Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
(a)
Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.
Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.
(b)
The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.
Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite. The carbon concentration for both ferrites is 0.022 wt% C.
(c)
(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.
(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.
(iii) Hardness: hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.
(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.
(d)
I would recommend hypereutectoid steel alloy to make a knife or ax blade
1- Hardness is required at the surface of the blades.
2- Ductility is not needed for such application.
3- Due to constant impact, the material will not easily yield to stress.
(e)
I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.
1- hypoeutectoid steel alloys have high machinability, hence better productivity
2- It will be used on softer metals, hence its fitness for the application
3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.
Explanation:
See all together above
a) The current that goes through a 100 mH inductor is given as
i(t) = 6 - 2e^-2t A t >= 0
Find the voltage v(t) across the inductor.
b) The voltage v(t) = 5sin(5t) V is applied across the terminals of a 200 mH inductor. The initial current through the inductor is i(0) = -10 A. Find the current i(t) through the inductor for t > 0.
Answer:
A) V(t) = 0.4e^-2t
B) i(t) = (25tsin5t+10) A for t>0
Explanation:
Formula for calculating voltage across an inductor is expressed as:
V = Ldi/dt
Given L = 100mH = 100×10^-3
If i(t) = 6 - 2e^-2t A t >= 0
di/dt = (-2)(-2)e^-2t
di/dt = 4e^-2t
If t ≥ 0
V(t) = 100×10^-3 × (4e^-2t)
V(t) = 0.1×4e^-2t
V(t) = 0.4e^-2t for t≥0
B) Applying the same formula as above
V = Ldi/dt
Vdt = Ldi
V/L dt = di
On integration
Vt/L = i + C
When t = 0, i = -10A
Substituting the values into the formula
V(0)/L = -10 + C
0 = -10+C
C = 10
To get the current i(t) through the inductor for t>0,
Since Vt/L = i + C
Given V(t) = 5sin5t Volts
L = 200mH = 200×10^-3H
C = 10
On substituting
(5sin5t)t/0.2 = i + 10
25tsin5t = i + 10
i(t) = (25tsin5t-10) A for t>0
An R-134a refrigeration system is operating with an evaporator pressure of 200 kPa. The refrigerant is 10% in vapor phase at the inlet of the evaporator. The enthalpy and temperature of the refrigerant at the exit of the compressor are 360 kJ/kg and 70 /C respectively. If this system (with the refrigerant flowing at 0.005 kg/s) is used to cool a 750 kg product (initial temperature = 80 /C and specific heat = 3,000 J/kg-K), what will be the temperature of the product after 6 hours?
Answer:
71.17°C
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its proportional limit, the specimen elongates by 0.3 mm. The total axial load is 20 kN. Determine the modulus of elasticity and the proportional limit.
Answer:
modulus of elasticity = 100.45 Gpa,
proportional limit = 150.68 N/mm^2.
Explanation:
We are given the following parameters or data in the question as;
=> "The original specimen = 200 mm long and has a diameter of 13 mm."
=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."
=> " The total axial load is 20 kN"
Step one: Calculate the area
Area = π/ 4 × c^2.
Area = π/ 4 × 13^2 = 132.73 mm^2.
Step two: determine the stress induced.
stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.
Step three: determine the strain rate:
The strain rate = change in length/original length = 0.3/ 200 = 0.0015.
Step four: determine the modulus of elasticity.
modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.
Step five: determine the proportional limit.
proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.
Answer:
Modulus of Elasticity = 100 GPa
Proportional limit = 0.15 GPa
Explanation:
Axial Load = 20 kN = 20000 N
Original length, L₀ = 200 mm = 0.2 m
diameter, d = 13 mm = 0.013 m
Elongation, ΔL = 0.3 mm = 0.0003 m
Area of the material:
[tex]A = \frac{\pi d^{2} }{4} \\A = \frac{\pi 0.013^{2} }{4}\\A = 0.000133 m[/tex]
Stress = Load / Area
Stress = 20000 / 0.000133
Stress = 150375940 N/m²
Stress = Proportional limit = 0.15 GPa
Modulus of Elasticity = Stress/Strain
Strain = ΔL / L₀
Strain = 0.0003 / 0.2
Strain = 0.0015
Modulus of Elasticity = 0.15 / 0.0015
Modulus of Elasticity = 100 GPa
A machine operates using air flow under steady conditions with the following inlet and exit flow parameters. At the inlet, the pressure is 1.75 MPa at a temperature of 300 K. The inlet velocity is 25 m/s. The inlet diameter is 20 cm. At the outlet, the pressure is decreased to 55 kPa at a temperature of 500 K. The outlet velocity is 100 m/s. The elevation increase between the inlet and outlet is 100 m. The machine includes a turbine operating at 5 kW
Determine mass flow rate through the machine.
