Problem 2. The length of a side of the square block is 4 in. Under the application of the load V, the top edge of the block displaces 1 16 in. (dashed lines show displacement). Determine the shear strain at corner A and the shear strain at angle COD.
Answers
Answer and Explanation:
The answer is attached below
Related Questions
In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variable, and terminates with a semicolon. It is possible to declare multiple variables separated by commas in one statement. The following statements present examples, float z; double z, w; The following partial grammar represents the specification for C++ style variable declaration. In this grammar, the letters z and w are terminals that represent two variable names. The non-terminal S is the start symbol. S=TV; V=CX X = , VIE T = float double C = z w 1 - Determine Nullable values for the LHS and RHS of all rules. Please note, your answer includes all Nullable functions for LHS and RHS, in addition to the resulting values. (25 points)
Answers
Answer:
The given grammar is :
S = T V ;
V = C X
X = , V | ε
T = float | double
C = z | w
1.
Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.
From the given grammar,
Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.
No other variables generate variable X or ε.
So, only variable X is nullable.
2.
First of nullable variable X is First (X ) = , and ε (epsilon).
L.H.S.
The first of other varibles are :
First (S) = {float, double }
First (T) = {float, double }
First (V) = {z, w}
First (C) = {z, w}
R.H.S.
First (T V ; ) = {float, double }
First ( C X ) = {z, w}
First (, V) = ,
First ( ε ) = ε
First (float) = float
First (double) = double
First (z) = z
First (w) = w
3.
Follow of nullable variable X is Follow (V).
Follow (S) = $
Follow (T) = {z, w}
Follow (V) = ;
Follow (X) = Follow (V) = ;
Follow (C) = , and ;
Explanation:
For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our website or in our book. You may discuss general ideas with other people. Assume A[1. . . n] is a heap, except that the element at index i might be too large. For the following parts, you should create a method that inputs A, n, and i, and makes A into a heap.
Answers
Answer:
(a)
(i) pseudo code :-
current = i
// assuming parent of root is -1
while A[parent] < A[current] && parent != -1 do,
if A[parent] < A[current] // if current element is bigger than parent then shift it up
swap(A[current],A[parent])
current = parent
(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))
(b)
(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.
If above condition is not true then we will check for parent node and if it is less than current node then swap these.
pseudo code :-
current = i
// assuming parent of root is -1
parent is parent node of current node
while A[parent] < A[current] && parent != -1 do,
if A[grandparent] < A[current] // if current element is bigger than parent then shift it up
swap(A[current],A[parent])
swap(A[grandparent],A[parent])
current = grandparent
else if A[parent] < A[current]
swap(A[parent],A[current])
current = parent
(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)
so order would be log(n)/2
(iii) C++ code :-
#include<bits/stdc++.h>
using namespace std;
// function to return index of parent node
int parent(int i)
{
if(i == 0)
return -1;
return (i-1)/2;
}
// function to return index of grandparent node
int grandparent(int i)
{
int p = parent(i);
if(p == -1)
return -1;
else
return parent(p);
}
void shift_up(int A[], int n, int ind)
{
int curr = ind-1; // because array is 0-indexed
while(parent(curr) != -1 && A[parent(curr)] < A[curr])
{
int g = grandparent(curr);
int p = parent(curr);
if(g != -1 && A[g] < A[curr])
{
swap(A[curr],A[p]);
swap(A[p],A[g]);
curr = g;
}
else if(A[p] < A[curr])
{
swap(A[p],A[curr]);
curr = p;
}
}
}
int main()
{
int n;
cout<<"enter the number of elements :-\n";
cin>>n;
int A[n];
cout<<"enter the elements of array :-\n";
for(int i=0;i<n;i++)
cin>>A[i];
int ind;
cout<<"enter the index value :- \n";
cin>>ind;
shift_up(A,n,ind);
cout<<"array after shift up :-\n";
for(int i=0;i<n;i++)
cout<<A[i]<<" ";
cout<<endl;
}
Explanation:
A vertical plate has a sharp-edged orifice at its center. A water jet of speed V strikes the plate concentrically. Obtain an expression for the external force needed to hold the plate in place, if the jet leaving the orifice also has speed V. Evaluate the force for V 5 15 ft/s, D 5 4 in., and d 5 1 in. Plot the required force as a function of diameter ratio for a suitable range of diameter d.
Answers
Answer:
184.077N
Explanation:
Please see attachment for step by step guide
Answer:
Answer is 184.077
Refer below for the explanation.
Explanation:
As per the question,
V 5 15 ft/s,
D 5 4 in,
d 5 1 in
Refer to the picture for detailed explanation.
Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street repair area of 1000 ft x 60 ft x 6 in. The aggregate in the stockpile contains 3.5% moisture. If the required compaction is 95% of the target, how many pounds of aggregate will be needed?
Answers
Answer:
total weight of the aggregate = 3594878.28 lbs
Explanation:
given data
density = 121.8 lb/ft³
area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft
moisture = 3.5 %
compaction = 95%
solution
we get here first volume of the space that is filled with the aggregate that is
volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft
now we get fill space with aggregate that compact to 95% of dry density.
so we fill space with aggregate of density that is = 95% of 121.8
= 115.71 lb/ cu ft
so now dry weight of aggregate is
dry weight of aggregate = 30,000 × 115.71 = 3471300 lb
when we assume that moisture percentage is by weight
then weight of the moisture in aggregate will be
weight of the moisture in aggregate = 3.56 % of 3471300 lb
weight of the moisture in aggregate = 123578.28 lbs
and
we get total weight of the aggregate to fill space that is
total weight of the aggregate = 3471300 lb +123578.28 lb
total weight of the aggregate = 3594878.28 lbs
A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradual contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions made.
Answers
Answer:
Depth in the contracted section = 2.896m
Velocity in the contracted section = 2.072m/s
Explanation:
Please see that attachment for the solving.
Assumptions:
1. Negligible head losses
2. Horizontal channel bottom
Define a function PyramidVolume with double parameters baseLength, baseWidth, and pyramidHeight, that returns as a double the volume of a pyramid with a rectangular base. Relevant geometry equations:Volume = base area x height x 1/3Base area = base length x base width.(Watch out for integer division).#include /* Your solution goes here */int main(void) {printf("Volume for 1.0, 1.0, 1.0 is: %.2f\n", PyramidVolume(1.0, 1.0, 1.0) );return 0;}
Answers
Answer / Explanation:
To answer this question, we first define the parameters which are,
Volume: This can be defined or refereed to the quantity of three-dimensional space enclosed by a closed surface. For the purpose of illustration, we can say that the space that a substance or shape occupies or contains. The SI used in measuring volume is mostly in cubic metre.
Therefore,
Volume, where Volume = base area x height x 1/3
where,
Base area = base length x base width
However, we also watch out for the division integer.
So moving forward to write the code solving the question, we have:
double Pyramid Volume (double baseLength, double baseWidth, double pyramid Height)
{
double baseArea = baseLength * baseWidth;
double vol = ((baseArea * pyramidHeight) * 1/3);
return vol;
}
int main() {
cout << "Volume for 1.0, 1.0, 1.0 is: " << PyramidVolume(1.0, 1.0, 1.0) <<
endl;
return 0;
}
Answer:
The problem demands a function PyramidVolume() in C language since printf() command exists in C language. Complete code, output and explanation is provided below.
Code in C Language:
#include <stdio.h>
double PyramidVolume(double baseLength, double baseWidth, double pyramidHeight)
{
return baseLength*baseWidth*pyramidHeight/3;
}
int main()
{
printf("Volume for 1.0, 1.0, 1.0 is: %.2f\n",PyramidVolume(1, 1, 1));
printf("Volume for 2.5, 5, 2.0 is: %.2f\n",PyramidVolume(2.5, 5, 2.0));
return 0;
}
Output:
Please also refer to the attached output results
Volume for 1.0, 1.0, 1.0 is: 0.33
Volume for 2.5, 5, 2.0 is: 8.33
Explanation:
A function PyramidVolume of type double is created which takes three inputs arguments of type double length, width and height and returns the volume of the pyramid as per the formula given in the question.
Then in the main function we called the PyramidVolume() function with inputs 1.0, 1.0, 1.0 and it returned correct output.
We again tested it with different inputs and again it returned correct output.
An AISI 1040 cold-drawn steel tube has an OD 5 50 mm and wall thickness 6 mm. Whatmaximum external pressure can this tube withstand if the largest principal normal stress is notto exceed 80 percent of the minimum yield strength of the material?
Answers
Calculating the maximum external pressure that a cold-drawn AISI 1040 steel tube can withstand involves using the formula for hoop stress and considering 80 percent of the steel's minimum yield strength. The formula relates the difference in pressure across the tube wall to its inner and outer radii.
Explanation:The question asks for the maximum external pressure a cold-drawn AISI 1040 steel tube can withstand, given that the largest principal normal stress should not exceed 80 percent of the steel's minimum yield strength. The tube's outer diameter (OD) is 50 mm, and it has a wall thickness of 6 mm. To solve this, we must utilize the formula for hoop stress (sigma) in a thin-walled cylinder under external pressure, which is σ = δP(r_o/(r_o - r_i)), where δP is the difference in internal and external pressure, r_o is the outer radius, and r_i is the inner radius. The yield strength of AISI 1040 steel must be considered, and 80 percent of this value is taken as the maximum allowable stress. Without knowing the exact yield strength, it cannot be directly calculated in this response. However, the process would involve deriving δP from the given conditions and substituting into the hoop stress formula to find the maximum external pressure.
