Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.
One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.
Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default initialize the artist's name to "None" and the years of birth and death to 0. Define the Artwork class in Artwork.py with a constructor to initialize an artwork's information. The constructor should by default initialize the title to "None", the year created to 0, and the artist to use the Artist default constructor parameter values. Add an import statement to import the Artist class. Add import statements to main.py to import the Artist and Artwork classes.
The codes for the files are:
File: Artist.py
This file will contain the Artist class with a constructor initializing the artist's name, birth year, and death year.
# Artist. py
class Artist:
def __init__(self, name="None", birth_year=0, death_year=0):
self.name = name
self. birth_year = birth_year
self. death_year = death_year
File: Artwork.py
This file will define the Artwork class and import the Artist class from Artist.py. It initializes the artwork's title, year of creation, and artist.
# Artwork. py
from Artist import Artist
class Artwork:
def __init__(self, title="None", year_created=0, artist=Artist()):
self. title = title
self. year_created = year_created
self. artist = artist
File: main. py
This file will import both Artist and Artwork classes and create instances of them to demonstrate their usage.
# main. py
from Artist import Artist
from Artwork import Artwork
# Creating an instance of Artist
artist1 = Artist("Vincent van Gogh", 1853, 1890)
# Creating an instance of Artwork using the artist instance
artwork1 = Artwork("Starry Night", 1889, artist1)
# Printing artist and artwork details to verify
print(f"Artist: {artwork1. artist. name}, Born: {artwork1. artist. birth_year}, Died: {artwork1. artist. death_year}")
print(f"Artwork: {artwork1. title}, Created in: {artwork1. year_created}")
Execution and Output
When you run the main.py script, it will create an Artist object for Vincent van Gogh and an Artwork object for one of his most famous pieces, "Starry Night". The script will then print the details of both the artist and the artwork.
Expected Output:
Artist: Vincent van Gogh, Born: 1853, Died: 1890
Artwork: Starry Night, Created in: 1889
If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relative velocity of block B with respect to C.
Answer:
Explanation:
The detailed steps and appropriate calculation with analysis is as shown in the attachment.
Determine the string's length.
[tex]S_A +2S_B +2S_C = Constant[/tex]
Calculate the equation in terms of time.
[tex]v_A+2v_B +2v_c =0 \\\\6 +2v_B +2(18)=0 \\\\2v_B = -42 \\\\V_B =-21 \frac{ft}{s}[/tex]
Calculate the relative velocity of B with respect to C.[tex]V_B = v_c + V_{\frac{B}{C}} \\\\-21=18+ V_{\frac{B}{C}} \\\\ V_{\frac{B}{C}}= -39 \ \frac{ft}{s} = 39 \ \frac{ft}{s} \uparrow[/tex]
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In many problems where the potential energy is considered between two objects, the potential energy is defined as zero when the particles are in infinite distance apart. Using the definition of potential energy, explain why the potential energy would be positive if the force between the particles is repulsive and negative if the force between the particles is attractive for noninfinite distances.
Answer:
potential energy=ε Qq/r. Putting values of charge along with sign and calculating the potential energy will tell whether the resultant force is attractive or repulsive.
Explanation:
potential energy=ε Qq/r
q and Q are charges on each particle
So in case the particles have attractive force between them, one particle will be negative and the other will be positiive. The result will be negative.
In case the particles have repulsive forces, the charge on both particles will be either positive or negative. The result will be positive
2. If you have 10 ft3 of an ideal gas stored at 75 F and at 12 psia that is expanded to 15 ft3 , what is the resulting pressure? 3. The atmospheric pressure is 14.3 psi. How many inches of mercury (s=13.6) does the National Weather Service report this pressure as? Pressure = __________________
Answer:
2. The decrease in pressure will be 8psia
3. The NWS will report the pressure to be 0.0657inchHg
Explanation:
2. This is an isothermal system because it was only the volume of the gas that was increased, and temperature was constant. To find the pressure, use the Boyle's law equation below.
P1V1 ÷ P2V2..............(1)
make P2 the subject of the formula
P2 = P1V1 ÷ V2 ..............(2)
P1 = 12psia
V1 = 10ft3
V2 = 15ft3
Using equation 2
P2 = (12 × 10) ÷ 15 = 8psia
Therefore, the resulting pressure of the gas will decrease to 8psia, which obeys Boyle's law that states that pressure is inversely proportion to volume in a given mass of gas. That means, as volume increases pressure should decrease.
3. The National weather Service report, are the national agency that forecast the weather, to report how the climate will change. To report the pressure of a weather, their use a pressure barometer, and reports the reading in the pressure barometer, which is the hight the Mercury in the barometer increased to.
Using a pressure barometer.
Patm = pgh..............(3)
Make h the subject of formula
h = Patm ÷ pg ..............(4)
Patm is the Atmospheric pressure = 4.3psi
p is the density of Mercury = 13.6
g is the force of gravity = 9.8
h is the hight of the Mercury in the barometer, which is the pressure that is measured.
Using equation 4
h = 4.3 ÷ (13.6 × 9.8) = 0.0322629psi
To convert Pound- force per square inch to inch of mecury (that's is psi to inchHg)
1psi = 2.03602inchHg
Therefore
0.0322629psi = 0.065689399inchHg
The NWS will report the pressure to be 0.0657inchHg
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps.
