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// Problem1 java code import java.util.Scanner; // Java code to print all possible strings of letter L and R class Problem1 { static void print(char set[], int length) { int setLength = set.length; printRecursion(set, \"\", setLength, length); } static void printRecursion(char set[], String prefixset, int setLength, int length) { // Base case: length is 0 if (length == 0) { System.out.println(prefixset); return; } for (int i = 0; i < setLength; ++i) { String newPrefixset = prefixset + set[i]; printRecursion(set, newPrefixset, setLength, length - 1); } } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter length: \"); int length = scan.nextInt(); char set[] = {\'L\', \'R\'}; print(set, length); } } /* output: Enter length: 3 LLL LLR LRL LRR RLL RLR RRL RRR */ // Problem2 java code import java.util.Scanner; // Java code to print all possible strings of letter L and R class Problem2 { static void print(char set[], int length) { int setLength = set.length; printRecursion(set, \"\", setLength, length); } static void printRecursion(char set[], String prefixset, int setLength, int length) { // Base case: length is 0 if (length == 0) { System.out.print(prefixset + \" \"); return; } for (int i = 0; i < setLength; ++i) { String newPrefixset = prefixset + set[i]; printRecursion(set, newPrefixset, setLength, length - 1); } } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter length: \"); int length = scan.nextInt(); char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'}; print(set, length); System.out.println(); } } /* output: Enter length: 3 111 113 115 117 119 131 133 135 137 139 151 153 155 157 159 171 173 175 177 179 191 193 195 197 199 311 313 315 317 319 331 333 335 337 339 351 353 355 357 359 371 373 375 377 379 391 393 395 397 399 511 513 515 517 519 531 533 535 537 539 551 553 555 557 559 571 573 575 577 579 591 593 595 597 599 711 713 715 717 719 731 733 735 737 739 751 753 755 757 759 771 773 775 777 779 791 793 795 797 799 911 913 915 917 919 931 933 935 937 939 951 953 955 957 959 971 973 975 977 979 991 993 995 997 999 */ // Problem3 java code import java.util.Scanner; // Java code to multiply 2 numbers class Problem3 { public static int multiply(int a, int b) { int temp = b; for (int i = 1; i < a; i++ ) { b = b + temp; } return b; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter a: \"); int a = scan.nextInt(); System.out.println(\"Enter b: \"); int b = scan.nextInt(); System.out.println(a + \"X\" + b + \" = \" + multiply(a,b)); } } /* output: Enter a: 7 Enter b: 6 7X6 = 42 */ // Problem4 java code import java.util.Scanner; // Java code to find gcd 2 numbers class Problem4 { public static int gcd(int a, int b) { if (b!=0) return gcd(b, a%b); else return a; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter a: \"); int a = scan.ne.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
anupamfootwear
All of the above Solution All of the above.
All of the aboveSolution All of the above.pdf
All of the aboveSolution All of the above.pdf
anupamfootwear
Yields a colorless solution and a white precipitate Solution Yields a colorless solution and a white precipitate.
Yields a colorless solution and a white precipita.pdf
Yields a colorless solution and a white precipita.pdf
anupamfootwear
ROund to one significant figures, THen, it\'s 2.2*10^8 Solution ROund to one significant figures, THen, it\'s 2.2*10^8.
ROund to one significant figures, THen, its 2..pdf
ROund to one significant figures, THen, its 2..pdf
anupamfootwear
P=7.7(0.92)^t growth rate is dP/dt differentiate p wrt t dP/dt = 7.7(0.92)^tlog(0.92) similarly P=15e^(-0.06t) dP/dt = 15.(-0.06)e^(-0.06t) dP/dt = -0.9e^(-0.06t) Solution P=7.7(0.92)^t growth rate is dP/dt differentiate p wrt t dP/dt = 7.7(0.92)^tlog(0.92) similarly P=15e^(-0.06t) dP/dt = 15.(-0.06)e^(-0.06t) dP/dt = -0.9e^(-0.06t).
