Dear,The answer is 3.SolutionDear,The answer is 3..pdf
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The document is a short note stating that the answer to an unknown question is the number 3. It provides no context or explanation for how the answer was determined. In just two sentences, it directly states that the answer is 3 and labels the following text as the solution.
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we know that there is an OH group present in phen.pdf
we know that there is an OH group present in phenols. now, in ortho phenols here
o-nitrophenols intramolecular bonding is possible due to the polar nature of OH bonds which can
result in the formation of hydrogen bonds with the same molecule . this makes it water soluble
and steam volatile which is not in the case of para-isomer. thus A mixture of p-nitrophenol and
o-nitrophenol can be separated by steam distillation.
Solution
we know that there is an OH group present in phenols. now, in ortho phenols here
o-nitrophenols intramolecular bonding is possible due to the polar nature of OH bonds which can
result in the formation of hydrogen bonds with the same molecule . this makes it water soluble
and steam volatile which is not in the case of para-isomer. thus A mixture of p-nitrophenol and
o-nitrophenol can be separated by steam distillation..
systems T and P are different from the reservoi.pdf
system\'s T and P are different from the reservoir.
Solution
system\'s T and P are different from the reservoir..
Since X represents the solid phase, Y represents .pdf
Since X represents the solid phase, Y represents the liquid phase,and Z is the vapor
phase: F to E = deposition C to D = boiling D to C = condensation E to F = sublimation E to C =
melting C to E = freezing A = triple point B = critical point
Solution
Since X represents the solid phase, Y represents the liquid phase,and Z is the vapor
phase: F to E = deposition C to D = boiling D to C = condensation E to F = sublimation E to C =
melting C to E = freezing A = triple point B = critical point.
P inversly proportional to 1V so answer is C .pdf
P inversly proportional to 1/V so answer is C
Solution
P inversly proportional to 1/V so answer is C.
nutrinos are the smallest part of an atom. NOTE-.pdf
nutrinos are the smallest part of an atom. NOTE:-Usually electrons are smallest but
they are a lot so quack occurs.
Solution
nutrinos are the smallest part of an atom. NOTE:-Usually electrons are smallest but
they are a lot so quack occurs..
Molecular solids are held together in single chai.pdf
Molecular solids are held together in single chains of bonds using either LDF forces
or dipole-dipole forces. These are weaker bonds than network covalent because instead of single
bonds, network covalent solids (such as sand, graphite and diamond) form webs of bonds that
interlock with one another.
Solution
Molecular solids are held together in single chains of bonds using either LDF forces
or dipole-dipole forces. These are weaker bonds than network covalent because instead of single
bonds, network covalent solids (such as sand, graphite and diamond) form webs of bonds that
interlock with one another..
Liquid crystals (LCs) are a state of matter that .pdf
Liquid crystals (LCs) are a state of matter that have properties between those of a
conventional liquid and those of a solid crystal.[1] For instance, an LC may flow like a liquid,
but its molecules may be oriented in a crystal-like way. There are many different types of LC
phases, which can be distinguished by their different optical properties (such as birefringence).
When viewed under a microscope using a polarized light source, different liquid crystal phases
will appear to have distinct textures. The contrasting areas in the textures correspond to domains
where the LC molecules are oriented in different directions. Within a domain, however, the
molecules are well ordered. LC materials may not always be in an LC phase (just as water may
turn into ice or steam). Liquid crystals can be divided into thermotropic, lyotropic and
metallotropic phases. Thermotropic and lyotropic LCs consist of organic molecules.
Thermotropic LCs exhibit a phase transition into the LC phase as temperature is changed.
Lyotropic LCs exhibit phase transitions as a function of both temperature and concentration of
the LC molecules in a solvent (typically water). Metallotropic LCs are composed of both organic
and inorganic molecules; their LC transition depends not only on temperature and concentration,
but also on the inorganic-organic composition ratio.
Solution
Liquid crystals (LCs) are a state of matter that have properties between those of a
conventional liquid and those of a solid crystal.[1] For instance, an LC may flow like a liquid,
but its molecules may be oriented in a crystal-like way. There are many different types of LC
phases, which can be distinguished by their different optical properties (such as birefringence).
When viewed under a microscope using a polarized light source, different liquid crystal phases
will appear to have distinct textures. The contrasting areas in the textures correspond to domains
where the LC molecules are oriented in different directions. Within a domain, however, the
molecules are well ordered. LC materials may not always be in an LC phase (just as water may
turn into ice or steam). Liquid crystals can be divided into thermotropic, lyotropic and
metallotropic phases. Thermotropic and lyotropic LCs consist of organic molecules.
Thermotropic LCs exhibit a phase transition into the LC phase as temperature is changed.
Lyotropic LCs exhibit phase transitions as a function of both temperature and concentration of
the LC molecules in a solvent (typically water). Metallotropic LCs are composed of both organic
and inorganic molecules; their LC transition depends not only on temperature and concentration,
but also on the inorganic-organic composition ratio..
Recommended
we know that there is an OH group present in phen.pdf
we know that there is an OH group present in phenols. now, in ortho phenols here
o-nitrophenols intramolecular bonding is possible due to the polar nature of OH bonds which can
result in the formation of hydrogen bonds with the same molecule . this makes it water soluble
and steam volatile which is not in the case of para-isomer. thus A mixture of p-nitrophenol and
o-nitrophenol can be separated by steam distillation.
Solution
we know that there is an OH group present in phenols. now, in ortho phenols here
o-nitrophenols intramolecular bonding is possible due to the polar nature of OH bonds which can
result in the formation of hydrogen bonds with the same molecule . this makes it water soluble
and steam volatile which is not in the case of para-isomer. thus A mixture of p-nitrophenol and
o-nitrophenol can be separated by steam distillation..
systems T and P are different from the reservoi.pdf
system\'s T and P are different from the reservoir.
Solution
system\'s T and P are different from the reservoir..
Since X represents the solid phase, Y represents .pdf
Since X represents the solid phase, Y represents the liquid phase,and Z is the vapor
phase: F to E = deposition C to D = boiling D to C = condensation E to F = sublimation E to C =
melting C to E = freezing A = triple point B = critical point
Solution
Since X represents the solid phase, Y represents the liquid phase,and Z is the vapor
phase: F to E = deposition C to D = boiling D to C = condensation E to F = sublimation E to C =
melting C to E = freezing A = triple point B = critical point.
P inversly proportional to 1V so answer is C .pdf
P inversly proportional to 1/V so answer is C
Solution
P inversly proportional to 1/V so answer is C.
nutrinos are the smallest part of an atom. NOTE-.pdf
nutrinos are the smallest part of an atom. NOTE:-Usually electrons are smallest but
they are a lot so quack occurs.
Solution
nutrinos are the smallest part of an atom. NOTE:-Usually electrons are smallest but
they are a lot so quack occurs..