Calculate:
a) lenghtb) exit velocity of the final slab
Answer:
L_f = 26.025 ft
v_f = 51.77 ft/min
Explanation:
Given:-
- The thickness of the slab initially, t_o = 2 in
- The width of the slab initially, w_o = 10 in
- The Length of the slab initially, L_o = 12.0 ft
- The reduction in thickness in each of three steps, r = 75%
- The widening of the slab in each of three steps , m = 3%
- The entry speed vi = 40 ft/min
- The roll speed remains the same
Find:-
a) length
b) exit velocity
Of the final slab
Solution:-
- The final thickness (t_f) after three passes is as follows:
t_f = ( r / 100 )^n * t_o
Where, n = number of passes.
t_f = ( 75 / 100 ) ^3 * ( 2.0 )
t_f = 0.844 in
- The final width (w_f) after three passes is as follows:
w_f = ( m / 100 + 1 )^n * w_o
Where, n = number of passes.
w_f = ( 3 / 100 + 1 ) ^3 * ( 10.0 )
w_f = 10.927 in
- Assuming the Volume of the slab remains the same. Zero material Loss. The final length of slab can be determined:
t_o*w_o*L_o = t_f*w_f*L_f
L_f = ( 2 * 10 * 12 ) / ( 0.844 * 10.927 )
L_f = 26.025 ft
- We can use the volume rate equation as the roll speed remains constant i.e change in rate of volume is zero. Hence, we can write the before and after the 3rd step formulation:
t_i*w_i*v_i = t_f*w_f*v_f
Where, v_i : The entry step speed
v_f : Third step exit speed.
(0.75)^2 * 2 * (1.03)^2 * 10 * 40 = (0.844)*(10.927)*v_f
v_f = 51.77 ft/min
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (10 pts) Calculate the von Mises stress and the factor of safety. 10. (10 pts) Of the two factors of safety computed, which one is more realistic? What failure theory should you use if you want to be conservative? 11. (10 pts) Suppose all the principal stresses are equal in magnitude and sign, and larger than Sy. What are the predicted safety factors by the maximum shear stress and distortion energy failure theories? Calculate your results and explain them. What do you think would happen in reality?
Answer:
Detailed solution is given below:
One-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q dot =1000 W/m^3 and is convectively cooled at x=±40m by an ambient fluid characterized by T [infinity] = 30 degrees C. If the steady-state temperature distribution within the wall is T(x) = a(L2-x2)+b where a = 15o C/m^2 and b=40oC, what is the thermal conductivityof the wall? What is the value of the convection heat transfer coefficient?
Answer:
Thermal Conductivity (K) = 33.33 W/m. ° C
The value of the convection heat transfer coefficient = 3 W/m².° C
Explanation:
The attached document file gives a detailed and clear explanation about the question.
A Diesel cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and cutoff ratio (rc) determine the efficiency and other values listed below.
Note: The gas constant for air is R=0.287 kJ/kg-K.
--Given Values--
T1 (K) = 306
P1 (kPa) = 140
r = 10
rc = 1.75
a) Determine the specific internal energy (kJ/kg) at state 1.
Your Answer =
b) Determine the relative specific volume at state 1.
Your Answer =
c) Determine the relative specific volume at state 2.
Your Answer =
d) Determine the temperature (K) at state 2.
Your Answer =
e) Determine the pressure (kPa) at state 2.
Your Answer =
f) Determine the specific enthalpy (kJ/kg) at state 2.
Your Answer =
g) Determine the temperature (K) at state 3.
Your Answer =
h) Determine the pressure (kPa) at state 3.
Your Answer =
i) Determine the specific enthalpy (kJ/kg) at state 3.
Your Answer =
j) Determine the relative specific volume at state 3.
Your Answer =
k) Determine the relative specific volume at state 4.
Your Answer =
l) Determine the temperature (K) at state 4.
Your Answer =
m) Determine the pressure (kPa) at state 4.
Your Answer =
n) Determine the specific internal energy (kJ/kg) at state 4.
Your Answer =
o) Determine the net work per cycle (kJ/kg) of the engine.
Your Answer =
p) Determine the heat addition per cycle (kJ/kg) of the engine.
Your Answer =
q) Determine the efficiency (%) of the engine.
Your Answer =
help A through Q
Answer:
Explanation: see attachment
Please write the following code in Python 3. Also please show all output(s) and share your code.
Below is a for loop that works. Underneath the for loop, rewrite the problem so that it does the same thing, but using a while loop instead of a for loop. Assign the accumulated total in the while loop code to the variable sum2. Once complete, sum2 should equal sum1.
sum1 = 0
lst = [65, 78, 21, 33]
for x in lst:
sum1 = sum1 + x
Explanation of rewriting code from for loop to while loop in Python.
Explanation:To rewrite the given code using a while loop instead of a for loop, you can iterate over the list indices and manually accumulate the total.
Here is the code:
sum1 = 0By using this code snippet, the sum accumulated using the while loop will be stored in the variable sum2, which should equal the sum1 calculated using the for loop.