A cylindrical specimen of a metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of 15,000 N (3370 lbf) produces a reduction in specimen diameter of 7 * 10-3 mm (2.8 * 10-4 in.). Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa (14.5 * 106 psi).
Answers
The Poisson's ratio is 0.36
Explanation:
Given-
Diameter = 10 mm = 10 X 10⁻³m
Force, F = 15000 N
Diameter reduction = 7 X 10⁻3 mm = 7 X 10⁻⁶ mm
The equation for Poisson's ratio:
ν = [tex]\frac{-E_{x} }{E_{z} }[/tex]
and
εₓ = Δd / d₀
εz = σ / E
= F / AE
We know,
Area of circle = π (d₀/2)²
[tex]E_{z}[/tex]= F / π (d₀/2)² E
εz = 4F / πd₀²E
Therefore, Poisson's ratio is
ν = - Δd ÷ d / 4F ÷ πd₀²E
ν = -d₀ΔDπE / 4F
ν = (10 X 10⁻³) (-7 X 10⁻⁶) (3.14) (100 X 10⁻⁹) / 4 (15000)
ν = 0.36
Therefore, the Poisson's ratio is 0.36
Design a circuit with output f and inputs x1, x0, y1, and y0. Let X = x1x0 and Y = y1y0 represent two 2-digit binary numbers. The output f should be 1 if the numbers represented by X and Y are equal. Otherwise, f should be 0.
Answers
Complete question
The complete question is shown on the first uploaded image
Answer:
a The table is shown on the second uploaded image
b The simplest possible sum product is shown on the fifth uploaded image
Explanation:
The explanation is shown on the third fourth and fifth uploaded image
Two bodies have heat capacities (at constant volume) c, = a and c2 = bT and are thermally isolated from the rest of the universe. Initial temperatures of the bodies are T_10 and T_20, with T_20, > T_10. The two bodies are brought into thermal equilibrium (keeping the volume constant) while delivering as much work as possible to a reversible work source.
(a) What is the final temperature T_f of the two bodies? (In case you are unable to solve for an explicit value of T_i, you will still get full credit if you explain in detail how to obtain the value).
(b) What is the maximum work delivered to the reversible work source? (You may express the answer in terms of T_e without having to explicitly solve for it).
Answers
Answer:
Explanation:
The answer to the above question is given in attached files.
Assume that a specific hard disk drive has an average access time of 16ms (i.e. the seek and rotational delay sums to 16ms) and a throughput or transfer rate of 185MBytes/s, where a megabyte is measured as 1024.
Required:
What is the average access time for a hard disk spinning at 360 revolutions per second with a seek time of 10 milliseconds?
Answers
Answer:
Average access time for a hard disk = 11.38 ms
Explanation:
See attached pictures for step by step explanation.
Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support for academic programs. Some IT personnel provide technology infrastructure s programs and technology infrastructure support. IT personnel are not professors. IT personne are required to take periodic training to retain their technical expertise. Tiny College tracks a IT personnel training by date, type, and results (completed vs. not completed). Given that information, create the complete ERD containing all primary keys, foreign keys, and main attributes.
Answers
Answer:
solution in the picture attached
Explanation:
Foreign keys help define the relationship between tables, which is what puts the "relational" in "relational database." They enable database developers to preserve referential integrity across their system.
What is a foreign key constraint?Foreign key constraints are the rules created when we add foreign keys to a table. Foreign key constraints in table A link to a column with unique values in table B and say that a value in A’s column is only valid if it also exists in B’s column.Foreign keys can be composite keys, so the foreign key for one column could be two or more columns in another table. In this article, for the sake of simplicity we’ll focus on linking a single column in one table to a single column in another.For example, imagine we’ve set up our orders table with the foreign keys we laid out earlier: orders.user id references users.user id and orders.product sku references books.product sku. These rules mean that:Users.user id must already contain any value inserted into orders.user id. : The orders table won't accept a new row or a row update if the value in orders.user id doesn't already exist in users.user id, which means that orders can only be placed by registered users.Books.product sku must already include any value inserted into orders.product sku : The orders table won't accept a new row or a row update if the value in orders.product sku doesn't already exist in books. product sku, which means that users can only order things that are already present in the database.To Learn more About Foreign keys refer to:
https://brainly.com/question/13437799
#SPJ2
A process involves the removal of oil and other liquid contaminants from metal parts using a heat-treat oven, which has a volume of 15,000 ft3. The oven is free of solvent vapors. The ventilation rate of the oven is 2,100 cfm, and the safety factor (K) is 3. The solvent used in the process evaporates at a rate of 0.6 cfm (cubic feet per minute). The operator would like to know how long it would take the concentration to reach 425 ppm.