Answer:
Mass flow rate = 15.96kg/s
Explanation:
From
P1V1=mRT1
V1=RT1/P1
V1=0.0492m³/kg
M=A(U)/V1
Where u is the velocity
A=πd²/4
M=π(0.2²)/4×25/0.0492
M=15.96kg/s
(USCS units) A deep drawing operation is performed on a sheet-metal blank that is 1/8 in thick. The height of the cup = 3.8 in and its diameter = 5.0 in (both inside dimensions).
(a) Assuming the punch radius = 0, compute the starting diameter of the blank to complete the operation with no material left in the flange.
(b) Is the operation feasible (ignoring the fact that the punch radius is too small)?
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
A converging–diverging nozzle is designed to produce a Mach number of 2.5 with air. (a) What operating pressure ratio (prec/pt inlet) will cause this nozzle to operate at the first, second, and third critical points? (b) If the inlet stagnation pressure is 150 psia, what receiver pressures represent operation at these critical points? (c) Suppose that the receiver pressure were fixed at 15 psia. What inlet pressures are necessary to cause operation at the critical points?
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator
Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
For what type of metal is high speed steel drill best suited?
Answer:
high speed steel I believe
Your company has been retained to start steel erection for a project. The steel will be delivered on an 18-wheel flat-bed truck and then lifted into place. You will have a crane on site, a Grove GMK 5175 all-terrain crane and you will be lifting with ½ steel chains. The crane is a 175 Ton telescoping crane and the crane will be sitting on Type B soil. The steel delivery consists of 48 pieces of W18x65 structural steel, Grade A 36, broken down as follows and delivered in 12 bundles of 4 pieces each.
13 pieces - 25’ long
17 pieces -50’ long
18 pieces – 15’ long
Based on the above, answer the following questions:
a. Neglecting the weight of the truck, what is the total weight of the steel load?
b. Assuming that the steel will be bundled and lifted as follows, calculate the weight of each of the lifts
4 pieces of 15 feet long, W18x65 Weight_______________________
4 pieces of 50 feet long, W18x65 Weight_______________________
4 pieces of 25 feet long, W18x65 Weight_______________________
c. With the soil conditions noted above,
Will the crane be able to safely lift the heaviest lift?
i. Yes
ii. No
d. What will be the maximum amount or load that can be lifted by this crane using the boom length of 155 feet?
A commonly used strategy to improve the efficiency of a power plant is to withdraw steam from the turbine before it fully expands, reheat it in the boiler and send it to another turbine. Consider a case where 1 kg/s steam enters a turbine at 5MPa, 500 °C, and is expanded to 1 MPa pressure. At this point, the steam is withdrawn from the turbine, reheated back to 500 °C at constant pressure, and then expanded in another turbine to 50 kPa pressure. Calculate (a) The temperature at which steam was sent back for reheating, (b) Heat added during the reheati
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
It is desired to obtain 500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor. What should be the angle of the AC to AC converter to be used? Calculate the THD of the current drawn from the mains (consider up to the 12th harmonic)?
Answer:
14.5° ; THD % = 3.873 × 100 = 387.3%.
Explanation:
Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;
(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".
(2). Consideration of up to 12th harmonic.
So, let us delve right into the solution to the question above;
Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;
Irms = reactive power /Vrms = 500/230 = 2.174 A.
Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.
Step two: Calculate the THD.
Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.
h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.
Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.
THD % = 3.873 × 100 = 387.3%.
Step four: angle AC - Ac converter
theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.
A deep drawing operation is performed on a sheet-metal blank that is 1/8 in thick. The height of the cup = 3.8 in and its diameter = 5.0 in (both inside dimensions). (a) Assuming the punch radius = 0, compute the starting diameter of the blank to complete the operation with no material left in the flange. (b) Is the operation feasible (ignoring the fact that the punch radius is too small)?
