B. The two primary components of a DAC are file and data ownership, .pdf
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B. The two primary components of a DAC are file and data ownership, and access rights and permissions. With file and data ownership, all objects within a system must have an owner. Objects without an owner will be left unprotected. Access rights and permissions control the access rights of an individual. Variation exists, but a basic access control list checks read, write, or execute privileges. Answers A, C, and D are incorrect. Solution B. The two primary components of a DAC are file and data ownership, and access rights and permissions. With file and data ownership, all objects within a system must have an owner. Objects without an owner will be left unprotected. Access rights and permissions control the access rights of an individual. Variation exists, but a basic access control list checks read, write, or execute privileges. Answers A, C, and D are incorrect..
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DnaAORC This protein is responsible for initiation of DNA replicat.pdf
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined.
Solution
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined..
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s
Solution
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s.
Answer Q1. Red mangroove is the last mangroove to colonise on the.pdf
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary
Solution
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary.
An Arrhenius acid produces H+ ions insolutionA Bronsted-Lowry acid.pdf
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq)
Solution
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq).
Ans. One of the main disadvantage is chromatic aberrasion. A more li.pdf
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information.
Solution
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information..
A coordination problem occurs in situations where there is no Nash E.pdf
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer.
Solution
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer..
1. When an integrated F plasmid leaves the chromosome it may take so.pdf
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens.
Solution
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens..
1. A DNA molecule contains two strands of poly deoxyribonucleotide s.pdf
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) .
Solution
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) ..
Recommended
DnaAORC This protein is responsible for initiation of DNA replicat.pdf
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined.
Solution
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined..
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s
Solution
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s.
Answer Q1. Red mangroove is the last mangroove to colonise on the.pdf
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary
Solution
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary.
An Arrhenius acid produces H+ ions insolutionA Bronsted-Lowry acid.pdf
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq)
Solution
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq).
Ans. One of the main disadvantage is chromatic aberrasion. A more li.pdf
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information.
Solution
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information..
A coordination problem occurs in situations where there is no Nash E.pdf
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer.
Solution
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer..
1. When an integrated F plasmid leaves the chromosome it may take so.pdf
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens.
Solution
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens..
1. A DNA molecule contains two strands of poly deoxyribonucleotide s.pdf
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) .
Solution
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) ..
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neut.pdf
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble.
Solution
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble..
draw a sphere and use raytracing on the sphere in OpenGL glut. .pdf
/* draw a sphere and use raytracing on the sphere in OpenGL glut. */
#include
#include
#include
#include
#include
#define screen_width 750
#define screen_height 750
#define true 1
#define false 0
#define perpendicular 0
int gridXsize = 20;
int gridZsize = 20;
float plane[] = {0.0, 1.0, 0.0, -50.0,};
float sphere[] = {250.0, 270.0, -100.0, 100.0};
float eye[] = {0.0, 400.0, 550.0};
float light[] = {250.0, 550.0, -200.0};
float dot(float *u, float *v)
{
return u[0]*v[0] + u[1]*v[1] + u[2]*v[2];
}
void norm(float *u)
{
float norm = sqrt(abs(dot(u,u)));
for (int i =0; i <3; i++)
{
u[i] = u[i]/norm;
}
}
float plane_intersect(float *u, float *pO)
{
float normt[3] = {plane[0], plane[1], plane[2]};
float s;
if (dot(u,normt) == 0)
{
s = -10;
}
else
{
s = (plane[3]-(dot(pO,normt)))/(dot(u,normt));
}
return s;
}
float sphere_intersect(float *u, float *pO)
{
float deltaP[3] = {sphere[0]-pO[0],sphere[1]-pO[1],sphere[2]-pO[2]};
float deltLen = sqrt(abs(dot(deltaP,deltaP)));
float t=0;
float answer;
float det;
if ((det =(abs(dot(u,deltaP)*dot(u,deltaP))- (deltLen*deltLen)+sphere[3]*sphere[3])) < 0)
{
answer = -10;
}
else
{
t =-1*dot(u,deltaP)- sqrt(det) ;
if (t>0)
{
answer = t;
}
else
{
answer = -10;
}
}
return answer;
}
void find_reflect(float *u, float s, float *pO)
{
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]- sphere[2]};
float l[3] = {s *u[0],s *u[1],s *u[2]};
u[0] =(2*dot(l,n)*n[0])-l[0];
u[1] = (2*dot(l,n)*n[1])-l[1];
u[2] = (2*dot(l,n)*n[2])-l[2];
}
float find_shade(float *u,float s, float *pO)
{
float answer;
float lightVec[3] = {light[0]-(pO[0]+s *u[0]), light[1]-(pO[1]+s *u[1]), light[2]-(pO[2]+s
*u[2])};
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]-sphere[2]};
answer = -1*dot(lightVec,n)/(sqrt(abs(dot(lightVec,lightVec)))*sqrt(abs(dot(n,n))));
return answer;
}
void init()
{
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluOrtho2D(0,screen_width,0,screen_height);
}
void display()
{
glClear(GL_COLOR_BUFFER_BIT| GL_DEPTH_BUFFER_BIT);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
for (int i=0; i < screen_width; i++)
{
for (int j=0; j < screen_height; j++)
{
float ray[3] = {1*(eye[0]-i),-1*(eye[1]-j),1*eye[2]};
float point[3] = {i,j,0};
norm(ray);
int plotted = false;
while (!plotted)
{
float s_plane = plane_intersect(ray, point);
float s_sphere = sphere_intersect(ray, point);
if (s_plane <= 0 && s_sphere <=0)
{
glColor3f(0,0,0);
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
else if (s_sphere >= 0 && (s_plane <=0 || s_sphere <= s_plane))
{
find_reflect(ray, s_sphere, point);
}
else if (s_plane >=0 && (s_sphere <=0 ||s_plane <= s_sphere))
{
float shade = find_shade(ray, s_plane, point);
float xx = s_plane*ray[0] + eye[0];
float z = s_plane*ray[2] + eye[2];
if (abs((int)xx/gridXsize)%2 == abs((int)z/gridZsize)%2)
{
glColor3f(shade,0,0);
}
else
{
glColor3f(shade,shade,shade);
}
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
}
}
}
glFlush();
}
int main(.
