Because ethanol and water are miscible and they w.pdf
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Because ethanol and water are miscible and they won\'t create two separate layers. Solution Because ethanol and water are miscible and they won\'t create two separate layers..
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object 1 sink object2 float as for object 2 weig.pdf
No.of moles of KOH , n = MOlarity * Volume in L = 0.0180 M * 0.032 L =
5.76*10^-4 moles HNO3 + KOH -----> KNO3 + H2O According to the balanced Equation , 1
mole of HNO3 reacts with 1 mole of KOH 5.76*10^-4 moles of HNO3 reacts with 5.76*10^-4
moles of KOH So Molarity of HNO3 , M = no.of moles / Volume in L = 5.76*10^-4 moles /
0.025 L = 0.02304 M
Solution
No.of moles of KOH , n = MOlarity * Volume in L = 0.0180 M * 0.032 L =
5.76*10^-4 moles HNO3 + KOH -----> KNO3 + H2O According to the balanced Equation , 1
mole of HNO3 reacts with 1 mole of KOH 5.76*10^-4 moles of HNO3 reacts with 5.76*10^-4
moles of KOH So Molarity of HNO3 , M = no.of moles / Volume in L = 5.76*10^-4 moles /
0.025 L = 0.02304 M.
Nitrogen Pollution Overuse or misuse of nitrogen .pdf
NaH then ethylene oxidide then PCC
Solution
NaH then ethylene oxidide then PCC.
many of compounds tend to be unstable and either .pdf
many of compounds tend to be unstable and either decompose or volatilize at higher
temperatures. the resultant peak will come with larger peak area so ans is b
Solution
many of compounds tend to be unstable and either decompose or volatilize at higher
temperatures. the resultant peak will come with larger peak area so ans is b.
Mendeleev,s law of periodic table .pdf
Mendeleev,s law of periodic table
Solution
Mendeleev,s law of periodic table.
It is very slightly soluble in in boiling water, .pdf
It is very slightly soluble in in boiling water, soluble in m-cresol and phenol, and is
degraded by acids.
Solution
It is very slightly soluble in in boiling water, soluble in m-cresol and phenol, and is
degraded by acids..
hydride shift is more favorable due to shearing e.pdf
hydride shift is more favorable due to shearing effect from surrounding atoms is
more in case of methanide .
Solution
hydride shift is more favorable due to shearing effect from surrounding atoms is
more in case of methanide ..
Highly conjugated organic compounds are normally .pdf
Highly conjugated organic compounds are normally colored or fluorescent. In order
to examine the nature of the color or the fluorescence we need lamda max, the maximum
wavelength at will a certain compound absorbs light. Once we find that wavelength we excite at
that wavelength to determine the fluorescence. With regard to metal complexes chemists use
UV-vis to determine the binding interaction between a metal and a ligand (this is called a metal
complex). This is normally done through UV-vis titrations. I can elaborate more but I think the
question is answered.
Solution
Highly conjugated organic compounds are normally colored or fluorescent. In order
to examine the nature of the color or the fluorescence we need lamda max, the maximum
wavelength at will a certain compound absorbs light. Once we find that wavelength we excite at
that wavelength to determine the fluorescence. With regard to metal complexes chemists use
UV-vis to determine the binding interaction between a metal and a ligand (this is called a metal
complex). This is normally done through UV-vis titrations. I can elaborate more but I think the
question is answered..
Recommended
object 1 sink object2 float as for object 2 weig.pdf
No.of moles of KOH , n = MOlarity * Volume in L = 0.0180 M * 0.032 L =
5.76*10^-4 moles HNO3 + KOH -----> KNO3 + H2O According to the balanced Equation , 1
mole of HNO3 reacts with 1 mole of KOH 5.76*10^-4 moles of HNO3 reacts with 5.76*10^-4
moles of KOH So Molarity of HNO3 , M = no.of moles / Volume in L = 5.76*10^-4 moles /
0.025 L = 0.02304 M
Solution
No.of moles of KOH , n = MOlarity * Volume in L = 0.0180 M * 0.032 L =
5.76*10^-4 moles HNO3 + KOH -----> KNO3 + H2O According to the balanced Equation , 1
mole of HNO3 reacts with 1 mole of KOH 5.76*10^-4 moles of HNO3 reacts with 5.76*10^-4
moles of KOH So Molarity of HNO3 , M = no.of moles / Volume in L = 5.76*10^-4 moles /
0.025 L = 0.02304 M.
Nitrogen Pollution Overuse or misuse of nitrogen .pdf
NaH then ethylene oxidide then PCC
Solution
NaH then ethylene oxidide then PCC.
many of compounds tend to be unstable and either .pdf
many of compounds tend to be unstable and either decompose or volatilize at higher
temperatures. the resultant peak will come with larger peak area so ans is b
Solution
many of compounds tend to be unstable and either decompose or volatilize at higher
temperatures. the resultant peak will come with larger peak area so ans is b.
Mendeleev,s law of periodic table .pdf
Mendeleev,s law of periodic table
Solution
Mendeleev,s law of periodic table.
It is very slightly soluble in in boiling water, .pdf
It is very slightly soluble in in boiling water, soluble in m-cresol and phenol, and is
degraded by acids.
Solution
It is very slightly soluble in in boiling water, soluble in m-cresol and phenol, and is
degraded by acids..
hydride shift is more favorable due to shearing e.pdf
hydride shift is more favorable due to shearing effect from surrounding atoms is
more in case of methanide .
Solution
hydride shift is more favorable due to shearing effect from surrounding atoms is
more in case of methanide ..
Highly conjugated organic compounds are normally .pdf
Highly conjugated organic compounds are normally colored or fluorescent. In order
to examine the nature of the color or the fluorescence we need lamda max, the maximum
wavelength at will a certain compound absorbs light. Once we find that wavelength we excite at
that wavelength to determine the fluorescence. With regard to metal complexes chemists use
UV-vis to determine the binding interaction between a metal and a ligand (this is called a metal
complex). This is normally done through UV-vis titrations. I can elaborate more but I think the
question is answered.