Answer:
a) Db = 10.05 in
b) The operation is not feasible
Explanation:
Let's begin by listing out the given variables:
Thickness = 1/8 in, Height (h) = 3.8 in,
Diameter (Dp) = 5 in ⇒ rp = Dp ÷ 2 = 5 ÷ 2 = 2.5 in
a) Area of cup = Area of wall + Area of base
A = 2πh(rp) + π(rp)²
A = (2π * 2.5 * 3.8) + (π * 2.5²)
A = 59.69 + 19.635 = 79.325 in²
A ≈ 79.33 in²
But π(rb)² = 79.33 ⇒ rb² = 79.33 ÷ π
rb² = 25.25
rb = 5.025 ⇒ Db = 2 * rb = 2 * 5.025 = 10.05 in
Db = 10.05 in
b) To calculate for feasibility, we use the formula, draw ratio equals diameter of the blank divided by diameter of the punch
Mathematically,
DR = Db ÷ Dp
DR = 10.05 ÷ 5 = 2.01
DR = 2.01
DR > 2 ⇒ the operation is not feasible
For an operation to be feasible, it must have a drawing ratio limit of 2 or lesser
Discuss the ethics of the circumstances that resulted in the Columbia shuttle disaster. Considering the predictions that were made years before the disaster, as well as the reliability of the Binomial distribution and its implications, what could or should the engineers associated with the program have done differently
Explanation:
This is not so much a mathematical issue as a case study, because the response will inevitably require us to test the special Columbic shuttle disaster scenario. I would suggest that you read this in detail and present the points accordingly. Here I give as many points as I think are relevant.
The failure of a space program is definitely a complex situation, more than a simple binomial distribution. It's definitely not as simple as repeating the flip of a coin. There are several coherent factors and situations that govern the overall coordination and execution of such an event. The problem is, those who are running a project like this are still making a trade off,It is never the case that they sealed the lid on any chance of failure between multiple parameters. You try to do something, but often, as is the case above, the potentially dangerous situation is impossible or uncontrollable. Since the root cause of failure, which is dried out tiles that can not withstand heat and water, it appears that owing to the constant use of the shuttle the head architects have not foreseen this.
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rate of 0.264 kg/s. The maximum absolute temperature in the cycle is 1.15 times the minimum absolute temperature, and the net power input to the cycle is 5 kW. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the ratio of the maximum to minimum pressures in the cycle.
Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
[tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]
Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]
Then, we should apply the values in the COP.
[tex]=\frac{1.15\;T_L}{1.15-1}[/tex]
[tex]=7.67[/tex]
The number of heat rejected by the heat pump must then be calculated.
[tex]Q_H=COP_{HP}\times W_{nst}[/tex]
[tex]=7.67\times5=38.35[/tex]
We must then calculate the refrigerant mass flow rate.
[tex]m=0.264\;kg/s[/tex]
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]=\frac{38.35}{0.264}=145.27[/tex]
The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
[tex]T_L=\frac{T_H}{1.15}[/tex]
[tex]=\frac{64+273}{1.15}=293.04[/tex]
[tex]=19.89\°C[/tex]
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
The amusement park ride consists of a fixed support near O, the 6-m arm OA, which rotates about the pivot at O, and the compartment, which remains horizontal by means of a mechanism at A. At a certain instant, β=30ο, 2 0.75 rad/s, and 0.5 rad/s , all clockwise. Determine the horizontal and vertical forces (F and N) exerted by the bench on the 75-kg rider at P. Compare your results with the static values of these forces. (Use x-y coordinate system and vector equations
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est
Question:
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.
Answer:
See explanation below
Explanation:
Given:
d = 2m = 2*10³ = 2000
thickness, t = 10 mm
Length of strain guage = 20 mm
i) Let's calculate d/t
[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]
Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.
For the minimum normal stress, we have:
[tex] \sigma max= \frac{pd}{4t} [/tex]
[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]
= 50p
For the minimum normal strain due to pressure, we have:
[tex] E_max= \frac{change in L}{L_g} [/tex]
[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]
The minimum normal stress for a thin pressure vessel is 0.
[tex] \sigma _min = 0 [/tex]
i) Let's use Hookes law to calculate the pressure causing this deformation.
