Two particles A and B of equal mass have velocities as shown in the f.pdf
•
0 likes•2 views
Two particles A and B of equal mass have velocities as shown in the figure to the right. If the particles collide, circle the situation(s) shown 111 the figures below that represent possible velocities for the particles after impact. The velocities in all figures are shown to scale. [0.5 mark for each correct answer] Solution Since thy have equal masses they will exchange velocities both for horizontal and vertical component. horizontal is head on collision so they will have horizontal velocity in opposite direction..
Report
Share
Report
Share
Download to read offline
![Two particles A and B of equal mass have velocities as shown in the figure to the right. If the
particles collide, circle the situation(s) shown 111 the figures below that represent possible
velocities for the particles after impact. The velocities in all figures are shown to scale. [0.5 mark
for each correct answer]
Solution
Since thy have equal masses they will exchange velocities both for horizontal and vertical
component.
horizontal is head on collision so they will have horizontal velocity in opposite direction.](https://image.slidesharecdn.com/twoparticlesaandbofequalmasshavevelocitiesasshowninthef-230702022652-a0d272a8/85/Two-particles-A-and-B-of-equal-mass-have-velocities-as-shown-in-the-f-pdf-1-320.jpg)
Recommended
Imagine a spherical balloon filled wild helium- the balloon itself is.pdf
Imagine a spherical balloon filled wild helium- the balloon itself is magically crafted to expand
and contract freely without exerting any restorative force Some weights (with negligible volume)
are hung from the balloon such that the total mass of helium + balloon + weights is m Assume
the balloon is surrounded by with constant density p_0 and constant pressure P_0, independent
of altitude.
Solution
Gas equation PV = nRT
As the pressure P is constant
P dV/dt = nR dT/dt
dV/dt =nR/P dT/dt
initial condition P0V0 = nRT0
dV/dt = (V0/T0) dT/dt
upward thrust on the ballon
F = Vg – mg
Differentiate with t , we have
dF/dt = gdV/dt
initial the balloon is at rest hence
V0 g – mg = 0 hence m = V0
mda/dt = gdV/dt
= g*(V0/T0)* dT/dt
da/dt = g/T0 * dT/dt
we can express rate of change of acceleration in terms of g, which has units of acceleration
da/dt = (1/T0)dT/dt
integrating it
a(t) = T(t)/T0
given T(t) = T0(2 - 1/t+1)
a(t) = 2 – 1/t+1
dv(t)/dt = 2 – 1/t+1
v(t) = 2t – ln(t+1).
If a substance has a pH of 4, is it acidic or basic. What type of io.pdf
If a substance has a pH of 4, is it acidic or basic. What type of ion is it going to give off when
dissolved in water?
Solution
If a substance has a pH of 4, it is acidic in nature.
An acid donates hydrogen ions. So,when this acid is dissolved in water (H2O), the balance
between hydrogen ions and hydroxide ions is shifted. There will be more hydrogen ions rather
than hydroxide ions in the solution. This kind of solution is called acidic solution.
H2O -----------------> H+ + OH-
Note:
If,
pH < 7 = Acidic in nature
pH = 7 = Neutral solutions
pH > 7 = Basic / Alkaline in nature.
Help with answering these questions for my Human Health Class Biolog.pdf
Help with answering these questions for my Human Health Class Biology 2020
INGREDIENTS: COOKED FETTUCCINI PASTA (WATER, SEMOLINA [DURUM
WHEAT, NIACIN, IRON, FERROUS SULFATE, THIAMINE MONONITRATE,
RIBOFLAVIN, FOLIC ACID], EGG WHITES), GARLIC BREAD (BLEACHED ENRICHED
FLOUR [MALTED BARLEY FLOUR, NIACIN, REDUCED IRON, THIAMINE
MONONITRATE, RIBOFLAVIN, FOLATE], WATER, VEGETABLE OIL [SOYBEAN
AND/OR COTTONSEED OIL], YEAST, SALT, HIGH FRUCTOSE CORN SYRUP,
ENRICHED WHEAT FLOUR [FLOUR, NIACIN, REDUCED IRON, THIAMINE
MONONITRATE, RIBOFLAVIN, FOLIC ACID], DATEM, VEGETABLE SHORTENING
[PARTIALLY HYDROGENATED SOYBEAN OIL, TBHQ ADDED TO PRESERVE
FRESHNESS], CONTAINS LESS THAN 2% OF THE FOLLOWING: ASCORBIC ACID, L-
CYSTEINE HYDROCHLORIDE, AZODICAR-BONAMIDE, ENZYMES, CALCIUM
PEROXIDE, CALCIUM PROPIONATE, ENZYME, TOPPINGS [LIQUID AND PARTIALLY
HYDROGENATED SOYBEAN OIL, WATER, SALT, LECITHIN, MONOGLYCERIDES,
SODIUM BENZOATE {PRESERVATIVE}, CITRIC ACID, NATURAL FLAVOR,
ARTIFICIAL FLAVOR, BETA CAROTENE, VITAMIN A PALMITATE, CALCIUM
DISODIUM EDTA {TO PROTECT FLAVOR}, PARSLEY FLAKES]), WATER, CARROTS,
BROCCOLI, HEAVY WHIPPING CREAM, SOYBEAN OIL, CREAM CHEESE
(PASTEURIZED MILK AND CREAM, CHEESE CULTURES, SALT, CAROB BEAN GUM),
ROMANO (FROM COW’S MILK) AND PARMESAN CHEESE (PART-SKIM MILK,
CHEESE CULTURES, SALT, ENZYMES, CELLULOSE POWDER [PREVENTS CAKING]),
HALF AND HALF (MILK, CREAM), CONTAINS 2% OR LESS OF: GARLIC SHERRY
WINE, STABILIZER (MODIFIED FOOD STARCH, SODIUM PHOSPHATE, MONO- AND
DIGLYCERIDES, DEHYDRATED GARLIC, XANTHAN AND GUAR GUMS, SALT,
PAPRIKA), SUGAR, SALT, BUTTER (CREAM, SALT), FLAVORINGS, BUTTER FLAVOR
(PARTIALLY HYDROGENATED SOYBEAN OIL, FLAVOR [BUTTER OIL, ENZYME
MODIFIED BUTTERFAT, WHEY POWDER, NONFAT DRY MILK POWDER, SOY
LECITHIN {EMULSIFIER}]), POLYGLYCEROL ESTERS OF FATTY ACIDS, SPICE,
PAPRIKA, BETA CAROTENE.
Nutrition Facts
Serving Size 1 Meal (397g)
Amount Per Serving
Calories 770
Calories from Fat 420
% Daily Value*
Total Fat 46g
?%
Saturated Fat 16g
?%
Trans Fat 0.5g
Cholesterol 30mg
?%
Sodium 129mg
?%
Total Carbohydrate 69g
?%
Dietary Fiber 11g
?%
Sugars 9g
Protein 19g
Vitamin A 80%
•
Vitamin C 0%
Calcium 35%
•
Iron 6%
* Percent Daily Values are based on a 2,000 calorie diet. Your Daily Values may be higher or
lower depending on your calorie needs.
Calories
2,000
2,500
Total Fat
Less than
65g
80g
Sat Fat
Less than
20g
25g
Cholesterol
Less than
300mg
300mg
Sodium
Less than
2400mg
2400mg
Total Carbohydrate
300g
375g
Dietary Fiber
25g
30g
Referring to the label above:
1. a. Is this product a good source of fiber?
b. Calculate the percentage of the Daily Value for fiber for a 2,500-calorie diet that is
contributed by this product.
c. Calculate the percentage of the DRI for a 2,500-calorie diet that is contributed by this
product.
2. Is this a nutrient-dense product? Why or why not?
3. What percentage of the Daily Value for total carbohydrate (for 2,000 calories) comes from
this product?
Label Analysis—Lipids (from chapter 5)
Instructions: Use t.
For each task, submit your source java code file.(1) Objective Im.pdf
For each task, submit your source java code file.
(1) Objective: Implement Link List
Write a class that maintains the scores for a game application. Implement the addition and
removal function to update the database. The gamescore.txt contains player’ name and score data
record fields separated by comma. For Removal function, uses the name field to select record to
remove the game score record.
Download – List.java, LList.java, Dlink.java, GameEntry.java, gamescore.txt
(a)Read gamescore.txt to initialize the Linked list in sorted order. (1 point)
(b)Provide Remove and Add function for user to update the sorted linked list. (1 point)
(2)Add a reverse function to the LList.java class to reverse the order of the linked list. (2 points).
---------------------------------------------------------------------------------------------------------------------
-----------------------------------------
//gamescore.txt
Mike,1105
Rob,750
Paul,720
Anna,660
Rose,590
Jack,510
--------------------------------------------------------------------------------------------------------------------
-----------------------------------------------------------------------
//DLink.java
/** Source code example for \"A Practical Introduction to Data
Structures and Algorithm Analysis, 3rd Edition (Java)\"
by Clifford A. Shaffer
Copyright 2008-2011 by Clifford A. Shaffer
*/
/** Doubly linked list node */
class DLink {
private E element; // Value for this node
private DLink next; // Pointer to next node in list
private DLink prev; // Pointer to previous node
/** Constructors */
DLink(E it, DLink p, DLink n)
{ element = it; prev = p; next = n; }
DLink(DLink p, DLink n) { prev = p; next = n; }
/** Get and set methods for the data members */
DLink next() { return next; }
DLink setNext(DLink nextval)
{ return next = nextval; }
DLink prev() { return prev; }
DLink setPrev(DLink prevval)
{ return prev = prevval; }
E element() { return element; }
E setElement(E it) { return element = it; }
}
--------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------
//GameEntry.java
public class GameEntry {
protected String name;
protected int score;
public GameEntry(String n, int s) {
name = n;
score = s;
}
public String getName() {return name;}
public int getScore() {return score;}
public String toString() {
return \"(\"+name+\",\"+score+\")\";
}
}
--------------------------------------------------------------------------------------------------------------------
----------------------------------
//List.java
/** Source code example for \"A Practical Introduction to Data
Structures and Algorithm Analysis, 3rd Edition (Java)\"
by Clifford A. Shaffer
Copyright 2008-2011 by Clifford A. Shaffer
*/
/** List ADT */
public interface List {
/** Remove all contents from the list, so it is once again
empty. Client is responsible for reclaiming storage
used by the list elements. */
p.