A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at the end of a randomly chosen red light. The engineer believes that the probability that X = x is proportional to (x + 1)(8 − x) for x = 0, . . . , 7 (the possible values of X), i.e. P(X = x) = ( C(x + 1)(8 − x) for x = 0, 1, . . . , 7 0 otherwise 1. Determine the value of C. 2. Find the probability that X will be at least 5.
Answer:
A) C = 1/120
B) P( x ≥ 5) = 1/3
Explanation:
I've attached explanations to this.
The value of C is: C = 1/126
The probability that X will be at least 5 is 1/21.
To determine the value of C, we need to make sure that the sum of the probabilities of all possible values of X is equal to 1. This means that the following equation must hold:
\sum_{x=0}^7 P(X = x) = 1
Substituting in the given expression for P(X = x), we get:
\sum_{x=0}^7 C(x + 1)(8 - x) = 1
Expanding the summation and simplifying, we get:
7C + 21C + 35C + 35C + 21C + 7C = 1
126C = 1
Therefore, the value of C is:
C = 1/126
To find the probability that X will be at least 5, we need to calculate the following sum:
P(X \ge 5) = P(X = 5) + P(X = 6) + P(X = 7)
Substituting in the expression for P(X = x), we get:
P(X \ge 5) = C(5 + 1)(8 - 5) + C(6 + 1)(8 - 6) + C(7 + 1)(8 - 7)
P(X \ge 5) = C(6)(3) + C(7)(2) + C(8)(1)
Substituting in the value of C, we get:
P(X \ge 5) = (1/126)(6)(3) + (1/126)(7)(2) + (1/126)(8)(1)
P(X \ge 5) = 1/21
Therefore, the probability that X will be at least 5 is 1/21.
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The structure supports a distributed load of w. The limiting stress in rod (1) is 370 MPa, and the limiting stress in each pin is 220 MPa. If the minimum factor of safety for the structure is 2.10, determine the maximum distributed load magnitude w that may be applied to the structure plus the stresses in the rod and pins at the maximum w.
In engineering, the maximum load a structure can withstand while maintaining a factor of safety can be calculated based on material properties and stress limits. Analyzing stress distributions in rods and pins under maximum loads is crucial for ensuring the structural integrity of a system.
Explanation:The maximum distributed load magnitude w that may be applied to the structure can be determined using the concept of factor of safety, which is the ratio of the materials' strength to the maximum stress it is subjected to. The factor of safety is given as 2.10, limiting stress in rod (1) as 370 MPa, and limiting stress in each pin as 220 MPa. By setting up equations based on these data, you can calculate the maximum load w.
To calculate the stresses in the rod and each pin at the maximum w, you would need to consider the equilibrium of forces acting on the structure with the maximum load applied. By analyzing the forces acting on the rod and pins, you can determine the stresses within them when the structure is subjected to the maximum load.
Example: Assuming the structure consists of multiple rods and pins, each experiencing different loads, you can analyze the stress distribution within the structure by considering the individual material properties and load distributions to ensure structural integrity.
A group of n Ghostbusters is battling n ghosts. Each Ghostbuster carries a proton pack, which shoots a stream at a ghost, eradicating it. A stream goes in a straight line and terminates when it hits the ghost. The Ghostbusters decide upon the following strategy. They will pair off with the ghosts, forming n Ghostbuster-ghost pairs, and then simultaneously each Ghostbuster will shoot a stream at his chosen ghost. As we all know, it is very dangerous to let streams cross, and so the Ghostbusters must choose pairings for which no streams will cross. Assume that the position of each Ghostbuster and each ghost is a fixed point in the plane and that no three positions are collinear.Give an O(n 2 lg n)-time algorithm to pair Ghostbusters with ghosts in such a way that no streams cross. Provide a step by step algorithm for this question.
Answer:
Using the above algorithm matches one pair of Ghostbuster and Ghost. On each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are the same, so use the algorithm recursively on each side of the line to find pairings. The worst case is when, after each iteration, one side of the line contains no Ghostbusters or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P([tex]n^{2} lg n[/tex])- time algorithm.
The described problem requires creating non-crossing pairings between points in a plane, utilizing a divide and conquer algorithm similar to finding the closest pair of points. The algorithm involves sorting, recursively pairing, checking for potential crossings, and merging pairs in O(n^2 lg n) time.
Explanation:The problem described is one of computational geometry, specifically related to the pairing of points (representing Ghostbusters and ghosts) in the plane so that the lines (streams) connecting the pairs do not cross. An O(n^2 lg n)-time algorithm to solve this can be designed by utilizing a divide and conquer strategy similar to the one used in the closest pair of points problem.
Sort all the points by their x-coordinates.Divide the set of points into two halves by drawing a vertical line through the median x-coordinate.Recursively pair off Ghostbusters and ghosts in each half. Ensure that each pair consists of one Ghostbuster and one ghost from the same half.Find potential cross-stream pairs by examining Ghostbusters and ghosts that are close to the dividing line. This step identifies pairs that may cause crossing streams after the recursive step.For each Ghostbuster on one side of the dividing line, pair with the closest ghost on the other side such that the pair does not cause a stream cross with already established pairs. Utilize a data structure to dynamically check for intersections while pairing.Repeat this for all unpaired Ghostbusters adjacent to the dividing line.Merge the pairs from both halves along with the Ghostbuster-ghost pairs across the divide.Steps 4 to 6 are critical in ensuring that no streams cross and contribute to the O(n lg n) complexity for pairing across the divide. The overall complexity is O(n^2 lg n) due to the recursive nature of the algorithm and the added complexity of checking for crossing streams.