SnO2 (ore) is initially washed and dressed. then .pdf
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form
Solution
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form.
d) Valine Note its side chain doesnt contain a.pdf
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens.
Solution
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens..
Amphiprotic. The preferred term is amphoteric. A .pdf
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ and HSO4-. The H+ will combine
with OH- to make water, and you\'re right back where you started. No. HSO4- is NOT
amphoteric in aqueous solution. ======= More follow up ========== \"Would it matter
about the physical state?\" No. For the substance to gain a proton, it will be in aqueous solution.
As soon as HCl interacts with water it will dissociate to form H+ and Cl-.
=============================== You may see that HSO4- is listed as an amphoteric
substance, but only because whoever wrote that didn\'t think it through.
Solution
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ a.
at equivalence point, Moles OH- added = moles H+.pdf
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases
Solution
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases.
We Know that Water is a Polar component as .pdf
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline.
Solution
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline..
molar mass of C2H5OH = 46.1 gmole moles of C2H5O.pdf
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules
Solution
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules.
What property of PN code makes them suitable for use in spread spect.pdf
What property of PN code makes them suitable for use in spread spectrum communication?
Solution
Pseudo random noise is used in some electronic musical instruments, either by itself or as an
input to subtractive synthesis, and in many white noise machines.
In spread-spectrum systems, the receiver correlates a locally generated signal with the received
signal. Such spread-spectrum systems require a set of one or more \"codes\" or \"sequences\"
such that
Like random noise, the local sequence has a very low correlation with any other sequence in the
set, or with the same sequence at a significantly different time offset, or with narrow band
interference, or with thermal noise.
Unlike random noise, it must be easy to generate exactly the same sequence at both the
transmitter and the receiver, so the receiver\'s locally generated sequence has a very high
correlation with the transmitted sequence.
In a direct-sequence spread spectrum system, each bit in the pseudorandom binary sequence is
known as a chip and the inverse of its period as chip rate; comparebit rate and symbol rate.
In a frequency-hopping spread spectrum sequence, each value in the pseudo random sequence is
known as a channel number and the inverse of its period as thehop rate. FCC Part 15 mandates at
least 50 different channels and at least a 2.5 Hz hop rate for narrow band frequency-hopping
systems.
GPS satellites broadcast data at a rate of 50 data bits per second – each satellite modulates its
data with one PN bit stream at 1.023 million chips per second and the same data with another PN
bit stream at 10.23 million chips per second. GPS receivers correlate the received PN bit stream
with a local reference to measure distance. GPS is a receive-only system that uses relative timing
measurements from several satellites (and the known positions of the satellites) to determine
receiver position.
Other range-finding applications involve two-way transmissions. A local station generates a
pseudo random bit sequence and transmits it to the remote location (using any modulation
technique). Some object at the remote location echoes this PN signal back to the location station
– either passively, as in some kinds of radar and sonar systems, or using an active transponder at
the remote location, as in the Apollo Unified S-band system.[4] By correlating a (delayed version
of) the transmitted signal with the received signal, a precise round trip time to the remote
location can be determined and thus the distance..
TLC can be used to help determine the number of components in a mixt.pdf
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container.
Solution
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container..
There may have some variables affect the model. So the error term wi.pdf
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation.
Solution
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation..
There are four major functional characteristics of skeletal muscle..pdf
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched.
Solution
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched..
The distinction is important due to the following reasons.The data.pdf
The distinction is important due to the following reasons.
The data and information are exchangable.
The data is real.
Data may be quantitative or qualitative
Information is a activity data.
To become a data as information ,we need to deposit the data into statements or condition.
Data is uncooked upto the time it is arranged.
Data may be a charecter,text,word,number....etc
i.e joe,smith,84084........
Data comes from the latin word \"datum\".
Data and information are involved binding with each other.
Data is incoherent with out information it means that it is irelevant.
The no of callers for a website by country is example of data.
With the help of formulas we can convert a data into information which can be understood by
humans.
Data can be described in terms of information .
\" DIKW pyramid helps to know the relation ship between data and information\".