Solution
Highly conjugated organic compounds are normally colored or fluorescent. In order
to examine the nature of the color or the fluorescence we need lamda max, the maximum
wavelength at will a certain compound absorbs light. Once we find that wavelength we excite at
that wavelength to determine the fluorescence. With regard to metal complexes chemists use
UV-vis to determine the binding interaction between a metal and a ligand (this is called a metal
complex). This is normally done through UV-vis titrations. I can elaborate more but I think the
question is answered..
What is question. i am ready to answerSolutionWhat is question.pdf
public final category Huffman non-public Huffman() ;
personal static category HuffmanNode
}
personal static category HuffManComparator implements Comparator<HuffmanNode> come
node1.frequency - node2.frequency;
}
}
/**
* Compresses the string victimization huffman formula.
* The huffman tree and also the huffman code area unit serialized to disk
*
* @param sentence The sentence to be serialized
* @throws FileNotFoundException If file isn\'t found
* @throws IOException If IO exception happens.
*/
public static void compress(String sentence) throws FileNotFoundException, IOException ought
to atleast have one character.\");
}
final Map<Character, Integer> charFreq = getCharFrequency(sentence);
final HuffmanNode root = buildTree(charFreq);
final Map<Character, String> charCode = generateCodes(charFreq.keySet(), root);
final String encodedMessage = encodeMessage(charCode, sentence);
serializeTree(root);
serializeMessage(encodedMessage);
}
personal static Map<Character, Integer> getCharFrequency(String sentence) else
}
come map;
}
/**
* Map<Character, Integer> map
* Some implementation of that treeSet is passed as parameter.
* @param map
*/
personal static HuffmanNode buildTree(Map<Character, Integer> map) whereas
(nodeQueue.size() > 1)
// take away it to forestall object leak.
come nodeQueue.remove();
}
personal static Queue<HuffmanNode> createNodeQueue(Map<Character, Integer> map)
return pq;
}
personal static Map<Character, String> generateCodes(Set<Character> chars, HuffmanNode
node) {
final Map<Character, String> map = new HashMap<Character, String>();
doGenerateCode(node, map, \"\");
come map;
}
personal static void doGenerateCode(HuffmanNode node, Map<Character, String> map, String
s)
doGenerateCode(node.left, map, s + \'0\');
doGenerateCode(node.right, map, s + \'1\' );
}
personal static String encodeMessage(Map<Character, String> charCode, String sentence)
come stringBuilder.toString();
}
personal static void serializeTree(HuffmanNode node) throws FileNotFoundException,
IOException attempt (ObjectOutputStream oosTree = new ObjectOutputStream(new
FileOutputStream(\"/Users/ap/Desktop/tree\"))) attempt (ObjectOutputStream oosChar = new
ObjectOutputStream(new FileOutputStream(\"/Users/ap/Desktop/char\"))) cushioned to mark
finish of bit set relevant for deserialization.
oosTree.writeObject(bitSet);
}
}
}
personal static category IntObject
/*
* Algo:
* 1. Access the node
* 2. Register the worth in bit set.
*
*
* here true and false dont correspond to left branch and right branch.
* there,
* - true means that \"a branch originates from leaf\"
* - false mens \"a branch originates from non-left\".
*
* conjointly since branches originate from some node, the foundation node should be provided
as supply
* or start line of initial branches.
*
* Diagram and the way associate degree bit set would look as a result.
* (source node)
* / \\
* true true
* / \\
* (leaf node) (leaf node)
* | |
* false false
* | |
*
* thus currently a little set sounds like [fals.
They are dependent since in order to be paired they must be correlat.pdf
Note that 1 - .05 = .95, 1 - .15 = .85, and 1 - .15 - .15 = .7
Thus, we have the transition matrix
.95 .05 .00
.15 .70 .15
.00 .15 .85
Thus, we solve for the eigenvector with eigenvalue 1 (This is the steady state solution.)
Let x = 1
Then,
.95 + .15y = 1
.05 + .7y + .15z = y
Solving,
.15y = 1 - .95
.15z - .3y = -.05
Then, .15y = .05
y = .05/.15 = 1/3
Then, .15z - .3(1/3) = -.05
.15z = -.05 + .1
.15z = .05
z = .05/.15 = 1/3
Thus, x = 1, y = 1/3, z = 1/3 is the eigenvector
Yet, as x + y + z = 1, then
(x, y, z) = c(1, 1/3, 1/3) and (1+1/3+1/3)c = 1
5/3c = 1
c = 3/5
Then, (x, y, z) = 3/5(1, 1/3, 1/3) = (3/5, 1/5, 1/5) or (.6, .2, .2) = (60%, 20%, 20%)
Poor = 60%
Satisfactory = 20%
Preferred = 20%
Solution
Note that 1 - .05 = .95, 1 - .15 = .85, and 1 - .15 - .15 = .7
Thus, we have the transition matrix
.95 .05 .00
.15 .70 .15
.00 .15 .85
Thus, we solve for the eigenvector with eigenvalue 1 (This is the steady state solution.)
Let x = 1
Then,
.95 + .15y = 1
.05 + .7y + .15z = y
Solving,
.15y = 1 - .95
.15z - .3y = -.05
Then, .15y = .05
y = .05/.15 = 1/3
Then, .15z - .3(1/3) = -.05
.15z = -.05 + .1
.15z = .05
z = .05/.15 = 1/3
Thus, x = 1, y = 1/3, z = 1/3 is the eigenvector
Yet, as x + y + z = 1, then
(x, y, z) = c(1, 1/3, 1/3) and (1+1/3+1/3)c = 1
5/3c = 1
c = 3/5
Then, (x, y, z) = 3/5(1, 1/3, 1/3) = (3/5, 1/5, 1/5) or (.6, .2, .2) = (60%, 20%, 20%)
Poor = 60%
Satisfactory = 20%
Preferred = 20%.
The correct answer is Synapomorphy.ReasonFlowering plants or an.pdf
Let T be the final temperature.
Heat gained by spoon = mass of spoon x S(Ag) x temperature change
= 40.00 x 0.235 x (T - 20.5)
Heat lost by coffee = mass of coffee x S(coffee) x temperature change
= 250.0 x 4.184 x (95 - T)
Heat gained = heat lost
40.00 x 0.235 x (T - 20.5) = 250.0 x 4.184 x (95 - T)
1055.4T = 99562.7
Final temperature T = 94.3oC 94oC
Solution
Let T be the final temperature.