This assignment covers the sequential circuit component: Register and ALU. In this assignment you are supposed to create your own storage component for two numbers using registers. Those two numbers are then passed into a custom ALU that calculates the result of one of four possible operations. Key aspect of this assignment is to understand how to control registers, how to route signals and how to design a custom ALU.
Answer:
The part I called command in the first diagram has been renamed to opcode, or operation code. This is a set of bits (a number) that will tell the ALU which action to perform. I can get the LC-3 opcodes for ADD and NOT and ADD from the book, so I'm not too worried.
Note the #? comment by the switch above opcode. This means I'm not sure how many switches I will need. How many bits do I need to perform all the operations I want? The textbook will tell me.
Materials
Now I make a list of all the materials you have accumulated so far. This list is just an example; yours may be different.
Two 4-bit inputs
One 4-bit output
Two keypads for 4-bit input
Three 7-segment displays (2 for input, 1 for output)
A bunch of switches for opcode (could use a keypad, I guess, but switches are so much more geeky)
A bunch of lights too
The "is zero" LED
One button for clock
One button for reset
One switch for carry-in
Include logic to perform a SUB instruction. That is, subtract the second operand from the first (out = in1 - in2). All three values -- both inputs and the output -- must be two's complement numbers (negative numbers must be represented). Your design may work in one (8 points) or two (4 points) clock cycles.
Explanation:
You are allowed to use the Logisim built-in registers.
The clear input of the register should not be used (do not connect anything to
them).
Custom ALU
Use the provided subcircuit in the template to implement your ALU. You do not have to create additional subcircuits to do this. The ALU has a total of three inputs: First number, second number and select operation input. And one output: Result. The first and second number are used as input for the operations the ALU performs. The select input decides which operation result will be on the single output of the ALU. The ALU is supposed to calculate: NumberA OPERATION NumberB. Register 1 of the storage contains NumberA and Register 2 contains NumberB. The ALU must be able to compute signals with a 4-bit width. Make sure to add labels to all inputs and outputs.
The following operations should be performed for each select input combination (s1s0): • 00: Logic Bitwise XOR
• 01: Multiplication
• 10: Division
• 11: Addition Notes:
You can change the inputs bit width / data bits of any gate to more that 1-bit.
The Logic Bitwise XOR operation can be done with a single XOR gate.
You are also allowed to use the built-in arithmetic logic components and multi- plexer provided by Logisim.
If the result is larger than 4 bits, it will be truncated (only 4 LSB will be shown). This behavior is intended for this assignment. Also, negative results do not have to be considered.
Once you have implemented the ALU circuit, connect the wires in the main circuit properly and test all four operations of your ALU in combination with the storage component.
1. A wood board is one of a dozen different parts in a homemade robot kit. The width, depth, and height dimensions of the board are 7.5 x 14 x 1.75 inches, respectively. The board is made from southern yellow pine, which has an air dry weight density of .025 lb/in.3. a. What is the volume of the wood board? Precision = 0.00
Answer:
183.75 cubic inches.
Explanation:
The volume of the wood board is determine by means of this expression:
[tex]V = w \cdot h \cdot l[/tex]
By replacing variables:
[tex]V = (7.5 in) \cdot (14 in) \cdot (1.75 in)\\V = 183.75 in^{3}[/tex]
The flow of a liquid in a 2 inch nominal diameter steel pipe produces a pressure drop due to friction of 78.86 kPa. The length of pipe is 40 m and the mean velocity is 3 m/s. if the density of the liquid is 1000 kg/m^3, then a) determine the reynolds number b) determind if the flow is laminar or tubulent c) compute viscosity of the liquid d) compute the mass flow rate (assume ?, equivalent roughness factor, for Steel pipe to be 45.7 x 10-6 m)
Answer:
Explanation:
The detailed steps and careful analysis is as shown in the attached file.
Based on the calculations, the Reynolds number is equal to [tex]2.9 \times 10^3[/tex]
Given the following data:
Length of pipe = 40 meters.Diameter of pipe = 2 inches to m = 0.0508 m.Pressure drop = 78.86 kPa.Mean velocity = 3 m/s.Density of liquid = 1000 [tex]kg/m^3[/tex].Roughness factor = [tex]45.7 \times 10^{-6}[/tex] m.How to calculate the Reynolds number.Reynolds number has a direct relationship with friction factor. Thus, we would determine the friction factor by using this formula:
[tex]f=\frac{2 \Delta P D}{ \rho Lu^2}[/tex]
Where:
D is the diameter.L is the length.[tex]\Delta P[/tex][tex]\DeltaP[/tex][tex]\DeltaP[/tex] is the pressure drop.u is the mean velocity.[tex]\rho[/tex] is the density.Substituting the given parameters into the formula, we have;
[tex]f=\frac{2 \times 78.86 \times 10^3 \times 0.0508}{ 1000 \times 40 \times 3^2}\\\\f=\frac{8012.176}{360000}[/tex]
f = 0.0223.