P=7.7(0.92)^t growth rate is dPdt differentiate .pdf
P=7.7(0.92)^t growth rate is dPdt differentiate .pdf
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option (D) ... reason posted in my previous answer Solution option (D) ... reason posted in my previous answer.
option (D) ... reason posted in my previous answe.pdf
option (D) ... reason posted in my previous answe.pdf
anupamfootwear
no.of moles of H2moles=28.8/2=14.4moles no.of moles of C2H6 formed=14.4moles Solution no.of moles of H2moles=28.8/2=14.4moles no.of moles of C2H6 formed=14.4moles.
no.of moles of H2moles=28.82=14.4moles no.of mol.pdf
no.of moles of H2moles=28.82=14.4moles no.of mol.pdf
anupamfootwear
Ionic Compounds: Think of an ionic compound as a sea of ions. All these ions are carefully arranged in some sort of 3D pattern with positive and negative ions in specific locations in the pattern. Take common table salt, Sodium Chloride, as an example. In Sodium Chloride, there aren\'t specific NaCl molecules, but a mix of Na+ ions and Cl- ions. Each Na+ ion is surrounded by Cl- ions, and each Cl- ion is surrounded by Na+ ions. The whole structure is held together by the attraction between the positive and negative ions. Molecular Substances: A molecule is a group of atoms which are sharing electrons with one another through covalent bonds. Because the sharing results in a lower energy state, the atoms are more stable together than alone. When a sample of a molecular substance is looked at, say water (H2O), within the substance each molecule is being attracted to other molecules. These attractive forces are called intermolecular forces and they are much weaker than covalent bonds or ionic bonds. For a substance to melt and then boil, the forces which hold the substance together must be broken. When water boils, the intermolecular forces between H2O molecules are broken and the molecules separate from one another. (Note that the covalent bonds are NOT broken, the gas is still made up of H2O molecules, they\'re just not \"sticking\" to one another anymore.) As you can imagine, the ionic forces holding an ionic compound together are MUCH stronger than the intermolecular forces holding a molecular substance together. A lot more energy is required to break the ionic bonds. So, ionic substances melt at very high temperatures and boil at even higher temperatures. Solution Ionic Compounds: Think of an ionic compound as a sea of ions. All these ions are carefully arranged in some sort of 3D pattern with positive and negative ions in specific locations in the pattern. Take common table salt, Sodium Chloride, as an example. In Sodium Chloride, there aren\'t specific NaCl molecules, but a mix of Na+ ions and Cl- ions. Each Na+ ion is surrounded by Cl- ions, and each Cl- ion is surrounded by Na+ ions. The whole structure is held together by the attraction between the positive and negative ions. Molecular Substances: A molecule is a group of atoms which are sharing electrons with one another through covalent bonds. Because the sharing results in a lower energy state, the atoms are more stable together than alone. When a sample of a molecular substance is looked at, say water (H2O), within the substance each molecule is being attracted to other molecules. These attractive forces are called intermolecular forces and they are much weaker than covalent bonds or ionic bonds. For a substance to melt and then boil, the forces which hold the substance together must be broken. When water boils, the intermolecular forces between H2O molecules are broken and the molecules separate from one another. (Note that the covalent bonds are NOT broken, the gas is still made u.