Molecular solids are held together in single chai.pdf
Molecular solids are held together in single chains of bonds using either LDF forces
or dipole-dipole forces. These are weaker bonds than network covalent because instead of single
bonds, network covalent solids (such as sand, graphite and diamond) form webs of bonds that
interlock with one another.
Solution
Molecular solids are held together in single chains of bonds using either LDF forces
or dipole-dipole forces. These are weaker bonds than network covalent because instead of single
bonds, network covalent solids (such as sand, graphite and diamond) form webs of bonds that
interlock with one another..
Liquid crystals (LCs) are a state of matter that .pdf
Liquid crystals (LCs) are a state of matter that have properties between those of a
conventional liquid and those of a solid crystal.[1] For instance, an LC may flow like a liquid,
but its molecules may be oriented in a crystal-like way. There are many different types of LC
phases, which can be distinguished by their different optical properties (such as birefringence).
When viewed under a microscope using a polarized light source, different liquid crystal phases
will appear to have distinct textures. The contrasting areas in the textures correspond to domains
where the LC molecules are oriented in different directions. Within a domain, however, the
molecules are well ordered. LC materials may not always be in an LC phase (just as water may
turn into ice or steam). Liquid crystals can be divided into thermotropic, lyotropic and
metallotropic phases. Thermotropic and lyotropic LCs consist of organic molecules.
Thermotropic LCs exhibit a phase transition into the LC phase as temperature is changed.
Lyotropic LCs exhibit phase transitions as a function of both temperature and concentration of
the LC molecules in a solvent (typically water). Metallotropic LCs are composed of both organic
and inorganic molecules; their LC transition depends not only on temperature and concentration,
but also on the inorganic-organic composition ratio.
Solution
Liquid crystals (LCs) are a state of matter that have properties between those of a
conventional liquid and those of a solid crystal.[1] For instance, an LC may flow like a liquid,
but its molecules may be oriented in a crystal-like way. There are many different types of LC
phases, which can be distinguished by their different optical properties (such as birefringence).
When viewed under a microscope using a polarized light source, different liquid crystal phases
will appear to have distinct textures. The contrasting areas in the textures correspond to domains
where the LC molecules are oriented in different directions. Within a domain, however, the
molecules are well ordered. LC materials may not always be in an LC phase (just as water may
turn into ice or steam). Liquid crystals can be divided into thermotropic, lyotropic and
metallotropic phases. Thermotropic and lyotropic LCs consist of organic molecules.
Thermotropic LCs exhibit a phase transition into the LC phase as temperature is changed.
Lyotropic LCs exhibit phase transitions as a function of both temperature and concentration of
the LC molecules in a solvent (typically water). Metallotropic LCs are composed of both organic
and inorganic molecules; their LC transition depends not only on temperature and concentration,
but also on the inorganic-organic composition ratio..
interstellar medium (gas) Solution .pdf
interstellar medium (gas)
Solution
interstellar medium (gas).
Hhyd = Hcrystallization + Hsolv Enthalpy of Solv.pdf
Hhyd = Hcrystallization + Hsolv Enthalpy of Solvation (Hsolv kJ/mol) of Some
Common Electrolytes Substanc Hsolv Substance Hsolv AlCl3(s) -373.63 H2SO4(l) -95.28
LiNO3(s) -2.51 LiCl(s) -37.03 NaNO3(s) 20.50 NaCl(s) 3.88 KNO3(s) 34.89 KCl(s) -17.22
NaOH(s) -44.51 NH4Cl(s) 14.77 use this i think may help u
Solution
Hhyd = Hcrystallization + Hsolv Enthalpy of Solvation (Hsolv kJ/mol) of Some
Common Electrolytes Substanc Hsolv Substance Hsolv AlCl3(s) -373.63 H2SO4(l) -95.28
LiNO3(s) -2.51 LiCl(s) -37.03 NaNO3(s) 20.50 NaCl(s) 3.88 KNO3(s) 34.89 KCl(s) -17.22
NaOH(s) -44.51 NH4Cl(s) 14.77 use this i think may help u.
The answer is (b) the condensation of steam to liiquid water.The e.pdf
The answer is (b) the condensation of steam to liiquid water.
The entropy decreases because the system becomes more ordered (free water molecules in steam
become less free in liquid water due to stronger bonding to neighbours).
Solution
The answer is (b) the condensation of steam to liiquid water.
The entropy decreases because the system becomes more ordered (free water molecules in steam
become less free in liquid water due to stronger bonding to neighbours)..
XML Extensiable Markup LanguageHTML Hypertext Markup LanguageD.pdf
XML: Extensiable Markup Language
HTML: Hypertext Markup Language
Difference between XML and HTML:
Solution
XML: Extensiable Markup Language
HTML: Hypertext Markup Language
Difference between XML and HTML:.
The Data recovery is of two types …One method is by using a softwa.pdf
The Data recovery is of two types …
One method is by using a software that can recover files.There are certain benifits and risks by
using this way.
\'File System has file records and each file record is stored in Master File Table(MFT).MFT
doesnt have file data but has only information about the file,which is used while retrieve the file.
Each file and folder stored in File System has its entry in Master File Table. Even deleted files
and folders also have there entry in Master File Table until it is reallocated to other file or folder.
We also know that $BITMAP metadata file has information about allocation status of disk
clusters. Using this information we have proposed the technique to recover deleted files from
File System partition which is given below:
Steps followed in deleted file recovery are:
1.Search Files MFT Entry
2. Process it’s $DATA attribute
3. If $DATA attribute is resident, file content is present in MFT Entry. Just copy it to external
location to complete recovery process.
4. If $DATA attribute is non-resident, file’s contents are present in external cluster. Then parse
the RunList present in the attribute.
5. Check if each cluster listed in Runlist is allocated or not. If all clusters have allocated status as
0, means that the files data is still present on disk. Therefore complete recovery is possible. Copy
file’s contents to external location to complete recovery process.
6. If some clusters have allocated status as 1, then the partial recovery is possible. Copy file’s
contents to external location with reallocated cluster’s values as zeros to complete partial
recovery process.
If all clusters have allocated status as 1, then the file’s contents are lost and recovery is
impossible. The above technique can handle the file’s data which is fragmented across number of
cluster runs so gives more granular results in recovery of deleted files. The technique does not
require any external storage or use of sorting therefore can increase the performance in recovery
process of deleted files.
As we are discussing the file recovery through software the recovery software starts by trying to
read and process the first copy of the info about files and folders. In some cases such as
accidental file deletion, this is the only step where we can recover the file entirely
If the first copy of the info about files and folders is severely damaged, the software scans the
disk for the second copy of the info about files and folders. It also attempts to glean additional
information about the folders and files structure that may be on the data part of the disk. Then, it
processes all this information to reconstruct the original folders and file structure.