Emperor penguins live on a diet of fish and crustaceans obtained fro.pdf
Emperor penguins live on a diet of fish and crustaceans obtained from cold Antarctic seawaters.
However, during their breeding cycle, they migrate to inland breeding grounds nearly 100 km
away from their source of food. As a result, male emperor penguins do not eat for the two
months that they protect and incubate the single egg laid by their mate. During this time, each of
the tens of thousands of male penguins can lose up to 50% of its body mass (by dry weight).
Where does the majority of this penguin mass go?A. The mass is converted to energy and used
up.B. The mass is converted to urine and feces and eliminated from the body.C. The mass is
converted to ATP molecules.D. The mass is released as CO2 and H2O.
Solution
DURING THIS PERIOD THE MASS WOULD BE USED UP IN FORM OF ENERGY
REQUIRED DURING SUCH PERIOD. OPTION A. The mass is converted to energy and used
up. is answer.
Even before the Ownership Society programs of Presidents Clinton.pdf
Even before the \\Ownership Society\" programs of Presidents Clinton and Bush, the federal
government was heavily involved in the housing market. In a study just prior to the foreclosure
crisis starting in 2008, the Urban Institute started about 80% of the estimated $200 billion
of federal housing subsidies consisted of tax breaks. Samples indicated that federal housing
benefits average $8,268 for those with incomes $200,000 and $500,000 and only $365 for those
with incomes of $40,000 to $50,000. The respective standard deviations were $2,100 and $150.
They were obtained from sample sizes of 150.
1. the statistical test obtained is 45.9741 and p value is 0.0000. Test whether the benefits
to the high income group are at least $7,500 more than the benefits to the low income
group using = 0:05. Make sure to state the null and the alternative hypotheses. Assume
bene ts are normally distributed for both income groups.
Solution.
Do the following picSolutionHere is the solution for the .pdf
Do the following pic:
Solution
Here is the solution for the first 2 problems in the pic:
#include
using namespace std;
template
class Dec
{
public:
//hasDuplicate function that returns true if the private data member has an item that appears
twice or more in the list.
bool hasDuplicate();
//eraseAll function to remove all items which are equal to the item specified in the function
parameter.
void eraseAll(ElementType element);
//isLessThan function to determine if another Dec object is greater than the Dec object itself.
//If the summation of all items from the Dec object itself is less than the summation of all
//items from another Dec object, return true, otherwise, return false.
bool isLessThan(Dec other);
//push_second function to add an item after the first item if there are more than one items.
void push_second(ElementType element);
private:
vector dec;
};
template
bool Dec::hasDuplicate()
{
for(vector::iterator ot = dec.begin(); ot != dec.end(); ot++)
for(vector::iterator it = dec.begin() + ot + 1; it != dec.end(); it++)
if(*it == *ot)
return true;
return false;
}
template
bool Dec::isLessThan(Dec other)
{
int sum1 = 0, sum2 = 0;
for(vector::iterator it = dec.begin(); it != dec.end(); it++)
sum1 += *it;
for(vector::iterator it = other.dec.begin(); it != other.dec.end(); it++)
sum2 += *it;
return sum1 < sum2;
}.
Can one mRNA be translated into more than one protein sequence Can .pdf
Can one mRNA be translated into more than one protein sequence? Can more than one type of
mRNA be transcribed in a single cell?
Solution
The prokaryotic mRNAs are called as polycistronic since a single mRNA codes for more than
one protein, whareas the eukaryotic mRNA is monocistronic, because it codes for single
polypeptide.hence the given statement true for prokaryotes. Second statement is also absolutely
true. Because in a cell meny genes are in the process of transcription or translation or both at the
same time in the same cell.hence one cell has more than one type of mRNA..
Recommended
Imagine a spherical balloon filled wild helium- the balloon itself is.pdf
Imagine a spherical balloon filled wild helium- the balloon itself is magically crafted to expand
and contract freely without exerting any restorative force Some weights (with negligible volume)
are hung from the balloon such that the total mass of helium + balloon + weights is m Assume
the balloon is surrounded by with constant density p_0 and constant pressure P_0, independent
of altitude.
Solution
Gas equation PV = nRT
As the pressure P is constant
P dV/dt = nR dT/dt
dV/dt =nR/P dT/dt
initial condition P0V0 = nRT0
dV/dt = (V0/T0) dT/dt
upward thrust on the ballon
F = Vg – mg
Differentiate with t , we have
dF/dt = gdV/dt
initial the balloon is at rest hence
V0 g – mg = 0 hence m = V0
mda/dt = gdV/dt
= g*(V0/T0)* dT/dt
da/dt = g/T0 * dT/dt
we can express rate of change of acceleration in terms of g, which has units of acceleration
da/dt = (1/T0)dT/dt
integrating it
a(t) = T(t)/T0
given T(t) = T0(2 - 1/t+1)
a(t) = 2 – 1/t+1
dv(t)/dt = 2 – 1/t+1
v(t) = 2t – ln(t+1).
If a substance has a pH of 4, is it acidic or basic. What type of io.pdf
If a substance has a pH of 4, is it acidic or basic. What type of ion is it going to give off when
dissolved in water?
Solution
If a substance has a pH of 4, it is acidic in nature.
An acid donates hydrogen ions. So,when this acid is dissolved in water (H2O), the balance
between hydrogen ions and hydroxide ions is shifted. There will be more hydrogen ions rather
than hydroxide ions in the solution. This kind of solution is called acidic solution.
H2O -----------------> H+ + OH-
Note:
If,
pH < 7 = Acidic in nature
pH = 7 = Neutral solutions
pH > 7 = Basic / Alkaline in nature.
Help with answering these questions for my Human Health Class Biolog.pdf
Help with answering these questions for my Human Health Class Biology 2020
INGREDIENTS: COOKED FETTUCCINI PASTA (WATER, SEMOLINA [DURUM
WHEAT, NIACIN, IRON, FERROUS SULFATE, THIAMINE MONONITRATE,
RIBOFLAVIN, FOLIC ACID], EGG WHITES), GARLIC BREAD (BLEACHED ENRICHED
FLOUR [MALTED BARLEY FLOUR, NIACIN, REDUCED IRON, THIAMINE
MONONITRATE, RIBOFLAVIN, FOLATE], WATER, VEGETABLE OIL [SOYBEAN
AND/OR COTTONSEED OIL], YEAST, SALT, HIGH FRUCTOSE CORN SYRUP,
ENRICHED WHEAT FLOUR [FLOUR, NIACIN, REDUCED IRON, THIAMINE
MONONITRATE, RIBOFLAVIN, FOLIC ACID], DATEM, VEGETABLE SHORTENING
[PARTIALLY HYDROGENATED SOYBEAN OIL, TBHQ ADDED TO PRESERVE
FRESHNESS], CONTAINS LESS THAN 2% OF THE FOLLOWING: ASCORBIC ACID, L-
CYSTEINE HYDROCHLORIDE, AZODICAR-BONAMIDE, ENZYMES, CALCIUM
PEROXIDE, CALCIUM PROPIONATE, ENZYME, TOPPINGS [LIQUID AND PARTIALLY
HYDROGENATED SOYBEAN OIL, WATER, SALT, LECITHIN, MONOGLYCERIDES,
SODIUM BENZOATE {PRESERVATIVE}, CITRIC ACID, NATURAL FLAVOR,
ARTIFICIAL FLAVOR, BETA CAROTENE, VITAMIN A PALMITATE, CALCIUM
DISODIUM EDTA {TO PROTECT FLAVOR}, PARSLEY FLAKES]), WATER, CARROTS,
BROCCOLI, HEAVY WHIPPING CREAM, SOYBEAN OIL, CREAM CHEESE
(PASTEURIZED MILK AND CREAM, CHEESE CULTURES, SALT, CAROB BEAN GUM),
ROMANO (FROM COW’S MILK) AND PARMESAN CHEESE (PART-SKIM MILK,
CHEESE CULTURES, SALT, ENZYMES, CELLULOSE POWDER [PREVENTS CAKING]),
HALF AND HALF (MILK, CREAM), CONTAINS 2% OR LESS OF: GARLIC SHERRY
WINE, STABILIZER (MODIFIED FOOD STARCH, SODIUM PHOSPHATE, MONO- AND
DIGLYCERIDES, DEHYDRATED GARLIC, XANTHAN AND GUAR GUMS, SALT,
PAPRIKA), SUGAR, SALT, BUTTER (CREAM, SALT), FLAVORINGS, BUTTER FLAVOR
(PARTIALLY HYDROGENATED SOYBEAN OIL, FLAVOR [BUTTER OIL, ENZYME
MODIFIED BUTTERFAT, WHEY POWDER, NONFAT DRY MILK POWDER, SOY
LECITHIN {EMULSIFIER}]), POLYGLYCEROL ESTERS OF FATTY ACIDS, SPICE,
PAPRIKA, BETA CAROTENE.
Nutrition Facts
Serving Size 1 Meal (397g)
Amount Per Serving
Calories 770
Calories from Fat 420
% Daily Value*
Total Fat 46g
?%
Saturated Fat 16g
?%
Trans Fat 0.5g
Cholesterol 30mg
?%
Sodium 129mg
?%
Total Carbohydrate 69g
?%
Dietary Fiber 11g
?%
Sugars 9g
Protein 19g
Vitamin A 80%
•
Vitamin C 0%
Calcium 35%
•
Iron 6%
* Percent Daily Values are based on a 2,000 calorie diet. Your Daily Values may be higher or
lower depending on your calorie needs.
Calories
2,000
2,500
Total Fat
Less than
65g
80g
Sat Fat
Less than
20g
25g
Cholesterol
Less than
300mg
300mg
Sodium
Less than
2400mg
2400mg
Total Carbohydrate
300g
375g
Dietary Fiber
25g
30g
Referring to the label above:
1. a. Is this product a good source of fiber?
b. Calculate the percentage of the Daily Value for fiber for a 2,500-calorie diet that is
contributed by this product.
c. Calculate the percentage of the DRI for a 2,500-calorie diet that is contributed by this
product.
2. Is this a nutrient-dense product? Why or why not?
3. What percentage of the Daily Value for total carbohydrate (for 2,000 calories) comes from
this product?
Label Analysis—Lipids (from chapter 5)
Instructions: Use t.
For each task, submit your source java code file.(1) Objective Im.pdf
For each task, submit your source java code file.