[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]
Substituting figures, we have:
[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]
[tex] 120 * 10^6 = 35p [/tex]
[tex] p = \frac{120*10^6}{35}[/tex]
[tex] p = 3.429 * 10^6 [/tex]
p = 3.4 MPa
ii) Calculating the maximum in-plane shear stress, we have:
[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]
[tex] = \frac{50p - 50p}{2} = 0 [/tex]
Max in plane shear stress = 0
iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:
[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]
[tex] = \frac{50p - 0}{2} = 25p [/tex]
since p = 3.429 MPa
25p = 25 * 3.4 MPa
= 85.71 ≈ 85.7 MPa
The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa
Tech A says that when checking tire pressure, the tire should be " cold." Tech B says that the tires should be driven more than 3 miles before checking tire presure. Who is correct?
Answer: Technician A is correct.
Explanation:
Technician A is correct because temperature of a tire will affect its pressure reading.
Tires attract heat because of their dark colour and then motion on the road generates heat. A car owner or technician should know that tire pressure is most accurate when the tire is cold (especially when atmospheric temperature is cool too).
If the tires are driven three miles first, their temperature will be high (due to the rubbing of the tires on the surface of the road). This higher temperature will result in higher per square inch (psi) readings.
Temperature has a great influence on the tire pressure.
Even if the tire is driven up to or more than 3 miles, it should still be left to cool, before tire pressure is checked.
The tire manufacturer's rating should be the maximum possible tire pressure. If an abnormal reading is gotten, the gauge should be properly checked.
A heat recovery device involves transferring energy from the hot flue gases passing through an annular region to pressurized water flowing through the inner tube of the annulus. The inner tube has inner and outer diameters of 24 and 30 mm and is connected by eight struts to an insulated outer tube of 60-mm diameter. Each strut is 3 mm thick and is integrally fabricated with the inner tube from carbon steel (k 50 W/m K). Consider conditions for which water at 300 K flows through the inner tube at 0.161 kg/s while flue gases at 800 K flow through the annulus, maintaining a convection coefficient of 100 W/m2 K on both the struts and the outer surface of the inner tube. What is the rate of heat transfer per unit length of tube from gas to the water?
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A large truck drives down the highway at 10 m/s hauling a rectangular trailer that is 6 m long, 2 m wide, and 2 m tall. The trailer contains frozen food and is therefore temperature-controlled. Its external surface can be approximated to be a consistent 15°C, while the outside air is at 20°C. Assume the heat transfer on the front, back, and bottom of the trailer is negligible.
a) How much cooling power must be provided to maintain the temperature controlled trailer? (i.e. What is the total heat transfer rate from the air to the trailer)?b) What is the minimum local heat transfer coefficient on the surface of the trailer? Where does the minimum occur?c) What percentage of the total heat transfer is occuring over a laminar boundary layer?d) If the cooling system can only provide 5 KW of cooling, what is the fastest speed that this truck can drive while still adequately maintaining the temperature within the trailer?
Answer:
3w/m²k
Explanation:
Base on the scenario been described in the question, the solution to the given problem solve in the file attached below
Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through some region of a crystal, the atomic arrangement of that region is disordered, Justify your answer - • True • False b. The process by which plastic deformation is produced by dislocation motion is called ____. • Interstitial Dislocation • Slip • Additional plain • Screw c. How does the theoretical strength of a solid material compare with its experimental strength? • Strength theoretical < Strength experimental • Strength theoretical = Strength experimental • Strength theoretical > Strength experimental d. The atoms surrounding a screw dislocation experience what kind(s) of strain(s)? • Shear strain • Compressive strain • Tensile strain e. For a particular crystal structure, the slip direction is that direction in the slip plane having the • highest linear density. • lowest linear density.
Answer:
a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line
b.Slip
C. Strength theoretical is greater than strength experimental
d. Shear stress
e. Highest linear density
A complex gear drawing done on a drawing sheet marked M-1 has many section views showing important interior details of the gear. One of the cutting-plane lines is marked at the ends with a callout in a circular bubble that says 7 above a line and M-3 below the line. To find this detail, you would
Answer:
The answer is "go to sheet M-3 and look for a detail labeled 7".
