1. E) The vapor pressure of a liquid.Increasing attractive intermo.pdf
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1. E) The vapor pressure of a liquid. Increasing attractive intermolecular forces causes molecules in the liquid to bond more strongly to each other. Thus it is more difficult for the molecules to escape into the gas phase and the vapor pressure decreases. All the other properties listed should increase in value. 2. A) Ba(OH)2 The light bulb glows because H2SO4 is an acid and strong electrolyte that ionizes to form H+ and SO42- ions that conduct electricity. Ba(OH)2 is a base that reacts with H2SO4 to form water and insoluble BaSO4: Ba(OH)2(aq) + H2SO4(aq) => BaSO4(s) + H2O(l) The removal of H+ and SO42-ions decreases the conductivity of the solution, less current flows and the bulb dims. 3. D) hydration Hydration is the process whereby water molecules form dipole-dipole interactions with the solute ions, releasing energy and causing the salt to fall apart. 4. E)All of these are correct. All the statements about light, matter and energy are true. 5. d) 780 kJ, 1,575 kJ, 3,220 kJ, 4,350 kJ 16,100 kJ (you may have mistyped this in your question but I have put the correct values here for you) Silicon has electronic configuration 1s2 2s2p6 3s2p2 The first 4 electrons removed are in the n = 3 shell and have similar (but slowly increasing) ionization energies. The 5th electron is from the n = 2 shell and is much more strongly bound to the nucleus, so the ionization energy shows a large increase. Solution 1. E) The vapor pressure of a liquid. Increasing attractive intermolecular forces causes molecules in the liquid to bond more strongly to each other. Thus it is more difficult for the molecules to escape into the gas phase and the vapor pressure decreases. All the other properties listed should increase in value. 2. A) Ba(OH)2 The light bulb glows because H2SO4 is an acid and strong electrolyte that ionizes to form H+ and SO42- ions that conduct electricity. Ba(OH)2 is a base that reacts with H2SO4 to form water and insoluble BaSO4: Ba(OH)2(aq) + H2SO4(aq) => BaSO4(s) + H2O(l) The removal of H+ and SO42-ions decreases the conductivity of the solution, less current flows and the bulb dims. 3. D) hydration Hydration is the process whereby water molecules form dipole-dipole interactions with the solute ions, releasing energy and causing the salt to fall apart. 4. E)All of these are correct. All the statements about light, matter and energy are true. 5. d) 780 kJ, 1,575 kJ, 3,220 kJ, 4,350 kJ 16,100 kJ (you may have mistyped this in your question but I have put the correct values here for you) Silicon has electronic configuration 1s2 2s2p6 3s2p2 The first 4 electrons removed are in the n = 3 shell and have similar (but slowly increasing) ionization energies. The 5th electron is from the n = 2 shell and is much more strongly bound to the nucleus, so the ionization energy shows a large increase..
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Light is a form of energy that can be released by.pdf
Light is a form of energy that can be released by an atom. It is made up of many
small particle-like packets that have energy and momentum but no mass. These particles, called
light photons, are the most basic units of light. Atoms release light photons when their electrons
become excited. When an atom gains or loses energy, the change is expressed by the movement
of electrons. When something passes energy on to an atom, an electron may be temporarily
boosted to a higher orbital (farther away from the nucleus). The electron only holds this position
for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its
original orbital. As it returns to its original orbital, the electron releases the extra energy in the
form of a photon, in some cases a light photon.
Solution
Light is a form of energy that can be released by an atom. It is made up of many
small particle-like packets that have energy and momentum but no mass. These particles, called
light photons, are the most basic units of light. Atoms release light photons when their electrons
become excited. When an atom gains or loses energy, the change is expressed by the movement
of electrons. When something passes energy on to an atom, an electron may be temporarily
boosted to a higher orbital (farther away from the nucleus). The electron only holds this position
for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its
original orbital. As it returns to its original orbital, the electron releases the extra energy in the
form of a photon, in some cases a light photon..
1) Presence of jaws with paired fins in fish helps to feed them. Fin.pdf
1) Presence of jaws with paired fins in fish helps to feed them. Fins helps them with increased
stability and maneuverability and jaws helps to capture the prey and seize it.
2) Fish with jaws need to regulate their body temperature so they need to feed themselves often.
