectively as Lewis acids in reactions with many molecules or ions that serve as electron donors or as Lewis bases. The formation of a complex ion tends to increase the solubility of a salt because the metal cations are removed from the solubility equilibria to form the complex ion. Copper ammonium complex-> Cu2+(aq) + 4 NH3(aq) <==> Cu(NH3)42+(aq) Nickel- ammonium complex ->Ni(NH3)62+.

1) When the two molecules are non polar there is a possibility of ex.pdf

aniyathikitchen

1) When the two molecules are non polar there is a possibility of existing dispersion forces. This is due to the arrangement of electrons around the central atom of the molecule. Ex: n-pentane, Neopentane, n-octane etc When the two molecules are polar there exists dipole - dipole forces. Ex: It can be seen in acetone molecules When there is a hydrogen atom attached to the electronegative atom then there is a possibility of existing hydrogen bonding. Ex: H-F-------H-F 2) O,N,F are electronegative in nature which brings a large positive charge on the hydrogen atom. Moreover being smaller in size a large interaction forces results. 3) In diamond C is in sp3 hybridization and all the C\'s are at the same distance. It is in polymeric tetrahedral structure. Hence it is very hard in nature. Solution 1) When the two molecules are non polar there is a possibility of existing dispersion forces. This is due to the arrangement of electrons around the central atom of the molecule. Ex: n-pentane, Neopentane, n-octane etc When the two molecules are polar there exists dipole - dipole forces. Ex: It can be seen in acetone molecules When there is a hydrogen atom attached to the electronegative atom then there is a possibility of existing hydrogen bonding. Ex: H-F-------H-F 2) O,N,F are electronegative in nature which brings a large positive charge on the hydrogen atom. Moreover being smaller in size a large interaction forces results. 3) In diamond C is in sp3 hybridization and all the C\'s are at the same distance. It is in polymeric tetrahedral structure. Hence it is very hard in nature..

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