Answer:
Explanation:
Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Strict schedule:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.
S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.
S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))
but T3 commits after T1. In a strict schedule T3 must read X after C1.
Cascadeless schedule:
Cascadeless schedule follows the below condition:
Tj reads X only? after Ti has written to X and terminated means aborted or committed.
S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.
But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,
T3 does not have to roll back if T1 is rolled back. The problem occurs because these
schedules are not serializable.
Recoverable schedule:
Schedule that follows the below condition:
-----Tj commits after Ti if Tj has?read any data item written by Ti.
Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is
recoverable one has to include abort operations. Thus in testing the recoverability abort
operations will have to used in place of commit one at a time. Also the strictest condition is
------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.
If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and
T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.
Strictest condition of schedule S3 is C3>C2.
If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]
Weight transfer is given as follows;
[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]
[tex]F_R[/tex] + [tex]F_F[/tex] = 18000
[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]
[tex]F_R[/tex] = 18000*17/30 = 10200 N
[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel [tex]F_R[/tex] = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel [tex]F_F[/tex] = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]
The least coefficient of friction, μ = 0.625.
WHAT IS A VACUOMETER?
Sometimes, steel studs may not be used on outside walls because they are?
Answer:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Explanation:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
When you do a vehicle check, what do you NOT need to keep an eye on?
A. Proper tire inflation
B. Cleanliness of windows and mirrors
C. Functioning indicator lights and headlights
D. Blindspot locations
Answer:
Blindspot Location
Explanation:
Just took the quiz
When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.
What is Blind spot location?A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.
While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.
Blind spot locations does not need to be checked.
Thus, the correct option is D.
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cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
Purely resistive loads of 24 kW, 18 kW, and 12 kW are connected between the neutral
and the red, yellow and blue phases respectively of a 3-0, four-wire system. The line
voltage is 415 V. Calculate:
i. the current in each line conductor (i.e., IR ,Iy and IB); and
ii. the current in the neutral conductor.
Answer:
(i) IR = 100.167 A Iy = 75.125∠-120 IB = 50.083 ∠+120 (ii) IN =43.374∠ -30°
Explanation:
Solution
Given that:
Three loads 24 kW, 18 kW, and 12 kW are connected between the neutral.
Voltage = 415V
Now,
(1)The current in each line conductor
Thus,
The Voltage Vpn = vL√3
Gives us, 415/√3 = 239.6 V
Then,
IR = 24 K/ Vpn ∠0°
24K/239.6 ∠0°= 100.167 A
For Iy
Iy = 18k/239. 6
= 75.125A
Thus,
Iy = 75.125∠-120 this is as a result of the 3- 0 system
Now,
IB = 12K /239.6
= 50.083 A
Thus,
IB is =50.083 ∠+120
(ii) We find the current in the neutral conductor
which is,
IN =Iy +IB +IR
= 75.125∠-120 + 50.083∠+120 +100.167
This will give us the following summation below:
-37.563 - j65.06 - 25.0415 +j 43.373 + 100.167
Thus,
IN = 37.563- j 21.687
Therefore,
IN =43.374∠ -30°
Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in computer applications. For example, an app that determines the winner of a sales contest would input the number of units sold by each salesperson. The sales person who sells the most units wins the contest. Write pseudocode, then a C# app that inputs a series of 10 integers, then determines and displays the largest integer. Your app should use at least the following three variables:
Counter: Acounter to count to 10 (i.e., to keep track of how many nimbers have been input and to determine when all 10 numbers have been processed).
Number: The integer most recently input by the user.
Largest: The largest number found so far.
Answer:
See Explanation
Explanation:
Required
- Pseudocode to determine the largest of 10 numbers
- C# program to determine the largest of 10 numbers
The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;
The largest of the 10 numbers is then saved in variable Largest and printed afterwards.