(1) Objective: Implement Link List
Write a class that maintains the scores for a game application. Implement the addition and
removal function to update the database. The gamescore.txt contains player’ name and score data
record fields separated by comma. For Removal function, uses the name field to select record to
remove the game score record.
Download – List.java, LList.java, Dlink.java, GameEntry.java, gamescore.txt
(a)Read gamescore.txt to initialize the Linked list in sorted order. (1 point)
(b)Provide Remove and Add function for user to update the sorted linked list. (1 point)
(2)Add a reverse function to the LList.java class to reverse the order of the linked list. (2 points).
---------------------------------------------------------------------------------------------------------------------
-----------------------------------------
//gamescore.txt
Mike,1105
Rob,750
Paul,720
Anna,660
Rose,590
Jack,510
--------------------------------------------------------------------------------------------------------------------
-----------------------------------------------------------------------
//DLink.java
/** Source code example for \"A Practical Introduction to Data
Structures and Algorithm Analysis, 3rd Edition (Java)\"
by Clifford A. Shaffer
Copyright 2008-2011 by Clifford A. Shaffer
*/
/** Doubly linked list node */
class DLink {
private E element; // Value for this node
private DLink next; // Pointer to next node in list
private DLink prev; // Pointer to previous node
/** Constructors */
DLink(E it, DLink p, DLink n)
{ element = it; prev = p; next = n; }
DLink(DLink p, DLink n) { prev = p; next = n; }
/** Get and set methods for the data members */
DLink next() { return next; }
DLink setNext(DLink nextval)
{ return next = nextval; }
DLink prev() { return prev; }
DLink setPrev(DLink prevval)
{ return prev = prevval; }
E element() { return element; }
E setElement(E it) { return element = it; }
}
--------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------
//GameEntry.java
public class GameEntry {
protected String name;
protected int score;
public GameEntry(String n, int s) {
name = n;
score = s;
}
public String getName() {return name;}
public int getScore() {return score;}
public String toString() {
return \"(\"+name+\",\"+score+\")\";
}
}
--------------------------------------------------------------------------------------------------------------------
----------------------------------
//List.java
/** Source code example for \"A Practical Introduction to Data
Structures and Algorithm Analysis, 3rd Edition (Java)\"
by Clifford A. Shaffer
Copyright 2008-2011 by Clifford A. Shaffer
*/
/** List ADT */
public interface List {
/** Remove all contents from the list, so it is once again
empty. Client is responsible for reclaiming storage
used by the list elements. */
p.
Emperor penguins live on a diet of fish and crustaceans obtained fro.pdf
Emperor penguins live on a diet of fish and crustaceans obtained from cold Antarctic seawaters.
However, during their breeding cycle, they migrate to inland breeding grounds nearly 100 km
away from their source of food. As a result, male emperor penguins do not eat for the two
months that they protect and incubate the single egg laid by their mate. During this time, each of
the tens of thousands of male penguins can lose up to 50% of its body mass (by dry weight).
Where does the majority of this penguin mass go?A. The mass is converted to energy and used
up.B. The mass is converted to urine and feces and eliminated from the body.C. The mass is
converted to ATP molecules.D. The mass is released as CO2 and H2O.
Solution
DURING THIS PERIOD THE MASS WOULD BE USED UP IN FORM OF ENERGY
REQUIRED DURING SUCH PERIOD. OPTION A. The mass is converted to energy and used
up. is answer.
Even before the Ownership Society programs of Presidents Clinton.pdf
Even before the \\Ownership Society\" programs of Presidents Clinton and Bush, the federal
government was heavily involved in the housing market. In a study just prior to the foreclosure
crisis starting in 2008, the Urban Institute started about 80% of the estimated $200 billion
of federal housing subsidies consisted of tax breaks. Samples indicated that federal housing
benefits average $8,268 for those with incomes $200,000 and $500,000 and only $365 for those
with incomes of $40,000 to $50,000. The respective standard deviations were $2,100 and $150.
They were obtained from sample sizes of 150.
1. the statistical test obtained is 45.9741 and p value is 0.0000. Test whether the benefits
to the high income group are at least $7,500 more than the benefits to the low income
group using = 0:05. Make sure to state the null and the alternative hypotheses. Assume
bene ts are normally distributed for both income groups.
Solution.
Do the following picSolutionHere is the solution for the .pdf
Do the following pic:
Solution
Here is the solution for the first 2 problems in the pic:
#include
using namespace std;
template
class Dec
{
public:
//hasDuplicate function that returns true if the private data member has an item that appears
twice or more in the list.
bool hasDuplicate();
//eraseAll function to remove all items which are equal to the item specified in the function
parameter.
void eraseAll(ElementType element);
//isLessThan function to determine if another Dec object is greater than the Dec object itself.
//If the summation of all items from the Dec object itself is less than the summation of all
//items from another Dec object, return true, otherwise, return false.
bool isLessThan(Dec other);
//push_second function to add an item after the first item if there are more than one items.
void push_second(ElementType element);
private:
vector dec;
};
template
bool Dec::hasDuplicate()
{
for(vector::iterator ot = dec.begin(); ot != dec.end(); ot++)
for(vector::iterator it = dec.begin() + ot + 1; it != dec.end(); it++)
if(*it == *ot)
return true;
return false;
}
template
bool Dec::isLessThan(Dec other)
{
int sum1 = 0, sum2 = 0;
for(vector::iterator it = dec.begin(); it != dec.end(); it++)
sum1 += *it;
for(vector::iterator it = other.dec.begin(); it != other.dec.end(); it++)
sum2 += *it;
return sum1 < sum2;
}.
Can one mRNA be translated into more than one protein sequence Can .pdf
Can one mRNA be translated into more than one protein sequence? Can more than one type of
mRNA be transcribed in a single cell?
Solution
The prokaryotic mRNAs are called as polycistronic since a single mRNA codes for more than
one protein, whareas the eukaryotic mRNA is monocistronic, because it codes for single
polypeptide.hence the given statement true for prokaryotes. Second statement is also absolutely
true. Because in a cell meny genes are in the process of transcription or translation or both at the
same time in the same cell.hence one cell has more than one type of mRNA..
why is there an uneven distribution of photosystem 1 and 2 on the th.pdf
why is there an uneven distribution of photosystem 1 and 2 on the thylakoid membrane?
Solution
Photosynthesis is carried out by the cyanobacteria, plants involved the absorption of light and the
conversion of its energy into chemical energy. The photosynthetic apparatus is made up of 4
multiprotein complexes, which are asymmetrically embedded in the thyalkoid memberane. It is
the high structural and functional coordination between these 4 complexes, which enables the
efficient coupling of photosynthetic electron transfer and photophosphorylation.
The 4 complexes - Photosystem II (PSII), cytochrome b6f complex, PS I and the proton-ATP
synthase, are unevenly distributed between the 2 regions of the thylakoids. The PS II complex is
situated in the appressed grana-lamella region, the PS I and the ATP synthase complex are
located in the non-appressed stroma-lamella region.
Light driven electron transfer is intiated with the oxidation of water mole. by the PS II complex
and the reduction of quinones associated with the PS II complex. The electrons are then
transferred across the memberane to plastocyanin via the Cyt b6f complex. Plastocyanin is a
copper protein facing the lumen of the thylakoids, is the secondary electron donor of the PS I
complex. In the PS I the electron are translocated across the memberane to ferredoxin which
further transfers them to NADP+. The membrane potental and proton gradient resulting from this
photochemical electron transport provide the motive force required for the formation of ATP by
the ATP synthase.
The transport of electron against the electrical potential and the formation of ATP and NADPH
are made possible due to the unique ability of the thylakoids to absorb light energy. The 2
pigment complexes (PS II and PS I) are responsible for harvesting the light and performing the
charge separation across the thylakoid memberanes.
Photosytem II (PS II):
1. PS II is located at the inner surface the grana thyalkoid memberane
2. The photocentre is P680
3. Pigments absorb shorter wavelengths of light (<680nm)
4. Participates only in non cyclic photophosphorylation
5. It is associated with photolysis of water
6. Main functions are ATP synthesis and hydrolysis of water
Photosystem I (PS I)
1. PS I is located at the outer surface the grana thylakoid memberane
2. The photocentre is P700
3. Pigments absorb longer wavelength of light (>680nm)
4. Participates in cyclic and non-cyclic photophosphorylation
5. It is not associated with photolysis of water
6. Main function is ATP synthesis
This is the purpose the two photosystems (PS) are unevely distributed in thyalkoid memberane..
Which of the following are examples of cell signaling Choose all tha.pdf
Which of the following are examples of cell signaling? Choose all that apply. As a response to
increasing densities, bacteria coordinate behaviors and form biofilms. Insulin hormone produced
by the pancreas triggers the intake of glucose by liver cells. Carbon dioxide diffuses across the
cell membrane into the blood plasma. Chlorophyll absorbs and transfers light energy during
photosynthesis. A neuron releases neurotransmitters into a synapse, triggering a target nerve
cell.
Solution
Cell signaling
Cell signaling is the communication between the cells through which they coordinate
certain activities.
What is the purpose of testing a program with different dataSol.pdf
What is the purpose of testing a program with different data?
Solution
The purpose of testing a program functions according to the expectation. The overall goal of
testing is to find a bug that might effect the functioning of the program. There are various types
of testing and each serve their own purpose. Different data is used for each type of testing. We
have to check the whether internal module is functioning properly, how the program responds to
invalid data, test how changing one module will effect the whole system etc.. There are many
automated tools which can be used for comprehensive testing..
what are the factors contributing to the rise in global media Will .pdf
what are the factors contributing to the rise in global media? Will global media eventually
replace national media?
Solution
Q1) Media is the mass communication system of letting people informed through broadcasting,
publishing, and the internet. It enriches public awareness and knowledge, which is very
important in today’s fast moving world.
Global media gives the news of whole world, suppose a person sitting in India can get the
happenings in the US.
Factors are as below:
Point 1) The urge of national media: There are so many small national media in various
countries. They want to get international news as well but can’t get because of their small
infrastructure. Therefore, they are now becoming tied up with big international media agencies in
order to get regular updating. This is the reason of rising trend of global media.
Point 2) Globalization of economy: Most of the countries have free-trade environment now.
They need cheaper rate of factors of production (land, labor, capital, and organization); the
global media is a great source of information of it.