Answers
Answer:
time = 4.89 min
Explanation:
given data
volume = 15,000 ft³
ventilation rate of oven = 2,100 cubic feet per minute
safety factor (K) = 3
evaporates at a rate = 0.6 cubic feet per minute
solution
we get here first solvent additional rate in oven that is
solvent additional rate = [tex]\frac{0.6}{1500}[/tex]
solvent additional rate = 4 × [tex]10^{-5}[/tex] min
solvent additional rate == 4 × [tex]10^{-5}[/tex] × [tex]10^{6}[/tex]
solvent additional rate = 40 ppm/min
and
solvent removal rate due to ventilation will be
removal rate = [tex]\frac{2100}{1500}[/tex] × concentration in ppm
removal rate = 0.14 C ppm/min
and
net additional rate is
net additional rate (c) = m - r
so net additional rate (c) is = 40 - 0.14C
so here
[tex]\frac{dc}{dt}[/tex] = 40 - 0.14C
so take integrate from 0 to t
[tex]\int\limits^t_o {dt}[/tex] = [tex]\int\limits^{425/3}_0 \frac{dc}{40-0.14C}[/tex] ....................1
here factor of safety is 3 so time taken is [tex]\frac{425}{3}[/tex]
solve it we get
time = [tex][\frac{-50}{7} \ log(40-\frac{7c}{50} ]^{425/3} _0[/tex]
time = 4.89 min
What are the molar volumes(cm3/mole) for the physically realistic stable root of ethylene gas at the given conditions? Note: Some mathematical solutions to the cubic may not be physically realistic.
Answers
Answer:
See the attached pictures.
Explanation:
See the attached pictures for explanation.
If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3 , find the junction built-in voltage. Assume ni = 1.5 × 1010/cm3 . Also, find the width of the depletion region (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 100 μm2 .
Answers
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
The value read at an analog input pin using analogRead() is returned as a binary number between 0 and the maximum value that can be stored in [X] bits. 1. This binary number is directly linearly proportional to the input voltage at the analog pin, with the smallest and largest numbers returned corresponding to the minimum and maximum ADC input values, respectively. At a Vcc of 3.3 V, analogRead(A0) returns a value of 1023. Approximately what voltage is present at the pin A0 on the MSP430F5529?
Answers
Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023
Explanation:
Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:
3.3 V ---------> 1023
X V------------>analogRead(A0)
Then [tex]X=\frac{3.3 \times analogRead(A0)}{1023} Volts[/tex]
Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).
Problem 6.3 7–20 Consider a hot automotive engine, which can be approximated as a 0.5-m-high, 0.40-m-wide, and 0.8-m-long rectangular block. The bottom surface of the block is at a temperature of 80°C and has an emissivity of 0.95. The ambient air is at 20°C, and the road surface is at 25°C. Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of 80 km/h. Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block
Answers
Answer:
Q_total = 1431 W
Explanation:
Given:-
- The dimension of the engine = ( 0.5 x 0.4 x 0.8 ) m
- The engine surface temperature T_s = 80°C
- The road surface temperature T_r = 25°C = 298 K
- The ambient air temperature T∞ = 20°C
- The emissivity of block has emissivity ε = 0.95
- The free stream velocity of air V∞ = 80 km/h
- The stefan boltzmann constant σ = 5.67*10^-8 W/ m^2 K^4
Find:-
Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation
Solution:-
- We will extract air properties at 1 atm from Table 15, assuming air to be an ideal gas. We have:
T_film = ( T_s + T∞ ) / 2 = ( 80 + 20 ) / 2
= 50°C = 323 K
k = 0.02808 W / m^2
v = 1.953*10^-5 m^2 /s
Pr = 0.705
- The air flows parallel to length of the block. The Reynold's number can be calculated as:
Re = V∞*L / v
= [ (80/3.6)*0.8 ] / [1.953*10^-5]
= 9.1028 * 10^5
- Even though the flow conditions are ( Laminar + Turbulent ). We are to assume Turbulent flow due to engine's agitation. For Turbulent conditions we will calculate Nusselt's number and convection coefficient with appropriate relation.
