a)PE + KE = TEFrom the figure from the left upto the top of the .pdf

•

0 likes•2 views

a) PE + KE = TE From the figure from the left upto the top of the dip, the PE remains constant,thus KE also remains constant (as TE is constant). As it goes into the dip of the curve, its PE decreses, so KE must increase so that TE remains constant. So, the body speeds up from the start of the dip to the bottom of the dip. After reaching the bottom, the PE increases,so KE must decrease and thus the body speeds down in this period. After completing the dip, again the KE remains constant and continues with the same speed afterwards b) Here the same thing occurs,. As the body goes inside the dip, its PE decreases and thus the KE increases . After reaching the bottom, it agains speeds down and finally comes to rest at +2.5m c) Yes it is reasonable to have TE to be -ve. This happens when the PE is exessively negative than KE as is the case above d) She would need 10 units of KE as is evident from the curve e) Same analysis as in part (b) f) For far away points, the PE is greater than the value is possible from the points under the thick line. So, some energy must be provided so that the TE increases and thus can be separated So, energy must be put in <---- answer Solution a) PE + KE = TE From the figure from the left upto the top of the dip, the PE remains constant,thus KE also remains constant (as TE is constant). As it goes into the dip of the curve, its PE decreses, so KE must increase so that TE remains constant. So, the body speeds up from the start of the dip to the bottom of the dip. After reaching the bottom, the PE increases,so KE must decrease and thus the body speeds down in this period. After completing the dip, again the KE remains constant and continues with the same speed afterwards b) Here the same thing occurs,. As the body goes inside the dip, its PE decreases and thus the KE increases . After reaching the bottom, it agains speeds down and finally comes to rest at +2.5m c) Yes it is reasonable to have TE to be -ve. This happens when the PE is exessively negative than KE as is the case above d) She would need 10 units of KE as is evident from the curve e) Same analysis as in part (b) f) For far away points, the PE is greater than the value is possible from the points under the thick line. So, some energy must be provided so that the TE increases and thus can be separated So, energy must be put in <---- answer.

Education

So, the body speeds up from the start of the dip to the bottom of the dip.
After reaching the bottom, the PE increases,so KE must decrease and thus the body speeds down
in this period.
After completing the dip, again the KE remains constant and continues with the same speed
afterwards
b)
Here the same thing occurs,. As the body goes inside the dip, its PE decreases and thus the KE
increases .
After reaching the bottom, it agains speeds down and finally comes to rest at +2.5m
c)
Yes it is reasonable to have TE to be -ve. This happens when the PE is exessively negative than
KE as is the case above
d)
She would need 10 units of KE as is evident from the curve
e)
Same analysis as in part (b)
f)
For far away points, the PE is greater than the value is possible from the points under the thick
line. So, some energy must be provided so that the TE increases and thus can be separated
So, energy must be put in <---- answer

More from Ankitchhabra28

the following statements are true Ubiquinone is a small, hydrophobic protein containing a metal group that acts as an electron carrier.Mitochondrial electron carriers with the highest redox potential generally contain copper ions and/or heme groups. Solution the following statements are true Ubiquinone is a small, hydrophobic protein containing a metal group that acts as an electron carrier.Mitochondrial electron carriers with the highest redox potential generally contain copper ions and/or heme groups..

the following statements are true Ubiquinone is a small, hydrophob.pdf

Ankitchhabra28

The evolution of one species is driven by the evolution of the other species. In this, two groups of organisms exert selective pressures on each other, which drives the co-evolution of genes, this may occur within the species or between the species. The co-evolution of certain insects and flowering plants is driven by their symbiotic relationship. Increased chances of pollination and seed dispersal by animals and other vectors, and all other factors that increased the fitness of plants. The co-evolved animals and plants share certain characteristics the benefit each other. Fruits are formed after pollination. The animal pollinator get some reward when it visit its coevolved plant. For example, the co- evolution of flowers with deep fragrance and UV-reflecting pigments. Bees have a keen sense of smell and they can also see UV light, so the flowers with these characteristics attract bees and thus, promotes pollination. In this case, the bees carry the pollen and nectar, pollen is transported to the other plants or used by bees and nectar is utilized as a food source by the bees (mutual benefit in coevolution). Solution The evolution of one species is driven by the evolution of the other species. In this, two groups of organisms exert selective pressures on each other, which drives the co-evolution of genes, this may occur within the species or between the species. The co-evolution of certain insects and flowering plants is driven by their symbiotic relationship. Increased chances of pollination and seed dispersal by animals and other vectors, and all other factors that increased the fitness of plants. The co-evolved animals and plants share certain characteristics the benefit each other. Fruits are formed after pollination. The animal pollinator get some reward when it visit its coevolved plant. For example, the co- evolution of flowers with deep fragrance and UV-reflecting pigments. Bees have a keen sense of smell and they can also see UV light, so the flowers with these characteristics attract bees and thus, promotes pollination. In this case, the bees carry the pollen and nectar, pollen is transported to the other plants or used by bees and nectar is utilized as a food source by the bees (mutual benefit in coevolution)..