Point 3) Technological development: The development of internet gives the platform of rising of
global media. A person sitting on computer can get to know the whole world.
Q2) To some extent it is true. Since national media getting tied up with big global firms, the
chance of being acquired may be high in future. The small national media could be acquired by
those global media..
WACC of Alibaba company WACC of Alibaba companySolutionTh.pdf
WACC of Alibaba company
WACC of Alibaba company
Solution
The Weighted Average Cost of Capital Alibaba company as of the latest quarter report is
18.28%. Alibaba company generates more return on investment than the cost of capital. WACC
is the rate that a company is expected to pay its stockholders and other lenders of money..
The forecast equation for mean wind in a turbulent flow is U_it + .pdf
The forecast equation for mean wind in a turbulent flow is: U_i/t + U_j U_i/x_j = delta_i3 g +
f_c epsilon_ij3 U_j - 1/p P/x_i + v ^2 U_i/x^2_j u\'_i u\'_j/x_j Name each term, and give its
physical interpretation.
Solution
solution
A is storage of mean momentum
B is advection of mean momentum by the mean wind
C is gravity to act in the vertical direction only
D is the Coriolis effect due to the earth’s rotation
E is the mean pressure gradient forces
F is the effect of viscous stress on the mean motions
G is the effect of Reynolds stress on the mean motions (also called the divergence of turbulent
momentum flux).
The image below is of a charged moving in a uniform magnetic field. I.pdf
The image below is of a charged moving in a uniform magnetic field. Is the particle positively
or negatively charged?
Solution
According to Fleming left hand rule,
When the thumb,first finger,middle finger are streatched perpendicular to each other and the first
finger represents the magnetic field direction ,middle finger represents the direction of motion of
positive charged particle then the thumb represents force on charged particle.
In figure , magnetic field into the paper and the particle deflect right.So, the charged particle is
negative nature.
The concept of a sampling distributionA-Is essential for drawing c.pdf
The concept of a sampling distribution
A-Is essential for drawing conclusions about a population from the results of a sample.
B- Is related to the use of probability theory in inferential statistics.
C-Helps us use statistics to estimate the corresponding true values in the population.
D-All of the above.
Solution
The concept of a sampling distribution
D-All of the above..
The diets common to some cultures emphasize meat. Meat, milk, and br.pdf
The diets common to some cultures emphasize meat. Meat, milk, and bread, but few fruits and
vegetables, might be a typical meal for populations in northern latitudes and in desert areas, for
example. What ecological factors seem relevant in explaining this human habit?
Solution
Various studies have shown that populations living in northern latitude and desert areas have
dietary patterns higher in animal fat such as meat and dairy products compared to those living in
southern latitudes who thrive majorly on fruits and vegetables. This may be because land in
northern and desert areas had been less fertile to grow seasonal vegeies, due to extremes of
environment and limited rainfalls. People living in these areas over generations have been mostly
hunter-gatherers. Such societies had depended for their economic survival primarily on hunting,
gathering (wild vegetable and fruits and small land fauna), and fishing. Thus they might have
developed the food habit of increased meat and milk consumption (as they are largely dependent
on herd of cattle/animals) along with staple diet of breads, over the generations.
On the other hand, southern populations residing on more fertile pieces of land relish a
vegetarian diet consisting of fresh fruits and vegetables. They are richer in biodiversity due to
moderate climates and rainfall distribution throughtout the year. These populations are typically
farmers, food growers over several generations and have developed food habits of vegetarian
diets..
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus.pdf
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus and at least one dominant
inant gene A pi R P A rose comb is produced when at least and two recessive genes are present
at the the first locus at pea comb is produced when two recessive genes are least one dominant
gene is at are present at the first and the second locas sNuces 2 for comb shape of the parents in
the fol walnut, 1 single, and 1 pea offspring. ngle comb ing cross: Rose 3/. The mutation (bard es
is inherited as a maternal effect recessive allele. What is the expected ratio of preho termale and
a bad bed male? in offspring cross between bcd normal: mutant 1 3 mutant normal c 3 mutant 1
normal 32 Could the characteristic followed in the pedigree be caused by an autosomal dominant
disease? not?
Solution
32............is pedigree explaining about autosomal dominant disease/.
t t Deg g olthe year.) 4.9 An engineering construction firm is .pdf
t t Deg g olthe year.) 4.9 An engineering construction firm is chartered to provide consulting
services, design, and a limited range of construction services. Its designs have developed into
registered patents Some designs are licensed by subcontractors, the firm not choosing to use
them itself; it does re- ceive income from that business arrangement. These accounts and their
balance are summa- rized below or a family-owned firm titled \"Eastwood s are in units of
$1000, but ignore r each account title, first identify if it is an asset, liability, net worth, income,
expense, Find the income statement and balance sheet f Engineering Services,\" Second Quarter.
(Hints: These value that. Fo or tax. Each account will have a spot in either the income or
balance-sheet statement.)
Solution
Income Statement Revenue in $ in $ Fees Recived 73 Sales of design 7 Equipment
Lease 6 Royality on Patent 40 Interest On owned Securities 126 Less Salary
Expense 28 Utilities Paid 5 Taxes 12 Office Rent 1 Travel 4 Interest On owned
Securities 2 Supplies Expense 9 61 Net Income after tax 65 Balance sheet
Liability Amount in $ Asset Amount in $ Opening bal of retained earning 27 Equipment Book
Value 58 Profit as per income statement 65 Receivable from project 55 Capital Stock 25
Account Payable 30 Patent Asset 11 Bond Owned 12 Cash in Hand 3.
Actual costs are determined by plugging the actual level of activity.pdf
Actual costs are determined by plugging the actual level of activity for the period into the cost
formulas used in flexible budgets.
Solution
False, Actual costs are not determined by plugging the actual level of activity for the period
into the cost formulas used in flexible budgets. Actual costs are the actual amounts paid for
expenses..
c++ program I need to sort arrays using an insertion sort and a mer.pdf
c++ program: I need to sort arrays using an insertion sort and a merge sort (I already have). I
need to use these sorting methods to sort input arrays in the form of randomly generated numbers
in the size of 20, 100 and 200. So I need 3 randomily generated arrays and sorted and outputed
by both sorting method. I currently have it as given arrays and need help changing them to
randomily generated arrays
Here is my code so far:
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
//Merges two sub-arrays. First sub-array is arr[l..m],second is arr[m+1..r]
void merge(int arr[], int l, int m, int r){
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
//Create temp arrays
int L[n1], R[n2];
//Copy data to temp arrays L[] and R[]
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2){
if (L[i] <= R[j]){
arr[k] = L[i];
i++;
}
else{
arr[k] = R[j];
j++;
}
k++;
}
//Copy the remaining elements of L[], if there are any
while (i < n1){
arr[k] = L[i];
i++;
k++;
}
//Copy the remaining elements of R[], if there are any
while (j < n2){
arr[k] = R[j];
j++;
k++;
}
}
//l is left index and r is right index of the sub-array of array to be sorted
void MergeSort(int arr[], int l, int r){
if (l < r){
// Same as (l+r)/2, but avoids overflow for large l and h
int m = l+(r-l)/2;
// Sort first and second halves
MergeSort(arr, l, m);
MergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
int array5[] = {100,99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by INSERTION SORT: \ \";
printArray(array1, size);
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
MergeSort(array2,0,size);
cout << \"Array 2 sorted by MERGE SORT: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by INSERTION SORT: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
MergeSort(array4,0,size);
cout << \"Array 4 sorted by MERGE SORT: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(arr.
Below is my program, I just have some issues when I want to check ou.pdf
Below is my program, I just have some issues when I want to check out a patient and when I
want to print a specific patient. Please help. I think there is a minor error, but I can\'t find it.
PATIENT ID MUST BE \"LASTNAMEFIRSTNAMEYEAR\" example PattersonMathew2005.
Please read instructions help me solve the problem.
In this assignment, you will write a struct plus two simple classes and then write a program
that uses them all. The first class is a Date class. You are given code for this at the end of this
assignment, but you must make some modifications. The other is a Patient class; you are given
the class definition and you must write the implementation. The program will keep track of
patients at a walk-in clinic. For each patient, the clinic keeps a record of all procedures done.
This is a good candidate for a class, but to simplify the assignment, we will make a procedure
into a struct. We will assume that the clinic has assigned all care providers (doctors, nurses, lab
technicians) with IDs and each procedure (“office visit”, “physical exam”, “blood sample taken”,
etc.) also has an ID. Each patient will also have an ID, which is formed by the last name
concatenated with the first name and the year of birth. All other IDs (care providers, procedures)
will be just integers.
When the user runs the program at the start of the day, it first tries to read in from a binary file
calledCurrentPatients.dat. It will contain the number of patients(in binary format) followed by
binary copies of records for the patients at the clinic . This information should be read and stored
in an array of Patient objects. There will be a separate array, initially empty at the start of the
program that will store patient records for all patients who are currently checked in at the clinic.
It then asks the user to enter the current date and reads it in. It should then presents the user with
a simple menu: the letter N to check in a new patient, R for checking in a returning patient, O to
check out a patient, I to print out information on a particular patient, P to print the list of patients
who have checked in, but not yet checked out, and Q for quitting the program.
The following tasks are done when the appropriate letter is chosen.
N: The new patient’s first name, last name, and birthdate are asked for and entered. The new
patient ID will be the last name concatenated with the first name and the year of birth. (Example:
SmithJohn1998) The primary doctor’s ID is also entered. The new patient object is placed in
both arrays: one keeping all patients, and one keeping the patients who checked in today.
R: The returning patient’s ID is asked for, and the array holding all patients is searched for the
patient object. If found, a copy of the patient object is placed in the array for all currently
checked in patients. If not found, the user is returned to the main menu after asking them to
either try R again, making sure the correct ID was entered, or choose N to enter th.
Alice uses a biased coin to create a one-time pad to encrypt a messa.pdf
Alice uses a biased coin to create a one-time pad to encrypt a message m of 1 bit. The coin
comes up 1 (Heads) 60% of the times and 0 (Tails) 40% of the times. Alice sends 0 as a message
75% of the times and 1 as a message 25% of the times. All of this is publicly known. Eve sees
cipher c = 1. What was the probability that m = 1?
Solution
An encryption scheme (E, D) is perfectly secret if there for every set M {0, 1} ` of plaintexts,
and for every strategy used by Eve, if we choose at random m M and a random key k {0, 1} n,
then the probability that Eve guesses m after seeing Ek(m) is at most 1/|M|.