Ionic Compounds Think of an ionic compound as a .pdf
Ionic Compounds Think of an ionic compound as a .pdf
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// Problem1 java code import java.util.Scanner; // Java code to print all possible strings of letter L and R class Problem1 { static void print(char set[], int length) { int setLength = set.length; printRecursion(set, \"\", setLength, length); } static void printRecursion(char set[], String prefixset, int setLength, int length) { // Base case: length is 0 if (length == 0) { System.out.println(prefixset); return; } for (int i = 0; i < setLength; ++i) { String newPrefixset = prefixset + set[i]; printRecursion(set, newPrefixset, setLength, length - 1); } } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter length: \"); int length = scan.nextInt(); char set[] = {\'L\', \'R\'}; print(set, length); } } /* output: Enter length: 3 LLL LLR LRL LRR RLL RLR RRL RRR */ // Problem2 java code import java.util.Scanner; // Java code to print all possible strings of letter L and R class Problem2 { static void print(char set[], int length) { int setLength = set.length; printRecursion(set, \"\", setLength, length); } static void printRecursion(char set[], String prefixset, int setLength, int length) { // Base case: length is 0 if (length == 0) { System.out.print(prefixset + \" \"); return; } for (int i = 0; i < setLength; ++i) { String newPrefixset = prefixset + set[i]; printRecursion(set, newPrefixset, setLength, length - 1); } } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter length: \"); int length = scan.nextInt(); char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'}; print(set, length); System.out.println(); } } /* output: Enter length: 3 111 113 115 117 119 131 133 135 137 139 151 153 155 157 159 171 173 175 177 179 191 193 195 197 199 311 313 315 317 319 331 333 335 337 339 351 353 355 357 359 371 373 375 377 379 391 393 395 397 399 511 513 515 517 519 531 533 535 537 539 551 553 555 557 559 571 573 575 577 579 591 593 595 597 599 711 713 715 717 719 731 733 735 737 739 751 753 755 757 759 771 773 775 777 779 791 793 795 797 799 911 913 915 917 919 931 933 935 937 939 951 953 955 957 959 971 973 975 977 979 991 993 995 997 999 */ // Problem3 java code import java.util.Scanner; // Java code to multiply 2 numbers class Problem3 { public static int multiply(int a, int b) { int temp = b; for (int i = 1; i < a; i++ ) { b = b + temp; } return b; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter a: \"); int a = scan.nextInt(); System.out.println(\"Enter b: \"); int b = scan.nextInt(); System.out.println(a + \"X\" + b + \" = \" + multiply(a,b)); } } /* output: Enter a: 7 Enter b: 6 7X6 = 42 */ // Problem4 java code import java.util.Scanner; // Java code to find gcd 2 numbers class Problem4 { public static int gcd(int a, int b) { if (b!=0) return gcd(b, a%b); else return a; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); System.out.println(\"Enter a: \"); int a = scan.ne.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
anupamfootwear
All of the above Solution All of the above.
All of the aboveSolution All of the above.pdf
All of the aboveSolution All of the above.pdf
anupamfootwear
Yields a colorless solution and a white precipitate Solution Yields a colorless solution and a white precipitate.
Yields a colorless solution and a white precipita.pdf
Yields a colorless solution and a white precipita.pdf
anupamfootwear
ROund to one significant figures, THen, it\'s 2.2*10^8 Solution ROund to one significant figures, THen, it\'s 2.2*10^8.
ROund to one significant figures, THen, its 2..pdf
ROund to one significant figures, THen, its 2..pdf
anupamfootwear
P=7.7(0.92)^t growth rate is dP/dt differentiate p wrt t dP/dt = 7.7(0.92)^tlog(0.92) similarly P=15e^(-0.06t) dP/dt = 15.(-0.06)e^(-0.06t) dP/dt = -0.9e^(-0.06t) Solution P=7.7(0.92)^t growth rate is dP/dt differentiate p wrt t dP/dt = 7.7(0.92)^tlog(0.92) similarly P=15e^(-0.06t) dP/dt = 15.(-0.06)e^(-0.06t) dP/dt = -0.9e^(-0.06t).
P=7.7(0.92)^t growth rate is dPdt differentiate .pdf
P=7.7(0.92)^t growth rate is dPdt differentiate .pdf
anupamfootwear
option (D) ... reason posted in my previous answer Solution option (D) ... reason posted in my previous answer.
option (D) ... reason posted in my previous answe.pdf
option (D) ... reason posted in my previous answe.pdf
anupamfootwear
no.of moles of H2moles=28.8/2=14.4moles no.of moles of C2H6 formed=14.4moles Solution no.of moles of H2moles=28.8/2=14.4moles no.of moles of C2H6 formed=14.4moles.