For the Reynolds number:
[tex]N_{Re}=\frac{64}{f} \\\\N_{Re}=\frac{64}{0.0223}[/tex]
Reynolds number = [tex]2.9 \times 10^3[/tex]
Note: Fluid flow is turbulent when Reynolds number is greater than 2000 ([tex]N_{Re} > 2000[/tex]) and it is laminar when it is lesser than 2000 ([tex]N_{Re} < 2000[/tex]).
b. The flow of this liquid is turbulent.
c. To determine the viscosity:
[tex]V=\frac{\rho uD}{N_{Re}} \\\\V=\frac{1000 \times 3 \times 0.0508}{2.9 \times 10^3} \\\\V=\frac{152.4}{2.9 \times 10^3}[/tex]
V = 0.0526 Kgm/s.
d. To determine the mass flow rate:
[tex]m=\rho A u=\rho u\frac{\pi}{4} D^2\\\\m=1000 \times 3 \times 0.7854 \times 0.0508^2[/tex]
m = 6.081 Kg/s.
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An equal-tangent vertical curve is to be constructed between grades of -2% (initial) and 1% (final). The PVI is at station 110 00 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 00 must be at 424.5 ft. Design the curve, and determine the elevations and stations for PVC and PVI.
Answer:
The curve length (L) will be = 1218 ft
The elevations and stations for PVC and PVI
a. station of PVC = 103 + 91.00
b. station of PVI = 116 + 09.00
c. elevation of PVC = 432.18ft
d. elevation of PVI = 426.09ft
Explanation:
First calculate for the length (L)
To calculate the length, use the formula of "elevation at any point".
where, elevation at any point = 424.5.
and ∴ PVC Elevation = (420 + 0.01L)
Then, calculate for Station of PVC and PVI and elevation of PVC and PVI
Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass flow rate of air through the cycle is 0.2 kg/s. Assuming variable specific heats for air, determine
(a) the rate of refrigeration, (b) the net power input, and (c) the coefficient of performance.
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Had to screenshot the solution check attached
A thick oak wall (rho = 545 kg/m3 , Cp = 2385 J/kgK, and k = 0.17 W/mK) initially at 25°C is suddenly exposed to combustion products at 800°C. Determine the time of exposure necessary for the surface to reach the ignition temperature of 400°C, assuming the convection heat transfer coefficient between the wall and the products to be 20 W/m2 K. At that time, what is the temperature 1 cm below the surface? (Note: use an appropriate equation for the semi-infinite wall case; compare equations 18.20 and 18.21 in the text).
Answer:
Explanation:
The detailed calculation and appropriate equation with substitution is as shown in the attached file.
The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.
Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200-g implant, determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.
Answer:
a) the velocity of the implant immediately after impact is 20 m/s
b) the average resistance of the implant is 40000 N
Explanation:
a) The impulse momentum is:
mv1 + ∑Imp(1---->2) = mv2
According the exercise:
v1=0
∑Imp(1---->2) = F(t2-t1)
m=0.2 kg
Replacing:
[tex]0+F(t_{2} -t_{1} )=0.2v_{2}[/tex]
if F=2 kN and t2-t1=2x10^-3 s. Replacing
[tex]0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s[/tex]
b) Work and energy in the system is:
T2 - U(2----->3) = T3
where T2 and T3 are the kinetic energy and U(2----->3) is the work.
[tex]T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x[/tex]
Replacing:
[tex]\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N[/tex]
randomFactory public static Shape randomFactory(int canvas_width, int canvas_height) Create a random shape. Create a random shape to fit on a canvas of the given size. The shape will be no larger than half the size of the canvas and will fit completely on it. Parameters: canvas_width - the width of the canvas being used. canvas_height - the height of the canvas being used. Returns: the generated shape.
Answer:
import java.awt.Color;
import java.awt.Canvas;
import java.awt.Button;
import java.awt.Image;
import java.awt.Graphics;
import java.awt.Frame;
import java.awt.event.*;
import java.util.*;
/**
* Check attached images for the continuation of the code
* 5000 characters exceeded
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*/
Air flows steadily between two sections in a long, straight portion of 10-cm inside diameter pipe. The uniformly distributed temperature and pressure at each section are given. The average air velocity at Sections (1) and (2) are 66 m/s and 300 m/s, respectively. Assume uniform velocity distributions. Determine the frictional force (Rx) exerted by the pipe wall on the airflow between the sections. At section (1), p1 = 7 MPa, T1 = 25°C, and V1 = 66 m/s. At section 2, p2 = 1.3 MPa, T1 = -20°C, and V2 = 300 m/s. (5 pt)
Answer:
solution attached below
Explanation:
The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The motor has an efficiency of ε = 0.65. Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 27 ft starting from rest.
The power that must be supplied to the motor is 136 hp
Explanation:
Given-
weight of the elevator, m = 1000 lb
Force on the table, F = 500 lb
Distance, s = 27 ft
Efficiency, ε = 0.65
Power = ?