In particular, if we encrypt either “Yes” or “No” with probability 1/2, then Eve won’t be able to
guess which one it is with probability better than half. In fact, that turns out to be the heart of the
matter:
Two to Many Theorem: An encryption scheme (E, D) is perfectly secret if and only if for every
two distinct plaintexts {m0, m1} {0, 1} ` and every strategy used by Eve, if we choose at
random b {0, 1} and a random key k {0, 1} n, then the probability that Eve guesses mb after
seeing Ek(mb) is at most 1/2.
Proof: The “only if” direction is obvious— this condition is a special case of the perfect secrecy
condition for a set m of size 2
The “if” direction is trickier. We need to show that if there is some set M (of size possibly much
larger than 2) and some strategy for Eve to guess (based on the ciphertext) a plaintext chosen
from M with probability larger than 1/|M|, then there is also some set M0 of size two and a
strategy Eve0 for Eve to guess a plaintext chosen from M0 with probability larger than 1/2.
Let’s fix the message m0 for example to be the all zeroes message. Since Eve(Ek(m0)) is a fixed
string, if we pick a random m1 from M then it holds that
Pr[Eve(Ek(m0)) = m1] 1/|M|
while under our assumption, on average it holds that
Pr[Eve(Ek(m1)) = m1] > 1/|M|
Thus in particular, due to linearity of expectation, there exists some m1 satisfying
Pr[Eve(Ek(m1)) = m1] > Pr[Eve(Ek(m0)) = m1] .
But this can be turned into an attacker Eve0 such that the probability that Eve0 (Ek(mb)) = mb is
larger than 1/2. Indeed, we can define Eve0 (c) to output m1 if Eve(c) = m1 and otherwise output
a random message in {m0, m1}. The probability that Eve0 (c) equals m1 is higher when c =
Ek(m1) than when c = Ek(m0), and since Eve0 outputs either m0 or m1, this means that the
probability that Eve0 (Ek(mb)) = mb is larger than 1/2 (Can you see why?) QED.
Exercise: Another equivalent condition for perfect secrecy is the following: (E, D) is perfectly
secret if for every plaintexts m, m0 {0, 1} ` , the two random variables {Ek(m)} and {Ek0 (m0
)} (for randomly chosen keys k and k 0 ) have precisely the same distribution.
So, perfect secrecy is a natural condition, and does not seem to be too weak for applications, but
can it actually be achieved? After all, the condition that two different plaintexts are mapped to
the same distribution seems somewhat at odds with th.
A paleontologist is studying the history of group of stratigraphic fo.pdf
A paleontologist is studying the history of group of stratigraphic fossils the arthropods. Use the
diagram below to answer the questions. The pictures of the organisms represent of two different
species that your research team has assigned to the various stratigraphic layers. Which of the
fossils has been assigned to the wrong layer? The paleontologist\'s graduate student was unsure
which rock layer was the most in important to date so she dated all of them with the results as
follows: A: 100 mya, B: 50 mya, C: 10 mya, D: 300 mya When did species 1 go extinct? When
did species 2 go extinct? If found a non-fossilized organic sample in the soil above the rock
layers, how might my approach to determining its age be different than for the fossilized
remains?
Solution
a) Species in the igneous rock has been assigned to the wrong layer.
Because, Igneous rocks do not contain fossils as they form from the molten rocks. These rocks
form when the molten rocks become cools and eventually solidify.
Instead of this, fossils are generally found in sedimentary rocks which is formed in layers by the
depositing and pressing of sediments on top of each other.
b) i) species 1 went extinct after 100 million years.
ii) Species 2 went extinct after 50 million years.
c) If I found a non-fossilized organic sample in the soil above the rock layers then I can use
carbon dating or radiometric dating. It is the process of determining the percentage of the
radioactive isotope in the sample and using the half life data to ultimately determine the age of
the sample.
Basic data of radiocarbon dating-
Carbon 14 is an isotope of carbon which is naturally formed in the atmosphere. All plants and
animals when they are alive, have regular intake carbon. When an animal or plant dies it no
longer takes in carbon of any form. That plant or animal most often decays and its constituent
parts break down and are eaten by scavengers, or they decay, or are washed away etc. Instead of
these events if the animal or plant is encased in some sediment, or frozen, or desiccated, or
otherwise preserved quickly so that its structure and constituent parts are preserved. Scientists
have ways of counting the carbon 14 atoms in a sample and then they find out the total number
of carbon atoms and figure out how old the material is.
In this way fairly new organic material’s age is also determined..
A negative stain is repelled by the negatively charged bacterial cel.pdf
A negative stain is repelled by the negatively charged bacterial cell and allows the observer to
see a clear picture of size, shape, and aggregation of the cell.
True
False
Solution
The statement is false. Negative stain means during the observation of bacteria or other microbes
under microscope, the background of them stained, not the microbes themselves. This allows the
microbes visible without staining them. If microbes stain, then it called as Positive stain..
1. The circulatory system of a whale is considered an organ system be.pdf
1. The circulatory system of a whale is considered an organ system because nis composed of
different A. cells. B. tissues. C. organs. E. heart. 2. Which of the following lists the terms from
simplest to most complex? A. cells, organs, tissues, organ systems, organism B, organs, organ
system, organism, cells, tissues C. tissues, organs, organ systems, organism, cells D, cells,
tissues, organs, organ systems, organism 3, The smallest unit of life is aln) A tissue. B. organ. C.
cell. D. species. transforms solar energy into chemical energy 4. The process of B, homeostasis
C. respiration D. photosynthesis 5. Which of the following is matched incorrectly? A. brain
organ B. neuron cell C skin organ D. gene organelle living things? 6. Which of the following is
not one of the most common elements in A, carbon B. oxygen C, hydrogen D. iron 7 Glucose,
CHizo. is best described as A) an element B) an a compound DJan ion isotope
Solution
Answer:
(1) C
The circulatory system of a whale is considered an organ system because it is composed of
different organs.
(2) D
Cells, tissues, organs, organ system, organism (Simplest to more complex).
(3) C
The smallest unit of life is cell.
(4) D
The process of photosynthesis transforms solar energy into chemical energy.
(5) D
Gene is not an organelle but is a short portion of DNA which encodes a particular protein.
(6) A
Carbon is the most abundant element in living organisms.
(7) C
Glucose is an organic compound composed of carbon, hydrogen and oxygen.
(8) A
An element that has electrical charge is called ion.
(9) B
Ionic bonds are formed by transfer of electrons between atoms.
(10) B
Oxygen is an element.
(11) C
Deoxyribose is a component of DNA.
(12) A
Digestion of sucrose results in 2 monomers.
(13) B
A fat is formed when three fatty acids join a glycerol by a dehydration reaction.
(14) B
Parts W and Z are hydrophilic since they lie outside the plasma membrane.
(15) B
Cell wall helps in maintaining cell shape.
(16) D
It represents cell wall (As evident from the image and its not very clear)
(17) B
Prokaryotic cell does not have a membrane bound nucleus.
(18) B
Mitochondria is the site of cellular respiration and produces ATP.
(19) A
Both chloroplast and mitochondria have their own DNA.
(20) C
Stack of thylakoids are called grana.
(21) A
Isotonic solutions have the same concentration.
(22) C
ATP releases energy when phosphate is removed.
(23) C
The cell wall protects the cell from bursting in a hypotonic solution.
(24) C
Solar energy is captured by chloroplasts for photosynthesis. So, solar energy, chloroplast,
mitochondria.
(25) B
The part of the enzyme into which the substrate fits is the active site..
Reliability and safety are related but distinct dependability attrib.pdf
Reliability and safety are related but distinct dependability attributes for software. Describe two
important distinctions between these attributes. Explain why it is possible for a reliable system to
be unsafe and vice versa.
Solution
Explain why it is possible for a reliable system to be unsafe and vice versa.?
ANS:------> Because If the specification does not explicitly exclude dangerous behavior then a
system can be reliable but unsafe.RELIABILITYSAFETY1) Reliability involves much more
than analyzing test results, estimating remaining faults, and modeling future failure
probabilities.1)safety optimizes system safety in the design, development, use, and maintenance
of software systems and their integration with safety-critical hardware systems in an operational
environment.2)Ensuring system reliability does not necessarily lead to system safety as
Reliability is concerned with meeting the system specification (the system \'shall\')2)Safety is
concerned with excluding the possibility of dangerous behavior (the system \'shall not\').
Please answer all questionsDefine1. Wild type –2. Epistasis –.pdf
Please answer all questions
Define
1. Wild type –
2. Epistasis –
Fill in the Blank
3. Two genes may interact effecting one trait and resulting in novel phenotypes with an F2
phenotypic ________________ ratio.
4. Under ________________ conditions, conditionally lethal mutants will die.
True or False
5. Penetrance is the degree or intensity with which a particular genotype is expressed in a
phenotype.
6. Continuous traits are usually controlled by multiple genes and may be influenced by the
environment.
Short Answer
7. Describe the inheritance pattern(s) of blood types.
8. When studying a newly discovered flowering plant in Costa Rica, you collect data on its petal
color in an effort to determine how this trait may be inherited. You find that 852 flowers are red,
331 are white and 389 are yellow. What are the possible inheritance patterns, and how could you
definitely conclude which of these possibilities is the true mode of inheritance?
Solution
Wild type- wild type phenotype refers to characteristics that occurs naturally in a breeding
population.
Epistasis - it refers to a phenomena when one gene’s expression is dependent on the presence of
another gene(s), called the modifier gene.
9:16
Let\'s take an example of comb varieties in chicken
We have two pure breeds : wyandotte whose phenotype is a rose comb and brahmas having the
phenotype of a pea comb, when crossed with each other, they create a completely new phenotype
of a walnut comb. Let\'s look at the ratio:
Parental cross: rose comb x pea comb
F1 : all walnut combs
F2 : 9 walnut : 3 rose: 3peas : 1 single
Hence the ratio is 9:16
Under the restrictive or nonpermissive condition conditions, conditionally lethal mutants will
die.
Conditionally lethal mutants are those which have an allele that will make the survival of the
species difficult only in a specific condition.
True. for example if a mutation in a gene gives the species a particular characterisitc which has
97%penetrance, only 97% of those with the mutation will develop that characteristic and 3 %
will not.