Heat gained by spoon = mass of spoon x S(Ag) x temperature change
= 40.00 x 0.235 x (T - 20.5)
Heat lost by coffee = mass of coffee x S(coffee) x temperature change
= 250.0 x 4.184 x (95 - T)
Heat gained = heat lost
40.00 x 0.235 x (T - 20.5) = 250.0 x 4.184 x (95 - T)
1055.4T = 99562.7
Final temperature T = 94.3oC 94oC.
Sufficient amount of water would absorbed by the carrot in the dish.pdf
Initial concentrations of H2 and I2 = moles/volume
= 1.00/0.5 = 2.00 M
H2 + I2 <=> 2 HI
I 2 2 0
C -a -a +2a
E 2-a 2-a 2a
Kc = [HI]2/[H2][I2]
= (2a)2/(2 - a)2 = 55.64
(2a)/(2 - a) = (55.64)1/2 = 7.459
a = 1.577
[H2] = 2 - a = 0.423 M
[I2] = 2 - a = 0.423 M
[HI] = 2a = 3.154 M
Solution
Initial concentrations of H2 and I2 = moles/volume
= 1.00/0.5 = 2.00 M
H2 + I2 <=> 2 HI
I 2 2 0
C -a -a +2a
E 2-a 2-a 2a
Kc = [HI]2/[H2][I2]
= (2a)2/(2 - a)2 = 55.64
(2a)/(2 - a) = (55.64)1/2 = 7.459
a = 1.577
[H2] = 2 - a = 0.423 M
[I2] = 2 - a = 0.423 M
[HI] = 2a = 3.154 M.
The answer is S2- has the largest radius.All the S species have t.pdf
Labor for excavation works:
Total area of footing = 1.2 x 1.2 = 1.44 m2
Depth of footing from floor line = 1.825 m
Total volume of soil to be excavated = 1.44 x 1.825 = 2.628 m3
Rate of excavation = 0.61 m3/hr
Time taken to excavate 2.628 m3 = 2.628/0.61 = 4.308 hrs
1 day = 8hrs
Time taken in days = 4.308 / 8 = 0.5385 day
SO, Quantity 0.5385 day, Cost P350/day, Total cost = 350 x .5385 = P 188.475
2) Rebar needed for spread footing-
with 7.5 cm covering the length of bar in both directions = 1.2 - .075x2 = 1.05 M
Total spacing of bars in one direction of length 1.05 M = 1.05 /.105 = 10
Total number of bars in one direction of length 1.05 M = 11
Total qty of 16mm rebar needed for spread footing of length 1.05 M in both direction = 22 nos
Qty of bar of 7.5 M (including 5 cm hook) = (1.15x 22) / 7.5 = 3.37 , say 4 nos
3) Total length of column = 5.37 m (elevation to top from floor line) + 1.825 (depth from floor
line to footing) = 7.195 m
Length of 16mm bar = 7.195 m + .3 (bend) = 7.495 m
Length of 8 rebars of 16 mm = 59.96 M, Quantity of 16mm rebar of 6 M length = 10 nos
Length of one lateral tie = (30 cm - 4x4 (cover))x4 + 3x3 (hook) = 94 cm
Let spacing of tie be 30 cm
Number of lateral tie = 26 nos
Quantity of 10mm rebar, 6M length = (0.94x26 nos)/6 = 5 nos
Solution
Labor for excavation works:
Total area of footing = 1.2 x 1.2 = 1.44 m2
Depth of footing from floor line = 1.825 m
Total volume of soil to be excavated = 1.44 x 1.825 = 2.628 m3
Rate of excavation = 0.61 m3/hr
Time taken to excavate 2.628 m3 = 2.628/0.61 = 4.308 hrs
1 day = 8hrs
Time taken in days = 4.308 / 8 = 0.5385 day
SO, Quantity 0.5385 day, Cost P350/day, Total cost = 350 x .5385 = P 188.475
2) Rebar needed for spread footing-
with 7.5 cm covering the length of bar in both directions = 1.2 - .075x2 = 1.05 M
Total spacing of bars in one direction of length 1.05 M = 1.05 /.105 = 10
Total number of bars in one direction of length 1.05 M = 11
Total qty of 16mm rebar needed for spread footing of length 1.05 M in both direction = 22 nos
Qty of bar of 7.5 M (including 5 cm hook) = (1.15x 22) / 7.5 = 3.37 , say 4 nos
3) Total length of column = 5.37 m (elevation to top from floor line) + 1.825 (depth from floor
line to footing) = 7.195 m
Length of 16mm bar = 7.195 m + .3 (bend) = 7.495 m
Length of 8 rebars of 16 mm = 59.96 M, Quantity of 16mm rebar of 6 M length = 10 nos
Length of one lateral tie = (30 cm - 4x4 (cover))x4 + 3x3 (hook) = 94 cm
Let spacing of tie be 30 cm
Number of lateral tie = 26 nos
Quantity of 10mm rebar, 6M length = (0.94x26 nos)/6 = 5 nos.
Sharing of electrons to form covalent bond one atom shares its one.pdf
Horizontal line is the line which passes through (-1,1) is parallel to x axis : Equation is y =1
vertical line is the line which passes through (-1,1) is parallel to Y axis : Equation is x =-1
So equations are : x =-1 and y =1
Solution
Horizontal line is the line which passes through (-1,1) is parallel to x axis : Equation is y =1
vertical line is the line which passes through (-1,1) is parallel to Y axis : Equation is x =-1
So equations are : x =-1 and y =1.
qweSolutionqwe.pdf
Hacker is a term used by some to mean \"a clever programmer\" and by others, especially those
in popular media, to mean \"someone who tries to break into computer systems.\"
A \"good hack\" is a clever solution to a programming problem and \"hacking\" is the act of
doing it. Raymond lists five possible characteristics that qualify one as a hacker, which we
paraphrase here:
Raymond deprecates the use of this term for someone who attempts to crack someone else\'s
system or otherwise uses programming or expert knowledge to act maliciously. He prefers the
term cracker for this meaning.