It capture the prey and send it to stomach for digestion and for temporary storage. Jawless fish
donot regulate their own body temperature so they don\'t need temporary storage like stomach
and also they have sucking mouth and sharp teeth which helps in digestion without the need for
pre digestion with gastric juice. So the fish which needs more food has jaw (Capture large prey)
with stomach (Temporary storage and pre digestion). The fish which needs less food has no jaw
and no stomach but with sucking mouth and also they have no fins. Jawless fish examples :
Lampreys and hagfish. Majority of the fishes are with jaws.
3) Two chambered heart have one atrium and one ventricle and three chambered heart have two
atrium and one ventricle. There are two types of circulatory system that is single circulation
system (fish) and double circulation system (reptiles, mammals)
In single circulation system, blood from the parts of the body enters heart through atrium and
then ventricle pumps the blood to gills where re-oxygenation takes place. From the gills, blood is
sent back to body parts. Blood enters the heart only once that is only deoxygenated blood enters.
In double circulation system, blood enters the heart twice that is from parts of the body to heart
then from heart to pulmonary circulation (Lungs) and from lungs to heart then heart to parts of
the body. Both deoxygenated and oxygenated blood enters the heart.
Solution
1) Presence of jaws with paired fins in fish helps to feed them. Fins helps them with increased
stability and maneuverability and jaws helps to capture the prey and seize it.
2) Fish with jaws need to regulate their body temperature so they need to feed themselves often.
It capture the prey and send it to stomach for digestion and for temporary storage. Jawless fish
donot regulate their own body temperature so they don\'t need temporary storage like stomach
and also they have sucking mouth and sharp teeth which helps in digestion without the need for
pre digestion with gastric juice. So the fish which needs more food has jaw (Capture large prey)
with stomach (Temporary storage and pre digestion). The fish which needs less food has no jaw
and no stomach but with sucking mouth and also they have no fins. Jawless fish examples :
Lampreys and hagfish. Majority of the fishes are with jaws.
3) Two chambered heart have one atrium and one ventricle and three chambered heart have two
atrium and one ventricle. There are two types of circulatory system that is single circulation
system (fish) and double circulation system (reptiles, mammals)
In single circulation system, blood from the parts of the body enters heart through atrium and
then ventricle pumps the blood to gills .
Youre looking for Ka, the dissociation constant.pdf
You\'re looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.
Solution
You\'re looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate..
Okay so dipole-dipole interactions would occur be.pdf
Okay so dipole-dipole interactions would occur between polar molecules such as
water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces)
would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar
part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient
charge among the different atoms. I hope that helps
Solution
Okay so dipole-dipole interactions would occur between polar molecules such as
water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces)
would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar
part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient
charge among the different atoms. I hope that helps.
remote operating system installationRemote Installation Services (.pdf
remote operating system installation
Remote Installation Services (RIS) ships as a part of the Windows® 2000 Server software
system. This document outlines the steps necessary to put in, configure, and use RIS.
RIS was designed to cut back the prices incurred by pre-installing or physically visiting every
consumer pc to put in the software system (OS). By combining RIS with alternative Windows
2000 IntelliMirror™ management technologies features—User knowledge Management,
software package Installation and Maintenance, and User Settings Management—companies get
pleasure from higher disaster recovery with easier OS and application management.
Prerequisites
See Appendix A below to confirm that each your server and consumer hardware meet the
remote installation hardware necessities.
The drive on the server wherever you decide on to put in RIS should be formatted with the
NTFS filing system. RIS needs a big quantity of space and can\'t be put in on constant drive or
partition on that Windows 2000 Server is put in. make sure that the chosen drive contains enough
free space for a minimum of one full Windows 2000 skilled compact disc—a minimum of
roughly 800 megabytes (MB)–1 GB (GB).
RIS needs many alternative services that conjointly ship as a part of Windows 2000 Server.
These services will be put in on individual servers, or all on one server, counting on your
network design:
Domain Name Service (DNS) Server. Remote installation depends on DNS for locating the
directory service and consumer machine accounts. you\'ll use any Windows 2000 Active
DirectoryTM service-compliant DNS server, otherwise you will use the DNS server given
Windows 2000 Server.
Dynamic Host Configuration Protocol (DHCP) Server. RIS needs a DHCP server to be gift and
active on the network. The remote boot-enabled consumer computers receive AN information
processing address from the DHCP server before contacting RIS.