Pseudocode (Number lines are used for indentation to illustrate the program flow)
1. Start:
2. Declare Number as 1 dimensional array of 10 integers
3. Initialize: counter = 0
4. Do:
4.1 Display “Enter Number ”+(counter + 1)
4.2 Accept input for Number[counter]
4.3 While counter < 10
5. Initialize: Largest = Number[0]
6. Loop: i = 0 to 10
6.1 if Largest < Number[i] Then
6.2 Largest = Number[i]
6.3 End Loop:
7. Display “The largest input is “+Largest
8. Stop
C# Program (Console)
Comments are used for explanatory purpose
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int[] Number = new int[10]; // Declare array of 10 elements
//Accept Input
int counter = 0;
while(counter<10)
{
Console.WriteLine("Enter Number " + (counter + 1)+": ");
string var = Console.ReadLine();
Number[counter] = Convert.ToInt32(var);
counter++;
}
//Initialize largest to first element of the array
int Largest = Number[0];
//Determine Largest
for(int i=0;i<10;i++)
{
if(Largest < Number[i])
{
Largest = Number[i];
}
}
//Print Largest
Console.WriteLine("The largest input is "+ Largest);
Console.ReadLine();
}
}
}
A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of 62 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have
Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
Choose two consumer services careers and research online to determine what kinds of professional organizations exist for these professions. Write a paragraph describing the purpose of the organization, the requirements for joining, and the benefits of membership.
Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.
Bank loan facilitator is a consumer service facilitator who ask and provide people loan in emergency, for the purpose of education, treatment, family events, and for other reasons. For bank loan facilitator the professional organizations should be banking and finance sector. The purpose of these organizations is to help people in financial matter seeking benefit by getting interest from customers. The requirements for joining of the employee must include strong convincing power for the employee, time management, strong and tactful communication skills. Benefits of membership of the customers can help them to seek loans on need basis on lower interest. Hospital emergency care specialist provides help to the staff and the customers in medical emergency. These professionals are necessary for the hospital, clinics, and rehabilitation centers. Purpose of the organization is to provide medical care to the patients. The requirements for joining of the employee includes ability to give information to patients and staff during emergency conditions, facilitating ambulance to rescue patients from their homes, and from other areas, providing medicine, medical equipment, and other facilities to the patients and other medical staff necessary for treatment. Benefits of membership in clinical or hospital settings can help the patient in frequent visits for treatment, concession in laboratory tests, and medication.Learn more about customer:
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Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel flow with a velocity of 12.5 m/sec. The plate has a length of 2.70 m (in the direction of the fluid flow), a width of 0.65 m, and is maintained at a constant temperature of 127.0°C. Determine the heat transfer rate from the top of the plate due to forced convection.
Answer:
Explanation:
Given that:
V = 12.5m/s
L= 2.70m
b= 0.65m
[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]
[tex]T_s= 127^0C = (127+273)= 400K[/tex]
P = 1atm
Film temperature
[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]
dynamic viscosity =
[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]
density = 0.9946kg/m³
Pr = 0.708564
K= 229.7984 * 10⁻³w/mk
Reynolds number,
[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]
[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]
we have,
[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]
we have,
heat transfer rate from top plate
[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]
Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most transmissions today use compound (multiple) planetary gear sets. Which technician is correct?
Answer:
Both technician A and technician B are correct
Explanation:
A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.
One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.
So, both technician A and technician B are correct.
what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the uniform magnetic field and name it. How can you represent this output as the periodic function?
Answer:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
Explanation:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field
B ( wave flux density ) = Bm sinwt and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec
You are standing at the edge of the roof of the engineering building, which is H meters high. You see Professor Murthy, who is h meters tall, jogging towards the building at a speed of v meters/second. You are holding an egg and want to release it so that it hits Prof Murthy squarely on top of his head. What formulas describes the distance from the building that Prof Murthy must be when you release the egg?
Answer:
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Explanation:
First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:
Vf = Vi + gt
where,
Vf = Velocity of egg at the time of hitting head of the Professor
Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)
g = acceleration due to gravity
t = time taken by egg to come down
Therefore,
Vf = 0 + gt
t = Vf/g ----- equation (1)
Now, we use 3rd euation of motion for Vf:
2gs = Vf² - Vi²
where,
s = height dropped = H - h
Therefore,
2g(H - h) = Vf²
Vf = √[2g(H - h)]
Therefore, equation (1) becomes:
t = √[2g(H - h)]/g
t = √[2(H - h)/g]
Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:
s = vt
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
Answer:
[tex] T_A_C = 6.296 kN [/tex]
[tex] R = 10.06 kN [/tex]
Explanation:
Given:
[tex] T_A_B = 8.7 kN[/tex]
Required:
Find the tension TAC and magnitude R of this downward force.