2) The term hacker is used in popular media to describe someone who attempts to break into
computer systems. Typically, this kind of hacker would be a proficient programmer or engineer
with sufficient technical knowledge to understand the weak points in a security system.
Solution
Hacker is a term used by some to mean \"a clever programmer\" and by others, especially those
in popular media, to mean \"someone who tries to break into computer systems.\"
A \"good hack\" is a clever solution to a programming problem and \"hacking\" is the act of
doing it. Raymond lists five possible characteristics that qualify one as a hacker, which we
paraphrase here:
Raymond deprecates the use of this term for someone who attempts to crack someone else\'s
system or otherwise uses programming or expert knowledge to act maliciously. He prefers the
term cracker for this meaning.
2) The term hacker is used in popular media to describe someone who attempts to break into
computer systems. Typically, this kind of hacker would be a proficient programmer or engineer
with sufficient technical knowledge to understand the weak points in a security system..
Please follow the data and description Interface ID In the for.pdf
From the kinetic theory of gases, the average speed of a molecule is given by:
v(average) = (8RT/M)1/2
where R is molar gas constant, T is temperature in K, M is molar mass of gas.
At a fixed temperature, the average molecular speed is highest for light molecules (small M) and
lowest for heavy molecules (large M).
Since M(H2O) = 18 < M(N2) = 28 < M(O2) = 32
Thus speed of H2O > N2 > O2
Solution
From the kinetic theory of gases, the average speed of a molecule is given by:
v(average) = (8RT/M)1/2
where R is molar gas constant, T is temperature in K, M is molar mass of gas.
At a fixed temperature, the average molecular speed is highest for light molecules (small M) and
lowest for heavy molecules (large M).
Since M(H2O) = 18 < M(N2) = 28 < M(O2) = 32
Thus speed of H2O > N2 > O2.
public final category Huffman non-public Huffman() ;personal stati.pdf
Given the assumptions of no friction and constant acceleration of 32 feet per second per second,
the height of an object is given by
[h=-1/2g t^2+v_0t+h_0] where [g=32] , [v_0=47] is the initial speed and [h_0=3] is the initial
height of the ball. Sub in these values to get:
[h=-16t^2+47t+3]
The height of the ball in feet after t seconds is [h=-16t^2+47t+3] .
Solution
Given the assumptions of no friction and constant acceleration of 32 feet per second per second,
the height of an object is given by
[h=-1/2g t^2+v_0t+h_0] where [g=32] , [v_0=47] is the initial speed and [h_0=3] is the initial
height of the ball. Sub in these values to get:
[h=-16t^2+47t+3]
The height of the ball in feet after t seconds is [h=-16t^2+47t+3] ..
C. is correct. note 1) aldehyde is more incline.pdf
C. is correct. note: 1) aldehyde is more inclined to hydrate than ketone due to less
steric hinderance. 2) electron-withdrawing groups such as F groups destabilize C=O and thus
benefit the hydration reaction.
Solution
C. is correct. note: 1) aldehyde is more inclined to hydrate than ketone due to less
steric hinderance. 2) electron-withdrawing groups such as F groups destabilize C=O and thus
benefit the hydration reaction..
Moral Standards - Morality is the disjuntion between right and wron.pdf
Do\'nt ask here
Solution
Do\'nt ask here.
m=Fr2GMSolutionm=Fr2GM.pdf
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4)
cos a cos b -sin a sin b =cos(a+b) is the formula
a=pi/3 , b=3pi/4
a+b =(pi/3) +(3pi/4)
a+b =(pi*4 +3*3pi)/(3*4)
a+b =(4pi +9pi)/(12)
a+b =(13pi)/(12)
a+b =(13pi/12)
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4) =cos(13pi/12)
Solution
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4)
cos a cos b -sin a sin b =cos(a+b) is the formula
a=pi/3 , b=3pi/4
a+b =(pi/3) +(3pi/4)
a+b =(pi*4 +3*3pi)/(3*4)
a+b =(4pi +9pi)/(12)
a+b =(13pi)/(12)
a+b =(13pi/12)
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4) =cos(13pi/12).
Misrepresentation is not considered a generic security issue in VoIP.pdf
Decay constant k = ln(2)/half life
= 0.6931/5730 = 1.20968 x 10-4 year-1
ln(A/Ao) = -kt where A is the activity at time t and Ao is initial activity
ln(7.10/15.3) = -1.20968 x 10-4 x t
Age = t = 6347 years old
Solution
Decay constant k = ln(2)/half life
= 0.6931/5730 = 1.20968 x 10-4 year-1
ln(A/Ao) = -kt where A is the activity at time t and Ao is initial activity
ln(7.10/15.3) = -1.20968 x 10-4 x t
Age = t = 6347 years old.
a). Poynting vectorPoynting vector, a quantity describing the mag.pdf
When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to
produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen
ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised. For
example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the
reverse reaction happens that we can write:
Solution
When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to
produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen
ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised. For
example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the
reverse reaction happens that we can write:.
A) At the center, the path travelled by the sounds waves from the tw.pdf
#include #include class shape { protected: double x,y; public: virtualvoid
get_data()=0; virtualvoid display_area()=0; }; class triangle : public shape { public:
void get_data(void) { cout<<\"\ \ =====Data Entry for Triangle=====\ \ \";
cout<<\"Enter base and height respectively : \"; cin>>x>>y; } void display_area(void)
{ cout<<\"\ \ =====Area of Triangle=====\ \ \"; double aot; aot = 0.5 * x * y;
cout<<\"Area of Triangle is \"<>x>>y; } void display_area(void) { cout<<\"\ \
=====Area of rectangle=====\ \ \"; double aor; aor = x * y; cout<<\"Area of
Rectangle is \"<>choice; switch(choice) { case 1 : list[0]->get_data(); list[0]-
>display_area(); getch(); break; case 2 : list[1]->get_data(); list[1]-
>display_area(); getch(); break; case 3 : goto end; default: cout<<\"\ \ Invalid
choice\ Try again\ \"; getch(); } } end: }
Solution
#include #include class shape { protected: double x,y; public: virtualvoid
get_data()=0; virtualvoid display_area()=0; }; class triangle : public shape { public:
void get_data(void) { cout<<\"\ \ =====Data Entry for Triangle=====\ \ \";
cout<<\"Enter base and height respectively : \"; cin>>x>>y; } void display_area(void)
{ cout<<\"\ \ =====Area of Triangle=====\ \ \"; double aot; aot = 0.5 * x * y;
cout<<\"Area of Triangle is \"<>x>>y; } void display_area(void) { cout<<\"\ \
=====Area of rectangle=====\ \ \"; double aor; aor = x * y; cout<<\"Area of
Rectangle is \"<>choice; switch(choice) { case 1 : list[0]->get_data(); list[0]-
>display_area(); getch(); break; case 2 : list[1]->get_data(); list[1]-
>display_area(); getch(); break; case 3 : goto end; default: cout<<\"\ \ Invalid
choice\ Try again\ \"; getch(); } } end: }.