True. multiple genes affect continuous traits. For such complex gene interactions, the traits are
influenced by environmental factors too. For example, height and skin color.
There are four basic blood types which are A, B, AB and O and together with the Rh factor,
these can be +ve or -ve.
A and B are dominant, while O is recessive.
Blood type
A
B
O
A
A
AB
A
B
AB
B
B
O
A
B
O
For Rh factor, Rh (+)ve is dominant.
So if the any one or both parents have Rh (+)ve factor, the progeny will be Rh(+)ve.
If both parents are Rh(-)ve, the progeny will be Rh (-)ve
Father
Mother
D
D
d
Dd
Dd
d
Dd
Dd
100% Rh (+)ve children
Father
Mother
D
d
d
Dd
dd
d
Dd
dd
50% Rh(+)ve children
Red: 852
White: 331
Yellow:389
According to the given data, red is dominant and white is recessive and yellow is a result of
incomplete dominance of both white and red phenotype.
It can\'t be codominance, because codominance involves both phenotypes being expressed in
di.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
More Related Content
More from dhavalbl38
why is there an uneven distribution of photosystem 1 and 2 on the th.pdf
why is there an uneven distribution of photosystem 1 and 2 on the thylakoid membrane?
Solution
Photosynthesis is carried out by the cyanobacteria, plants involved the absorption of light and the
conversion of its energy into chemical energy. The photosynthetic apparatus is made up of 4
multiprotein complexes, which are asymmetrically embedded in the thyalkoid memberane. It is
the high structural and functional coordination between these 4 complexes, which enables the
efficient coupling of photosynthetic electron transfer and photophosphorylation.
The 4 complexes - Photosystem II (PSII), cytochrome b6f complex, PS I and the proton-ATP
synthase, are unevenly distributed between the 2 regions of the thylakoids. The PS II complex is
situated in the appressed grana-lamella region, the PS I and the ATP synthase complex are
located in the non-appressed stroma-lamella region.
Light driven electron transfer is intiated with the oxidation of water mole. by the PS II complex
and the reduction of quinones associated with the PS II complex. The electrons are then
transferred across the memberane to plastocyanin via the Cyt b6f complex. Plastocyanin is a
copper protein facing the lumen of the thylakoids, is the secondary electron donor of the PS I
complex. In the PS I the electron are translocated across the memberane to ferredoxin which
further transfers them to NADP+. The membrane potental and proton gradient resulting from this
photochemical electron transport provide the motive force required for the formation of ATP by
the ATP synthase.
The transport of electron against the electrical potential and the formation of ATP and NADPH
are made possible due to the unique ability of the thylakoids to absorb light energy. The 2
pigment complexes (PS II and PS I) are responsible for harvesting the light and performing the
charge separation across the thylakoid memberanes.
Photosytem II (PS II):
1. PS II is located at the inner surface the grana thyalkoid memberane
2. The photocentre is P680
3. Pigments absorb shorter wavelengths of light (<680nm)
4. Participates only in non cyclic photophosphorylation
5. It is associated with photolysis of water
6. Main functions are ATP synthesis and hydrolysis of water
Photosystem I (PS I)
1. PS I is located at the outer surface the grana thylakoid memberane
2. The photocentre is P700
3. Pigments absorb longer wavelength of light (>680nm)
4. Participates in cyclic and non-cyclic photophosphorylation
5. It is not associated with photolysis of water
6. Main function is ATP synthesis
This is the purpose the two photosystems (PS) are unevely distributed in thyalkoid memberane..
Which of the following are examples of cell signaling Choose all tha.pdf
Which of the following are examples of cell signaling? Choose all that apply. As a response to
increasing densities, bacteria coordinate behaviors and form biofilms. Insulin hormone produced
by the pancreas triggers the intake of glucose by liver cells. Carbon dioxide diffuses across the
cell membrane into the blood plasma. Chlorophyll absorbs and transfers light energy during
photosynthesis. A neuron releases neurotransmitters into a synapse, triggering a target nerve
cell.
Solution
Cell signaling
Cell signaling is the communication between the cells through which they coordinate
certain activities.
What is the purpose of testing a program with different dataSol.pdf
What is the purpose of testing a program with different data?
Solution
The purpose of testing a program functions according to the expectation. The overall goal of
testing is to find a bug that might effect the functioning of the program. There are various types
of testing and each serve their own purpose. Different data is used for each type of testing. We
have to check the whether internal module is functioning properly, how the program responds to
invalid data, test how changing one module will effect the whole system etc.. There are many
automated tools which can be used for comprehensive testing..
what are the factors contributing to the rise in global media Will .pdf
what are the factors contributing to the rise in global media? Will global media eventually
replace national media?
Solution
Q1) Media is the mass communication system of letting people informed through broadcasting,
publishing, and the internet. It enriches public awareness and knowledge, which is very
important in today’s fast moving world.
Global media gives the news of whole world, suppose a person sitting in India can get the
happenings in the US.
Factors are as below:
Point 1) The urge of national media: There are so many small national media in various
countries. They want to get international news as well but can’t get because of their small
infrastructure. Therefore, they are now becoming tied up with big international media agencies in
order to get regular updating. This is the reason of rising trend of global media.
Point 2) Globalization of economy: Most of the countries have free-trade environment now.
They need cheaper rate of factors of production (land, labor, capital, and organization); the
global media is a great source of information of it.
Point 3) Technological development: The development of internet gives the platform of rising of
global media. A person sitting on computer can get to know the whole world.
Q2) To some extent it is true. Since national media getting tied up with big global firms, the
chance of being acquired may be high in future. The small national media could be acquired by
those global media..
WACC of Alibaba company WACC of Alibaba companySolutionTh.pdf
WACC of Alibaba company
WACC of Alibaba company
Solution
The Weighted Average Cost of Capital Alibaba company as of the latest quarter report is
18.28%. Alibaba company generates more return on investment than the cost of capital. WACC
is the rate that a company is expected to pay its stockholders and other lenders of money..
The forecast equation for mean wind in a turbulent flow is U_it + .pdf
The forecast equation for mean wind in a turbulent flow is: U_i/t + U_j U_i/x_j = delta_i3 g +
f_c epsilon_ij3 U_j - 1/p P/x_i + v ^2 U_i/x^2_j u\'_i u\'_j/x_j Name each term, and give its
physical interpretation.
Solution
solution
A is storage of mean momentum
B is advection of mean momentum by the mean wind
C is gravity to act in the vertical direction only
D is the Coriolis effect due to the earth’s rotation
E is the mean pressure gradient forces
F is the effect of viscous stress on the mean motions
G is the effect of Reynolds stress on the mean motions (also called the divergence of turbulent
momentum flux).
The image below is of a charged moving in a uniform magnetic field. I.pdf
The image below is of a charged moving in a uniform magnetic field. Is the particle positively
or negatively charged?
Solution
According to Fleming left hand rule,
When the thumb,first finger,middle finger are streatched perpendicular to each other and the first
finger represents the magnetic field direction ,middle finger represents the direction of motion of
positive charged particle then the thumb represents force on charged particle.
In figure , magnetic field into the paper and the particle deflect right.So, the charged particle is
negative nature.
The concept of a sampling distributionA-Is essential for drawing c.pdf
The concept of a sampling distribution
A-Is essential for drawing conclusions about a population from the results of a sample.
B- Is related to the use of probability theory in inferential statistics.
C-Helps us use statistics to estimate the corresponding true values in the population.
D-All of the above.
Solution
The concept of a sampling distribution
D-All of the above..
The diets common to some cultures emphasize meat. Meat, milk, and br.pdf
The diets common to some cultures emphasize meat. Meat, milk, and bread, but few fruits and
vegetables, might be a typical meal for populations in northern latitudes and in desert areas, for
example. What ecological factors seem relevant in explaining this human habit?
Solution
Various studies have shown that populations living in northern latitude and desert areas have
dietary patterns higher in animal fat such as meat and dairy products compared to those living in
southern latitudes who thrive majorly on fruits and vegetables. This may be because land in
northern and desert areas had been less fertile to grow seasonal vegeies, due to extremes of
environment and limited rainfalls. People living in these areas over generations have been mostly
hunter-gatherers. Such societies had depended for their economic survival primarily on hunting,
gathering (wild vegetable and fruits and small land fauna), and fishing. Thus they might have
developed the food habit of increased meat and milk consumption (as they are largely dependent
on herd of cattle/animals) along with staple diet of breads, over the generations.
On the other hand, southern populations residing on more fertile pieces of land relish a
vegetarian diet consisting of fresh fruits and vegetables. They are richer in biodiversity due to
moderate climates and rainfall distribution throughtout the year. These populations are typically
farmers, food growers over several generations and have developed food habits of vegetarian
diets..
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus.pdf
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus and at least one dominant
inant gene A pi R P A rose comb is produced when at least and two recessive genes are present
at the the first locus at pea comb is produced when two recessive genes are least one dominant
gene is at are present at the first and the second locas sNuces 2 for comb shape of the parents in
the fol walnut, 1 single, and 1 pea offspring. ngle comb ing cross: Rose 3/. The mutation (bard es
is inherited as a maternal effect recessive allele. What is the expected ratio of preho termale and
a bad bed male? in offspring cross between bcd normal: mutant 1 3 mutant normal c 3 mutant 1
normal 32 Could the characteristic followed in the pedigree be caused by an autosomal dominant
disease? not?
Solution
32............is pedigree explaining about autosomal dominant disease/.
t t Deg g olthe year.) 4.9 An engineering construction firm is .pdf
t t Deg g olthe year.) 4.9 An engineering construction firm is chartered to provide consulting
services, design, and a limited range of construction services. Its designs have developed into
registered patents Some designs are licensed by subcontractors, the firm not choosing to use
them itself; it does re- ceive income from that business arrangement. These accounts and their
balance are summa- rized below or a family-owned firm titled \"Eastwood s are in units of
$1000, but ignore r each account title, first identify if it is an asset, liability, net worth, income,
expense, Find the income statement and balance sheet f Engineering Services,\" Second Quarter.
(Hints: These value that. Fo or tax. Each account will have a spot in either the income or
balance-sheet statement.)