If the file system on the disk isn\'t severely damaged, it is often possible to recover the entire file
and folder structure.
If the file system on the disk is severely damaged, this recovery method cannot recreate the
entire folder structure.
Then these files are stored in the unkonwn folders..
Thanks anon101. D is correctSolutionThanks anon101. D is corre.pdf
Thanks anon101. D is correct
Solution
Thanks anon101. D is correct.
StudentData.javaimport java.util.Scanner; public class Student.pdf
StudentData.java
import java.util.Scanner;
public class StudentData {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int totalStudents = 0;
int maleCount = 0, femaleCount = 0, minorCount = 0, adultCount = 0, seniorCount = 0;
while(true){
System.out.println(\"Do you ant to add a student (n to stop):\");
char ch = scan.next().charAt(0);
if(ch == \'n\' || ch == \'N\'){
break;
}
else{
System.out.println(\"Enter the student gender (male or female)\");
String gender = scan.next();
if(gender.equalsIgnoreCase(\"male\")){
maleCount++;
}
else{
femaleCount++;
}
System.out.println(\"Enter student age:\");
int age = scan.nextInt();
if(age <= 18){
minorCount++;
}
else if(age >=19 && age <= 65){
adultCount++;
}
else{
seniorCount++;
}
}
}
totalStudents = maleCount + femaleCount;
System.out.println(\"Total Students: \"+totalStudents);
System.out.println(\"Male: \"+maleCount);
System.out.println(\"Female: \"+femaleCount);
System.out.println(\"Minor: \"+minorCount);
System.out.println(\"Adult: \"+adultCount);
System.out.println(\"Senior: \"+seniorCount);
String s = (maleCount >= femaleCount)? \"Students are mre males\":\"Students are more
females\";
System.out.println(s);
}
}
Output:
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
44
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
77
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
12
Do you ant to add a student (n to stop):
yy
Enter the student gender (male or female)
female
Enter student age:
55
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
female
Enter student age:
33
Do you ant to add a student (n to stop):
n
Total Students: 5
Male: 3
Female: 2
Minor: 1
Adult: 3
Senior: 1
Students are mre males
Solution
StudentData.java
import java.util.Scanner;
public class StudentData {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int totalStudents = 0;
int maleCount = 0, femaleCount = 0, minorCount = 0, adultCount = 0, seniorCount = 0;
while(true){
System.out.println(\"Do you ant to add a student (n to stop):\");
char ch = scan.next().charAt(0);
if(ch == \'n\' || ch == \'N\'){
break;
}
else{
System.out.println(\"Enter the student gender (male or female)\");
String gender = scan.next();
if(gender.equalsIgnoreCase(\"male\")){
maleCount++;
}
else{
femaleCount++;
}
System.out.println(\"Enter student age:\");
int age = scan.nextInt();
if(age <= 18){
minorCount++;
}
else if(age >=19 && age <= 65){
adultCount++;
}
else{
seniorCount++;
}
}
}
totalStudents = maleCount + femaleCount;
System.out.println(\"Total Students: \"+totalStudents);
System.out.println(\"Male: \"+maleCount);
System.out.println(\"Female: \"+femaleCount);
System.out.println(\"Minor: \"+minorCount);
System.out.println(\"Adult: \"+adultCount);
System.out.println(\"Senior: \"+seni.
Reverse EngineeringReverse engineering is also a back engineering .pdf
Reverse Engineering
Reverse engineering is also a back engineering which is extracting knowledge from manmade
and reproducing it from extracted information. Reverse engineering convert’s binary code of the
program and recreating it back to the original source code is known as reverse engineering.
Reverse engineering is common in both hardware and software. It convert’s the code written in
high level language into a low level language without changing the original program is also a
reverse engineering The process of recreating a design by analyzing a final product of a software
is also known as reverse engineering.
Reverse Engineering is used in many fields of Information Technology in form of Malware
Analysis, Network Analysis, debugging, and improvising existing algorithms, software
reusability
Reverse engineering is used in software design for to add new features to the existing software
with or without knowing the source code.
It help’s to study the virus and malware codeIt makes the system does not have any major
.vulnerabilities and security flaws.Some of the reverse engineering tools are debugger and
disassemblers.Reverse engineering is also beneficial for business owners as they can incorporate
advanced features into their software to meet the demands of the growing markets.
Solution
Reverse Engineering
Reverse engineering is also a back engineering which is extracting knowledge from manmade
and reproducing it from extracted information. Reverse engineering convert’s binary code of the
program and recreating it back to the original source code is known as reverse engineering.
Reverse engineering is common in both hardware and software. It convert’s the code written in
high level language into a low level language without changing the original program is also a
reverse engineering The process of recreating a design by analyzing a final product of a software
is also known as reverse engineering.
Reverse Engineering is used in many fields of Information Technology in form of Malware
Analysis, Network Analysis, debugging, and improvising existing algorithms, software
reusability
Reverse engineering is used in software design for to add new features to the existing software
with or without knowing the source code.
It help’s to study the virus and malware codeIt makes the system does not have any major
.vulnerabilities and security flaws.Some of the reverse engineering tools are debugger and
disassemblers.Reverse engineering is also beneficial for business owners as they can incorporate
advanced features into their software to meet the demands of the growing markets..
Quantum mechanics predicts the existence of what are usually called .pdf
Quantum mechanics predicts the existence of what are usually called \'\'zero-point\'\' energies for
the strong, the weak and the electromagnetic interactions, where \'\'zero-point\'\' refers to the
energy of the system at temperature T=0, or the lowest quantized energy level of a quantum
mechanical system. Although the term \'\'zero-point energy\'\' applies to all three of these
interactions in nature, customarily (and hereafter in this article) it is used in reference only to the
electromagnetic case.
In conventional quantum physics, the origin of zero-point energy is the Heisenberg uncertainty
principle, which states that, for a moving particle such as an electron, the more precisely one
measures the position, the less exact the best possible measurement of its momentum (mass
times velocity), and vice versa. The least possible uncertainty of position times momentum is
specified by Planck\'s constant, h. A parallel uncertainty exists between measurements involving
time and energy (and other so-called conjugate variables in quantum mechanics). This minimum
uncertainty is not due to any correctable flaws in measurement, but rather reflects an intrinsic
quantum fuzziness in the very nature of energy and matter springing from the wave nature of the
various quantum fields. This leads to the concept of zero-point energy.
Zero-point energy is the energy that remains when all other energy is removed from a system.
This behaviour is demonstrated by, for example, liquid helium. As the temperature is lowered to
absolute zero, helium remains a liquid, rather than freezing to a solid, owing to the irremovable
zero-point energy of its atomic motions. (Increasing the pressure to 25 atmospheres will cause
helium to freeze.)