According to the equation of motion:
F = ma
[tex]3(500) - 1000 = \frac{1000}{32.2} * a[/tex]
a = 16.1 ft/s²
We know,
[tex]v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s[/tex]
To calculate the output power:
Pout = F. v
Pout = 3 (500) * 29.48
Pout = 44220 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = Pout / Pin
0.65 = 44220 / Pin
Pin = 68030.8 lb.ft/s
Pin = 68030.8 / 500 hp
= 136 hp
Therefore, the power that must be supplied to the motor is 136 hp
To calculate the power supplied to the motor, we use the work done by the motor and the motor efficiency. However, without the time, it takes to lift the elevator, a crucial piece of information is missing, thus preventing an exact calculation.
Explanation:To determine the power that must be supplied to the motor at the instant the 1000-lb elevator has been hoisted 27 ft from rest, we must first calculate the useful power output. Since the motor has an efficiency of 0.65 and exerts a constant force of 500 lb, we can use the work-energy principle along with the efficiency to find the required power.
The work done by the motor on the elevator is the force applied times the distance moved. The useful work is W = force × distance = 500 lb × 27 ft.
The useful power output, or work done per unit time, can be found by P = W / t. However, the time is not provided directly in this scenario; we assume this process occurs instantaneously, which is not realistic for real-world scenarios.
Since power supplied is equal to the useful power output divided by the efficiency, we have Power supplied = Useful power / ε.
Assumptions in the CalculationWe would normally need time to calculate power, which is work done or energy converted per unit time.The efficiency of the motor is used to calculate the actual power supplied versus the useful power output.The actual lifting process takes time, and an instantaneous lift is not realistic.Due to missing time information, the exact numerical answer cannot be provided without assumptions or additional data concerning how quickly the elevator is lifted.
The surface energy of a single crystal depends on crystallographic orientation. Does this surface energy increase or decrease with an increase in planar density
Answer:
The surface energy of a single crystal depends on the planar density (i.e., degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds. As the planar density increases, the number of nearest atoms in the plane increases, which results in an increase in the number of satisfied atomic bonds in the plane, and a decrease in the number of unsatisfied bonds. Since the number of unsatisfied bonds diminishes, so also does the surface energy decrease. (That is, surface energy decreases with an increase in planar density.)
Explanation:
Which has the capability to produce the most work in a closed system;
1 kg of steam at 800 kPa and 180°C or 1 kg of R–134a at 800 kPa and 180°C? Take T0 = 25°C and P0 = 100 kPa.
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 amount of work produced by 1 kg of steam and 1 kg of R–134a would be the same in a closed system at the given conditions.
Explanation:In a closed system, the amount of work produced depends on the change in internal energy of the system. The change in internal energy is given by the formula ΔU = Q - W, where Q is the heat added to the system and W is the work done by the system. Since both 1 kg of steam and 1 kg of R–134a are at the same temperature and pressure, their internal energy changes would be the same for the same amount of heat added. Therefore, the amount of work produced by both substances would also be the same in a closed system.
Learn more about work produced in a closed system here:https://brainly.com/question/34701684
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The elastic cords used for bungee jumping are designed to endure large strains. Consider a bungee cord that stretches to a maximum length 3.85 times the original length. There are different ways to report this extensional deformation. Calculate how 'wrong' the engineering strain is compared to the true strain by evaluating the ratio:
εtrue / εengr = __________
Answer:
(εtrue/εengr) = (1.3481/2.85) = 0.473
This shows that the engineering strain is truly a bit far off the true strain.
Explanation:
εengr
Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.
ε(engineering) = ΔL/L₀
Lf = final length = 3.85 L₀
L₀ = original length = L₀
ΔL = Lf - L₀ = 3.85 L₀ - L₀ = 2.85 L₀
ε(engineering) = ΔL/L₀ = (2.85L₀)/L₀ = 2.85
εtrue
True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and Running them up. Hence,
ε(true) = In (Lf/L₀)
Lf = 3.85L₀
L₀ = L₀
ε(true) = In (Lf/L₀) = In (3.85L₀/L₀) = In 3.85 = 1.3481
(εtrue/εengr) = (1.3481/2.85) = 0.473
Consider the animation code from the end of Chapter 4. Modify the ShapeIcon class so that it aggregates a collection of objects implementing the MovableShape interface and its paintIcon() method takes care of painting all the given MovableShape objects. Modify the animation program (AnimationTester.main) to display 5 cars moving horizontally, each starting from a different vertical coordinate, so that the car with index i moves twice as fast as the car with index i - 1.
Answer:
import java.awt.*;
import java.util.*;
import javax.swing.*;
/**
An icon that contains a moveable shape.
*/
public class ShapeIcon implements Icon
{
private int width;
private int height;
private MoveableShape shape;
private ArrayList <MoveableShape> cars;
public ShapeIcon(MoveableShape shape,
int width, int height)
{
this.shape = shape;
this.width = width;
this.height = height;
cars = new ArrayList<MoveableShape>();
}
public int getIconWidth()
{
return width;
}
public int getIconHeight()
{
return height;
}
public void paintIcon(Component c, Graphics g, int x, int y)
{
Graphics2D g2 = (Graphics2D) g;
shape.draw(g2);
}
}
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
/**
This program implements an animation that moves
a car shape.