First calculate [tex] \alpha, \beta, \gamma [/tex]
[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]
[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]
[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]
To Find tension in AC and magnitude R, use sine rule.
[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]
Substitute values:
[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]
Solve for T_A_C:
[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]
[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]
Solve for R.
[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]
[tex] R = 8.7 * 1.156 [/tex]
R = 10.06 kN
Tension AC = 6.296kN
Magnitude,R = 10.06 kN
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2. The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2. (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile
Answer:
A.) Time = 17.13 seconds
B.) Distance = 31.9 m
C.) V = 11.18 m/s
D.) V = 7.1 m/s
Explanation:
The initial velocity U of the automobile is 15.65 m/s.
At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.
For the automobile, let us use first equation of motion
V = U - at.
Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And
V = 0.
Substitute U and a into the formula
0 = 15.65 - 3.05t
15.65 = 3.05t
t = 15.65/3.05
t = 5.13 seconds
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².
The total time required for the police car to overtake the automobile will be
12 + 5.13 = 17.13 seconds.
b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²
V^2 = U^2 + 2aS
Where S = distance travelled.
Substitute V and a into the formula
11.18^2 = 0 + 2 × 1.96 ×S
124.99 = 3.92S
S = 124.99/3.92
S = 31.88 m
c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s
d.) That will be the final velocity V of the automobile car.
We will use third equation of motion to solve that.
V^2 = U^2 + 2as
V^2 = 15.65^2 - 2 × 3.05 × 31.88
V^2 = 244.9225 - 194.468
V = sqrt( 50.4545)
V = 7.1 m/s
A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the material is 240 MPa. A factor of safety of 2.5 is desired. The maximum permissible internal pressure is most nearly: 90 kPa 230 kPa 430 kPa D. 570 kPa csauteol psotolem here Pcr 8. A structural steel tube with a 203 mm x 203 mm square cross section has an average wall thickness of 6.35 mm. The tube resists a torque of 8 N m. The average shear flow is most nearly
A. 100 N/m
B. 200 N/m
C. 400 N/m
D. 800 N/m
Answer:
1) 2304 kPa
2) B. 200 N/m
Explanation:
The internal pressure of the of the tank can be found from the following relations;
Resisting wall force F = p×(1/4·π·D²)
σ×A = p×(1/4·π·D²)
Where:
σ = Allowable stress of the tank
A = Area of the wall of the tank = π·D·t
t = Thickness of the tank = 15 mm. = 0.015 m
D = Diameter of the tank = 25 m
p = Maximum permissible internal pressure pressure
∴ σ×π·D·t = p×(1/4·π·D²)
p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa
With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa
2) The formula for average shear flow is given as follows;
[tex]q = \dfrac{T}{2 \times A_m}[/tex]
Where:
q = Average shear flow
T = Torque = 8 N·m
[tex]A_m[/tex] = Average area enclosed within tube
t = Thickness of tube = 6.35 mm = 0.00635 m
Side length of the square cross sectioned tube, s = 203 mm = 0.203 m
Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²
[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]
Hence the average shear flow is most nearly 200 N/m.
Following are the solution to the given question:
Calculating the allowable stress:
[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]
[tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]
Calculating the Thickness:
[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]
The stress in a spherical tank is defined as
[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]
[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]
Calculating the shear flow:
[tex]\to q=\frac{T}{2A}[/tex]
[tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]
[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]
Therefore, the final answer is "".
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Solid spherical particles having a diameter of 0.090 mm and a density of 2002 kg/m3 are settling in a solution of water at 26.7C. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity and the Reynolds number.
Answer:
Settling Velocity (Up)= 2.048*10^-5 m/s
Reynolds number Re = 2.159*10^-3
Explanation:
We proceed as follows;
Diameter of Particle = 0.09 mm = 0.09*10^-3 m
Solid Particle Density = 2002 kg/m3
Solid Fraction, θ= 0.45
Temperature = 26.7°C
Viscosity of water = 0.8509*10^-3 kg/ms
Density of water at 26.7 °C = 996.67 kg/m3
The velocity between the interface, i.e between the suspension and clear water is given by,
U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]
D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)
D = 2.147
U = 0.0003m/s (n = 4.49)
Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s
Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3
A piston–cylinder device contains 0.85 kg of refrigerant- 134a at 2108C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 158C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.