A harmonic oscillator is a useful conceptual tool in physics. Classically a harmonic oscillator,
such as a mass on a spring, can always be brought to rest. However a quantum harmonic
oscillator does not permit this. A residual motion will always remain due to the requirements of
the Heisenberg uncertainty principle, resulting in a zero-point energy, equal to 1/2 hf, where f is
the oscillation frequency.
Electromagnetic radiation can be pictured as waves flowing through space at the speed of light.
The waves are not waves of anything substantive, but are ripples in a state of a theoretically
defined field. However these waves do carry energy (and momentum), and each wave has a
specific direction, frequency and polarization state. Each wave represents a \'\'propagating mode
of the electromagnetic field.\'\'
Each mode is equivalent to a harmonic oscillator and is thus subject to the Heisenberg
uncertainty principle. From this analogy, every mode of the field must have 1/2 hf as its average
minimum energy. That is a tiny amount of energy in each mode, but the number of modes is
enormous, and indeed increases per unit frequency interval as the square of the frequency. The
spectral energy density is determined by the density of modes times the.
Program Educational Objectives • To create competencies and opportun.pdf
Program Educational Objectives • To create competencies and opportunities for Higher
Education; • To create professional manpower skilled for the IT Industry; • To provide
laboratory practices with advanced FOSS Tools; • To provide inter-disciplinary opportunities; •
To provide opportunities of developing technical documents and presentation skills. • to provide
opportunities of industry-Institute interactions; • To develop opportunities to promote
Entrepreneurship and start-ups; • To nurture professional and social ethics. Program Objectives •
To expose students to the Systems and Applications Programming, Organizations and
Architectures; • To provide conceptual knowledge in the Computing domain; • To provide
interdisciplinary knowledge; • To expose students with advanced tools used in industry; • To
develop written and soft-skill competencies; • To develop team work experience of professionals
skills for IT Industry. Program Outcomes • To test applications with concepts and skills in the
domain subjects; • To demonstrate skills in programming techniques and technologies; • To
demonstrate oral and written skills for technical presentations and documentation; • To
demonstrate IT project as team work; • To demonstrate social and Professional ethical practices;
2 SAVITRIBAI PHULE PUNE UNIVERSITY BE (COMPUTER ENGINEERING)- 2012
COURSE STRUCTURE Term-I Subject Subject Teaching Scheme Examination Scheme Total
Code Marks Lect Tut Pract In PR/ OR/ End Sem TW TW Sem Asmnt Asmnt 410441 Design &
Analysis 03 — — 30 — — 70 100 of Algorithms 410442 Principles of Modern 04 — — 30 —
— 70 100 Compiler Design 410443 Smart System Design 03 — — 30 — — 70 100 and
Applications 410444 Elective-I 03 — — 30 — — 70 100 410445 Elective-II 03 — — 30 — —
70 100 410446 Computer — — 04 — 50 50 — 100 laboratory-I 410447 Computer — — 04 —
50 50 — 100 Laboratory-II 410448 Project — 02 — — 50 — — 50 Total 16 02 08 150 150 100
350 750 Term-II 410449 Software Design 03 — — 30 — — 70 100 Methodologies & Testing
410450 High 03 — — 30 — — 70 100 Performance Computing 410451 Elective-III 03 — — 30
— — 70 100 410452 Elective-IV 03 — — 30 — — 70 100 Open Elective 410453 Computer —
— 04 — 50 50 — 100 laboratory-III 410454 Computer — — 04 — 50 50 — 100 Laboratory-IV
410455 Project — 06 — — 50 100 — 150 Total 12 06 08 120 150 200 280 750
Solution
Program Educational Objectives • To create competencies and opportunities for Higher
Education; • To create professional manpower skilled for the IT Industry; • To provide
laboratory practices with advanced FOSS Tools; • To provide inter-disciplinary opportunities; •
To provide opportunities of developing technical documents and presentation skills. • to provide
opportunities of industry-Institute interactions; • To develop opportunities to promote
Entrepreneurship and start-ups; • To nurture professional and social ethics. Program Objectives •
To expose students to the Systems and Applications Programming, Organizations and
Architectures; • To.
P2O5 removes H2O from Alcohol to form Alkene O of Alcohol gives it.pdf
P2O5 removes H2O from Alcohol to form Alkene
O of Alcohol gives its lone pair to Higly e- defficient P of P2O5
O makes bond to P and breaks its bond to C atom
P also breaks its Double bond with O which now works as a base and abstract H atom from C
resulting in formation of an alkene
Solution
P2O5 removes H2O from Alcohol to form Alkene
O of Alcohol gives its lone pair to Higly e- defficient P of P2O5
O makes bond to P and breaks its bond to C atom
P also breaks its Double bond with O which now works as a base and abstract H atom from C
resulting in formation of an alkene.
Japanese colonial design in china korea and mongolia create an image.pdf
Japanese colonial design in china korea and mongolia create an image of utopian republic and
discards local context.
During the 1930s, Japanese colonial designed utopias in the newly conquered lands of China,
Korea, Mongolia, Thailand, Vietnam, Laos, Burma, the Philippines, and Indonesia.
Solution
Japanese colonial design in china korea and mongolia create an image of utopian republic and
discards local context.
During the 1930s, Japanese colonial designed utopias in the newly conquered lands of China,
Korea, Mongolia, Thailand, Vietnam, Laos, Burma, the Philippines, and Indonesia..
Is the measurement of the drug water solubility.Partitioning means.pdf
Is the measurement of the drug water solubility.
Partitioning means that the drug will be divided in parts between water and oil layer.
LogP is for hydrophilic or lipophilic. Lipophilicity tells about the compounds ability to dissolve
into lipohilic (non-aqueous) solutions. Lipophilicity is needed for the compounds to permeate
through the various biological membrane. Lipophilicity is typically measured as the compounds
distribution between non-aqueous (octanol) and aqueous (water) phase and the result is
expressed as a 10-base logarithm of the concentration ratios between these phases (partition
coefficient), logP.
P = [Co ]/[Cw]
LogP = Log[Co ]/[Cw].
LogP > 2 lipophilic drug.
LogP < 2 hydrophilic drug
LogP only applied for neutral compound
Low logP….. Low penetration to CNS
High logP….. Low water solubility…. Not suitable for oral administration
Solution
Is the measurement of the drug water solubility.
Partitioning means that the drug will be divided in parts between water and oil layer.
LogP is for hydrophilic or lipophilic. Lipophilicity tells about the compounds ability to dissolve
into lipohilic (non-aqueous) solutions. Lipophilicity is needed for the compounds to permeate
through the various biological membrane. Lipophilicity is typically measured as the compounds
distribution between non-aqueous (octanol) and aqueous (water) phase and the result is
expressed as a 10-base logarithm of the concentration ratios between these phases (partition
coefficient), logP.