Joints innervated with the nerves necessary for movement, it lacks b.pdf
Budget Deficit = Government\'s Total Expenditures Government\'s Total Income
Government\'s Total Income = 10% of actual GDP
= 10% * $5 trillion = $0.5 trillion
Government\'s Total Expenditures = $1 trillion
Actual Budget Deficit =1-0.5 => 0.5 Trillion
so answer is More than $.2 trillion
Solution
Budget Deficit = Government\'s Total Expenditures Government\'s Total Income
Government\'s Total Income = 10% of actual GDP
= 10% * $5 trillion = $0.5 trillion
Government\'s Total Expenditures = $1 trillion
Actual Budget Deficit =1-0.5 => 0.5 Trillion
so answer is More than $.2 trillion.
Freezing point depression = i x Kf x molality= 1.88 x 1.86 x 0.50.pdf
a. hyperbola
b parellel
c.left side
Solution
a. hyperbola
b parellel
c.left side.
ExplanationAldosterone stimulates kidneys to reabsorb sodium and .pdf
a) =hf=hc/v
5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v
v= 3.51 x10-7 m or 351 nm
b) =hf
=(6.626 x10-34Js2)(1.24 x1015 Hz)
= 8.28 x10-19 J
c) first, calculate the work function that 289 nm gives:
=hf=hc/v
= (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m)
= 6.88 x10-19
comparing this with the values of given for each metal, you can conclude that electrons will be
ejected from cadmium, manganese, and calcium because their respective work functions are
lower that 6.88 x10-19
Solution
a) =hf=hc/v
5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v
v= 3.51 x10-7 m or 351 nm
b) =hf
=(6.626 x10-34Js2)(1.24 x1015 Hz)
= 8.28 x10-19 J
c) first, calculate the work function that 289 nm gives:
=hf=hc/v
= (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m)
= 6.88 x10-19
comparing this with the values of given for each metal, you can conclude that electrons will be
ejected from cadmium, manganese, and calcium because their respective work functions are
lower that 6.88 x10-19.
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.
More Related Content
More from apposekitchenpvd
What is question. i am ready to answerSolutionWhat is question.pdf
public final category Huffman non-public Huffman() ;
personal static category HuffmanNode
}
personal static category HuffManComparator implements Comparator<HuffmanNode> come
node1.frequency - node2.frequency;
}
}
/**
* Compresses the string victimization huffman formula.
* The huffman tree and also the huffman code area unit serialized to disk
*
* @param sentence The sentence to be serialized
* @throws FileNotFoundException If file isn\'t found
* @throws IOException If IO exception happens.
*/
public static void compress(String sentence) throws FileNotFoundException, IOException ought
to atleast have one character.\");
}
final Map<Character, Integer> charFreq = getCharFrequency(sentence);
final HuffmanNode root = buildTree(charFreq);
final Map<Character, String> charCode = generateCodes(charFreq.keySet(), root);
final String encodedMessage = encodeMessage(charCode, sentence);
serializeTree(root);
serializeMessage(encodedMessage);
}
personal static Map<Character, Integer> getCharFrequency(String sentence) else
}
come map;
}
/**
* Map<Character, Integer> map
* Some implementation of that treeSet is passed as parameter.
* @param map
*/
personal static HuffmanNode buildTree(Map<Character, Integer> map) whereas
(nodeQueue.size() > 1)
// take away it to forestall object leak.
come nodeQueue.remove();
}
personal static Queue<HuffmanNode> createNodeQueue(Map<Character, Integer> map)
return pq;
}
personal static Map<Character, String> generateCodes(Set<Character> chars, HuffmanNode
node) {
final Map<Character, String> map = new HashMap<Character, String>();
doGenerateCode(node, map, \"\");
come map;
}
personal static void doGenerateCode(HuffmanNode node, Map<Character, String> map, String
s)
doGenerateCode(node.left, map, s + \'0\');
doGenerateCode(node.right, map, s + \'1\' );
}
personal static String encodeMessage(Map<Character, String> charCode, String sentence)
come stringBuilder.toString();
}
personal static void serializeTree(HuffmanNode node) throws FileNotFoundException,
IOException attempt (ObjectOutputStream oosTree = new ObjectOutputStream(new
FileOutputStream(\"/Users/ap/Desktop/tree\"))) attempt (ObjectOutputStream oosChar = new
ObjectOutputStream(new FileOutputStream(\"/Users/ap/Desktop/char\"))) cushioned to mark
finish of bit set relevant for deserialization.
oosTree.writeObject(bitSet);
}
}
}
personal static category IntObject
/*
* Algo:
* 1. Access the node
* 2. Register the worth in bit set.
*
*
* here true and false dont correspond to left branch and right branch.
* there,
* - true means that \"a branch originates from leaf\"
* - false mens \"a branch originates from non-left\".
*
* conjointly since branches originate from some node, the foundation node should be provided
as supply
* or start line of initial branches.
*
* Diagram and the way associate degree bit set would look as a result.
* (source node)
* / \\
* true true
* / \\
* (leaf node) (leaf node)
* | |
* false false
* | |
*
* thus currently a little set sounds like [fals.
They are dependent since in order to be paired they must be correlat.pdf
Note that 1 - .05 = .95, 1 - .15 = .85, and 1 - .15 - .15 = .7
Thus, we have the transition matrix
.95 .05 .00
.15 .70 .15
.00 .15 .85
Thus, we solve for the eigenvector with eigenvalue 1 (This is the steady state solution.)