Solution
The distinction is important due to the following reasons.
The data and information are exchangable.
The data is real.
Data may be quantitative or qualitative
Information is a activity data.
To become a data as information ,we need to deposit the data into statements or condition.
Data is uncooked upto the time it is arranged.
Data may be a charecter,text,word,number....etc
i.e joe,smith,84084........
Data comes from the latin word \"datum\".
Data and information are involved binding with each other.
Data is incoherent with out information it means that it is irelevant.
The no of callers for a website by country is example of data.
With the help of formulas we can convert a data into information which can be understood by
humans.
Data can be described in terms of information .
\" DIKW pyramid helps to know the relation ship between data and information\"..
sequence Fnconverges to L hence by definition for given € 0 there.pdf
sequence Fnconverges to L hence by definition for given € 0 there exist N such that
|Fn-L| < €1/3 whenever (nN)
|F3n - L3 | |Fn - L|3 by Cauhy Schwartz inequality.
|F3n - L3 | |Fn - L|3 < € whenever (nN)
Hece by definition F3 converges to L3.
Solution
sequence Fnconverges to L hence by definition for given € 0 there exist N such that
|Fn-L| < €1/3 whenever (nN)
|F3n - L3 | |Fn - L|3 by Cauhy Schwartz inequality.
|F3n - L3 | |Fn - L|3 < € whenever (nN)
Hece by definition F3 converges to L3..
Project .javaimport java.util.;public class Project impleme.pdf
//Project .java
import java.util.*;
public class Project implements Comparable, Comparator {
int projectId;
boolean active;
String projectName;
String projectDescription;
public Project()
{
}
public Project(int id, String stat, String name,String des)
{
projectId=id;
if(stat.equals(\"A\"))
active=true;
else
active=false;
projectName=name;
projectDescription=des;
}
public int getProjectID()
{
return projectId;
}
public String getProjectName()
{
return projectName;
}
public String getProjectDescription()
{
return projectDescription;
}
public boolean getActiveFlag()
{
return active;
}
public void display()
{
System.out.println(\"Project ID: \"+projectId);
if(active==true)
System.out.println(\"Active flag: \"+\"A\");
else
System.out.println(\"Active flag: \"+\"I\");
System.out.println(\"Project Name: \"+projectName);
System.out.println(\"Project Description: \"+projectDescription);
}
@Override
public int compareTo(Project p) {
return (this.getProjectName().compareTo(p.getProjectName()));
}
@Override
public int compare(Project p1, Project p2) {
return p1.projectId - p2.projectId;
}
}
////////////////////////////////////////////////////////////////
//ProjectDAO.java
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.*;
public class ProjectDAO {
String fileName;
List projectList=new ArrayList<>();
public ProjectDAO(String file) throws FileNotFoundException
{
fileName=file;
Scanner infile=new Scanner(new FileReader(file));
while(infile.hasNextLine())
{
String str[]=infile.nextLine().split(\"|\");
projectList.add(new Project(Integer.parseInt(str[0]),str[1],str[2],str[3]));
}
}
public Project getProject(int id)
{
return projectList.get(id);
}
public void sortByProjectID(List projectList) {
Collections.sort(projectList, new Project());
}
public void sortByProjectName(List projectList) {
Collections.sort(projectList);
}
public void PrintProjectDetails()
{
for(int i=0;i
Solution
//Project .java
import java.util.*;
public class Project implements Comparable, Comparator {
int projectId;
boolean active;
String projectName;
String projectDescription;
public Project()
{
}
public Project(int id, String stat, String name,String des)
{
projectId=id;
if(stat.equals(\"A\"))
active=true;
else
active=false;
projectName=name;
projectDescription=des;
}
public int getProjectID()
{
return projectId;
}
public String getProjectName()
{
return projectName;
}
public String getProjectDescription()
{
return projectDescription;
}
public boolean getActiveFlag()
{
return active;
}
public void display()
{
System.out.println(\"Project ID: \"+projectId);
if(active==true)
System.out.println(\"Active flag: \"+\"A\");
else
System.out.println(\"Active flag: \"+\"I\");
System.out.println(\"Project Name: \"+projectName);
System.out.println(\"Project Description: \"+projectDescription);
}
@Override
public int compareTo(Project p) {
return (this.getProjectName().compareTo(p.getProjectName()));
}
@Override
public int compare(Project p1, Project p2) {
return p1..
Our immune system will respond and protect our body from pathogens b.pdf
Our immune system will respond and protect our body from pathogens by three lines of
defenses.
First Line of Defense :-
It is the non specific defense mechanism, in which pathogen or antigen restrict to enter into the
body, this includes the skin, mucous membranes(lining of respiratory tracts, genito urinary
tracts) cilia of the respiratory tract, nasal hairs,urine, perspiration, saliva, sebum, sweat from
sweat pores, gastric juices are included. These plays an vital role and protect from the pathogens
at entry level,
Second line of defense :-
When the pathogen enters into the body by crossing the first line of defense which manifest the
symptomatic disease, this is due to second line of defense, the immunogen triggers immune
response and shows immunity in order to decrease the multiplication rate. This is innate immune
response which prevents disease manifestation this includes, inflammatory response,
phagocytosis( WBC cells, NK cells, Neutrophils, Redness,( Dolar,Rubor, Colar,Tumor), fever
response.