The evolution of one species is driven by the evolution of the other.pdf

Ankitchhabra28

Student choosed wrong subject. Question was not showning.Solutio.pdf

Ankitchhabra28

S(sulfur) atom Glucose molecule and compound Oxygen atom DN.pdf

Ankitchhabra28

Question not visible..please post the imageSolutionQuestion no.pdf

Ankitchhabra28

Program 1 public class QueueUsingArray { private static final i.pdf

Ankitchhabra28

PrimeRange.java import java.util.Scanner; public class PrimeRange { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter low range:\"); int low = scan.nextInt(); System.out.println(\"Enter high range:\"); int high = scan.nextInt(); for(int i=low; i<=high; i++){ if(isPrime(i)) System.out.print(i+\" \"); } } public static boolean isPrime(int n){ for (int f = 2; f < n / 2; f++) { if (n % f == 0) { return false; } } return true; } } Output: Enter low range: 10 Enter high range: 30 11 13 17 19 23 29 Solution PrimeRange.java import java.util.Scanner; public class PrimeRange { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter low range:\"); int low = scan.nextInt(); System.out.println(\"Enter high range:\"); int high = scan.nextInt(); for(int i=low; i<=high; i++){ if(isPrime(i)) System.out.print(i+\" \"); } } public static boolean isPrime(int n){ for (int f = 2; f < n / 2; f++) { if (n % f == 0) { return false; } } return true; } } Output: Enter low range: 10 Enter high range: 30 11 13 17 19 23 29.

PrimeRange.java import java.util.Scanner;public class PrimeRan.pdf

Ankitchhabra28

opps no format is there...Solutionopps no format is there....pdf

Ankitchhabra28

Note: Given Code modified as required and required methods added. Code: // class node class Node { // declare the required variable String data; Node next; // constructor Node(){} // constructor with arguments Node(String data) { this.data = data; next = null; } // method to get the data public String getData() { return data; } // method to set the data public void setData(String data) { this.data = data; } // method to get the next data public Node getNext() { return next; } // method to set the next data public void setNext(Node next) { this.next = next; } } // class for a linked list class MyLInkedList { Node head;alue into a list // method to add the v public void add(String num) { if (head == null) { head = new Node(num); } else{ Node temp = new Node(num); Node current = head; if (current != null) { while (current.getNext() != null) { current = current.getNext(); } current.setNext(temp); } } } // method to search a value from a list public void search(String num) { Node current = head; boolean flag=true; if (current != null) { while (current != null) { if(current.getData() == num) { System.out.println(\"Search value \"+ current.getData()+\" is available\"); flag=false; } current = current.getNext(); } if (flag==true) { System.out.println(\"Search value \"+ current.getData()+\" is not available\"); } } else { System.out.println(\"Linked list is empty\"); } } // method to delete a value from a list public void delete(String num) { //Node current = head; boolean flag=true; if (head == null) { System.out.println(\"Can\'t Delete, Linked list is empty\"); return; } if(head.getData() == num) { head=head.getNext(); return; } Node temp = new Node(num); Node cr = head; Node prev=null; while ((cr != null) && !(cr.getData() == num)) { prev=cr; cr = cr.getNext(); } prev.setNext(cr.getNext()); } // method to print the list public void print() { Node current = head; if (current != null) { while (current != null) { System.out.println(current.getData()); current = current.getNext(); } } } } // method to test the class public class MyLInkedListTester { // main method to perform the linked list operatrions public static void main(String args[]) { MyLInkedList llist = new MyLInkedList(); // insert the values into list llist.add(\"1\"); llist.add(\"2\"); llist.add(\"3\"); llist.add(\"4\"); // print the list values llist.print(); // sear the values in a list llist.search(\"1\"); llist.print(); // delete the value from a list llist.delete(\"4\"); System.out.println(\"List after deletion:\"); llist.print(); } } Result: 1 2 3 4 Search value 1 is available 1 2 3 4 List after deletion: 1 2 3 Solution Note: Given Code modified as required and required methods added. Code: // class node class Node { // declare the required variable String data; Node next; // constructor Node(){} // constructor with arguments Node(String data) { this.data = data; next = null; } // method to get the data public String getData() { return data; } // method to set the data public void se.

Note Given Code modified as required and required met.pdf

Ankitchhabra28

No, weight is not the measure of quantity of matter.Quantity of ma.pdf

Ankitchhabra28

Two. The pyruvate with carbons 4-6 will go throu.pdf

Ankitchhabra28

Listing.javaimport java.util.Scanner;public class Listing { .pdf

Ankitchhabra28

is rightSolutionis right.pdf

Ankitchhabra28

Information managementThe term information management is used .pdf

Ankitchhabra28

Hi,I have implemented increment() method. Please find the below up.pdf

Ankitchhabra28

P belongs to the V A group of the periodic table so its valance is 5 ( having 5 electrons around it). It combines with single valenced Cl atoms makes one single bond.When it combines with 3 Cl atoms forms 3 single bonds ie., 3 bonding pairs ( there is one lone pair on P). Solution P belongs to the V A group of the periodic table so its valance is 5 ( having 5 electrons around it). It combines with single valenced Cl atoms makes one single bond.When it combines with 3 Cl atoms forms 3 single bonds ie., 3 bonding pairs ( there is one lone pair on P)..

P belongs to the V A group of the periodic table .pdf

Ankitchhabra28

only (2) can have equilibrium because all the re.pdf

Ankitchhabra28

Mass Spectrum is 2-Methyl-2-pentene. However I ca.pdf

Ankitchhabra28

Intermolecular forces, also called van der Waals forces are attractions between molecules. Between all molecules are London dispersion forces. These are present even if there are other types of van der Waals forces present. We also see dipole-dipole attractions and hydrogen bonding. SO4^2- == a nonpolar polyatomic ion, London dispersion forces. Since it is charged it is attracted to water and positively charged ions. Solution Intermolecular forces, also called van der Waals forces are attractions between molecules. Between all molecules are London dispersion forces. These are present even if there are other types of van der Waals forces present. We also see dipole-dipole attractions and hydrogen bonding. SO4^2- == a nonpolar polyatomic ion, London dispersion forces. Since it is charged it is attracted to water and positively charged ions..

Intermolecular forces, also called van der Waals .pdf

Ankitchhabra28

I get C but im really not sure .pdf

Ankitchhabra28

More from Ankitchhabra28 (20)