Let x = 1
Then,
.95 + .15y = 1
.05 + .7y + .15z = y
Solving,
.15y = 1 - .95
.15z - .3y = -.05
Then, .15y = .05
y = .05/.15 = 1/3
Then, .15z - .3(1/3) = -.05
.15z = -.05 + .1
.15z = .05
z = .05/.15 = 1/3
Thus, x = 1, y = 1/3, z = 1/3 is the eigenvector
Yet, as x + y + z = 1, then
(x, y, z) = c(1, 1/3, 1/3) and (1+1/3+1/3)c = 1
5/3c = 1
c = 3/5
Then, (x, y, z) = 3/5(1, 1/3, 1/3) = (3/5, 1/5, 1/5) or (.6, .2, .2) = (60%, 20%, 20%)
Poor = 60%
Satisfactory = 20%
Preferred = 20%
Solution
Note that 1 - .05 = .95, 1 - .15 = .85, and 1 - .15 - .15 = .7
Thus, we have the transition matrix
.95 .05 .00
.15 .70 .15
.00 .15 .85
Thus, we solve for the eigenvector with eigenvalue 1 (This is the steady state solution.)
Let x = 1
Then,
.95 + .15y = 1
.05 + .7y + .15z = y
Solving,
.15y = 1 - .95
.15z - .3y = -.05
Then, .15y = .05
y = .05/.15 = 1/3
Then, .15z - .3(1/3) = -.05
.15z = -.05 + .1
.15z = .05
z = .05/.15 = 1/3
Thus, x = 1, y = 1/3, z = 1/3 is the eigenvector
Yet, as x + y + z = 1, then
(x, y, z) = c(1, 1/3, 1/3) and (1+1/3+1/3)c = 1
5/3c = 1
c = 3/5
Then, (x, y, z) = 3/5(1, 1/3, 1/3) = (3/5, 1/5, 1/5) or (.6, .2, .2) = (60%, 20%, 20%)
Poor = 60%
Satisfactory = 20%
Preferred = 20%.
The correct answer is Synapomorphy.ReasonFlowering plants or an.pdf
Let T be the final temperature.
Heat gained by spoon = mass of spoon x S(Ag) x temperature change
= 40.00 x 0.235 x (T - 20.5)
Heat lost by coffee = mass of coffee x S(coffee) x temperature change
= 250.0 x 4.184 x (95 - T)
Heat gained = heat lost
40.00 x 0.235 x (T - 20.5) = 250.0 x 4.184 x (95 - T)
1055.4T = 99562.7
Final temperature T = 94.3oC 94oC
Solution
Let T be the final temperature.
Heat gained by spoon = mass of spoon x S(Ag) x temperature change
= 40.00 x 0.235 x (T - 20.5)
Heat lost by coffee = mass of coffee x S(coffee) x temperature change
= 250.0 x 4.184 x (95 - T)
Heat gained = heat lost
40.00 x 0.235 x (T - 20.5) = 250.0 x 4.184 x (95 - T)
1055.4T = 99562.7
Final temperature T = 94.3oC 94oC.
Sufficient amount of water would absorbed by the carrot in the dish.pdf
Initial concentrations of H2 and I2 = moles/volume
= 1.00/0.5 = 2.00 M
H2 + I2 <=> 2 HI
I 2 2 0
C -a -a +2a
E 2-a 2-a 2a
Kc = [HI]2/[H2][I2]
= (2a)2/(2 - a)2 = 55.64
(2a)/(2 - a) = (55.64)1/2 = 7.459
a = 1.577
[H2] = 2 - a = 0.423 M
[I2] = 2 - a = 0.423 M
[HI] = 2a = 3.154 M
Solution
Initial concentrations of H2 and I2 = moles/volume
= 1.00/0.5 = 2.00 M
H2 + I2 <=> 2 HI
I 2 2 0
C -a -a +2a
E 2-a 2-a 2a
Kc = [HI]2/[H2][I2]
= (2a)2/(2 - a)2 = 55.64
(2a)/(2 - a) = (55.64)1/2 = 7.459
a = 1.577
[H2] = 2 - a = 0.423 M
[I2] = 2 - a = 0.423 M
[HI] = 2a = 3.154 M.
The answer is S2- has the largest radius.All the S species have t.pdf
Labor for excavation works:
Total area of footing = 1.2 x 1.2 = 1.44 m2
Depth of footing from floor line = 1.825 m
Total volume of soil to be excavated = 1.44 x 1.825 = 2.628 m3
Rate of excavation = 0.61 m3/hr
Time taken to excavate 2.628 m3 = 2.628/0.61 = 4.308 hrs
1 day = 8hrs
Time taken in days = 4.308 / 8 = 0.5385 day
SO, Quantity 0.5385 day, Cost P350/day, Total cost = 350 x .5385 = P 188.475
2) Rebar needed for spread footing-
with 7.5 cm covering the length of bar in both directions = 1.2 - .075x2 = 1.05 M
Total spacing of bars in one direction of length 1.05 M = 1.05 /.105 = 10
Total number of bars in one direction of length 1.05 M = 11
Total qty of 16mm rebar needed for spread footing of length 1.05 M in both direction = 22 nos
Qty of bar of 7.5 M (including 5 cm hook) = (1.15x 22) / 7.5 = 3.37 , say 4 nos
3) Total length of column = 5.37 m (elevation to top from floor line) + 1.825 (depth from floor
line to footing) = 7.195 m
Length of 16mm bar = 7.195 m + .3 (bend) = 7.495 m
Length of 8 rebars of 16 mm = 59.96 M, Quantity of 16mm rebar of 6 M length = 10 nos
Length of one lateral tie = (30 cm - 4x4 (cover))x4 + 3x3 (hook) = 94 cm
Let spacing of tie be 30 cm
Number of lateral tie = 26 nos
Quantity of 10mm rebar, 6M length = (0.94x26 nos)/6 = 5 nos
Solution
Labor for excavation works:
Total area of footing = 1.2 x 1.2 = 1.44 m2
Depth of footing from floor line = 1.825 m
Total volume of soil to be excavated = 1.44 x 1.825 = 2.628 m3
Rate of excavation = 0.61 m3/hr
Time taken to excavate 2.628 m3 = 2.628/0.61 = 4.308 hrs
1 day = 8hrs
Time taken in days = 4.308 / 8 = 0.5385 day
SO, Quantity 0.5385 day, Cost P350/day, Total cost = 350 x .5385 = P 188.475
2) Rebar needed for spread footing-
with 7.5 cm covering the length of bar in both directions = 1.2 - .075x2 = 1.05 M
Total spacing of bars in one direction of length 1.05 M = 1.05 /.105 = 10
Total number of bars in one direction of length 1.05 M = 11
Total qty of 16mm rebar needed for spread footing of length 1.05 M in both direction = 22 nos
Qty of bar of 7.5 M (including 5 cm hook) = (1.15x 22) / 7.5 = 3.37 , say 4 nos
3) Total length of column = 5.37 m (elevation to top from floor line) + 1.825 (depth from floor
line to footing) = 7.195 m
Length of 16mm bar = 7.195 m + .3 (bend) = 7.495 m
Length of 8 rebars of 16 mm = 59.96 M, Quantity of 16mm rebar of 6 M length = 10 nos
Length of one lateral tie = (30 cm - 4x4 (cover))x4 + 3x3 (hook) = 94 cm
Let spacing of tie be 30 cm
Number of lateral tie = 26 nos
Quantity of 10mm rebar, 6M length = (0.94x26 nos)/6 = 5 nos.