Nu = 0.037*Re^0.8 * Pr^(1/3)
= 0.037*(9.1028 * 10^5)^0.8 * 0.705^(1/3)
= 1927.3
h = k*Nu / L
= (0.02808*1927.3) / 0.8
= 67.65 W/m^2 °C
- The heat transfer by convection is given by:
Q_convec = A_s*h*( T_s - T∞ )
= 0.8*0.4*67.65*(80-20)
= 1299 W
- The heat transfer by radiation we have:
Q_rad = A_s*ε*σ*( T_s - T∞ )
= 0.8*0.4*0.95*(5.67*10^-8)* (353^4 - 298^4)
= 131.711 W
- The total heat transfer from the engine block bottom surface is given by:
Q_total = Q_convec + Q_rad
Q_total = 1299 + 131.711
Q_total = 1431 W
Following are the calculation to the given question:
Calculating the average film temperature:
[tex]\to T_f = \frac{T_s + T_{\infty}}{2} =\frac{80+20}{2}= 50^{\circ}\ C\\\\[/tex]
At [tex]T_f = 50^{\circ}\ C[/tex], obtain the properties of air:
[tex]\to k=0.02735\ \frac{W}{m \cdot K} \\\\\to v=1.798 \times 10^{-5} \ \frac{m^2}{s}\\\\ \to Pr =0.7228 \\\\\to Re_{L} =\frac{VL}{v} =\frac{80 (\frac{5}{18}) \times 0.8}{1.798 \times 10^{-5}} = 988752.935 \\\\[/tex]
Assume the engine block's bottom surface is a flat plate.
For
[tex]\to 5\times 10^{5} \leq Re_{L} \leq {10^7}, 0.6 \leq Pr \leq 60 \\\\[/tex]
[tex]\to Nu=0.0296 \ \ Re_{L}_{0.8} Pr^{\frac{1}{3}}\\\\[/tex]
[tex]= 0.0296(988752.935)^{0.8} (0.7228)^{\frac{1}{3}}\\\\ = 1660.99[/tex]
But,
[tex]\to Nu=\frac{hL}{k} \\\\ \to h= Nu \frac{k}{L}\\\\[/tex]
[tex]=\frac{1660.99 \times 0.02735}{0.8}\\\\ = 56.7850 \frac{W}{m^2 . K}\\\\[/tex]
[tex]\to A_s= L \times w\\\\[/tex]
[tex]= 0.8 \times 0.4\\\\ =0.32\ m^2\\\\[/tex]
[tex]\to Q_{com} = h A_s (T_s -T_{\infty})\\\\[/tex]
[tex]= 56.7850 \times 0.32 \times (80-20)\\\\ = 1090.272\ W\\\\[/tex]
[tex]\to Q_{rad} = \varepsilon A_s \sigma (T_{s}^{4} -T_{surr}^4) \\\\[/tex]
[tex]= (0.95)(0.32 )(5.67 \times 10^{-8}) [(80+273)^4-(25+273)^4]\\\\ = 131.710\ W\\\\[/tex]
Then the total rate of heat transfer is
[tex]\to Q_{total} = Q_{corw} + Q_{rad}\\\\[/tex]
[tex]=1090.272 +131.710 \\\\= 1221.982\ W\\\\[/tex]
Learn more:
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214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle, as represented by the nuclear equation above. Which of the following quantities plotted versus time will produce a straight line?(A) [Bi](B) [Po](C) ln[Bi](D) 1/[Bi]
Answers
Answer:
(C) ln [Bi]
Explanation:
Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.
Answer:
Option C. ln[Bi]
Explanation:
The nuclear equation of first-order radioactive decay of Bi to Po is:
²¹⁴Bi₈₃ → ²¹⁴Po₈₄ + e⁻
The radioactive decay is expressed by the following equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex] (1)
where Nt: is the number of particles at time t, No: is the initial number of particles, and λ: is the decay constant.
To plot the variation of the quantities in function of time, we need to solve equation (1) for t:
[tex] Ln(\frac{N_{t}}{N_{0}}) = -\lambda t [/tex] (2)
Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:
[tex] [Bi] = \frac {N particles}{N_{A} * V} [/tex] (3)
[tex] [Bi]_{0} = \frac {N_{0} particles}{N_{A} * V} [/tex] (4)
where [tex]N_{A}[/tex]: si the Avogadro constant and V is the volume.
Now, introducing equations (3) and (4) into (2), we have:
[tex] Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t [/tex]
[tex] Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t [/tex] (5)
Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:
[tex] Ln ([Bi]) = -\lambda t + Ln([Bi]_{0}) [/tex]
Therefore, the correct answer is option C. ln[Bi].
I hope it helps you!