no.of moles of H2moles=28.82=14.4moles no.of mol.pdf
no.of moles of H2moles=28.82=14.4moles no.of mol.pdf
anupamfootwear
Ionic Compounds: Think of an ionic compound as a sea of ions. All these ions are carefully arranged in some sort of 3D pattern with positive and negative ions in specific locations in the pattern. Take common table salt, Sodium Chloride, as an example. In Sodium Chloride, there aren\'t specific NaCl molecules, but a mix of Na+ ions and Cl- ions. Each Na+ ion is surrounded by Cl- ions, and each Cl- ion is surrounded by Na+ ions. The whole structure is held together by the attraction between the positive and negative ions. Molecular Substances: A molecule is a group of atoms which are sharing electrons with one another through covalent bonds. Because the sharing results in a lower energy state, the atoms are more stable together than alone. When a sample of a molecular substance is looked at, say water (H2O), within the substance each molecule is being attracted to other molecules. These attractive forces are called intermolecular forces and they are much weaker than covalent bonds or ionic bonds. For a substance to melt and then boil, the forces which hold the substance together must be broken. When water boils, the intermolecular forces between H2O molecules are broken and the molecules separate from one another. (Note that the covalent bonds are NOT broken, the gas is still made up of H2O molecules, they\'re just not \"sticking\" to one another anymore.) As you can imagine, the ionic forces holding an ionic compound together are MUCH stronger than the intermolecular forces holding a molecular substance together. A lot more energy is required to break the ionic bonds. So, ionic substances melt at very high temperatures and boil at even higher temperatures. Solution Ionic Compounds: Think of an ionic compound as a sea of ions. All these ions are carefully arranged in some sort of 3D pattern with positive and negative ions in specific locations in the pattern. Take common table salt, Sodium Chloride, as an example. In Sodium Chloride, there aren\'t specific NaCl molecules, but a mix of Na+ ions and Cl- ions. Each Na+ ion is surrounded by Cl- ions, and each Cl- ion is surrounded by Na+ ions. The whole structure is held together by the attraction between the positive and negative ions. Molecular Substances: A molecule is a group of atoms which are sharing electrons with one another through covalent bonds. Because the sharing results in a lower energy state, the atoms are more stable together than alone. When a sample of a molecular substance is looked at, say water (H2O), within the substance each molecule is being attracted to other molecules. These attractive forces are called intermolecular forces and they are much weaker than covalent bonds or ionic bonds. For a substance to melt and then boil, the forces which hold the substance together must be broken. When water boils, the intermolecular forces between H2O molecules are broken and the molecules separate from one another. (Note that the covalent bonds are NOT broken, the gas is still made u.
Ionic Compounds Think of an ionic compound as a .pdf
Ionic Compounds Think of an ionic compound as a .pdf
anupamfootwear
In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-. Solution In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-..
In the First molecule there will be Resonance whi.pdf
In the First molecule there will be Resonance whi.pdf
anupamfootwear
He down the group, IE decreases. Solution He down the group, IE decreases..
He down the group, IE decreases. Solution.pdf
He down the group, IE decreases. Solution.pdf
anupamfootwear
Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potential energy of the proton gradient will be seen losing as mere heat. Research has showed when 2,4-dintrophenol is added to the reaction mixture, it has uncoupled electron transfer from that of ATP synthesis and as a result the oxygen is completely seen to be consumed in the absence of ADP. Solution Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potenti.
Uncouple agents will never disrupt the electron transport (they will.pdf
Uncouple agents will never disrupt the electron transport (they will.pdf
anupamfootwear
The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O. Solution The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O..
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
anupamfootwear
d.Addition of sulfuric acid to copper(II) oxide produces water as a by-product Solution d.Addition of sulfuric acid to copper(II) oxide produces water as a by-product.
d.Addition of sulfuric acid to copper(II) oxide p.pdf
d.Addition of sulfuric acid to copper(II) oxide p.pdf
anupamfootwear
The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them. Solution The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them..
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
anupamfootwear
S2F6 Solution S2F6.
S2F6SolutionS2F6.pdf
S2F6SolutionS2F6.pdf
anupamfootwear
Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid element or right element is encountered . 3) Compare list1[i] and list[j]. i) If list1[i]<=list2[j] c[k++]=list1[i++] else c[k++]=list2[j++] count = count + mid-i; 4) add rest elements of list1 and list2 in c. 5) copy sorted list c back to original list. 6) return count. void quickSort(int arr[], int left, int right) { int i = left, j = right; int tmp; int pivot = arr[(left + right) / 2]; /* partition */ while (i <= j) { while (arr[i] < pivot) i++; while (arr[j] > pivot) j--; if (i <= j) { tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; i++; j--; } }; /* recursion */ if (left < j) quickSort(arr, left, j); if (i < right) quickSort(arr, i, right); } Solution Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid elem.