Question:
A piston–cylinder device contains 0.85 kg of refrigerant- 134a at -10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.
Answer:
a) 90.4 kPa
b) 0.0205 m³
c) 17.4 kJ/kg
Explanation:
Given:
Mass, m = 0.85 kg
a) The final pressure here is equal to the initial pressure. Let's use the formula:
[tex] P_2 = P_1 = P_a_t_m + \frac{mg}{\pi D^2 / 4}[/tex]
[tex] = 88*10^3 + \frac{12kg * 9.81}{\pi (0.25)^2 / 4} [/tex]
= 90398 Pa
≈ 90.4 KPa
Final pressure = 90.4 kPa
b) Change in volume of the cylinder:
To find the initial and final volume, let's use the values from the A-13 table for refrigerant-134a, at initial values of 90.4 kPa and -10°C and final values of 90.4 kPa and 15°C
v1 = 0.2302m³/kg
h1 = 247.76 kJ/kg
v2 = 0.2544 m³/kg
h2 = 268.2 kJ/kg
Change in volume is calculated as:
Δv = m(v2 - v1)
Δv = 0.85(0.2544 - 0.2302)
= 0.0205 m³
Change in volume = 0.0205 m³
c) Change in enthalpy
Let's use the formula:
Δh = m(h2 - h1)
= 0.85(268.2 - 247.76)
= 17.4 kJ/kg
Change in enthalpy = 17.4 kJ/kg
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 4200 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assume uniformly accelerated motion. determine the number of revolutions that the motor executes
(a) in reaching its rated speed,
(b) in coating to rest.
Answer:
a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]
Explanation:
a) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]
[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 175\,rev[/tex]
b) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]
[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 2450\,rev[/tex]
The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of kO = 0.3 m.
Image of wheel is missing, so i attached it.
Answer:
ω = 14.95 rad/s
Explanation:
We are given;
Mass of wheel; m = 20kg
T = 20 N
k_o = 0.3 m
Since the wheel starts from rest, T1 = 0.
The mass moment of inertia of the wheel about point O is;
I_o = m(k_o)²
I_o = 20 * (0.3)²
I_o = 1.8 kg.m²
So, T2 = ½•I_o•ω²
T2 = ½ × 1.8 × ω²
T2 = 0.9ω²
Looking at the image of the wheel, it's clear that only T does the work.
Thus, distance is;
s_t = θr
Since 4 revolutions,
s_t = 4(2π) × 0.4
s_t = 3.2π
So, Energy expended = Force x Distance
Wt = T x s_t = 20 × 3.2π = 64π J
Using principle of work-energy, we have;
T1 + W = T2
Plugging in the relevant values, we have;
0 + 64π = 0.9ω²
0.9ω² = 64π
ω² = 64π/0.9
ω = √64π/0.9
ω = 14.95 rad/s
A cylinder of metal that is originally 450 mm tall and 50 mm in diameter is to be open-die upset forged to a final height of 100 mm. The strength coefficient is 230 MPa and the work hardening exponent is 0.15 while the coefficient of friction of the metal against the tool is 0.1. If the maximum force that the forging hammer can deliver is 3 MN, can the forging be completed
Answer:
Yes, the forging can be completed
Explanation:
Given h = 100 mm, ε = ㏑(450/100) = 1.504
[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]
V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³
At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²
D = √(4·A/π) = 106.07 mm
[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042
F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN
Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.
When an electrical signal travels through a conductive wire, it produces an electromagnetic (EM) field. Likewise, when an EM field encounters a conductive wire, it produces a proportional electrical current.
A. True
B. False
Answer:
A. True
Explanation:
When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.
The lower half of a 7-m-high cylindrical container is filled with water (rho = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and the bottom of the cylinder. (Round the final answer to one decimal place.)
Answer:
Pressure difference (ΔP) = 63,519.75 kpa
Explanation:
Given:
ρ = 1,000 kg/m³
Height of cylindrical container used (h) = 7m / 2 = 3.5m
Specific gravity (sg) = 0.85
Find:
Pressure difference (ΔP).