Active Directory. RIS depends on Windows 2000 Active Directory for locating existing
consumer machines further as existing RIS servers. RIS should be put in on a Windows 2000-
based server that has access to Active Directory. this will be a site controller or a server that\'s a
member of a site with access to the Active Directory.
To ensure a in installation, you want to install and tack together the extra services delineate
higher than so as for RIS to perform. additionally, certify that you just have each the Windows
2000 Server and skilled CDs on the market.
Before starting this gradual guide, you wish to create the common infrastructure, that specifies a
specific hardware and software package configuration. The common infrastructure is roofed
within the Common Infrastructure gradual guide, \"Part 1: putting in a Windows 2000 Server as
a site Controller\"
http://www.microsoft.com/windows2000/techinfo/planning/server/serversteps.asp. If you\'re not
victimization the common infrastructure, you wish to create the suitable changes to the current
instruction set.
Alth.
Title of this process is The Project Life Cycle (Phases)The pr.pdf
Title of this process is \"The Project Life Cycle (Phases)\"
The project manager and project team have one shared goal: to carry out the work of the project
for the purpose of meeting the project’s objectives. Every project has a beginning, a middle
period during which activities move the project toward completion, and an ending.
A standard project typically has the following four major phases: initiation, planning,
implementation, and closure. Taken together, these phases represent the path a project takes from
the beginning to its end and are generally referred to as the project “life cycle.”
Initiation Phase
During the first phases, the initiation phase, the project objective is identified; this can be a
business problem or opportunity. An appropriate response to the objective of the project is
documented with recommended solution options. A feasibility study is conducted to investigate
whether each option addresses the project objective and a final recommended solution is
determined. Issues of feasibility (“can we do the project?”) and justification (“should we do the
project?”) are addressed.
Once the recommended solution is approved, a project is initiated to deliver the approved
solution and a project manager is appointed. Approval is then sought by the project manager to
move onto the detailed planning phase.
Planning Phase
The planning phase, is where the project solution is further developed in as much detail as
possible and the steps necessary to meet the project’s objective are planned. In this step, the team
identifies all of the work to be done. The project’s tasks and resource requirements are identified,
along with the strategy for producing them.
A project plan is created outlining the activities, tasks, dependencies, and timeframes. The
project manager coordinates the preparation of a project budget by providing cost estimates for
the labor, equipment, and materials costs. The budget is used to monitor and control cost
expenditures during project implementation.
Implementation (Execution) Phase
During the third phase, the implementation phase, the project plan is put into motion and the
work of the project is performed. It is important to maintain control and communicate as needed
during implementation. Progress is continuously monitored and appropriate adjustments are
made and recorded as variances from the original plan.
Throughout this step, project sponsors and other key stakeholders should be kept informed of the
project’s status according to the agreed-on frequency and format of communication. The plan
should be updated and published on a regular basis.
Status reports should always emphasize the anticipated end point in terms of cost, schedule, and
quality of deliverables. Each project deliverable produced should be reviewed for quality and
measured against the acceptance criteria. Once all of the deliverables have been produced and
the customer has accepted the final solution, the project is ready for closure.
Closing.
The answer is A) Pol II, with twelve subunits on its own, is capable.pdf
The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription.
because it is composed with 12 subunits and due to incapable of specific promoter recognition it
can\'t initiate transcription.
Solution
The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription.
because it is composed with 12 subunits and due to incapable of specific promoter recognition it
can\'t initiate transcription..
solutionA= 1 1 0 4 3 1 2 0 .pdf
solution:
A=
1
1
0
4
3
1
2
0
5
Finding the determinant of the matrix ?
det A = -3
The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse
matrix was calculated the minors and cofactors of the matrix ?