Third line of defense :-
When the pathogen load is high then immunogen presented to the cell mediated immune
response, which relies on T cells and B cell immunity, in which antibodies, interleukins,
interferon are produced and kills the pathogen. These are antigen specific, acquired immune
response, and memory cells are developed.
Fever is the second line of defense in which body temperature is abnormally raised and halt the
metabolic rate, in which pathogen are stunted their growth to avail our immune system to
eradicate pathogen from the body. Body uses up the metabolic energy to raise and maintain
temperature higher than normal. So infections always manifest the fever as symptom,
Solution
Our immune system will respond and protect our body from pathogens by three lines of
defenses.
First Line of Defense :-
It is the non specific defense mechanism, in which pathogen or antigen restrict to enter into the
body, this includes the skin, mucous membranes(lining of respiratory tracts, genito urinary
tracts) cilia of the respiratory tract, nasal hairs,urine, perspiration, saliva, sebum, sweat from
sweat pores, gastric juices are included. These plays an vital role and protect from the pathogens
at entry level,
Second line of defense :-
When the pathogen enters into the body by crossing the first line of defense which manifest the
symptomatic disease, this is due to second line of defense, the immunogen triggers immune
response and shows immunity in order to decrease the multiplication rate. This is innate immune
response which prevents disease manifestation this includes, inflammatory response,
phagocytosis( WBC cells, NK cells, Neutrophils, Redness,( Dolar,Rubor, Colar,Tumor), fever
response.
Third line of defense :-
When the pathogen load is high then immunogen presented to the cell mediated immune
response, which relies on T cells and B cell immunity, in which antibodies, interleukins,
interferon are produced and kills the pathogen. These are anti.
Different thermal conductivity of different materials causes damage .pdf
Different thermal conductivity of different materials causes damage of the instruments as
weather conditions changes bring change in temperature over which various material properties
are dependent.
Solution
Different thermal conductivity of different materials causes damage of the instruments as
weather conditions changes bring change in temperature over which various material properties
are dependent..
#include node1.h #include list.h #include cassert P.pdf
#include \"node1.h\"
#include \"list.h\"
#include // Provides assert
#include
using namespace std;
value_type head(list lst)
{
list p;
p=list_locate(lst,1);
return p->data();
}
list tail(list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_remove(head);
return head;
}
bool list_is_empty(list lst)
{
size_t n= list_length(lst);
if(n==0)
return true;
else
return false;
}
list cons(value_type val, const list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_insert(head, val);
return head;
}
Solution
#include \"node1.h\"
#include \"list.h\"
#include // Provides assert
#include
using namespace std;
value_type head(list lst)
{
list p;
p=list_locate(lst,1);
return p->data();
}
list tail(list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_remove(head);
return head;
}
bool list_is_empty(list lst)
{
size_t n= list_length(lst);
if(n==0)
return true;
else
return false;
}
list cons(value_type val, const list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_insert(head, val);
return head;
}.
D is correct. Dual stack uses side-by-side protocol stacks on the de.pdf
D is correct. Dual stack uses side-by-side protocol stacks on the device.
A, B, C, and E are incorrect. A and C are incorrect because they do not maintain side-by-side
protocol stacks for IPv4 and IPv6. B and E are incorrect because neither is a transition method.
Solution
D is correct. Dual stack uses side-by-side protocol stacks on the device.
A, B, C, and E are incorrect. A and C are incorrect because they do not maintain side-by-side
protocol stacks for IPv4 and IPv6. B and E are incorrect because neither is a transition method..
More than one hundred radio-isotopes have been used to assist in a w.pdf
More than one hundred radio-isotopes have been used to assist in a wide variety of diagnostic
tests and treatment protocols. The use of radio-isotopes is convenient because they can emit
gamma radiation, have a short half-life, and are eliminated from the body shortly after the
completion of the diagnostic test or treatment. Radiodiagnostic protocols include:
1. Radioactive tracers are added to the blood stream to enable us to monitor the movement of
biochemical components in the human body.
2. Iodine 131, a beta emitter, is used for the treatment of overactive thyroids and also for
detecting internal hemorrhages.
3. Diagnostic techniques:
Diagnostic techniques in nuclear medicine use radioactive tracers which emit gamma rays from
within the body. These tracers are generally short-lived isotopes linked to chemical compounds
which permit specific physiological processes to be scrutinized. They can be given by injection,
inhalation, or orally.
An instrument called a gamma camera produces an image of the distribution of radioactivity in
an organ and by using this in conjunction with a computer, the functioning of the organ can be
closely monitored.
4. In Positron Emission Tomography (PET) scan, which is \"a more precise and sophisticated
technique\" for detecting cancerous growth in the human body:
A positron-emitting radionuclide is introduced, usually by injection, and accumulates in the
target tissue. As it decays it emits a positron, which promptly combines with a nearby electron
resulting in the simultaneous emission of two identifiable gamma rays in opposite directions.
These are detected by a PET camera and give very precise indication of their origin. PET\'s most
important clinical role is in oncology, with fluorine-18 as the tracer, since it has proven to be the
most accurate non-invasive method of detecting and evaluating most cancers. It is also well used
in cardiac and brain imaging.