*/
public class AnimationTester
{
private static final int ICON_WIDTH = 400;
private static final int ICON_HEIGHT = 100;
private static final int CAR_WIDTH = 100;
public static void main(String[] args)
{
JFrame frame = new JFrame();
final MoveableShape shape
= new CarShape(0, 0, CAR_WIDTH);
ShapeIcon icon = new ShapeIcon(shape,
ICON_WIDTH, ICON_HEIGHT);
final JLabel label = new JLabel(icon);
frame.setLayout(new FlowLayout());
frame.add(label);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setVisible(true);
final int DELAY = 10; //100
// Milliseconds between timer ticks
Timer t = new Timer(DELAY, new
ActionListener()
{
public void actionPerformed(ActionEvent event)
{
shape.translate(1, 0);
label.repaint();
}
});
t.start();
}
}
import java.awt.*;
import java.awt.geom.*;
import java.util.*;
/**
A car that can be moved around.
*/
public class CarShape implements MoveableShape
{
private int x;
private int y;
private int width;
/**
Constructs a car item.
@param x the left of the bounding rectangle
@param y the top of the bounding rectangle
@param width the width of the bounding rectangle
*/
public CarShape(int x, int y, int width)
{
this.x = x;
this.y = y;
this.width = width;
}
public void translate(int dx, int dy)
{
// x += dx;
// y += dy;
if (x <= 400)
x += dx;
else
x = 0;
y += dy;
}
public void draw(Graphics2D g2)
{
Rectangle2D.Double body
= new Rectangle2D.Double(x, y + width / 6,
width - 1, width / 6);
Ellipse2D.Double frontTire
= new Ellipse2D.Double(x + width / 6, y + width / 3,
width / 6, width / 6);
Ellipse2D.Double rearTire
= new Ellipse2D.Double(x + width * 2 / 3, y + width / 3,
width / 6, width / 6);
// The bottom of the front windshield
Point2D.Double r1
= new Point2D.Double(x + width / 6, y + width / 6);
// The front of the roof
Point2D.Double r2
= new Point2D.Double(x + width / 3, y);
// The rear of the roof
Point2D.Double r3
= new Point2D.Double(x + width * 2 / 3, y);
// The bottom of the rear windshield
Point2D.Double r4
= new Point2D.Double(x + width * 5 / 6, y + width / 6);
Line2D.Double frontWindshield
= new Line2D.Double(r1, r2);
Line2D.Double roofTop
= new Line2D.Double(r2, r3);
Line2D.Double rearWindshield
= new Line2D.Double(r3, r4);
g2.draw(body);
g2.draw(frontTire);
g2.draw(rearTire);
g2.draw(frontWindshield);
g2.draw(roofTop);
g2.draw(rearWindshield);
Explanation:
piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?
Answer: boundary work is 1.33kJ and heat lost is 200kJ
Explanation: mass of water m = 5kg
Molar mass of water mm = 18kg/mol
No of moles n = 5/18 = 0.28
Pressure P = 500x10^3
Temperature T = 300°c = 573K
R = 8.314pa.m3/mol/k a constant.
Using PV = nRT
V = nRT/P
V = (0.28x8.314x573)÷(500x10^3)
V = 0.00266m3
Work on boundary w = PV
W = 500x10^3 x 0.00266 = 1330 J
= 1.33kJ
Heat lost on cooling is equal to electrical heat Q used to heat the water initially
Q =Ivt
I = current = 4A
v = voltage = 100
T = time = 500sec
Q = 4x100x500 = 200000 = 200kJ
Water at 158C (r 5 999.1 kg/m3 and m 5 1.138 3 1023 kg/m·s) is flowing steadily in a 30-m-long and 5-cm-diameter horizontal pipe made of stainless steel at a rate of 9 L/s. Determine (a) the pressure drop, (b) the head loss, and (c) the pumping power requirement to overcome this pressure drop.