P = [Co ]/[Cw]
LogP = Log[Co ]/[Cw].
LogP > 2 lipophilic drug.
LogP < 2 hydrophilic drug
LogP only applied for neutral compound
Low logP….. Low penetration to CNS
High logP….. Low water solubility…. Not suitable for oral administration.
Information systems Information systems are the combination of Hard.pdf
Information systems: Information systems are the combination of Hardware, Software and
telecommunication networks that people build and use to collect, create and distribute useful
data, typically in organizational settings.
Hardware refers to computer devices such as monitor, cpu etc.
Software refers to programming instructions to perform some activity
Telecommunication networks refers to set of equipment used to interconnect two or more
computers
People in organizations use information systems to process sales transactions, manage loan
applications, or help financial analysts decide where, when and how the investments can be
made. Product manager also use them to help decide when, where and how to market their
products and related serives. Production managers use them to help decide when and how to
manufacture products. Information systems also enable us in performing account transactions
through ATMs, communicate by live video with people living in other parts of the world and
purchasing anything in the global market.
Globalization: The integration of economies through out the world, enabled by technological
progress is called globalization. effects of globalization are in many ways such as greater
international movement of commodities, money, information and labour, as well as the
development of technologies, standards and processes to facilitate this movement. Specifically, a
more global and competitive world includes visible economic, cultural and technological
changes.
Changes in MIS because of Globalization
Globalization affects information systems in a lot of aspects like the use of internet by general
public in the world, global e-mail providers like hotmail and yahoo connects the whole world
together. Information systems has big role in globalization by influencing different cultures
through internet, where big economies and developed countries benefit the most out of this.
Globalization has revolutionized internal management. It has also made easier the interaction
between countries, regions and continents, thus contributing to profitability. It is the private
sector\'s philosophy that propelled efforts to utilize every means, including information
technology, to make companies survive, even the biggest and the most powerful company in the
world.
The role of the MIS in an organization can be compared to the role of heart in the body. The
information is the blood and MIS is the heart. In the body the heart plays the role of supplying
pure blood to all the elements of the body including the brain. The heart work faster and supplies
more blood when needed. It regulates and controls the incoming impure blood, processed it and
sends it to the destination in the quantity needed. It fulfills the needs of blood supply to human
body in normal course and also in crisis.
The MIS plays exactly the same role in the organization. The system ensures that an appropriate
data is collected from the various sources, processed and send further to all.
Hi,I have modified the Point.java file as per your requirement.P.pdf
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
@Test public void testXY()
{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}
Solution
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
@Test public void testXY()
{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}.
four types of eolian landforms are1)erosinal landforms2)depositi.pdf
four types of eolian landforms are
1)erosinal landforms
2)depositional landforms
3)sand dune landforms
4)desert dunes
erosional landforms:whenever wind force is concentrated over a area it will create a hollow pit
namely called deflation hollow.it will range a diameter of hundred meters.these broad shallow
depressions are called as pans among which qattara is the largest pan present in lybian desert of
egypt.
deposional landforms:these landforms are caused by the deposition of the sediment.these types
of largest dune fields are found in middle east and north africa.
sand dunes:this will occur in the form of deposition of sand.
desert dunes:this will occur in amazing diversity of forms.
Solution
four types of eolian landforms are
1)erosinal landforms
2)depositional landforms
3)sand dune landforms
4)desert dunes
erosional landforms:whenever wind force is concentrated over a area it will create a hollow pit
namely called deflation hollow.it will range a diameter of hundred meters.these broad shallow
depressions are called as pans among which qattara is the largest pan present in lybian desert of
egypt.
deposional landforms:these landforms are caused by the deposition of the sediment.these types
of largest dune fields are found in middle east and north africa.
sand dunes:this will occur in the form of deposition of sand.
desert dunes:this will occur in amazing diversity of forms..
As Al+3 has noble gas configuration. the IE3 of A.pdf
As Al+3 has noble gas configuration. the IE3 of Al is very low As Li has 3 e- only.
Li+3 will be having no e-.hence a higgh amount of energy is needed. lowest= al highest=Li
Solution
As Al+3 has noble gas configuration. the IE3 of Al is very low As Li has 3 e- only.
Li+3 will be having no e-.hence a higgh amount of energy is needed. lowest= al highest=Li.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
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More from anokhijew
interstellar medium (gas) Solution .pdf
interstellar medium (gas)
Solution
interstellar medium (gas).
Hhyd = Hcrystallization + Hsolv Enthalpy of Solv.pdf
Hhyd = Hcrystallization + Hsolv Enthalpy of Solvation (Hsolv kJ/mol) of Some
Common Electrolytes Substanc Hsolv Substance Hsolv AlCl3(s) -373.63 H2SO4(l) -95.28
LiNO3(s) -2.51 LiCl(s) -37.03 NaNO3(s) 20.50 NaCl(s) 3.88 KNO3(s) 34.89 KCl(s) -17.22
NaOH(s) -44.51 NH4Cl(s) 14.77 use this i think may help u
Solution
Hhyd = Hcrystallization + Hsolv Enthalpy of Solvation (Hsolv kJ/mol) of Some
Common Electrolytes Substanc Hsolv Substance Hsolv AlCl3(s) -373.63 H2SO4(l) -95.28
LiNO3(s) -2.51 LiCl(s) -37.03 NaNO3(s) 20.50 NaCl(s) 3.88 KNO3(s) 34.89 KCl(s) -17.22
NaOH(s) -44.51 NH4Cl(s) 14.77 use this i think may help u.
The answer is (b) the condensation of steam to liiquid water.The e.pdf
The answer is (b) the condensation of steam to liiquid water.
The entropy decreases because the system becomes more ordered (free water molecules in steam
become less free in liquid water due to stronger bonding to neighbours).
Solution
The answer is (b) the condensation of steam to liiquid water.
The entropy decreases because the system becomes more ordered (free water molecules in steam
become less free in liquid water due to stronger bonding to neighbours)..
XML Extensiable Markup LanguageHTML Hypertext Markup LanguageD.pdf
XML: Extensiable Markup Language
HTML: Hypertext Markup Language
Difference between XML and HTML:
Solution
XML: Extensiable Markup Language
HTML: Hypertext Markup Language
Difference between XML and HTML:.
The Data recovery is of two types …One method is by using a softwa.pdf
The Data recovery is of two types …
One method is by using a software that can recover files.There are certain benifits and risks by
using this way.
\'File System has file records and each file record is stored in Master File Table(MFT).MFT
doesnt have file data but has only information about the file,which is used while retrieve the file.
Each file and folder stored in File System has its entry in Master File Table. Even deleted files
and folders also have there entry in Master File Table until it is reallocated to other file or folder.