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
anupamfootwear
Pictures are not legible, could you plz post it again. Solution Pictures are not legible, could you plz post it again..
Pictures are not legible, could you plz post it again.Solution.pdf
Pictures are not legible, could you plz post it again.Solution.pdf
anupamfootwear
Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat. Solution Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat..
Carbonic acid leaves the soda solution as CO2 (ca.pdf
Carbonic acid leaves the soda solution as CO2 (ca.pdf
anupamfootwear
C. III Solution C. III.
C. III S.pdf
C. III S.pdf
anupamfootwear
mass= density *volume= 2.33*0.1*0.1*0.01 = 2.33*10-4gms atomic mass of Si = 28.1 amu moles of Si =2.33*10-4/28.1 = 8.29*10-6 number of atoms =8.29*10-6*6.023*1023= 5*1018 atoms Solution mass= density *volume= 2.33*0.1*0.1*0.01 = 2.33*10-4gms atomic mass of Si = 28.1 amu moles of Si =2.33*10-4/28.1 = 8.29*10-6 number of atoms =8.29*10-6*6.023*1023= 5*1018 atoms.
mass= density volume= 2.330.10.10.01 = 2.3310-4gmsatomic .pdf
mass= density volume= 2.330.10.10.01 = 2.3310-4gmsatomic .pdf
anupamfootwear
John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body. Solution John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body..
John is suffering from fifth disease.It is caused by an airborne v.pdf
John is suffering from fifth disease.It is caused by an airborne v.pdf
anupamfootwear
Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M Solution Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M.
Initial concentration of IO4- = moles of NaIO4final volume= 0.025.pdf
Initial concentration of IO4- = moles of NaIO4final volume= 0.025.pdf
anupamfootwear
atomic elements, Hg, Au, molecular elements, H2, F2 molecular compounds, CO2, HCl or ionic compounds Al2S3, KCl Solution atomic elements, Hg, Au, molecular elements, H2, F2 molecular compounds, CO2, HCl or ionic compounds Al2S3, KCl.
atomic elements, Hg, Au, molecular elements, H2,.pdf
atomic elements, Hg, Au, molecular elements, H2,.pdf
anupamfootwear
In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein. Solution In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein..
In Drosophila the gene Tra encodes a protein that is expressed only .pdf
In Drosophila the gene Tra encodes a protein that is expressed only .pdf
anupamfootwear
Ecell = E0 - (0.0591/n)log[H+] 0.207 = 0.0591*log[H+]/1 Therefore - log[H+] = pH = 3.5 Solution Ecell = E0 - (0.0591/n)log[H+] 0.207 = 0.0591*log[H+]/1 Therefore - log[H+] = pH = 3.5.
Ecell = E0 - (0.0591n)log[H+]0.207 = 0.0591log[H+]1Therefore.pdf
Ecell = E0 - (0.0591n)log[H+]0.207 = 0.0591log[H+]1Therefore.pdf
anupamfootwear
Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane. Solution Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane..
Explanation- The processes that occur within the nucleus areRepli.pdf
Explanation- The processes that occur within the nucleus areRepli.pdf
anupamfootwear
Dominant strategy for Jane: Early Dominant strategy for Bob: Early Therefore, Nash equilibrium: (Early, Early) Socially optimal solution: (Early, Early) Solution Dominant strategy for Jane: Early Dominant strategy for Bob: Early Therefore, Nash equilibrium: (Early, Early) Socially optimal solution: (Early, Early).
Dominant strategy for Jane EarlyDominant strategy for Bob Early.pdf
Dominant strategy for Jane EarlyDominant strategy for Bob Early.pdf
anupamfootwear
Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : is a concept that involves the use and management of information and communication technology in avocation of a competitive advantage over an opposing party. Ping sweep :ping sweep is a method which establishes a range of IP addresses that can map to live hosts. The tool used for ping sweeps is fping Port scan : It is basically a attack in which the attacker sends packets to your machine, varying the destination port and find out the sevices you are running. Script kiddies : person who uses existing computer scripts or codes to hack into computers, as he lacks the expertise to write his own code or scripts Structured threat : Explained above in highly structured threat Unstructured threat : these are basically an Organized efforts to attack a un specific target , it can be a organization specific or related to anything but not specific. Solution Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : .
Critical infrastructures Compose the asset or systems or network r.pdf
Critical infrastructures Compose the asset or systems or network r.pdf
anupamfootwear
Skills of introducing the lesson presents by Mrs. Amanpreet Kaur, Assistant Professor Khalsa College of Education, G.T. Road Amritsar
SKILL OF INTRODUCING THE LESSON MICRO SKILLS.pptx
SKILL OF INTRODUCING THE LESSON MICRO SKILLS.pptx
Amanpreet Kaur
Wednesday 20 March 2024, 09:30-15:30.
Jamworks pilot and AI at Jisc (20/03/2024)
Jamworks pilot and AI at Jisc (20/03/2024)
Jisc
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In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-. Solution In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-..
In the First molecule there will be Resonance whi.pdf
In the First molecule there will be Resonance whi.pdf
anupamfootwear
He down the group, IE decreases. Solution He down the group, IE decreases..
He down the group, IE decreases. Solution.pdf
He down the group, IE decreases. Solution.pdf
anupamfootwear
Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potential energy of the proton gradient will be seen losing as mere heat. Research has showed when 2,4-dintrophenol is added to the reaction mixture, it has uncoupled electron transfer from that of ATP synthesis and as a result the oxygen is completely seen to be consumed in the absence of ADP. Solution Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potenti.
Uncouple agents will never disrupt the electron transport (they will.pdf
Uncouple agents will never disrupt the electron transport (they will.pdf
anupamfootwear
The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O. Solution The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O..
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
anupamfootwear
d.Addition of sulfuric acid to copper(II) oxide produces water as a by-product Solution d.Addition of sulfuric acid to copper(II) oxide produces water as a by-product.
d.Addition of sulfuric acid to copper(II) oxide p.pdf
d.Addition of sulfuric acid to copper(II) oxide p.pdf
anupamfootwear
The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them. Solution The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them..
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
anupamfootwear
S2F6 Solution S2F6.
S2F6SolutionS2F6.pdf
S2F6SolutionS2F6.pdf
anupamfootwear
Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid element or right element is encountered . 3) Compare list1[i] and list[j]. i) If list1[i]<=list2[j] c[k++]=list1[i++] else c[k++]=list2[j++] count = count + mid-i; 4) add rest elements of list1 and list2 in c. 5) copy sorted list c back to original list. 6) return count. void quickSort(int arr[], int left, int right) { int i = left, j = right; int tmp; int pivot = arr[(left + right) / 2]; /* partition */ while (i <= j) { while (arr[i] < pivot) i++; while (arr[j] > pivot) j--; if (i <= j) { tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; i++; j--; } }; /* recursion */ if (left < j) quickSort(arr, left, j); if (i < right) quickSort(arr, i, right); } Solution Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid elem.
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
anupamfootwear
Pictures are not legible, could you plz post it again. Solution Pictures are not legible, could you plz post it again..
Pictures are not legible, could you plz post it again.Solution.pdf
Pictures are not legible, could you plz post it again.Solution.pdf
anupamfootwear
Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat. Solution Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat..
Carbonic acid leaves the soda solution as CO2 (ca.pdf
Carbonic acid leaves the soda solution as CO2 (ca.pdf
anupamfootwear
C. III Solution C. III.
C. III S.pdf
C. III S.pdf
anupamfootwear
mass= density *volume= 2.33*0.1*0.1*0.01 = 2.33*10-4gms atomic mass of Si = 28.1 amu moles of Si =2.33*10-4/28.1 = 8.29*10-6 number of atoms =8.29*10-6*6.023*1023= 5*1018 atoms Solution mass= density *volume= 2.33*0.1*0.1*0.01 = 2.33*10-4gms atomic mass of Si = 28.1 amu moles of Si =2.33*10-4/28.1 = 8.29*10-6 number of atoms =8.29*10-6*6.023*1023= 5*1018 atoms.
mass= density volume= 2.330.10.10.01 = 2.3310-4gmsatomic .pdf
mass= density volume= 2.330.10.10.01 = 2.3310-4gmsatomic .pdf
anupamfootwear
John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body. Solution John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body..