Solution
Income Statement Revenue in $ in $ Fees Recived 73 Sales of design 7 Equipment
Lease 6 Royality on Patent 40 Interest On owned Securities 126 Less Salary
Expense 28 Utilities Paid 5 Taxes 12 Office Rent 1 Travel 4 Interest On owned
Securities 2 Supplies Expense 9 61 Net Income after tax 65 Balance sheet
Liability Amount in $ Asset Amount in $ Opening bal of retained earning 27 Equipment Book
Value 58 Profit as per income statement 65 Receivable from project 55 Capital Stock 25
Account Payable 30 Patent Asset 11 Bond Owned 12 Cash in Hand 3.
Actual costs are determined by plugging the actual level of activity.pdf
Actual costs are determined by plugging the actual level of activity for the period into the cost
formulas used in flexible budgets.
Solution
False, Actual costs are not determined by plugging the actual level of activity for the period
into the cost formulas used in flexible budgets. Actual costs are the actual amounts paid for
expenses..
c++ program I need to sort arrays using an insertion sort and a mer.pdf
c++ program: I need to sort arrays using an insertion sort and a merge sort (I already have). I
need to use these sorting methods to sort input arrays in the form of randomly generated numbers
in the size of 20, 100 and 200. So I need 3 randomily generated arrays and sorted and outputed
by both sorting method. I currently have it as given arrays and need help changing them to
randomily generated arrays
Here is my code so far:
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
//Merges two sub-arrays. First sub-array is arr[l..m],second is arr[m+1..r]
void merge(int arr[], int l, int m, int r){
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
//Create temp arrays
int L[n1], R[n2];
//Copy data to temp arrays L[] and R[]
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2){
if (L[i] <= R[j]){
arr[k] = L[i];
i++;
}
else{
arr[k] = R[j];
j++;
}
k++;
}
//Copy the remaining elements of L[], if there are any
while (i < n1){
arr[k] = L[i];
i++;
k++;
}
//Copy the remaining elements of R[], if there are any
while (j < n2){
arr[k] = R[j];
j++;
k++;
}
}
//l is left index and r is right index of the sub-array of array to be sorted
void MergeSort(int arr[], int l, int r){
if (l < r){
// Same as (l+r)/2, but avoids overflow for large l and h
int m = l+(r-l)/2;
// Sort first and second halves
MergeSort(arr, l, m);
MergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
int array5[] = {100,99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by INSERTION SORT: \ \";
printArray(array1, size);
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
MergeSort(array2,0,size);
cout << \"Array 2 sorted by MERGE SORT: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by INSERTION SORT: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
MergeSort(array4,0,size);
cout << \"Array 4 sorted by MERGE SORT: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(arr.
Below is my program, I just have some issues when I want to check ou.pdf
Below is my program, I just have some issues when I want to check out a patient and when I
want to print a specific patient. Please help. I think there is a minor error, but I can\'t find it.
PATIENT ID MUST BE \"LASTNAMEFIRSTNAMEYEAR\" example PattersonMathew2005.
Please read instructions help me solve the problem.
In this assignment, you will write a struct plus two simple classes and then write a program
that uses them all. The first class is a Date class. You are given code for this at the end of this
assignment, but you must make some modifications. The other is a Patient class; you are given
the class definition and you must write the implementation. The program will keep track of
patients at a walk-in clinic. For each patient, the clinic keeps a record of all procedures done.
This is a good candidate for a class, but to simplify the assignment, we will make a procedure
into a struct. We will assume that the clinic has assigned all care providers (doctors, nurses, lab
technicians) with IDs and each procedure (“office visit”, “physical exam”, “blood sample taken”,
etc.) also has an ID. Each patient will also have an ID, which is formed by the last name
concatenated with the first name and the year of birth. All other IDs (care providers, procedures)
will be just integers.
When the user runs the program at the start of the day, it first tries to read in from a binary file
calledCurrentPatients.dat. It will contain the number of patients(in binary format) followed by
binary copies of records for the patients at the clinic . This information should be read and stored
in an array of Patient objects. There will be a separate array, initially empty at the start of the
program that will store patient records for all patients who are currently checked in at the clinic.
It then asks the user to enter the current date and reads it in. It should then presents the user with
a simple menu: the letter N to check in a new patient, R for checking in a returning patient, O to
check out a patient, I to print out information on a particular patient, P to print the list of patients
who have checked in, but not yet checked out, and Q for quitting the program.
The following tasks are done when the appropriate letter is chosen.
N: The new patient’s first name, last name, and birthdate are asked for and entered. The new
patient ID will be the last name concatenated with the first name and the year of birth. (Example:
SmithJohn1998) The primary doctor’s ID is also entered. The new patient object is placed in
both arrays: one keeping all patients, and one keeping the patients who checked in today.
R: The returning patient’s ID is asked for, and the array holding all patients is searched for the
patient object. If found, a copy of the patient object is placed in the array for all currently
checked in patients. If not found, the user is returned to the main menu after asking them to
either try R again, making sure the correct ID was entered, or choose N to enter th.
Alice uses a biased coin to create a one-time pad to encrypt a messa.pdf
Alice uses a biased coin to create a one-time pad to encrypt a message m of 1 bit. The coin
comes up 1 (Heads) 60% of the times and 0 (Tails) 40% of the times. Alice sends 0 as a message
75% of the times and 1 as a message 25% of the times. All of this is publicly known. Eve sees
cipher c = 1. What was the probability that m = 1?
Solution
An encryption scheme (E, D) is perfectly secret if there for every set M {0, 1} ` of plaintexts,
and for every strategy used by Eve, if we choose at random m M and a random key k {0, 1} n,
then the probability that Eve guesses m after seeing Ek(m) is at most 1/|M|.
In particular, if we encrypt either “Yes” or “No” with probability 1/2, then Eve won’t be able to
guess which one it is with probability better than half. In fact, that turns out to be the heart of the
matter:
Two to Many Theorem: An encryption scheme (E, D) is perfectly secret if and only if for every
two distinct plaintexts {m0, m1} {0, 1} ` and every strategy used by Eve, if we choose at
random b {0, 1} and a random key k {0, 1} n, then the probability that Eve guesses mb after
seeing Ek(mb) is at most 1/2.
Proof: The “only if” direction is obvious— this condition is a special case of the perfect secrecy
condition for a set m of size 2
The “if” direction is trickier. We need to show that if there is some set M (of size possibly much
larger than 2) and some strategy for Eve to guess (based on the ciphertext) a plaintext chosen
from M with probability larger than 1/|M|, then there is also some set M0 of size two and a
strategy Eve0 for Eve to guess a plaintext chosen from M0 with probability larger than 1/2.
Let’s fix the message m0 for example to be the all zeroes message. Since Eve(Ek(m0)) is a fixed
string, if we pick a random m1 from M then it holds that
Pr[Eve(Ek(m0)) = m1] 1/|M|
while under our assumption, on average it holds that
Pr[Eve(Ek(m1)) = m1] > 1/|M|
Thus in particular, due to linearity of expectation, there exists some m1 satisfying
Pr[Eve(Ek(m1)) = m1] > Pr[Eve(Ek(m0)) = m1] .
But this can be turned into an attacker Eve0 such that the probability that Eve0 (Ek(mb)) = mb is
larger than 1/2. Indeed, we can define Eve0 (c) to output m1 if Eve(c) = m1 and otherwise output
a random message in {m0, m1}. The probability that Eve0 (c) equals m1 is higher when c =
Ek(m1) than when c = Ek(m0), and since Eve0 outputs either m0 or m1, this means that the
probability that Eve0 (Ek(mb)) = mb is larger than 1/2 (Can you see why?) QED.
Exercise: Another equivalent condition for perfect secrecy is the following: (E, D) is perfectly
secret if for every plaintexts m, m0 {0, 1} ` , the two random variables {Ek(m)} and {Ek0 (m0
)} (for randomly chosen keys k and k 0 ) have precisely the same distribution.
So, perfect secrecy is a natural condition, and does not seem to be too weak for applications, but
can it actually be achieved? After all, the condition that two different plaintexts are mapped to
the same distribution seems somewhat at odds with th.
A paleontologist is studying the history of group of stratigraphic fo.pdf
A paleontologist is studying the history of group of stratigraphic fossils the arthropods. Use the
diagram below to answer the questions. The pictures of the organisms represent of two different
species that your research team has assigned to the various stratigraphic layers. Which of the
fossils has been assigned to the wrong layer? The paleontologist\'s graduate student was unsure
which rock layer was the most in important to date so she dated all of them with the results as
follows: A: 100 mya, B: 50 mya, C: 10 mya, D: 300 mya When did species 1 go extinct? When
did species 2 go extinct? If found a non-fossilized organic sample in the soil above the rock
layers, how might my approach to determining its age be different than for the fossilized
remains?
Solution
a) Species in the igneous rock has been assigned to the wrong layer.
Because, Igneous rocks do not contain fossils as they form from the molten rocks. These rocks
form when the molten rocks become cools and eventually solidify.
Instead of this, fossils are generally found in sedimentary rocks which is formed in layers by the
depositing and pressing of sediments on top of each other.
b) i) species 1 went extinct after 100 million years.
ii) Species 2 went extinct after 50 million years.
c) If I found a non-fossilized organic sample in the soil above the rock layers then I can use
carbon dating or radiometric dating. It is the process of determining the percentage of the
radioactive isotope in the sample and using the half life data to ultimately determine the age of
the sample.
Basic data of radiocarbon dating-
Carbon 14 is an isotope of carbon which is naturally formed in the atmosphere. All plants and
animals when they are alive, have regular intake carbon. When an animal or plant dies it no
longer takes in carbon of any form. That plant or animal most often decays and its constituent
parts break down and are eaten by scavengers, or they decay, or are washed away etc. Instead of
these events if the animal or plant is encased in some sediment, or frozen, or desiccated, or
otherwise preserved quickly so that its structure and constituent parts are preserved. Scientists
have ways of counting the carbon 14 atoms in a sample and then they find out the total number
of carbon atoms and figure out how old the material is.
In this way fairly new organic material’s age is also determined..
A negative stain is repelled by the negatively charged bacterial cel.pdf
A negative stain is repelled by the negatively charged bacterial cell and allows the observer to
see a clear picture of size, shape, and aggregation of the cell.
True
False
Solution
The statement is false. Negative stain means during the observation of bacteria or other microbes
under microscope, the background of them stained, not the microbes themselves. This allows the
microbes visible without staining them. If microbes stain, then it called as Positive stain..