We also know that $BITMAP metadata file has information about allocation status of disk
clusters. Using this information we have proposed the technique to recover deleted files from
File System partition which is given below:
Steps followed in deleted file recovery are:
1.Search Files MFT Entry
2. Process it’s $DATA attribute
3. If $DATA attribute is resident, file content is present in MFT Entry. Just copy it to external
location to complete recovery process.
4. If $DATA attribute is non-resident, file’s contents are present in external cluster. Then parse
the RunList present in the attribute.
5. Check if each cluster listed in Runlist is allocated or not. If all clusters have allocated status as
0, means that the files data is still present on disk. Therefore complete recovery is possible. Copy
file’s contents to external location to complete recovery process.
6. If some clusters have allocated status as 1, then the partial recovery is possible. Copy file’s
contents to external location with reallocated cluster’s values as zeros to complete partial
recovery process.
If all clusters have allocated status as 1, then the file’s contents are lost and recovery is
impossible. The above technique can handle the file’s data which is fragmented across number of
cluster runs so gives more granular results in recovery of deleted files. The technique does not
require any external storage or use of sorting therefore can increase the performance in recovery
process of deleted files.
As we are discussing the file recovery through software the recovery software starts by trying to
read and process the first copy of the info about files and folders. In some cases such as
accidental file deletion, this is the only step where we can recover the file entirely
If the first copy of the info about files and folders is severely damaged, the software scans the
disk for the second copy of the info about files and folders. It also attempts to glean additional
information about the folders and files structure that may be on the data part of the disk. Then, it
processes all this information to reconstruct the original folders and file structure.
If the file system on the disk isn\'t severely damaged, it is often possible to recover the entire file
and folder structure.
If the file system on the disk is severely damaged, this recovery method cannot recreate the
entire folder structure.
Then these files are stored in the unkonwn folders..
Thanks anon101. D is correctSolutionThanks anon101. D is corre.pdf
Thanks anon101. D is correct
Solution
Thanks anon101. D is correct.
StudentData.javaimport java.util.Scanner; public class Student.pdf
StudentData.java
import java.util.Scanner;
public class StudentData {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int totalStudents = 0;
int maleCount = 0, femaleCount = 0, minorCount = 0, adultCount = 0, seniorCount = 0;
while(true){
System.out.println(\"Do you ant to add a student (n to stop):\");
char ch = scan.next().charAt(0);
if(ch == \'n\' || ch == \'N\'){
break;
}
else{
System.out.println(\"Enter the student gender (male or female)\");
String gender = scan.next();
if(gender.equalsIgnoreCase(\"male\")){
maleCount++;
}
else{
femaleCount++;
}
System.out.println(\"Enter student age:\");
int age = scan.nextInt();
if(age <= 18){
minorCount++;
}
else if(age >=19 && age <= 65){
adultCount++;
}
else{
seniorCount++;
}
}
}
totalStudents = maleCount + femaleCount;
System.out.println(\"Total Students: \"+totalStudents);
System.out.println(\"Male: \"+maleCount);
System.out.println(\"Female: \"+femaleCount);
System.out.println(\"Minor: \"+minorCount);
System.out.println(\"Adult: \"+adultCount);
System.out.println(\"Senior: \"+seniorCount);
String s = (maleCount >= femaleCount)? \"Students are mre males\":\"Students are more
females\";
System.out.println(s);
}
}
Output:
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
44
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
77
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
male
Enter student age:
12
Do you ant to add a student (n to stop):
yy
Enter the student gender (male or female)
female
Enter student age:
55
Do you ant to add a student (n to stop):
y
Enter the student gender (male or female)
female
Enter student age:
33
Do you ant to add a student (n to stop):
n
Total Students: 5
Male: 3
Female: 2
Minor: 1
Adult: 3
Senior: 1
Students are mre males
Solution
StudentData.java
import java.util.Scanner;
public class StudentData {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int totalStudents = 0;
int maleCount = 0, femaleCount = 0, minorCount = 0, adultCount = 0, seniorCount = 0;
while(true){
System.out.println(\"Do you ant to add a student (n to stop):\");
char ch = scan.next().charAt(0);
if(ch == \'n\' || ch == \'N\'){
break;
}
else{
System.out.println(\"Enter the student gender (male or female)\");
String gender = scan.next();
if(gender.equalsIgnoreCase(\"male\")){
maleCount++;
}
else{
femaleCount++;
}
System.out.println(\"Enter student age:\");
int age = scan.nextInt();
if(age <= 18){
minorCount++;
}
else if(age >=19 && age <= 65){
adultCount++;
}
else{
seniorCount++;
}
}
}
totalStudents = maleCount + femaleCount;
System.out.println(\"Total Students: \"+totalStudents);
System.out.println(\"Male: \"+maleCount);
System.out.println(\"Female: \"+femaleCount);
System.out.println(\"Minor: \"+minorCount);
System.out.println(\"Adult: \"+adultCount);
System.out.println(\"Senior: \"+seni.
Reverse EngineeringReverse engineering is also a back engineering .pdf
Reverse Engineering
Reverse engineering is also a back engineering which is extracting knowledge from manmade
and reproducing it from extracted information. Reverse engineering convert’s binary code of the
program and recreating it back to the original source code is known as reverse engineering.
Reverse engineering is common in both hardware and software. It convert’s the code written in
high level language into a low level language without changing the original program is also a
reverse engineering The process of recreating a design by analyzing a final product of a software
is also known as reverse engineering.
Reverse Engineering is used in many fields of Information Technology in form of Malware
Analysis, Network Analysis, debugging, and improvising existing algorithms, software
reusability
Reverse engineering is used in software design for to add new features to the existing software
with or without knowing the source code.
It help’s to study the virus and malware codeIt makes the system does not have any major
.vulnerabilities and security flaws.Some of the reverse engineering tools are debugger and
disassemblers.Reverse engineering is also beneficial for business owners as they can incorporate
advanced features into their software to meet the demands of the growing markets.
Solution
Reverse Engineering
Reverse engineering is also a back engineering which is extracting knowledge from manmade
and reproducing it from extracted information. Reverse engineering convert’s binary code of the
program and recreating it back to the original source code is known as reverse engineering.
Reverse engineering is common in both hardware and software. It convert’s the code written in
high level language into a low level language without changing the original program is also a
reverse engineering The process of recreating a design by analyzing a final product of a software
is also known as reverse engineering.
Reverse Engineering is used in many fields of Information Technology in form of Malware
Analysis, Network Analysis, debugging, and improvising existing algorithms, software
reusability
Reverse engineering is used in software design for to add new features to the existing software
with or without knowing the source code.
It help’s to study the virus and malware codeIt makes the system does not have any major
.vulnerabilities and security flaws.Some of the reverse engineering tools are debugger and
disassemblers.Reverse engineering is also beneficial for business owners as they can incorporate
advanced features into their software to meet the demands of the growing markets..