M1,1 =
3
1
0
5
=
15
C1,1 = (-1)1+1M1,1 =
15
M1,2 =
4
1
2
5
=
18
C1,2 = (-1)1+2M1,2 =
-18
M1,3 =
4
3
2
0
=
-6
C1,3 = (-1)1+3M1,3 =
-6
M2,1 =
1
0
0
5
=
5
C2,1 = (-1)2+1M2,1 =
-5
M2,2 =
1
0
2
5
=
5
C2,2 = (-1)2+2M2,2 =
5
M2,3 =
1
1
2
0
=
-2
C2,3 = (-1)2+3M2,3 =
2
M3,1 =
1
0
3
1
=
1
C3,1 = (-1)3+1M3,1 =
1
M3,2 =
1
0
4
1
=
1
C3,2 = (-1)3+2M3,2 =
-1
M3,3 =
1
1
4
3
=
-1
C3,3 = (-1)3+3M3,3 =
-1
Matrix of cofactors:
C =
15
-18
-6
-5
5
2
1
-1
-1
Transpose of cofactor matrix:
CT =
15
-5
1
-18
5
-1
-6
2
-1
Finding the inverse matrix.
A-1 =
CT
=
det A
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
B)
First, let\'s add the Identity Matrix to the right of our matrix
1
1
0
1
0
0
4
3
1
0
1
0
2
0
5
0
0
1
Now, let\'s do Gauss-Jordan Elimination on our new matrix...
Add (-4 * row1) to row2
1
1
0
1
0
0
0
-1
1
-4
1
0
2
0
5
0
0
1
Add (-2 * row1) to row3
1
1
0
1
0
0
0
-1
1
-4
1
0
0
-2
5
-2
0
1
Divide row2 by -1
1
1
0
1
0
0
0
1
-1
4
-1
0
0
-2
5
-2
0
1
Add (2 * row2) to row3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
3
6
-2
1
Divide row3 by 3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
1
2
-2/3
1/3
Add (1 * row3) to row2
1
1
0
1
0
0
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
Add (-1 * row2) to row1
1
0
0
-5
5/3
-1/3
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix
Here is the inverse matrix:
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
A=
1
1
0
4
3
1
2
0
5
Solution
solution:
A=
1
1
0
4
3
1
2
0
5
Finding the determinant of the matrix ?
det A = -3
The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse
matrix was calculated the minors and cofactors of the matrix ?
M1,1 =
3
1
0
5
=
15
C1,1 = (-1)1+1M1,1 =
15
M1,2 =
4
1
2
5
=
18
C1,2 = (-1)1+2M1,2 =
-18
M1,3 =
4
3
2
0
=
-6
C1,3 = (-1)1+3M1,3 =
-6
M2,1 =
1
0
0
5
=
5
C2,1 = (-1)2+1M2,1 =
-5
M2,2 =
1
0
2
5
=
5
C2,2 = (-1)2+2M2,2 =
5
M2,3 =
1
1
2
0
=
-2
C2,3 = (-1)2+3M2,3 =
2
M3,1 =
1
0
3
1
=
1
C3,1 = (-1)3+1M3,1 =
1
M3,2 =
1
0
4
1
=
1
C3,2 = (-1)3+2M3,2 =
-1
M3,3 =
1
1
4
3
=
-1
C3,3 = (-1)3+3M3,3 =
-1
Matrix of cofactors:
C =
15
-18
-6
-5
5
2
1
-1
-1
Transpose of cofactor matrix:
CT =
15
-5
1
-18
5
-1
-6
2
-1
Finding the inverse matrix.
A-1 =
CT
=
det A
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
B)
First, let\'s add the Identity Matrix to the right of our matrix
1
1
0
1
0
0
4
3
1
0
1
0
2
0
5
0
0
1
Now, let\'s do Gauss-Jordan Elimination on our new matrix...
Add (-4 * row1) to row2
1
1
0
1
0
0
0
-1
1
-4
1
0
2
0
5
0
0
1
Add (-2 * row1) to row3
1
1
0
1
0
0
0
-1
1
-4
1
0
0
-2
5
-2
0
1
Divide row2 by -1
1
1
0
1
0
0
0
1
-1
4
-1
0
0
-2
5
-2
0
1
Add (2 * row2) to row3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
3
6
-2
1
Divide row3 by 3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
1
2
-2/3
1/3
Add (1 * row3) to row2
1
1
0
1
0
0
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
Add (-1 * row2) to row1
1.
Q1). Gene therapy is an experimental approach to treat the disease b.pdf
Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes
or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these
therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated
gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular
dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune
deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active
investigation.
All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy
include,
1). Diseases with known genetic defects or mutations
2). Diseases with point mutations on one or more genes
3). The possibility of delivering genes to the affected tissues
4). If the substitution of defective gene correct the disease status
Solution
Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes
or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these
therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated
gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular
dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune
deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active
investigation.