Computation:
⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ] ∵ [ g = 9.81]
⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [1,850]
⇒ Pressure difference (ΔP) = 63,519.75 kpa
Participating in extracurricular activities in high school helps:
Answer:
Develop social skills
Explanation:
Answer:
strengthen your college applications
Explanation:
Describe what you have been taught about the relationship between basic science research, and technological innovation before this class. Have you been told that it is similar to the linear model? Is your view of this relationship different after studying this unit's lectures and readings? Explain why in 3-4 sentences
Answer:
With the Breakthrough of Technology, the rate at which things are done are becoming much more easy. but without basic science, innovation towards technology cannot occur, so the both work hand in hand in the world of technology today.
Explanation:
Technological innovation and Basic science research plays a major role in the world of science and technology today, while we all want technology innovation the more, without basic science, innovation cannot come in place,
Just as we are going further in technology, breakthroughs and growth are been made which helps on the long run in science research which in turn has made things to be done much better and easily.
Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an instrument used to determine density) indicates that the stream has a density of 0.810 g/mL. Using specific tractors from Table B.1, estimate the mass and volumetric feed rates of the two hydrocarbons to the mixing vessel (in U.S. customary units). State at least two assumptions required to obtain the estimate from the recommended date.
Use a delta-star conversion to simplify the delta BCD (40 , 16 , and 8 ) in the
bridge network in Fig. f and find the equivalent resistance that replaces the network
between terminals A and B, and hence find the current I if the source voltage is 52 V.
Answer:
Current, I = 4A
Explanation:
Since the connection is in delta, let's convert to star.
Simplify BCD:
[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]
[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]
[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]
From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.
Simplify:
[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]
Req = 10 ohms + 3 ohms
Req = 13 ohms
To find the current, use ohms law.
V = IR
Where, V = 52volts and I = 13 ohms
Solve for I,
[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]
Current, I = 4 A
Suppose you used the pipette to make 10 additions to a flask, and suppose the pipette had a 10% random error in the amount delivered with each delivery. Use equation 1 on page 25 to calculate the percent error in the total volume delivered to the flask using the number of clicks you were permitted to make. Report that total percentage below.
Here is the equation: random error of average= error in one measurement/n^1/2
Answer:
The total percentage is 3.16237%
Explanation:
Solution
Now,
We have to know what a random error is.
A random error is an error in measured caused by factors or elements which varies from one measurement to another.
The random error is shown as follows:
The average random error is = the error found in one measurement/n^1/2
Where
n =Number ( how many times the experiment was done)
Now that we added 10 times we have that,
n → 10
Thus,
The error in one measurement = 10%
So,
The average random error = 10 %/(10)^1/2
= (10)^1/2 %
√10%
The total percentage is = 3.16237%
The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Projected Sales Units Price T1 60,000 $165 T2 40,000 $250
Inventories in Units January 1, 20XX December 31, 20XX T1 20,000 25,000 T2 8,000 9,000
The following direct materials are used in the two products: Amount used per unit Direct Material Unit T1 T2 A pound 4 5 B pound 2 3 C each 0 1
Anticipated Inventories Direct Material Purchase Price January 1, 2012 December 31, 2012 A $12 32,000 lb. 36,000 lb. B $5 29,000 lb. 32,000 lb. C $3 6,000 units 7,000 units
Projected direct manufacturing labor requirements and rates for 20XX are as follows: Hours per Unit Rate per Hour T1 2 $12 T2 3 $16
4
Manufacturing overhead is allocated at the rate of $20 per direct manufacturing labor-hour. Marketing and distribution costs are projected to be $100,000 and $ 300,000, respectively.
a. What is the total expected revenue (in dollars) for 20XX? b. What is the expected production level (in units) both for T1 and T2? c. What is the total direct material purchases (in dollars) for each type of direct material? d. What is the total direct manufacturing labor cost (in dollars)? e. What is the total overhead cost (in dollars)? f. What is the total cost of goods sold (in dollars)? g. What is the total expected operating income (in dollars) for 20XX?