Sharing of electrons to form covalent bond one atom shares its one.pdf
Horizontal line is the line which passes through (-1,1) is parallel to x axis : Equation is y =1
vertical line is the line which passes through (-1,1) is parallel to Y axis : Equation is x =-1
So equations are : x =-1 and y =1
Solution
Horizontal line is the line which passes through (-1,1) is parallel to x axis : Equation is y =1
vertical line is the line which passes through (-1,1) is parallel to Y axis : Equation is x =-1
So equations are : x =-1 and y =1.
qweSolutionqwe.pdf
Hacker is a term used by some to mean \"a clever programmer\" and by others, especially those
in popular media, to mean \"someone who tries to break into computer systems.\"
A \"good hack\" is a clever solution to a programming problem and \"hacking\" is the act of
doing it. Raymond lists five possible characteristics that qualify one as a hacker, which we
paraphrase here:
Raymond deprecates the use of this term for someone who attempts to crack someone else\'s
system or otherwise uses programming or expert knowledge to act maliciously. He prefers the
term cracker for this meaning.
2) The term hacker is used in popular media to describe someone who attempts to break into
computer systems. Typically, this kind of hacker would be a proficient programmer or engineer
with sufficient technical knowledge to understand the weak points in a security system.
Solution
Hacker is a term used by some to mean \"a clever programmer\" and by others, especially those
in popular media, to mean \"someone who tries to break into computer systems.\"
A \"good hack\" is a clever solution to a programming problem and \"hacking\" is the act of
doing it. Raymond lists five possible characteristics that qualify one as a hacker, which we
paraphrase here:
Raymond deprecates the use of this term for someone who attempts to crack someone else\'s
system or otherwise uses programming or expert knowledge to act maliciously. He prefers the
term cracker for this meaning.
2) The term hacker is used in popular media to describe someone who attempts to break into
computer systems. Typically, this kind of hacker would be a proficient programmer or engineer
with sufficient technical knowledge to understand the weak points in a security system..
Please follow the data and description Interface ID In the for.pdf
From the kinetic theory of gases, the average speed of a molecule is given by:
v(average) = (8RT/M)1/2
where R is molar gas constant, T is temperature in K, M is molar mass of gas.
At a fixed temperature, the average molecular speed is highest for light molecules (small M) and
lowest for heavy molecules (large M).
Since M(H2O) = 18 < M(N2) = 28 < M(O2) = 32
Thus speed of H2O > N2 > O2
Solution
From the kinetic theory of gases, the average speed of a molecule is given by:
v(average) = (8RT/M)1/2
where R is molar gas constant, T is temperature in K, M is molar mass of gas.
At a fixed temperature, the average molecular speed is highest for light molecules (small M) and
lowest for heavy molecules (large M).
Since M(H2O) = 18 < M(N2) = 28 < M(O2) = 32
Thus speed of H2O > N2 > O2.
public final category Huffman non-public Huffman() ;personal stati.pdf
Given the assumptions of no friction and constant acceleration of 32 feet per second per second,
the height of an object is given by
[h=-1/2g t^2+v_0t+h_0] where [g=32] , [v_0=47] is the initial speed and [h_0=3] is the initial
height of the ball. Sub in these values to get:
[h=-16t^2+47t+3]
The height of the ball in feet after t seconds is [h=-16t^2+47t+3] .
Solution
Given the assumptions of no friction and constant acceleration of 32 feet per second per second,
the height of an object is given by
[h=-1/2g t^2+v_0t+h_0] where [g=32] , [v_0=47] is the initial speed and [h_0=3] is the initial
height of the ball. Sub in these values to get:
[h=-16t^2+47t+3]
The height of the ball in feet after t seconds is [h=-16t^2+47t+3] ..
C. is correct. note 1) aldehyde is more incline.pdf
C. is correct. note: 1) aldehyde is more inclined to hydrate than ketone due to less
steric hinderance. 2) electron-withdrawing groups such as F groups destabilize C=O and thus
benefit the hydration reaction.
Solution
C. is correct. note: 1) aldehyde is more inclined to hydrate than ketone due to less
steric hinderance. 2) electron-withdrawing groups such as F groups destabilize C=O and thus
benefit the hydration reaction..