All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy
include,
1). Diseases with known genetic defects or mutations
2). Diseases with point mutations on one or more genes
3). The possibility of delivering genes to the affected tissues
4). If the substitution of defective gene correct the disease status.
Solution Three modes of DNA replication 1) Semi conservative.pdf
Solution
:
Three modes of DNA replication
1) Semi conservative:
=> Each daughter strand is composed of one old DNA strand and one new DNA strand.
2) Conservative:
=> The original double helix remains as complete unit.
=> The new DNA double helix is produced as a single unit.
=> The old DNA is completly conserved.
3) Dispersive mode:
=>Separation of the individual covalent phosphodiester bonds is required for this mode of
replication..
package s3; Copy paste this Java Template and save it as Emer.pdf
package s3;
// Copy paste this Java Template and save it as \"EmergencyRoom.java\"
import java.util.*;
import java.io.*;
// write your matric number here:
// write your name here:
// write list of collaborators here:
// year 2016 hash code: XAbyuzR78fXeaMHBdLan (do NOT delete this line)
class patientlist {
private String PatientName;
private int emergencyLvl;
public patientlist(String PatientName, int emergencyLvl) {
this.PatientName = PatientName;
this.emergencyLvl = emergencyLvl;
}
public String getPatientName() {
return PatientName;
}
public void setPatientName(String PatientName) {
this.PatientName = PatientName;
}
public int getEmergencyLvl() {
return emergencyLvl;
}
public void setEmergencyLvl(int emergencyLvl) {
this.emergencyLvl = emergencyLvl;
}
}
class EmergencyRoom {
// if needed, declare a private data structure here that
// is accessible to all methods in this class
private List patientlist;
public EmergencyRoom() {
// Write necessary code during construction
patientlist = new ArrayList();
}
void ArriveAtHospital(String patientName, int emergencyLvl) {
// You have to insert the information (patientName, emergencyLvl)
// into your chosen data structure
if (patientName.length() > 15 || patientName.length() < 1) {
System.out.println(\"patient name is either too long or too short.\ Please enter a name between 1
to 15 characters.\");
} else if (emergencyLvl > 100 || emergencyLvl < 30) {
System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid level
between 30 to 100.\");
} else {
int i;
for (i = 0; i < patientlist.size(); i++) {
if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) {
System.out.println(\"Patient already admitted\");
break;
}
}
if (i == patientlist.size()) {
patientlist p = new patientlist(patientName.toUpperCase(), emergencyLvl);
patientlist.add(p);
}
}
}
void UpdateEmergencyLvl(String patientName, int incEmergencyLvl) {
// You have to update the emergencyLvl of patientName to
// emergencyLvl += incEmergencyLvl
// and modify your chosen data structure (if needed)
int i;
for (i = 0; i < patientlist.size(); i++) {
if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) {
//System.out.println(\"Patient already admitted\");
//break;
if (incEmergencyLvl > 70 || incEmergencyLvl < 0) {
System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid
increment between 0 to 70.\");
break;
} else {
patientlist.get(i).setEmergencyLvl(patientlist.get(i).getEmergencyLvl() + incEmergencyLvl);
}
}
}
}
void Treat(String patientName) {
// This patientName is treated by the doctor
// remove him/her from your chosen data structure
int i;
for (i = 0; i < patientlist.size(); i++) {
if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) {
patientlist.remove(i);
}
}
}
String Query() {
String ans = \"The emergency room is empty\";
int i;
if(patientlist.isEmpty())
return ans;
else{
int max=patientlist.get(0).getEme.
Our body is having two line defence system against pathogens.Pathoge.pdf
Our body is having two line defence system against pathogens.Pathogens are micro-organisms
such like virus,bactria,parasite,fungi.
First line of defence system:- It is a outside defence system which protect body from the outside
,it includes skin ,tear,mucus,cilia,stomach acid,urine flow.
Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to
your body unless your skin get brokken.The pathogens can get enter through your nose
,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop
pathogen and break down their cell wall because they contain enzyme.those that are not killed
immediately are trapped in mucus swallowed.
cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine
flow help to flush out pathogens from the bladder.stomach acid is present in stomach and helps
to kill pathogens.this is the fine defence.