Answer:
The ABC Corporation
a) Total Expected Revenue (in dollars) for 20XX:
Revenue from T1 = 60,000 x $165 = $26,400,000
Revenue from T2 = 40,000 x $250 = $10,000,000
Total Revenue from T1 and T2 = $36,400,000
b) Production Level (in units) for T1 and T2
T1 T2
Total Units sold 160,000 40,000
Add Closing Inventory 25,000 9,000
Units Available for sale 185,000 49,000
less opening inventory 20,000 8,000
Production Level 165,000 units 41,000 units
c) Total Direct Material Purchases (in dollars):
Cost of direct materials used T1 T2
A: (165,000 x 4 x $12) $7,920,000 $2,460,000 (41,000 x 5 x $12)
B: (165,000 x 2 x $5) 1,650,000 615,000 (41,000 x 3 x $5)
C: 0 123,000 (41,000 x 1 x$3)
Total cost $9,570,000 $3,198,000 Total = $12,768,000
Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2
Cost of direct materials used for production $12,768,000
Cost of closing direct materials:
A (36,000 x $12) $432,000
B (32,000 x $5) 160,000
C (7,000 x $3) 21,000 $613,000
Cost of direct materials available for prodn $13,381,000
Less cost of beginning direct materials:
A (32,000 x $12) $384,000
B (29,000 x $5) 145,000
C (6,000 x $3) 18,000 $547,000
Cost of direct materials purchases $12,834,000
d) The Total Direct Manufacturing Labor Cost (in dollars):
T1 T2
Direct labor per unit 2 hours 3 hours
Direct labor rate per hour $12 $16
Direct labor cost per unit $24 $48
Production level 165,000 units 41,000 units
Labor Cost ($) $3,960,000 $1,968,000
Total labor cost $5,928,000 ($3,960,000 + $1,968,000)
e) Total Overhead cost (in dollars):
Overhead rate = $20 per labor hour
Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)
T1 overhead = $20 x 2 x 165,000) = $6,600,000
T2 overhead = $20 x 3 x 41,000) = $2,460,000
Total Overhead cost = $9,060,000
Cost of goods produced:
Cost of opening inventory of materials = $547,000
Purchases of directials materials 12,834,000
less closing inventory of materials = $613,000
Cost of materials used for production 12,768,000
add Labor cost 5,928,000
add Overhead cost 9,060,000
Total production cost $27,756,000
f) Total cost of goods sold (in dollars):
Cost of opening inventory = $3,928,000
Total Production cost = $27,756,000
Cost of goods available for sale $31,684,000
Less cost of closing inventory $4,724,000
Total cost of goods sold $26,960,000
g) Total expected operating income (in dollars)
Sales Revenue: T1 and T2 $36,400,000
Cost of goods sold 26,960,000
Gross profit $9,440,000
less marketing & distribution 400,000
Total Expected Operating Income = $9,040,000
Explanation:
a) Cost of beginning inventory of finished goods:
T1, (Direct materials + Labor + Overhead) X inventory units =
T1 = 20,000 x ($58 + 24 + 40) = $2,440,000
T2 = 8,000 ($78 + 48 + 60) = $1,488,000
Total cost of beginning inventory = $3,928,000
b) Cost of closing Inventory of finished goods:
T1 = 25,000 x ($58 + 24 + 40) = $3,050,000
T2 = 9,000 ($78 + 48 + 60) = $1,674,000
Total cost of closing inventory = $4,724,000
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P
Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]
Where:
[tex]n[/tex] - Safety factor, dimensionless.
[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])
[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]
[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]
[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]
Where:
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
[tex]V[/tex] - Shear force, measured in kilonewtons.
[tex]A[/tex] - Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
[tex]V = \frac{P}{5}[/tex]
[tex]V = \frac{450\,kN}{5}[/tex]
[tex]V = 90\,kN[/tex]
The minimum allowable cross section area is cleared in the shearing stress equation:
[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]
If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:
[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]
[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]
The diameter is now cleared and computed:
[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]
[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]
[tex]D = 0.0457\,m[/tex]
[tex]D = 45.7\,mm[/tex]
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
We have that the minimum allowable bolt diameter is mathematically given as
d = 26.65mmFrom the question we are told
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of Assuming P to be P = 425 kN.DiameterGenerally the equation for the stress is mathematically given as
[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]
Therefore
Force = Stress * area
Force = P/2
F= 425,000 N / 2 = 212,500 N
Hence area of each bolt is given as
212,500 = 76.190*( 5* area of each bolt)
area of each bolt = 557.815
Since
area of each bolt=\pi*d^2/4
\pi*d^2/4 = 557.815
d = 26.65mmFor more information on diameter visit
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