Quantum mechanics predicts the existence of what are usually called .pdf
Quantum mechanics predicts the existence of what are usually called \'\'zero-point\'\' energies for
the strong, the weak and the electromagnetic interactions, where \'\'zero-point\'\' refers to the
energy of the system at temperature T=0, or the lowest quantized energy level of a quantum
mechanical system. Although the term \'\'zero-point energy\'\' applies to all three of these
interactions in nature, customarily (and hereafter in this article) it is used in reference only to the
electromagnetic case.
In conventional quantum physics, the origin of zero-point energy is the Heisenberg uncertainty
principle, which states that, for a moving particle such as an electron, the more precisely one
measures the position, the less exact the best possible measurement of its momentum (mass
times velocity), and vice versa. The least possible uncertainty of position times momentum is
specified by Planck\'s constant, h. A parallel uncertainty exists between measurements involving
time and energy (and other so-called conjugate variables in quantum mechanics). This minimum
uncertainty is not due to any correctable flaws in measurement, but rather reflects an intrinsic
quantum fuzziness in the very nature of energy and matter springing from the wave nature of the
various quantum fields. This leads to the concept of zero-point energy.
Zero-point energy is the energy that remains when all other energy is removed from a system.
This behaviour is demonstrated by, for example, liquid helium. As the temperature is lowered to
absolute zero, helium remains a liquid, rather than freezing to a solid, owing to the irremovable
zero-point energy of its atomic motions. (Increasing the pressure to 25 atmospheres will cause
helium to freeze.)
A harmonic oscillator is a useful conceptual tool in physics. Classically a harmonic oscillator,
such as a mass on a spring, can always be brought to rest. However a quantum harmonic
oscillator does not permit this. A residual motion will always remain due to the requirements of
the Heisenberg uncertainty principle, resulting in a zero-point energy, equal to 1/2 hf, where f is
the oscillation frequency.
Electromagnetic radiation can be pictured as waves flowing through space at the speed of light.
The waves are not waves of anything substantive, but are ripples in a state of a theoretically
defined field. However these waves do carry energy (and momentum), and each wave has a
specific direction, frequency and polarization state. Each wave represents a \'\'propagating mode
of the electromagnetic field.\'\'
Each mode is equivalent to a harmonic oscillator and is thus subject to the Heisenberg
uncertainty principle. From this analogy, every mode of the field must have 1/2 hf as its average
minimum energy. That is a tiny amount of energy in each mode, but the number of modes is
enormous, and indeed increases per unit frequency interval as the square of the frequency. The
spectral energy density is determined by the density of modes times the.
Program Educational Objectives • To create competencies and opportun.pdf
Program Educational Objectives • To create competencies and opportunities for Higher
Education; • To create professional manpower skilled for the IT Industry; • To provide
laboratory practices with advanced FOSS Tools; • To provide inter-disciplinary opportunities; •
To provide opportunities of developing technical documents and presentation skills. • to provide
opportunities of industry-Institute interactions; • To develop opportunities to promote
Entrepreneurship and start-ups; • To nurture professional and social ethics. Program Objectives •
To expose students to the Systems and Applications Programming, Organizations and
Architectures; • To provide conceptual knowledge in the Computing domain; • To provide
interdisciplinary knowledge; • To expose students with advanced tools used in industry; • To
develop written and soft-skill competencies; • To develop team work experience of professionals
skills for IT Industry. Program Outcomes • To test applications with concepts and skills in the
domain subjects; • To demonstrate skills in programming techniques and technologies; • To
demonstrate oral and written skills for technical presentations and documentation; • To
demonstrate IT project as team work; • To demonstrate social and Professional ethical practices;
2 SAVITRIBAI PHULE PUNE UNIVERSITY BE (COMPUTER ENGINEERING)- 2012
COURSE STRUCTURE Term-I Subject Subject Teaching Scheme Examination Scheme Total
Code Marks Lect Tut Pract In PR/ OR/ End Sem TW TW Sem Asmnt Asmnt 410441 Design &
Analysis 03 — — 30 — — 70 100 of Algorithms 410442 Principles of Modern 04 — — 30 —
— 70 100 Compiler Design 410443 Smart System Design 03 — — 30 — — 70 100 and
Applications 410444 Elective-I 03 — — 30 — — 70 100 410445 Elective-II 03 — — 30 — —
70 100 410446 Computer — — 04 — 50 50 — 100 laboratory-I 410447 Computer — — 04 —
50 50 — 100 Laboratory-II 410448 Project — 02 — — 50 — — 50 Total 16 02 08 150 150 100
350 750 Term-II 410449 Software Design 03 — — 30 — — 70 100 Methodologies & Testing
410450 High 03 — — 30 — — 70 100 Performance Computing 410451 Elective-III 03 — — 30
— — 70 100 410452 Elective-IV 03 — — 30 — — 70 100 Open Elective 410453 Computer —
— 04 — 50 50 — 100 laboratory-III 410454 Computer — — 04 — 50 50 — 100 Laboratory-IV
410455 Project — 06 — — 50 100 — 150 Total 12 06 08 120 150 200 280 750
Solution
Program Educational Objectives • To create competencies and opportunities for Higher
Education; • To create professional manpower skilled for the IT Industry; • To provide
laboratory practices with advanced FOSS Tools; • To provide inter-disciplinary opportunities; •
To provide opportunities of developing technical documents and presentation skills. • to provide
opportunities of industry-Institute interactions; • To develop opportunities to promote
Entrepreneurship and start-ups; • To nurture professional and social ethics. Program Objectives •
To expose students to the Systems and Applications Programming, Organizations and
Architectures; • To.
P2O5 removes H2O from Alcohol to form Alkene O of Alcohol gives it.pdf
P2O5 removes H2O from Alcohol to form Alkene
O of Alcohol gives its lone pair to Higly e- defficient P of P2O5
O makes bond to P and breaks its bond to C atom
P also breaks its Double bond with O which now works as a base and abstract H atom from C
resulting in formation of an alkene
Solution
P2O5 removes H2O from Alcohol to form Alkene
O of Alcohol gives its lone pair to Higly e- defficient P of P2O5
O makes bond to P and breaks its bond to C atom
P also breaks its Double bond with O which now works as a base and abstract H atom from C
resulting in formation of an alkene.
Japanese colonial design in china korea and mongolia create an image.pdf
Japanese colonial design in china korea and mongolia create an image of utopian republic and
discards local context.
During the 1930s, Japanese colonial designed utopias in the newly conquered lands of China,
Korea, Mongolia, Thailand, Vietnam, Laos, Burma, the Philippines, and Indonesia.
Solution
Japanese colonial design in china korea and mongolia create an image of utopian republic and
discards local context.