. A 10W light bulb connected to a series of batteries may produce a brighter lightthan a 250W light bulb connected to the same batteries. Why? Explain.
Answers
Answer:
Explanation:
From the equation:
Power dissipated= square of voltage supplied by battery ÷ Resistance of the load
i.e P= V^2/R
It means that at constant voltage, the the power consumed is inversely related to the resistance. Therefore the 10W bulb which has a higher resistance will consume less power using the sufficiently excess power dissipated to glow brighter than the 250W bulb which has a low resistance. The power dissipated will partly be used to overcome this low resistance making less power available for heating up the 250W bulb .
ANSWER:
A 10W bulb may shine brighter than a 250W bulb, when the two are connected to the same battery. This can happen when the battery is low or not fully charged. Because the 10W bulb has a high resistance more than a 250W bulb, it makes the 10W bulb to be more efficient than the 250W bulb.
When a low current is passed through the fillament of the two bulbs, the 10W bulb which has high resistance, uses almost all the current that enters the filament to emit more light, more than the 250W bulb which has a low resistance and will converts almost all the current that enters the filament into heat energy. In the 250W bulb only about 8% of the current are used to light up the bulb and the rest are converted to heat energy. This explains why a bulb which it's watt is higher will be more hotter when lighted up, than a bulb which watt is lower.
Another reason while the 10W bulb will shine brighter is when the fillament in it are coiled tight round itself ( example is a fluorescent bulb). It will emit more light than a 250W watt bulb that the filament are not coiled (example is an incandescent bulb).
NOTE : On a normal circumstances, a 250W bulb will shine brighter than a 10W bulb, because the higher the electric energy a bulb consumes the brighter light the filament will produce. Watt is the amount of electric energy the bulb can consume, therefore a bulb with 250W is assumed to produce more light than a bulb with 10W.
Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2 2P2 1 1P3 8 4P4 4 2P5 5 3The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0.(a) Draw four Gantt charts that illustrate the execution of these processes using the following scheduling algorithms: FCFS, SJF, nonpreemptive priority (a larger priority number implies a higher priority), and RR (quantum = 2).
Answers
Answer and Explanation:
The answer is attached below
Blocks A and B are able to slide on a smooth surface. Block A has a mass of 30 kg. Block B has a mass of 60 kg. At a given instant, block A is moving to the right at 0.5 m/s, block B is moving to the left at 0.3 m/s, and the spring connected between them is stretched 1.5 m. Determine the speed of both blocks at the instant the spring becomes unstretched. 9.
Answers
Answer and Explanation:
The answer is attached below
Steam enters an adiabatic nozzle at 2.5 MPa and 450oC with a velocity of 55 m/s and exits at 1 MPa and 390 m/s. If the nozzle has an inlet area of 6 cm2 , determine (a) the exit temperature. (b) the rate of entropy generation for this process.
Answers
To answer the student's question on the exit temperature and rate of entropy generation in an adiabatic nozzle, we need more context-specific information or the assumption that steam behaves as an ideal gas.
Explanation:The student's question involves determining the exit temperature and the rate of entropy generation in an adiabatic nozzle process, which falls under the subject of thermodynamics, a branch of Physics. Calculating these properties requires knowledge of the first and second laws of thermodynamics, as well as the ideal gas law and flow equations.
For part (a), the exit temperature can be estimated using the ideal gas law and the conservation of energy. For part (b), the Clausius–Clapeyron relation and the concept of entropy would be applied. However, to accurately determine these values, more information would be needed, such as specific heat capacities, or the assumption that the steam behaves as an ideal gas.
Given the provided problem statements refer to various examples, which could lead to confusion, the correct answer would be provided if the precise context and details specific to the student's actual question were known.
StackOfStrings s = new StackOfStrings(); while (!StdIn.isEmpty()) { String item = StdIn.readString(); if (!item.equals("-")) s.push(item); else if (s.isEmpty()) StdOut.println("BAD INPUT"); else StdOut.print(s.pop() + " "); }
Answers
Explanation
Question 1:
!StdIn.isEmpty() is false.
So, it does not run the loop body.
So, It does not print anything.