Major therapeutic uses of radio-isotopes are:
1. Cobalt 60, an intense gamma emitter is often used in cancer therapy through direct irradiation.
2. Phosphorous 32, another beta emitter is often injected intravenously for treatment of leukemia.
3. Plutonium 238, an alpha emitter is used as a long lasting low level power source for
pacemakers to regulate heart action.
4. Caesium 137, encased in a biologically inert material container, is inserted into body cavities
like cervix or rectum to treat cancers in these regions effectively. Intense gamma radiation from
the radioisotope kills cancer cells locally, while the beta emissions are shielded by the container
walls.
5. Gold 198 or Iridium 192 wires can be inserted as implant for treatment of cancer in the tongue.
The list seems endless. A more detailed account can be obtained in the reference link given here.
The future of radiotherapy is probably the tagging of a monoclonal antibody for a specific cell
followed by the injection of the radioactive complex into the blood. The monoclonal antibodies
then bind.
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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Ethyl 4-aminobenzoate is a basic organic compound. In basic and neut.pdf
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble.
Solution
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble..
draw a sphere and use raytracing on the sphere in OpenGL glut. .pdf
/* draw a sphere and use raytracing on the sphere in OpenGL glut. */
#include
#include
#include
#include
#include
#define screen_width 750
#define screen_height 750
#define true 1
#define false 0
#define perpendicular 0
int gridXsize = 20;
int gridZsize = 20;
float plane[] = {0.0, 1.0, 0.0, -50.0,};
float sphere[] = {250.0, 270.0, -100.0, 100.0};
float eye[] = {0.0, 400.0, 550.0};
float light[] = {250.0, 550.0, -200.0};
float dot(float *u, float *v)
{
return u[0]*v[0] + u[1]*v[1] + u[2]*v[2];
}
void norm(float *u)
{
float norm = sqrt(abs(dot(u,u)));
for (int i =0; i <3; i++)
{
u[i] = u[i]/norm;
}
}
float plane_intersect(float *u, float *pO)
{
float normt[3] = {plane[0], plane[1], plane[2]};
float s;
if (dot(u,normt) == 0)
{
s = -10;
}
else
{
s = (plane[3]-(dot(pO,normt)))/(dot(u,normt));
}
return s;
}
float sphere_intersect(float *u, float *pO)
{
float deltaP[3] = {sphere[0]-pO[0],sphere[1]-pO[1],sphere[2]-pO[2]};
float deltLen = sqrt(abs(dot(deltaP,deltaP)));
float t=0;
float answer;
float det;
if ((det =(abs(dot(u,deltaP)*dot(u,deltaP))- (deltLen*deltLen)+sphere[3]*sphere[3])) < 0)
{
answer = -10;
}
else
{
t =-1*dot(u,deltaP)- sqrt(det) ;
if (t>0)
{
answer = t;
}
else
{
answer = -10;
}
}
return answer;
}
void find_reflect(float *u, float s, float *pO)
{
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]- sphere[2]};
float l[3] = {s *u[0],s *u[1],s *u[2]};
u[0] =(2*dot(l,n)*n[0])-l[0];
u[1] = (2*dot(l,n)*n[1])-l[1];
u[2] = (2*dot(l,n)*n[2])-l[2];
}
float find_shade(float *u,float s, float *pO)
{
float answer;
float lightVec[3] = {light[0]-(pO[0]+s *u[0]), light[1]-(pO[1]+s *u[1]), light[2]-(pO[2]+s
*u[2])};
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]-sphere[2]};
answer = -1*dot(lightVec,n)/(sqrt(abs(dot(lightVec,lightVec)))*sqrt(abs(dot(n,n))));
return answer;
}
void init()
{
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluOrtho2D(0,screen_width,0,screen_height);
}
void display()
{
glClear(GL_COLOR_BUFFER_BIT| GL_DEPTH_BUFFER_BIT);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
for (int i=0; i < screen_width; i++)
{
for (int j=0; j < screen_height; j++)
{
float ray[3] = {1*(eye[0]-i),-1*(eye[1]-j),1*eye[2]};
float point[3] = {i,j,0};
norm(ray);
int plotted = false;
while (!plotted)
{
float s_plane = plane_intersect(ray, point);
float s_sphere = sphere_intersect(ray, point);
if (s_plane <= 0 && s_sphere <=0)
{
glColor3f(0,0,0);
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
else if (s_sphere >= 0 && (s_plane <=0 || s_sphere <= s_plane))
{
find_reflect(ray, s_sphere, point);
}
else if (s_plane >=0 && (s_sphere <=0 ||s_plane <= s_sphere))
{
float shade = find_shade(ray, s_plane, point);
float xx = s_plane*ray[0] + eye[0];
float z = s_plane*ray[2] + eye[2];
if (abs((int)xx/gridXsize)%2 == abs((int)z/gridZsize)%2)
{
glColor3f(shade,0,0);
}
else
{
glColor3f(shade,shade,shade);
}
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
}
}
}
glFlush();
}
int main(.