Moral Standards - Morality is the disjuntion between right and wron.pdf
Do\'nt ask here
Solution
Do\'nt ask here.
m=Fr2GMSolutionm=Fr2GM.pdf
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4)
cos a cos b -sin a sin b =cos(a+b) is the formula
a=pi/3 , b=3pi/4
a+b =(pi/3) +(3pi/4)
a+b =(pi*4 +3*3pi)/(3*4)
a+b =(4pi +9pi)/(12)
a+b =(13pi)/(12)
a+b =(13pi/12)
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4) =cos(13pi/12)
Solution
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4)
cos a cos b -sin a sin b =cos(a+b) is the formula
a=pi/3 , b=3pi/4
a+b =(pi/3) +(3pi/4)
a+b =(pi*4 +3*3pi)/(3*4)
a+b =(4pi +9pi)/(12)
a+b =(13pi)/(12)
a+b =(13pi/12)
cos(pi/3)*cos(3pi/4) - sin(pi/3)*sin(3pi/4) =cos(13pi/12).
Misrepresentation is not considered a generic security issue in VoIP.pdf
Decay constant k = ln(2)/half life
= 0.6931/5730 = 1.20968 x 10-4 year-1
ln(A/Ao) = -kt where A is the activity at time t and Ao is initial activity
ln(7.10/15.3) = -1.20968 x 10-4 x t
Age = t = 6347 years old
Solution
Decay constant k = ln(2)/half life
= 0.6931/5730 = 1.20968 x 10-4 year-1
ln(A/Ao) = -kt where A is the activity at time t and Ao is initial activity
ln(7.10/15.3) = -1.20968 x 10-4 x t
Age = t = 6347 years old.
a). Poynting vectorPoynting vector, a quantity describing the mag.pdf
When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to
produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen
ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised. For
example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the
reverse reaction happens that we can write:
Solution
When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to
produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen
ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised. For
example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the
reverse reaction happens that we can write:.
A) At the center, the path travelled by the sounds waves from the tw.pdf
#include #include class shape { protected: double x,y; public: virtualvoid
get_data()=0; virtualvoid display_area()=0; }; class triangle : public shape { public:
void get_data(void) { cout<<\"\ \ =====Data Entry for Triangle=====\ \ \";
cout<<\"Enter base and height respectively : \"; cin>>x>>y; } void display_area(void)
{ cout<<\"\ \ =====Area of Triangle=====\ \ \"; double aot; aot = 0.5 * x * y;
cout<<\"Area of Triangle is \"<>x>>y; } void display_area(void) { cout<<\"\ \
=====Area of rectangle=====\ \ \"; double aor; aor = x * y; cout<<\"Area of
Rectangle is \"<>choice; switch(choice) { case 1 : list[0]->get_data(); list[0]-
>display_area(); getch(); break; case 2 : list[1]->get_data(); list[1]-
>display_area(); getch(); break; case 3 : goto end; default: cout<<\"\ \ Invalid
choice\ Try again\ \"; getch(); } } end: }
Solution
#include #include class shape { protected: double x,y; public: virtualvoid
get_data()=0; virtualvoid display_area()=0; }; class triangle : public shape { public:
void get_data(void) { cout<<\"\ \ =====Data Entry for Triangle=====\ \ \";
cout<<\"Enter base and height respectively : \"; cin>>x>>y; } void display_area(void)
{ cout<<\"\ \ =====Area of Triangle=====\ \ \"; double aot; aot = 0.5 * x * y;
cout<<\"Area of Triangle is \"<>x>>y; } void display_area(void) { cout<<\"\ \
=====Area of rectangle=====\ \ \"; double aor; aor = x * y; cout<<\"Area of
Rectangle is \"<>choice; switch(choice) { case 1 : list[0]->get_data(); list[0]-
>display_area(); getch(); break; case 2 : list[1]->get_data(); list[1]-
>display_area(); getch(); break; case 3 : goto end; default: cout<<\"\ \ Invalid
choice\ Try again\ \"; getch(); } } end: }.
Joints innervated with the nerves necessary for movement, it lacks b.pdf
Budget Deficit = Government\'s Total Expenditures Government\'s Total Income
Government\'s Total Income = 10% of actual GDP
= 10% * $5 trillion = $0.5 trillion
Government\'s Total Expenditures = $1 trillion
Actual Budget Deficit =1-0.5 => 0.5 Trillion
so answer is More than $.2 trillion
Solution
Budget Deficit = Government\'s Total Expenditures Government\'s Total Income
Government\'s Total Income = 10% of actual GDP
= 10% * $5 trillion = $0.5 trillion
Government\'s Total Expenditures = $1 trillion
Actual Budget Deficit =1-0.5 => 0.5 Trillion
so answer is More than $.2 trillion.
Freezing point depression = i x Kf x molality= 1.88 x 1.86 x 0.50.pdf
a. hyperbola
b parellel
c.left side
Solution
a. hyperbola
b parellel
c.left side.
ExplanationAldosterone stimulates kidneys to reabsorb sodium and .pdf
a) =hf=hc/v
5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v
v= 3.51 x10-7 m or 351 nm
b) =hf
=(6.626 x10-34Js2)(1.24 x1015 Hz)
= 8.28 x10-19 J
c) first, calculate the work function that 289 nm gives:
=hf=hc/v
= (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m)
= 6.88 x10-19
comparing this with the values of given for each metal, you can conclude that electrons will be
ejected from cadmium, manganese, and calcium because their respective work functions are
lower that 6.88 x10-19
Solution
a) =hf=hc/v
5.66 x10-19 J = (6.626 x10-34 Js2)(3.0 x108 m/s2)/v
v= 3.51 x10-7 m or 351 nm
b) =hf
=(6.626 x10-34Js2)(1.24 x1015 Hz)
= 8.28 x10-19 J
c) first, calculate the work function that 289 nm gives:
=hf=hc/v
= (6.626 x10-34 Js2)(3.0 x108 m/s2)/ (289 x10-9 m)
= 6.88 x10-19
comparing this with the values of given for each metal, you can conclude that electrons will be
ejected from cadmium, manganese, and calcium because their respective work functions are
lower that 6.88 x10-19.
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A) At the center, the path travelled by the sounds waves from the tw.pdf
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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.
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/
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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.
Best Digital Marketing Institute In NOIDA
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Pavilions
NGV Architecture Commission Competition
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Other Pavilion Designs
Normal Labour/ Stages of Labour/ Mechanism of Labour
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
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.
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Home assignment II on Spectroscopy 2024 Answers.pdf
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Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
1.4 modern child centered education - mahatma gandhi-2.pptx
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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.
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Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
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Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Investigating Pavilion Designs
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