They are dependent since in order to be paired they must be correlat.pdf
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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%.
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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.
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..
Because ethanol and water are miscible and they w.pdf
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..
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.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
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.
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..
Because ethanol and water are miscible and they w.pdf
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..
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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What is question. i am ready to answerSolutionWhat is question.pdf
What is question. i am ready to answerSolutionWhat is question.pdf
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The correct answer is Synapomorphy.ReasonFlowering plants or an.pdf
Sufficient amount of water would absorbed by the carrot in the dish.pdf
Sufficient amount of water would absorbed by the carrot in the dish.pdf
The answer is S2- has the largest radius.All the S species have t.pdf
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Please follow the data and description Interface ID In the for.pdf
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public final category Huffman non-public Huffman() ;personal stati.pdf
C. is correct. note 1) aldehyde is more incline.pdf
C. is correct. note 1) aldehyde is more incline.pdf
Because ethanol and water are miscible and they w.pdf
Because ethanol and water are miscible and they w.pdf
Moral Standards - Morality is the disjuntion between right and wron.pdf
Moral Standards - Morality is the disjuntion between right and wron.pdf
Misrepresentation is not considered a generic security issue in VoIP.pdf
Misrepresentation is not considered a generic security issue in VoIP.pdf
a). Poynting vectorPoynting vector, a quantity describing the mag.pdf
a). Poynting vectorPoynting vector, a quantity describing the mag.pdf
A) At the center, the path travelled by the sounds waves from the tw.pdf
A) At the center, the path travelled by the sounds waves from the tw.pdf
Joints innervated with the nerves necessary for movement, it lacks b.pdf
Joints innervated with the nerves necessary for movement, it lacks b.pdf
Freezing point depression = i x Kf x molality= 1.88 x 1.86 x 0.50.pdf
Freezing point depression = i x Kf x molality= 1.88 x 1.86 x 0.50.pdf
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