SnO2 (ore) is initially washed and dressed. then .pdf
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form
Solution
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form.
d) Valine Note its side chain doesnt contain a.pdf
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens.
Solution
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens..
Amphiprotic. The preferred term is amphoteric. A .pdf
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ and HSO4-. The H+ will combine
with OH- to make water, and you\'re right back where you started. No. HSO4- is NOT
amphoteric in aqueous solution. ======= More follow up ========== \"Would it matter
about the physical state?\" No. For the substance to gain a proton, it will be in aqueous solution.
As soon as HCl interacts with water it will dissociate to form H+ and Cl-.
=============================== You may see that HSO4- is listed as an amphoteric
substance, but only because whoever wrote that didn\'t think it through.
Solution
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ a.
at equivalence point, Moles OH- added = moles H+.pdf
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases
Solution
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases.
We Know that Water is a Polar component as .pdf
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline.
Solution
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline..
molar mass of C2H5OH = 46.1 gmole moles of C2H5O.pdf
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules
Solution
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules.
What property of PN code makes them suitable for use in spread spect.pdf
What property of PN code makes them suitable for use in spread spectrum communication?
Solution
Pseudo random noise is used in some electronic musical instruments, either by itself or as an
input to subtractive synthesis, and in many white noise machines.
In spread-spectrum systems, the receiver correlates a locally generated signal with the received
signal. Such spread-spectrum systems require a set of one or more \"codes\" or \"sequences\"
such that
Like random noise, the local sequence has a very low correlation with any other sequence in the
set, or with the same sequence at a significantly different time offset, or with narrow band
interference, or with thermal noise.
Unlike random noise, it must be easy to generate exactly the same sequence at both the
transmitter and the receiver, so the receiver\'s locally generated sequence has a very high
correlation with the transmitted sequence.
In a direct-sequence spread spectrum system, each bit in the pseudorandom binary sequence is
known as a chip and the inverse of its period as chip rate; comparebit rate and symbol rate.
In a frequency-hopping spread spectrum sequence, each value in the pseudo random sequence is
known as a channel number and the inverse of its period as thehop rate. FCC Part 15 mandates at
least 50 different channels and at least a 2.5 Hz hop rate for narrow band frequency-hopping
systems.
GPS satellites broadcast data at a rate of 50 data bits per second – each satellite modulates its
data with one PN bit stream at 1.023 million chips per second and the same data with another PN
bit stream at 10.23 million chips per second. GPS receivers correlate the received PN bit stream
with a local reference to measure distance. GPS is a receive-only system that uses relative timing
measurements from several satellites (and the known positions of the satellites) to determine
receiver position.
Other range-finding applications involve two-way transmissions. A local station generates a
pseudo random bit sequence and transmits it to the remote location (using any modulation
technique). Some object at the remote location echoes this PN signal back to the location station
– either passively, as in some kinds of radar and sonar systems, or using an active transponder at
the remote location, as in the Apollo Unified S-band system.[4] By correlating a (delayed version
of) the transmitted signal with the received signal, a precise round trip time to the remote
location can be determined and thus the distance..
TLC can be used to help determine the number of components in a mixt.pdf
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container.
Solution
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container..
There may have some variables affect the model. So the error term wi.pdf
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation.
Solution
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation..
There are four major functional characteristics of skeletal muscle..pdf
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched.
Solution
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched..
The distinction is important due to the following reasons.The data.pdf
The distinction is important due to the following reasons.
The data and information are exchangable.
The data is real.
Data may be quantitative or qualitative
Information is a activity data.
To become a data as information ,we need to deposit the data into statements or condition.
Data is uncooked upto the time it is arranged.
Data may be a charecter,text,word,number....etc
i.e joe,smith,84084........
Data comes from the latin word \"datum\".
Data and information are involved binding with each other.
Data is incoherent with out information it means that it is irelevant.
The no of callers for a website by country is example of data.
With the help of formulas we can convert a data into information which can be understood by
humans.
Data can be described in terms of information .
\" DIKW pyramid helps to know the relation ship between data and information\".
Solution
The distinction is important due to the following reasons.
The data and information are exchangable.
The data is real.
Data may be quantitative or qualitative
Information is a activity data.
To become a data as information ,we need to deposit the data into statements or condition.
Data is uncooked upto the time it is arranged.
Data may be a charecter,text,word,number....etc
i.e joe,smith,84084........
Data comes from the latin word \"datum\".
Data and information are involved binding with each other.
Data is incoherent with out information it means that it is irelevant.
The no of callers for a website by country is example of data.
With the help of formulas we can convert a data into information which can be understood by
humans.
Data can be described in terms of information .
\" DIKW pyramid helps to know the relation ship between data and information\"..
sequence Fnconverges to L hence by definition for given € 0 there.pdf
sequence Fnconverges to L hence by definition for given € 0 there exist N such that
|Fn-L| < €1/3 whenever (nN)
|F3n - L3 | |Fn - L|3 by Cauhy Schwartz inequality.
|F3n - L3 | |Fn - L|3 < € whenever (nN)
Hece by definition F3 converges to L3.