Second line of defence system :- once pathogen cross your first line of defence second line of
defence active for example ,through the cut pathogen enter into body and develope infection then
second line defence active .Through a sequence of steps called immune response ,the immune
system attacks these pathogens.
In the second line of deffence system is a group of cells ,tissues,organs work together to protect
the body .This is called immune system.
The cells involved in the immune system called white blood cells is helps to destoys the desease
causing organisms.there are different types of white blood cells but function is same that protect
you from the organisms.
They are neutrophiles,T-helper cells,Cytotoxic cells,marcophases,B-cells.
These cells work together and kill the micro organism.
Tissues and Organs :- This are involved in the immune system are the lymphatic system ,lymph
nodes and lymp fluid.This are having specific function.
Lymphatic system is a system of thin tube present through out the body .These tubes are called
lymph vessels.they contain lymph.
Lymph is a fluid that contains white blood cells. lymph helps to protect body form micro
organisms.
This is a second line of defence system.
Solution
Our body is having two line defence system against pathogens.Pathogens are micro-organisms
such like virus,bactria,parasite,fungi.
First line of defence system:- It is a outside defence system which protect body from the outside
,it includes skin ,tear,mucus,cilia,stomach acid,urine flow.
Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to
your body unless your skin get brokken.The pathogens can get enter through your nose
,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop
pathogen and break down their cell wall because they contain enzyme.those that are not killed
immediately are trapped in mucus swallowed.
cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine
flow help to flush out.
Once neurons are produced, they migrate and modify to form six layer.pdf
Once neurons are produced, they migrate and modify to form six layered structure of cortex. The
earliest formed neurons form the preplate which is then split into marginal plate and subplate by
invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and
ganglionic eminences. The corticular neurons exhibit radial migration and form projection
neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both
types arrange in inside out sequence and the early neurons form the deep layers while the late
born neuron form the outer layers of cortex.
Solution
Once neurons are produced, they migrate and modify to form six layered structure of cortex. The
earliest formed neurons form the preplate which is then split into marginal plate and subplate by
invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and
ganglionic eminences. The corticular neurons exhibit radial migration and form projection
neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both
types arrange in inside out sequence and the early neurons form the deep layers while the late
born neuron form the outer layers of cortex..
PDU is called Protocol Data Unit.It consists of user data and protoc.pdf
PDU is called Protocol Data Unit.It consists of user data and protocol control information. PDU
at layer 4 i.e Networks layer is called Segment.
Solution
PDU is called Protocol Data Unit.It consists of user data and protocol control information. PDU
at layer 4 i.e Networks layer is called Segment..
1.A hole is the absence of an electron in a particular place in an a.pdf
1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not
move as rapidly through the solid materials as electrons do in a vaccum.The holes move more
slowly than the electrons.The holes and electrons move in opposite direction.
2.Here if the electric current moving from left to right ,than the holes will be moving in opposite
direction that is from right to left.
Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both
electrons and holes contribute to total current flow,altough they are moving in opposite
directions,they are of opposite charge.
Solution
1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not
move as rapidly through the solid materials as electrons do in a vaccum.The holes move more
slowly than the electrons.The holes and electrons move in opposite direction.
2.Here if the electric current moving from left to right ,than the holes will be moving in opposite
direction that is from right to left.
Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both
electrons and holes contribute to total current flow,altough they are moving in opposite
directions,they are of opposite charge..
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- 2. Ba(OH)2(aq) + H2SO4(aq) => BaSO4(s) + H2O(l) The removal of H+ and SO42-ions decreases the conductivity of the solution, less current flows and the bulb dims. 3. D) hydration Hydration is the process whereby water molecules form dipole-dipole interactions with the solute ions, releasing energy and causing the salt to fall apart. 4. E)All of these are correct. All the statements about light, matter and energy are true. 5. d) 780 kJ, 1,575 kJ, 3,220 kJ, 4,350 kJ 16,100 kJ (you may have mistyped this in your question but I have put the correct values here for you) Silicon has electronic configuration 1s2 2s2p6 3s2p2 The first 4 electrons removed are in the n = 3 shell and have similar (but slowly increasing) ionization energies. The 5th electron is from the n = 2 shell and is much more strongly bound to the nucleus, so the ionization energy shows a large increase.