Answer:
none of these
Question 2:
If we print the list starting at x, the result would be: 2 3
Answer:
none of these
We would like to measure the density (p) of an ideal gas. We know the ideal gas law provides p= , where P represents pressure, R is a constant equal to 287.058 J/kg-K and I represents temperature. We use a pressure transducer to take 15 measurements, all under the same conditions. The sample mean and standard deviation obtained from these 15 pressure measurements are 120,300 Pa and 6,600 Pa respectively. The pressure transducer specifications sheet reports an accuracy of 0.6% of the full-scale output (FSO) and a sensitivity error of 0.3% of the FSO. The FSO for this instrument is 180,000 Pa. In addition, we use a thermocouple to measure the gas temperature. We take 20 temperature measurements and obtain a sample mean and standard deviation of 340 K and 8 K, respectively. The accuracy of the thermocouple is 0.25% of the reading and its hysteresis uncertainty is + 2 K. Calculate the density of the ideal gas and its total uncertainty for a 95% probability. (Sol: 1.2325 +0.1578 kg/m3)
Answers
Answer: =
Explanation:
= P / (R * T) P- Pressure, R=287.058, T- temperature
From the given that
Sample mean(pressure) = 120300 Pa
Standard deviation (pressure) = 6600 Pa
Sample mean(temperature) = 340K
Standard deviation(temperature) = 8K
To calculate the Density;
Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa
Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa
Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K
Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K
So now to calculate the density:
Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331
Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138
Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345
cheers i hope this helps
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and the strain-hardening exponent in the flow curve equation.
Answers
Answer:
The strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex]
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find [tex](K)[/tex] and the [tex](n)[/tex].
[tex]\sigma =K\epsilon^n[/tex]
We will plug the values in the formula.
[tex]250=K\times (0.12)^n\\350=K\times (0.26)^n[/tex]
We will solve these equation.
[tex]K=\frac{250}{(0.12)^n}[/tex] plug this value in [tex]350=K\times (0.26)^n[/tex]
[tex]350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n[/tex]
Taking a natural log both sides we get.
[tex]ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435[/tex]
Now, we will find value of [tex]K[/tex]
[tex]K=\frac{250}{(0.12)^n}[/tex]
[tex]K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625[/tex]
So, the strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex].
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upper surface: Cp constant at – 0.8 from the leading edge to 60% chord, then increasing linearly to +0.1 at the trailing edge. Lower surface: Cp constant at – 0.4 from the leading edge to 60% chord, then increasing linearly to + 0.1 at the trailing edge. Estimate the lift coefficient and the pitching moment coefficient about the leading edge due to lift.
Answers
Answer:
The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
Explanation:
The Upper Surface Cp is given as
[tex]Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1[/tex]
The Lower Surface Cp is given as
[tex]Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1[/tex]
The difference of the Cp over the airfoil is given as
[tex]\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32[/tex]
Now the Lift Coefficient is given as
[tex]C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192[/tex]
Now the coefficient of moment about the leading edge is given as
[tex]C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306[/tex]
So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
The lift coefficient and the pitching moment coefficient about the leading edge due to lift are respectively; 0.3192 and -0.13
Aerodynamics engineeringWe are given;
Distance of upper surface from leading edge to percentage of chord = -0.8
Percentage of chord for both surfaces = 60% = 0.6
Rate of increase at trailing edge for both surfaces = +0.1
Distance of lower surface from leading edge to percentage of chord = -0.6
angle of incidence; α_i = 4° = 4π/180 rad
Let us first calculate the Cp constant for both the upper and lower surface.
Cp for upper surface is;
Cp_u = (-0.8 × 0.6) - 0.1∫¹₀.₆ dx
Solving this integral gives;
Cp_u = (-0.8 × 0.6) - (0.1 × 0.4)
Cp_u = -0.52
Cp for lower surface is;
Cp_l = (-0.4 × 0.6) + 0.1∫¹₀.₆ dx
Solving this integral gives;
Cp_l = (-0.4 × 0.6) + (0.1 × 0.4)
Cp_l = -0.2
Change in Cp across the foil is;
ΔCp = Cp_l - Cp_u
ΔCp = -0.2 - (-0.52)
ΔCp = 0.32
Formula for the lift coefficient is;
C_L = ΔCp * cosα_i
C_L = 0.32 * cos (4π/180)
C_L = 0.3192
Formula for the pitching moment coefficient is;
(-0.3 * 0.4 * 0.6) - ((0.6 + (0.4/3)) * 0.2 * 0.4)
C_m,p = -0.072 - 0.059
C_mp ≈ -0.13
Read more about aerodynamics at; https://brainly.com/question/6672263
Janelle Heinke, the owner of Ha�Peppas!, is considering a new oven in which to bake the firm�s signature dish, vegetarian pizza. Oven type A can handle 20 pizzas an hour. The fixed costs associated with oven A are $20,000 and the variable costs are $2.00 per pizza. Oven B is larger and can handle 40 pizzas an hour. The fixed costs associated with oven B are $30,000 and the variable costs are $1.25 per pizza. The pizzas sell for $14 each.
a) What is the break-even point for each oven?
b) If the owner expects to sell 9,000 pizzas, which oven should she purchase?
c) If the owner expects to sell 12,000 pizzas, which oven should she purchase?
d) At what volume should Janelle switch ovens?