Solution
sequence Fnconverges to L hence by definition for given € 0 there exist N such that
|Fn-L| < €1/3 whenever (nN)
|F3n - L3 | |Fn - L|3 by Cauhy Schwartz inequality.
|F3n - L3 | |Fn - L|3 < € whenever (nN)
Hece by definition F3 converges to L3..
Project .javaimport java.util.;public class Project impleme.pdf
//Project .java
import java.util.*;
public class Project implements Comparable, Comparator {
int projectId;
boolean active;
String projectName;
String projectDescription;
public Project()
{
}
public Project(int id, String stat, String name,String des)
{
projectId=id;
if(stat.equals(\"A\"))
active=true;
else
active=false;
projectName=name;
projectDescription=des;
}
public int getProjectID()
{
return projectId;
}
public String getProjectName()
{
return projectName;
}
public String getProjectDescription()
{
return projectDescription;
}
public boolean getActiveFlag()
{
return active;
}
public void display()
{
System.out.println(\"Project ID: \"+projectId);
if(active==true)
System.out.println(\"Active flag: \"+\"A\");
else
System.out.println(\"Active flag: \"+\"I\");
System.out.println(\"Project Name: \"+projectName);
System.out.println(\"Project Description: \"+projectDescription);
}
@Override
public int compareTo(Project p) {
return (this.getProjectName().compareTo(p.getProjectName()));
}
@Override
public int compare(Project p1, Project p2) {
return p1.projectId - p2.projectId;
}
}
////////////////////////////////////////////////////////////////
//ProjectDAO.java
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.*;
public class ProjectDAO {
String fileName;
List projectList=new ArrayList<>();
public ProjectDAO(String file) throws FileNotFoundException
{
fileName=file;
Scanner infile=new Scanner(new FileReader(file));
while(infile.hasNextLine())
{
String str[]=infile.nextLine().split(\"|\");
projectList.add(new Project(Integer.parseInt(str[0]),str[1],str[2],str[3]));
}
}
public Project getProject(int id)
{
return projectList.get(id);
}
public void sortByProjectID(List projectList) {
Collections.sort(projectList, new Project());
}
public void sortByProjectName(List projectList) {
Collections.sort(projectList);
}
public void PrintProjectDetails()
{
for(int i=0;i
Solution
//Project .java
import java.util.*;
public class Project implements Comparable, Comparator {
int projectId;
boolean active;
String projectName;
String projectDescription;
public Project()
{
}
public Project(int id, String stat, String name,String des)
{
projectId=id;
if(stat.equals(\"A\"))
active=true;
else
active=false;
projectName=name;
projectDescription=des;
}
public int getProjectID()
{
return projectId;
}
public String getProjectName()
{
return projectName;
}
public String getProjectDescription()
{
return projectDescription;
}
public boolean getActiveFlag()
{
return active;
}
public void display()
{
System.out.println(\"Project ID: \"+projectId);
if(active==true)
System.out.println(\"Active flag: \"+\"A\");
else
System.out.println(\"Active flag: \"+\"I\");
System.out.println(\"Project Name: \"+projectName);
System.out.println(\"Project Description: \"+projectDescription);
}
@Override
public int compareTo(Project p) {
return (this.getProjectName().compareTo(p.getProjectName()));
}
@Override
public int compare(Project p1, Project p2) {
return p1..
Our immune system will respond and protect our body from pathogens b.pdf
Our immune system will respond and protect our body from pathogens by three lines of
defenses.
First Line of Defense :-
It is the non specific defense mechanism, in which pathogen or antigen restrict to enter into the
body, this includes the skin, mucous membranes(lining of respiratory tracts, genito urinary
tracts) cilia of the respiratory tract, nasal hairs,urine, perspiration, saliva, sebum, sweat from
sweat pores, gastric juices are included. These plays an vital role and protect from the pathogens
at entry level,
Second line of defense :-
When the pathogen enters into the body by crossing the first line of defense which manifest the
symptomatic disease, this is due to second line of defense, the immunogen triggers immune
response and shows immunity in order to decrease the multiplication rate. This is innate immune
response which prevents disease manifestation this includes, inflammatory response,
phagocytosis( WBC cells, NK cells, Neutrophils, Redness,( Dolar,Rubor, Colar,Tumor), fever
response.
Third line of defense :-
When the pathogen load is high then immunogen presented to the cell mediated immune
response, which relies on T cells and B cell immunity, in which antibodies, interleukins,
interferon are produced and kills the pathogen. These are antigen specific, acquired immune
response, and memory cells are developed.
Fever is the second line of defense in which body temperature is abnormally raised and halt the
metabolic rate, in which pathogen are stunted their growth to avail our immune system to
eradicate pathogen from the body. Body uses up the metabolic energy to raise and maintain
temperature higher than normal. So infections always manifest the fever as symptom,
Solution
Our immune system will respond and protect our body from pathogens by three lines of
defenses.