Complete Question
The complete question is shown on the first uploaded image
Answer:
(a) the pressure drop is [tex]\Delta P_L = 100.185\ kPa[/tex]
(b) the head loss is the [tex]h_L =10.22\ m[/tex]
(c) the pumping power requirement to overcome this pressure drop [tex]W_p = 901.665\ W[/tex]
Explanation:
So the question we are told that the water is at a temperature of 15°C
And also we are told that the density of the water is [tex]\rho=999.1\ kg/m^3[/tex]
We are also told that the dynamic viscosity of the water is [tex]\mu = 1.138 *10^{-3} kg/m \cdot s[/tex]
From the diagram the length of the pipe is [tex]L=[/tex] 30 m
The diameter is given as [tex]D=[/tex] 5 cm [tex]\frac{5}{100} m = 0.05m[/tex]
The volumetric flow rate is given as [tex]Q=[/tex] 9 L/s [tex]= \frac{9}{1000} m^3/ s = 0.009\ m^3/s[/tex]
Now the objective of this solution is to obtain
i the pressure drop, ii the head loss iii the pumping power requirement to overcome this pressure drop
to obtain this we need to get the cross-sectional area of the pipe which will help when looking at the flow analysis
So mathematically the cross-sectional area is
[tex]A_s =\frac{\pi}{4} D^2[/tex]
[tex]= \frac{\pi}{4} (5*10^{-2})[/tex]
[tex]=1.963*10^{-3} m^2[/tex]
Next thing to do is to obtain average flow velocity which is mathematically represented as
[tex]v = \frac{Q}{A_s}[/tex]
[tex]v =\frac{9*10^{-3}}{1.963*10^{-3}}[/tex]
[tex]=4.585 \ m/s[/tex]
Now to determine the type of flow we have i.e to know whether it a laminar flow , a turbulent flow or an intermediary flow
We use the Reynolds number
if it is below [tex]4000[/tex] then it is a laminar flow but if it is higher then it is a turbulent flow ,now when it is exactly the value then it is an intermediary flow
This Reynolds number is mathematically represented as
[tex]Re = \frac{\rho vD}{\mu}[/tex]
[tex]=\frac{(999.1)(4.585)(0.05)}{1.138*10^{-3}}[/tex]
[tex]=2.0127*10^5[/tex]
Now since this number is greater than 4000 the flow is turbulent
So we are going to be analyse the flow using the Colebrook's equation which is mathematically represented as
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{\epsilon/D_h}{3.7} +\frac{2.51}{Re\sqrt{f} } ][/tex]
Where f is the friction factor , [tex]\epsilon[/tex] is the surface roughness ,
Now generally the surface roughness for stainless steel is
[tex]\epsilon = 0.002 mm = 2*10^{-6} m[/tex]
Now substituting the values into the equation we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{2*10^{-6}/5*10^{-2}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
So solving to obtain f we have
[tex]\frac{1}{\sqrt{f} } =-2.0\ log [\frac{4*10^{-5}}{3.7} +\frac{2.51}{(2.0127*10^{5})\sqrt{f} } ][/tex]
[tex]= -2.0log[1.0811 *10^{-5} +\frac{1.2421*10^{-5}}{\sqrt{f} } ][/tex]
[tex]f = 0.0159[/tex]
Generally the pressure drop is mathematically represented as
[tex]\Delta P_L =f\ \frac{L}{D} \ \frac{\rho v^2}{2}[/tex]
Now substituting values into the equation
[tex]= 0.0159 [\frac{30}{5*10^{-5}} ][\frac{(999.1)(4.585)}{2} ][/tex]
[tex]=100.185 *10^3 Pa[/tex]
[tex]=100.185 kPa[/tex]
Generally the head loss in the pipe is mathematically represented as
[tex]h_L =\frac{\Delta P_L}{\rho g}[/tex]
[tex]= \frac{100.185 *10^3}{(999.1)(9.81)}[/tex]
[tex]=10.22m[/tex]
Generally the power input required to overcome this pressure drop is mathematically represented as
[tex]W_p = Q \Delta P_L[/tex]
[tex]=(9*10^{-3}(100.185*10^3))[/tex]
[tex]= 901.665\ W[/tex]
The contents of a tank are to be mixed with a turbine impeller that has six flat blades. The diameter of the impeller is 3 m. If the temperature is 20°Cand the impeller is rotated at 30 rpm (rev/min), what will be the power consumption? Use power number (Np) of 3.5
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).
An online music platform, S record, is planning to implement a database to enhance its data management practice and ultimately advance its business operations. The initial planning analysis phases have revealed the following system requirements:
Each album has a unique Album ID as well as the following attributes: Album Title, Album Price, and Release Date. An album contains at least one song or more songs. Songs are identified by Song ID. Each song can be contained in more than one album or not contained in any of them at all and has a Song Title and Play Time. Each song belongs to at least one genre or multiple genres. Songs are written by at least an artist or multiple artists. Each artist has a unique Artist ID, and an artist writes at least one song or multiple songs, to be recorded in the database. Data held by each artist includes Artist Name and Debut Date.
Each customer must sign up as a member to make a purchase on the platform. The customer membership information includes Customer ID, Customer Name, Address (consisting of City, State, Postal Code), Phone Number, Birthday, Registration Date. Customers place orders to purchase at least one album or more albums. They can purchase multiple quantities of the same album, which should be recorded as Quantities Ordered. Each order is identified by an Order IDand has Order Date, Total Price, Payment Method, and Delivery Option.
Q1. Draw an ER diagram for Statement 1 (You can add notes to your diagram to explain additional assumptions, if necessary).
File format: asgmt1_q1_ lastname_firstname.pdf
Answer:
Explanation:
We would be taking a breakdown of the following entities.
online music platform database can have the following entities:
1. Artist
has the following attributes:
Artists Name
Artist ID
Debut Date
2. Albums
has the following attributes
:
Album ID
Title
Release Date
Price
3. Song
has the following attributes
:
Song ID
Play Time
Title
genres
4. Items
This represents the items the customers purchased with several albums and quantity.
this shows the following attributes
Album ID
Qty
5. Orders
has the following attributes
The Order ID
The Order Date
Total Price
Payment Method
Delivery Option
6. Customer
has the following attributes
Customer ID
Customer Name
Address - City, State, Postal code
Phone Number
Birthday
Registration Date
NB. the uploaded image shows the ER Diagram.
cheers i hope this helps.