Answers
Answer:
a) A = 1667 and B = 2353
b) Oven A
c) Oven A
d) Below 13,333 pizza: Oven A
Above 13,334 pizza: Oven B
Explanation:
We have the following data:
Oven A: Oven B:
Capacity 20 p/hr 40p/hr
Fixed Cost $20,000 $30,000
Variable Cost $2.00/p $1.25/p
Selling Price: $14
a) Break-even point → Cost = Revenue
([tex]x[/tex] refers to the number of pizza sold)
Oven A:
20000 + 2[tex]x[/tex] = 14[tex]x[/tex]
20000 = 14[tex]x[/tex] - 2[tex]x[/tex]
[tex]x[/tex] = 20000/ 12
[tex]x[/tex] = 1666.67 ≈ 1667 pizza
Oven B:
30000 + 1.25[tex]x[/tex] = 14[tex]x[/tex]
30000 = 14[tex]x[/tex] - 1.25[tex]x[/tex]
[tex]x[/tex] = 30000/ 12.75
[tex]x[/tex] = 2352.9 ≈ 2353 pizza
b) Comparing both oven for 9,000 pizza
Profit = Selling Price - Cost Price
Oven A:
Profit = (9000 x 14) - (20,000 + 2 x 9000)
Profit = 126000 - 38000
Profit = 88000
Oven B:
Profit = (9000 x 14) - (30,000 + 1.25 x 9000)
Profit = 126000 - 41250
Profit = 84750
Oven A is more profitable.
c)
Oven A:
Profit = (12000 x 14) - (20,000 + 2 x 12000)
Profit = 168000 - 44000
Profit = 124000
Oven B:
Profit = (12000 x 14) - (30,000 + 1.25 x 12000)
Profit = 168000 - 45000
Profit = 123000
Oven A is more profitable.
d) Using the equation formed in a):
20,000 - 12[tex]x[/tex] < 30,000 - 12.75[tex]x[/tex]
12.75[tex]x[/tex] - 12[tex]x[/tex] < 30000 - 20000
0.75[tex]x[/tex] < 10000
[tex]x[/tex] < 10000/0.75
[tex]x[/tex] < 13333.3
Hence, if the production is below 13,333 Oven A is beneficial.
For production of 13,334 and above, Oven B is beneficial.
Answer:
a. Find the break even points in units for each oven.
Breakeven for type A pizza x = = 1,666.6 units of pizza need to be sold in order to obtain breakeven for Type A
Breakeven for type B pizza x = = 2,352.9 units of pizza need to be sold in order to obtain breakeven for Type B
b. If the owner expects to sell 9000 pizzas, which oven should she purchase?
Type B: because the profit will be twice what will be obtainable from type A considering the fact that it produces pizza at the ration of TypeB:TypeA, 40:20 or 2:1
Profit for type a = 9000/20 x 14 = 6,300 – 1,666,6units ($23, 3332) = 4366.4 units
Profit for type B = 10,247.1 units of pizza - which makes it justifiable
Explanation:
Air at 620 kPa and 500 K enters an adiabatic nozzle that has an inlet-to-exit area ratio of 2:1 with a velocity of 120 m/s and leaves with a velocity of 380 m/s. The enthalpy of air at the inlet temperature of 500 K is h1 = 503.02 kJ/kg
Answers
Answer:
therefore the exit pressure of air is 331.2 kPa
Explanation:
When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a high-pressure region in front of the piston. It takes more work to move the piston against this high-pressure region. Hence, a non-quasi-equilibrium compression process requires a larger work input than the corresponding quasi-equilibrium one.a. trueb. false
Answers
Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer:
True
Explanation:
Because in quasi static process, process occurs very slowly so it is treated as reversible process.
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.
Answers
Answer:
critical stress required is 18.92 MPa
Explanation:
given data
specific surface energy = 1.0 J/m²
modulus of elasticity = 225 GPa
internal crack of length = 0.8 mm
solution
we get here one half length of internal crack that is
2a = 0.8 mm
so a = 0.4 mm = 0.4 × [tex]10^{-3}[/tex] m
so we get here critical stress that is
[tex]\sigma _c = \sqrt{\frac{2E \gamma }{\pi a}}[/tex] ...............1
put here value we get
[tex]\sigma _c[/tex] = [tex]\sqrt{\frac{2\times 225\times 10^9 \times 1 }{\pi \times 0.4\times 10^{-3}}}[/tex]
[tex]\sigma _c[/tex] = 18923493.9151 N/m²
[tex]\sigma _c[/tex] = 18.92 MPa