During the 1930s, Japanese colonial designed utopias in the newly conquered lands of China,
Korea, Mongolia, Thailand, Vietnam, Laos, Burma, the Philippines, and Indonesia..
Is the measurement of the drug water solubility.Partitioning means.pdf
Is the measurement of the drug water solubility.
Partitioning means that the drug will be divided in parts between water and oil layer.
LogP is for hydrophilic or lipophilic. Lipophilicity tells about the compounds ability to dissolve
into lipohilic (non-aqueous) solutions. Lipophilicity is needed for the compounds to permeate
through the various biological membrane. Lipophilicity is typically measured as the compounds
distribution between non-aqueous (octanol) and aqueous (water) phase and the result is
expressed as a 10-base logarithm of the concentration ratios between these phases (partition
coefficient), logP.
P = [Co ]/[Cw]
LogP = Log[Co ]/[Cw].
LogP > 2 lipophilic drug.
LogP < 2 hydrophilic drug
LogP only applied for neutral compound
Low logP….. Low penetration to CNS
High logP….. Low water solubility…. Not suitable for oral administration
Solution
Is the measurement of the drug water solubility.
Partitioning means that the drug will be divided in parts between water and oil layer.
LogP is for hydrophilic or lipophilic. Lipophilicity tells about the compounds ability to dissolve
into lipohilic (non-aqueous) solutions. Lipophilicity is needed for the compounds to permeate
through the various biological membrane. Lipophilicity is typically measured as the compounds
distribution between non-aqueous (octanol) and aqueous (water) phase and the result is
expressed as a 10-base logarithm of the concentration ratios between these phases (partition
coefficient), logP.
P = [Co ]/[Cw]
LogP = Log[Co ]/[Cw].
LogP > 2 lipophilic drug.
LogP < 2 hydrophilic drug
LogP only applied for neutral compound
Low logP….. Low penetration to CNS
High logP….. Low water solubility…. Not suitable for oral administration.
Information systems Information systems are the combination of Hard.pdf
Information systems: Information systems are the combination of Hardware, Software and
telecommunication networks that people build and use to collect, create and distribute useful
data, typically in organizational settings.
Hardware refers to computer devices such as monitor, cpu etc.
Software refers to programming instructions to perform some activity
Telecommunication networks refers to set of equipment used to interconnect two or more
computers
People in organizations use information systems to process sales transactions, manage loan
applications, or help financial analysts decide where, when and how the investments can be
made. Product manager also use them to help decide when, where and how to market their
products and related serives. Production managers use them to help decide when and how to
manufacture products. Information systems also enable us in performing account transactions
through ATMs, communicate by live video with people living in other parts of the world and
purchasing anything in the global market.
Globalization: The integration of economies through out the world, enabled by technological
progress is called globalization. effects of globalization are in many ways such as greater
international movement of commodities, money, information and labour, as well as the
development of technologies, standards and processes to facilitate this movement. Specifically, a
more global and competitive world includes visible economic, cultural and technological
changes.
Changes in MIS because of Globalization
Globalization affects information systems in a lot of aspects like the use of internet by general
public in the world, global e-mail providers like hotmail and yahoo connects the whole world
together. Information systems has big role in globalization by influencing different cultures
through internet, where big economies and developed countries benefit the most out of this.
Globalization has revolutionized internal management. It has also made easier the interaction
between countries, regions and continents, thus contributing to profitability. It is the private
sector\'s philosophy that propelled efforts to utilize every means, including information
technology, to make companies survive, even the biggest and the most powerful company in the
world.
The role of the MIS in an organization can be compared to the role of heart in the body. The
information is the blood and MIS is the heart. In the body the heart plays the role of supplying
pure blood to all the elements of the body including the brain. The heart work faster and supplies
more blood when needed. It regulates and controls the incoming impure blood, processed it and
sends it to the destination in the quantity needed. It fulfills the needs of blood supply to human
body in normal course and also in crisis.
The MIS plays exactly the same role in the organization. The system ensures that an appropriate
data is collected from the various sources, processed and send further to all.
Hi,I have modified the Point.java file as per your requirement.P.pdf
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
@Test public void testXY()
{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}
Solution
Hi,
I have modified the Point.java file as per your requirement.
Point.java
public class Point
{
double x, y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
public double getRadius() { return Math.sqrt(x * x + y * y); }
public double getAngle() { return Math.atan2(y, x); }
}
import org.junit.*;
import static org.junit.Assert.*;
public class PointTest
{
private static final double TOLERANCE = 1E-12;
@Test public void testNoArg()
{
Point p = new Point();
assertEquals(0, p.getX(), TOLERANCE);
assertEquals(0, p.getY(), TOLERANCE);
assertEquals(0, p.getRadius(), TOLERANCE);
assertEquals(0, p.getAngle(), TOLERANCE);
}
@Test public void testXY()
{
Point p = new Point(3, 4);
assertEquals(3, p.getX(), TOLERANCE);
assertEquals(4, p.getY(), TOLERANCE);
assertEquals(5, p.getRadius(), TOLERANCE);
assertEquals(0.9272952180016122, p.getAngle(), TOLERANCE);
}
}.
four types of eolian landforms are1)erosinal landforms2)depositi.pdf
four types of eolian landforms are
1)erosinal landforms
2)depositional landforms
3)sand dune landforms
4)desert dunes
erosional landforms:whenever wind force is concentrated over a area it will create a hollow pit
namely called deflation hollow.it will range a diameter of hundred meters.these broad shallow
depressions are called as pans among which qattara is the largest pan present in lybian desert of
egypt.
deposional landforms:these landforms are caused by the deposition of the sediment.these types
of largest dune fields are found in middle east and north africa.
sand dunes:this will occur in the form of deposition of sand.
desert dunes:this will occur in amazing diversity of forms.
Solution
four types of eolian landforms are
1)erosinal landforms
2)depositional landforms
3)sand dune landforms
4)desert dunes
erosional landforms:whenever wind force is concentrated over a area it will create a hollow pit
namely called deflation hollow.it will range a diameter of hundred meters.these broad shallow
depressions are called as pans among which qattara is the largest pan present in lybian desert of
egypt.
deposional landforms:these landforms are caused by the deposition of the sediment.these types
of largest dune fields are found in middle east and north africa.
sand dunes:this will occur in the form of deposition of sand.
desert dunes:this will occur in amazing diversity of forms..
As Al+3 has noble gas configuration. the IE3 of A.pdf
As Al+3 has noble gas configuration. the IE3 of Al is very low As Li has 3 e- only.
Li+3 will be having no e-.hence a higgh amount of energy is needed. lowest= al highest=Li
Solution
As Al+3 has noble gas configuration. the IE3 of Al is very low As Li has 3 e- only.
Li+3 will be having no e-.hence a higgh amount of energy is needed. lowest= al highest=Li.
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