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..
object 1 sink object2 float as for object 2 weig.pdfapposekitchenpvd
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.
many of compounds tend to be unstable and either .pdfapposekitchenpvd
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.
hydride shift is more favorable due to shearing e.pdfapposekitchenpvd
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 .pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.
object 1 sink object2 float as for object 2 weig.pdfapposekitchenpvd
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.
many of compounds tend to be unstable and either .pdfapposekitchenpvd
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.
hydride shift is more favorable due to shearing e.pdfapposekitchenpvd
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 .pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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..
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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:.
Joints innervated with the nerves necessary for movement, it lacks b.pdfapposekitchenpvd
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.
ExplanationAldosterone stimulates kidneys to reabsorb sodium and .pdfapposekitchenpvd
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.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
The correct answer is Synapomorphy.ReasonFlowering plants or an.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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..
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.pdfapposekitchenpvd
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.pdfapposekitchenpvd
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:.
Joints innervated with the nerves necessary for movement, it lacks b.pdfapposekitchenpvd
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.
ExplanationAldosterone stimulates kidneys to reabsorb sodium and .pdfapposekitchenpvd
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.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
How to Make a Field invisible in Odoo 17Celine George
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.
Embracing GenAI - A Strategic ImperativePeter Windle
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.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
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
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
It is very slightly soluble in in boiling water, .pdf
1. 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.