1. The circulatory system of a whale is considered an organ system be.pdf
1. The circulatory system of a whale is considered an organ system because nis composed of
different A. cells. B. tissues. C. organs. E. heart. 2. Which of the following lists the terms from
simplest to most complex? A. cells, organs, tissues, organ systems, organism B, organs, organ
system, organism, cells, tissues C. tissues, organs, organ systems, organism, cells D, cells,
tissues, organs, organ systems, organism 3, The smallest unit of life is aln) A tissue. B. organ. C.
cell. D. species. transforms solar energy into chemical energy 4. The process of B, homeostasis
C. respiration D. photosynthesis 5. Which of the following is matched incorrectly? A. brain
organ B. neuron cell C skin organ D. gene organelle living things? 6. Which of the following is
not one of the most common elements in A, carbon B. oxygen C, hydrogen D. iron 7 Glucose,
CHizo. is best described as A) an element B) an a compound DJan ion isotope
Solution
Answer:
(1) C
The circulatory system of a whale is considered an organ system because it is composed of
different organs.
(2) D
Cells, tissues, organs, organ system, organism (Simplest to more complex).
(3) C
The smallest unit of life is cell.
(4) D
The process of photosynthesis transforms solar energy into chemical energy.
(5) D
Gene is not an organelle but is a short portion of DNA which encodes a particular protein.
(6) A
Carbon is the most abundant element in living organisms.
(7) C
Glucose is an organic compound composed of carbon, hydrogen and oxygen.
(8) A
An element that has electrical charge is called ion.
(9) B
Ionic bonds are formed by transfer of electrons between atoms.
(10) B
Oxygen is an element.
(11) C
Deoxyribose is a component of DNA.
(12) A
Digestion of sucrose results in 2 monomers.
(13) B
A fat is formed when three fatty acids join a glycerol by a dehydration reaction.
(14) B
Parts W and Z are hydrophilic since they lie outside the plasma membrane.
(15) B
Cell wall helps in maintaining cell shape.
(16) D
It represents cell wall (As evident from the image and its not very clear)
(17) B
Prokaryotic cell does not have a membrane bound nucleus.
(18) B
Mitochondria is the site of cellular respiration and produces ATP.
(19) A
Both chloroplast and mitochondria have their own DNA.
(20) C
Stack of thylakoids are called grana.
(21) A
Isotonic solutions have the same concentration.
(22) C
ATP releases energy when phosphate is removed.
(23) C
The cell wall protects the cell from bursting in a hypotonic solution.
(24) C
Solar energy is captured by chloroplasts for photosynthesis. So, solar energy, chloroplast,
mitochondria.
(25) B
The part of the enzyme into which the substrate fits is the active site..
Reliability and safety are related but distinct dependability attrib.pdf
Reliability and safety are related but distinct dependability attributes for software. Describe two
important distinctions between these attributes. Explain why it is possible for a reliable system to
be unsafe and vice versa.
Solution
Explain why it is possible for a reliable system to be unsafe and vice versa.?
ANS:------> Because If the specification does not explicitly exclude dangerous behavior then a
system can be reliable but unsafe.RELIABILITYSAFETY1) Reliability involves much more
than analyzing test results, estimating remaining faults, and modeling future failure
probabilities.1)safety optimizes system safety in the design, development, use, and maintenance
of software systems and their integration with safety-critical hardware systems in an operational
environment.2)Ensuring system reliability does not necessarily lead to system safety as
Reliability is concerned with meeting the system specification (the system \'shall\')2)Safety is
concerned with excluding the possibility of dangerous behavior (the system \'shall not\').
Please answer all questionsDefine1. Wild type –2. Epistasis –.pdf
Please answer all questions
Define
1. Wild type –
2. Epistasis –
Fill in the Blank
3. Two genes may interact effecting one trait and resulting in novel phenotypes with an F2
phenotypic ________________ ratio.
4. Under ________________ conditions, conditionally lethal mutants will die.
True or False
5. Penetrance is the degree or intensity with which a particular genotype is expressed in a
phenotype.
6. Continuous traits are usually controlled by multiple genes and may be influenced by the
environment.
Short Answer
7. Describe the inheritance pattern(s) of blood types.
8. When studying a newly discovered flowering plant in Costa Rica, you collect data on its petal
color in an effort to determine how this trait may be inherited. You find that 852 flowers are red,
331 are white and 389 are yellow. What are the possible inheritance patterns, and how could you
definitely conclude which of these possibilities is the true mode of inheritance?
Solution
Wild type- wild type phenotype refers to characteristics that occurs naturally in a breeding
population.
Epistasis - it refers to a phenomena when one gene’s expression is dependent on the presence of
another gene(s), called the modifier gene.
9:16
Let\'s take an example of comb varieties in chicken
We have two pure breeds : wyandotte whose phenotype is a rose comb and brahmas having the
phenotype of a pea comb, when crossed with each other, they create a completely new phenotype
of a walnut comb. Let\'s look at the ratio:
Parental cross: rose comb x pea comb
F1 : all walnut combs
F2 : 9 walnut : 3 rose: 3peas : 1 single
Hence the ratio is 9:16
Under the restrictive or nonpermissive condition conditions, conditionally lethal mutants will
die.
Conditionally lethal mutants are those which have an allele that will make the survival of the
species difficult only in a specific condition.
True. for example if a mutation in a gene gives the species a particular characterisitc which has
97%penetrance, only 97% of those with the mutation will develop that characteristic and 3 %
will not.
True. multiple genes affect continuous traits. For such complex gene interactions, the traits are
influenced by environmental factors too. For example, height and skin color.
There are four basic blood types which are A, B, AB and O and together with the Rh factor,
these can be +ve or -ve.
A and B are dominant, while O is recessive.
Blood type
A
B
O
A
A
AB
A
B
AB
B
B
O
A
B
O
For Rh factor, Rh (+)ve is dominant.
So if the any one or both parents have Rh (+)ve factor, the progeny will be Rh(+)ve.
If both parents are Rh(-)ve, the progeny will be Rh (-)ve
Father
Mother
D
D
d
Dd
Dd
d
Dd
Dd
100% Rh (+)ve children
Father
Mother
D
d
d
Dd
dd
d
Dd
dd
50% Rh(+)ve children
Red: 852
White: 331
Yellow:389
According to the given data, red is dominant and white is recessive and yellow is a result of
incomplete dominance of both white and red phenotype.
It can\'t be codominance, because codominance involves both phenotypes being expressed in
di.
More from dhavalbl38 (20)
why is there an uneven distribution of photosystem 1 and 2 on the th.pdf
why is there an uneven distribution of photosystem 1 and 2 on the th.pdf
Which of the following are examples of cell signaling Choose all tha.pdf
Which of the following are examples of cell signaling Choose all tha.pdf
What is the purpose of testing a program with different dataSol.pdf
What is the purpose of testing a program with different dataSol.pdf
what are the factors contributing to the rise in global media Will .pdf
what are the factors contributing to the rise in global media Will .pdf
WACC of Alibaba company WACC of Alibaba companySolutionTh.pdf
WACC of Alibaba company WACC of Alibaba companySolutionTh.pdf
The forecast equation for mean wind in a turbulent flow is U_it + .pdf
The forecast equation for mean wind in a turbulent flow is U_it + .pdf
The image below is of a charged moving in a uniform magnetic field. I.pdf
The image below is of a charged moving in a uniform magnetic field. I.pdf
The concept of a sampling distributionA-Is essential for drawing c.pdf
The concept of a sampling distributionA-Is essential for drawing c.pdf
The diets common to some cultures emphasize meat. Meat, milk, and br.pdf
The diets common to some cultures emphasize meat. Meat, milk, and br.pdf
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus.pdf
aed by genes at two loci R. rand P. p). A walnut comb is at one Tocus.pdf
t t Deg g olthe year.) 4.9 An engineering construction firm is .pdf
t t Deg g olthe year.) 4.9 An engineering construction firm is .pdf
Actual costs are determined by plugging the actual level of activity.pdf
Actual costs are determined by plugging the actual level of activity.pdf
c++ program I need to sort arrays using an insertion sort and a mer.pdf
c++ program I need to sort arrays using an insertion sort and a mer.pdf
Below is my program, I just have some issues when I want to check ou.pdf
Below is my program, I just have some issues when I want to check ou.pdf
Alice uses a biased coin to create a one-time pad to encrypt a messa.pdf
Alice uses a biased coin to create a one-time pad to encrypt a messa.pdf
A paleontologist is studying the history of group of stratigraphic fo.pdf
A paleontologist is studying the history of group of stratigraphic fo.pdf
A negative stain is repelled by the negatively charged bacterial cel.pdf
A negative stain is repelled by the negatively charged bacterial cel.pdf
1. The circulatory system of a whale is considered an organ system be.pdf
1. The circulatory system of a whale is considered an organ system be.pdf
Reliability and safety are related but distinct dependability attrib.pdf
Reliability and safety are related but distinct dependability attrib.pdf
Please answer all questionsDefine1. Wild type –2. Epistasis –.pdf
Please answer all questionsDefine1. Wild type –2. Epistasis –.pdf
Recently uploaded
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
678020731-Sumas-y-Restas-Para-Colorear.pdf
KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Class 11 CBSE Business Studies Project ( AIDS TO TRADE - INSURANCE)
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
Recently uploaded (20)
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
The French Revolution Class 9 Study Material pdf free download
The French Revolution Class 9 Study Material pdf free download
Home assignment II on Spectroscopy 2024 Answers.pdf
Home assignment II on Spectroscopy 2024 Answers.pdf
Adversarial Attention Modeling for Multi-dimensional Emotion Regression.pdf
Adversarial Attention Modeling for Multi-dimensional Emotion Regression.pdf
1.4 modern child centered education - mahatma gandhi-2.pptx
1.4 modern child centered education - mahatma gandhi-2.pptx
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI fund...
Overview on Edible Vaccine: Pros & Cons with Mechanism
Overview on Edible Vaccine: Pros & Cons with Mechanism
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
Two particles A and B of equal mass have velocities as shown in the f.pdf
- 1. Two particles A and B of equal mass have velocities as shown in the figure to the right. If the particles collide, circle the situation(s) shown 111 the figures below that represent possible velocities for the particles after impact. The velocities in all figures are shown to scale. [0.5 mark for each correct answer] Solution Since thy have equal masses they will exchange velocities both for horizontal and vertical component. horizontal is head on collision so they will have horizontal velocity in opposite direction.