Explanation:
In the given question some information is missing, that is choices so, the correct choice can be described as follows:
In gear drawing, we use equipment that sorts a very important technical reference necessary for machinery design. If a manufacturer wants a tool in the production of a new computer, two choices are available to design the new equipment itself. To use standard features that have already been developed. In this gear drawing to find the details we go to sheet in M-3 and for the detailed labeled 7.Determine the drag on a small circular disk of 0.02-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.89 and a viscosity 10000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
Answer:
33.3%
Explanation:
Given that:
specific gravity (SG) = 0.89
Diameter (D) = 0.01 ft/s
Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]
Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]
Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]
the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]
Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
3. A storage tank is connected to a pond (at atmospheric pressure!) by a length of 4" pipe and a gate valve. From previous operating experience, it has been found that when the tank is at a pressure of 3 atm the flow through the pipe is 35 m3/h when the gate valve is fully open. If the pressure in the tank increases to 5 atm, what will be the maximum discharge rate from the tank?
Answer:
V2= 21m³/h
Explanation:
According to Boyle's law, pressure and flow rate of gas can be calculated from the following equation:
P1V1=P2V2
3 × 35 = 5 × V2
V2= 21m³/h
Suppose you have a coworker who is a high Mach in your workplace. What could you do to counter the behavior of that individual? Put the high Mach individual in charge of a project by himself, and don’t let others work with him. Set up work projects for teams, rather than working one on one with the high Mach person. Work with the high Mach individual one on one, rather than in a team setting. Explain to the high Mach individual what is expected of him and ask him to agree to your terms.
Answer:
To counter the behavior of a high Mach individual in my workplace, I could put the individual in charge of a project by himself, and don't let others work with him.
Explanation:
A high Mach individual is one who exhibits a manipulative and self-centered behavior. The personality trait is characterized by the use of manipulation and persuasion to achieve power and results. But, such individuals are hard to be persuaded. They do not function well in team settings and asking them to agree to terms is very difficult. "The presence of Machiavellianism in an organisation has been positively correlated with counterproductive workplace behaviour and workplace deviance," according to wikipedia.com.
Mach is an abbreviation for Machiavellianism. Machiavellianism is referred to in psychology as a personality trait which sees a person so focused on their own interests that they will manipulate, deceive, and exploit others to achieve their selfish goals. It is one of the Dark Triad traits. The others are narcissism and psychopathy, which are very dangerous behaviors.
Two Electric field vectors E1 and E2 are perpendicular to each other; obtain its base
vectors.
Answer:
<E1, E2>.
Explanation:
So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;
E = |E| cos θi + |E| sin θj. -------------------(1).
Therefore, if we make use of the components division rule we will have something like what we have below;
x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ
Therefore, we will now have;
E = x |E2| i + y |E1| j.
The base vectors is then Given as <E1, E2>.
Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-ton sluice gate of a dam. Quality of construction and maintenance (including lubrication) are good, resulting in an estimated friction coefficient of only 0.1 for the screw. Collar friction can be neglected, as ball thrust bearings are used. Assuming that, because of gate friction, each screw must provide a lifting force of 26 tons, what power is required to drive each screw when the gate is being raised at the rate of 3 ft/min
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountainous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF=0.90.
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remain the same but the peak-hour factor increases to 0.95?
Process: (1) determine passenger car equivalent for trucks and buses; (2) determine passenger car equivalent for recreational vehicles; (3) calculate heavy vehicle factor; (4) determine 15-min passenger equivalent flow rate for current conditions; (5) determine 15-min passenger equivalent flow rate at full capacity; (6) calculate the volume for current and capacity conditions; (7) take the difference of the two volumes to determine how many vehicles were added
Answer:
The number of vehicles added to this highway before the capacity is reached is 1,511 vehicles.
Explanation:
see attached image
How does a car batteray NOT die?
Answer:
bye hooking plugs up to it to amp it up
Have you ever had an ice cream headache that’s when a painful sensation resonates in your head after eating something cold usually ice cream on a hot day this pain is produced by the dilation of a nerve center in the roof of your mouth the nerve center is overreacting to the cold by trying to hit your brain ice cream headaches have turned many smiles to frowns identify the structure
Answer:
Cause and effect
Explanation:
pls mark brainliest
The structure that makes or turned many smiles to frowns can be regarded as compare/contrast.
What is compare contrast?The term compare/contrast is a common terms. The act of comparing is known to be depicting the similarities, and contrasting is said to be showing differences that exist between two things.
Conclusively, from the above, we can see that it is a compare/contrast scenario as it talks about the effects of taking ice cream. It went from smiles to frowns.
See option below
cause/effect
descriptive
compare/contrast
sequence/process
Learn more about compare/contrast from
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