John is suffering from fifth disease.It is caused by an airborne v.pdf
John is suffering from fifth disease.It is caused by an airborne v.pdf
anupamfootwear
Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M Solution Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M.
Initial concentration of IO4- = moles of NaIO4final volume= 0.025.pdf
Initial concentration of IO4- = moles of NaIO4final volume= 0.025.pdf
anupamfootwear
atomic elements, Hg, Au, molecular elements, H2, F2 molecular compounds, CO2, HCl or ionic compounds Al2S3, KCl Solution atomic elements, Hg, Au, molecular elements, H2, F2 molecular compounds, CO2, HCl or ionic compounds Al2S3, KCl.
atomic elements, Hg, Au, molecular elements, H2,.pdf
atomic elements, Hg, Au, molecular elements, H2,.pdf
anupamfootwear
In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein. Solution In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein..
In Drosophila the gene Tra encodes a protein that is expressed only .pdf
In Drosophila the gene Tra encodes a protein that is expressed only .pdf
anupamfootwear
Ecell = E0 - (0.0591/n)log[H+] 0.207 = 0.0591*log[H+]/1 Therefore - log[H+] = pH = 3.5 Solution Ecell = E0 - (0.0591/n)log[H+] 0.207 = 0.0591*log[H+]/1 Therefore - log[H+] = pH = 3.5.
Ecell = E0 - (0.0591n)log[H+]0.207 = 0.0591log[H+]1Therefore.pdf
Ecell = E0 - (0.0591n)log[H+]0.207 = 0.0591log[H+]1Therefore.pdf
anupamfootwear
Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane. Solution Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane..
Explanation- The processes that occur within the nucleus areRepli.pdf
Explanation- The processes that occur within the nucleus areRepli.pdf
anupamfootwear
Dominant strategy for Jane: Early Dominant strategy for Bob: Early Therefore, Nash equilibrium: (Early, Early) Socially optimal solution: (Early, Early) Solution Dominant strategy for Jane: Early Dominant strategy for Bob: Early Therefore, Nash equilibrium: (Early, Early) Socially optimal solution: (Early, Early).
Dominant strategy for Jane EarlyDominant strategy for Bob Early.pdf
Dominant strategy for Jane EarlyDominant strategy for Bob Early.pdf
anupamfootwear
Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : is a concept that involves the use and management of information and communication technology in avocation of a competitive advantage over an opposing party. Ping sweep :ping sweep is a method which establishes a range of IP addresses that can map to live hosts. The tool used for ping sweeps is fping Port scan : It is basically a attack in which the attacker sends packets to your machine, varying the destination port and find out the sevices you are running. Script kiddies : person who uses existing computer scripts or codes to hack into computers, as he lacks the expertise to write his own code or scripts Structured threat : Explained above in highly structured threat Unstructured threat : these are basically an Organized efforts to attack a un specific target , it can be a organization specific or related to anything but not specific. Solution Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : .
Critical infrastructures Compose the asset or systems or network r.pdf
Critical infrastructures Compose the asset or systems or network r.pdf
anupamfootwear
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In the First molecule there will be Resonance whi.pdf
He down the group, IE decreases. Solution.pdf
He down the group, IE decreases. Solution.pdf
Uncouple agents will never disrupt the electron transport (they will.pdf
Uncouple agents will never disrupt the electron transport (they will.pdf
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf
d.Addition of sulfuric acid to copper(II) oxide p.pdf
d.Addition of sulfuric acid to copper(II) oxide p.pdf
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
The 2nd statement and 3rd statement follow the seed and soil theory .pdf
S2F6SolutionS2F6.pdf
S2F6SolutionS2F6.pdf
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf
Pictures are not legible, could you plz post it again.Solution.pdf
Pictures are not legible, could you plz post it again.Solution.pdf
Carbonic acid leaves the soda solution as CO2 (ca.pdf
Carbonic acid leaves the soda solution as CO2 (ca.pdf
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110 answersSolution110 answers.pdf
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