First Line of Defense :-
It is the non specific defense mechanism, in which pathogen or antigen restrict to enter into the
body, this includes the skin, mucous membranes(lining of respiratory tracts, genito urinary
tracts) cilia of the respiratory tract, nasal hairs,urine, perspiration, saliva, sebum, sweat from
sweat pores, gastric juices are included. These plays an vital role and protect from the pathogens
at entry level,
Second line of defense :-
When the pathogen enters into the body by crossing the first line of defense which manifest the
symptomatic disease, this is due to second line of defense, the immunogen triggers immune
response and shows immunity in order to decrease the multiplication rate. This is innate immune
response which prevents disease manifestation this includes, inflammatory response,
phagocytosis( WBC cells, NK cells, Neutrophils, Redness,( Dolar,Rubor, Colar,Tumor), fever
response.
Third line of defense :-
When the pathogen load is high then immunogen presented to the cell mediated immune
response, which relies on T cells and B cell immunity, in which antibodies, interleukins,
interferon are produced and kills the pathogen. These are anti.
Different thermal conductivity of different materials causes damage .pdf
Different thermal conductivity of different materials causes damage of the instruments as
weather conditions changes bring change in temperature over which various material properties
are dependent.
Solution
Different thermal conductivity of different materials causes damage of the instruments as
weather conditions changes bring change in temperature over which various material properties
are dependent..
#include node1.h #include list.h #include cassert P.pdf
#include \"node1.h\"
#include \"list.h\"
#include // Provides assert
#include
using namespace std;
value_type head(list lst)
{
list p;
p=list_locate(lst,1);
return p->data();
}
list tail(list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_remove(head);
return head;
}
bool list_is_empty(list lst)
{
size_t n= list_length(lst);
if(n==0)
return true;
else
return false;
}
list cons(value_type val, const list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_insert(head, val);
return head;
}
Solution
#include \"node1.h\"
#include \"list.h\"
#include // Provides assert
#include
using namespace std;
value_type head(list lst)
{
list p;
p=list_locate(lst,1);
return p->data();
}
list tail(list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_remove(head);
return head;
}
bool list_is_empty(list lst)
{
size_t n= list_length(lst);
if(n==0)
return true;
else
return false;
}
list cons(value_type val, const list lst)
{
node *head=NULL;
node *tail=NULL;
list_copy(lst, head, tail);
list_head_insert(head, val);
return head;
}.
D is correct. Dual stack uses side-by-side protocol stacks on the de.pdf
D is correct. Dual stack uses side-by-side protocol stacks on the device.
A, B, C, and E are incorrect. A and C are incorrect because they do not maintain side-by-side
protocol stacks for IPv4 and IPv6. B and E are incorrect because neither is a transition method.
Solution
D is correct. Dual stack uses side-by-side protocol stacks on the device.
A, B, C, and E are incorrect. A and C are incorrect because they do not maintain side-by-side
protocol stacks for IPv4 and IPv6. B and E are incorrect because neither is a transition method..
More than one hundred radio-isotopes have been used to assist in a w.pdf
More than one hundred radio-isotopes have been used to assist in a wide variety of diagnostic
tests and treatment protocols. The use of radio-isotopes is convenient because they can emit
gamma radiation, have a short half-life, and are eliminated from the body shortly after the
completion of the diagnostic test or treatment. Radiodiagnostic protocols include:
1. Radioactive tracers are added to the blood stream to enable us to monitor the movement of
biochemical components in the human body.
2. Iodine 131, a beta emitter, is used for the treatment of overactive thyroids and also for
detecting internal hemorrhages.
3. Diagnostic techniques:
Diagnostic techniques in nuclear medicine use radioactive tracers which emit gamma rays from
within the body. These tracers are generally short-lived isotopes linked to chemical compounds
which permit specific physiological processes to be scrutinized. They can be given by injection,
inhalation, or orally.
An instrument called a gamma camera produces an image of the distribution of radioactivity in
an organ and by using this in conjunction with a computer, the functioning of the organ can be
closely monitored.
4. In Positron Emission Tomography (PET) scan, which is \"a more precise and sophisticated
technique\" for detecting cancerous growth in the human body:
A positron-emitting radionuclide is introduced, usually by injection, and accumulates in the
target tissue. As it decays it emits a positron, which promptly combines with a nearby electron
resulting in the simultaneous emission of two identifiable gamma rays in opposite directions.
These are detected by a PET camera and give very precise indication of their origin. PET\'s most
important clinical role is in oncology, with fluorine-18 as the tracer, since it has proven to be the
most accurate non-invasive method of detecting and evaluating most cancers. It is also well used
in cardiac and brain imaging.
Major therapeutic uses of radio-isotopes are:
1. Cobalt 60, an intense gamma emitter is often used in cancer therapy through direct irradiation.
2. Phosphorous 32, another beta emitter is often injected intravenously for treatment of leukemia.
3. Plutonium 238, an alpha emitter is used as a long lasting low level power source for
pacemakers to regulate heart action.
4. Caesium 137, encased in a biologically inert material container, is inserted into body cavities
like cervix or rectum to treat cancers in these regions effectively. Intense gamma radiation from
the radioisotope kills cancer cells locally, while the beta emissions are shielded by the container
walls.
5. Gold 198 or Iridium 192 wires can be inserted as implant for treatment of cancer in the tongue.
The list seems endless. A more detailed account can be obtained in the reference link given here.
The future of radiotherapy is probably the tagging of a monoclonal antibody for a specific cell
followed by the injection of the radioactive complex into the blood. The monoclonal antibodies
then bind.
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