Although many countries have issues with soil erosion due to deforestation, some of the most serious effects are seen
in which country?
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel is 200 GPa and 75 GPa for aluminium. The allowable normal stress in steel is 220 MPa and 100 MPa in aluminium. Determine the largest permissible bending moment when the composite bar is bent about horizontal axis. a
Answer:
1.933 KN-M
Explanation:
Determine the largest permissible bending moment when the composite bar is bent horizontally
Given data :
modulus of elasticity of steel = 200 GPa
modulus of elasticity of aluminum = 75 GPa
Allowable stress for steel = 220 MPa
Allowable stress for Aluminum = 100 MPa
a = 10 mm
First step
determine moment of resistance when steel reaches its max permissible stress
next : determine moment of resistance when Aluminum reaches its max permissible stress
Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M
attached below is a detailed solution
When choosing a respirator for your job, you must conduct a
test.
In a metal-oxide-semiconductor (MOS) device, a thin layer of SiO2 (density = 2.20 Mg/m3) is grown on a single crystal chip of silicon. How many Si atoms and how many O atoms are present per square millimeter of the oxide layer? Assume that the layer thickness is 160 nm.
Answer:
3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O
Explanation:
It is all about unit conversions.
The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.
160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm
By the way, now we can determine the volume of MOS, in order to work with density.
1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³
But density is mg/m³, so we convert mm³ to m³
1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³
Now, we apply density to determine the mass of MOS
Density = mass /volume → Density . volume = mass
1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg
To make more easier the calculate, we convert mg to g.
3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g
To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol
We need to know moles of Si and O₂ in the MOS
Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles
1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:
5.86×10⁻¹⁸ mol of SiO₂ may have:
(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si
(5.86×10⁻¹⁸ .2) /1 = 1.17×10⁻¹⁷ moles of O₂
Let's count the atoms (1 mol of anything contain NA particles)
5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si
1.17×10⁻¹⁷ mol of O₂ . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O
The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O
What is the number of atoms present?We are given;
Area of MOS device; A = 1 mm²
Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm
Formula for volume is;
V = Area * thickness
V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³
Converting volume to m³ gives;
V = 1.6 × 10⁻¹³ m³
Now, to get the mass of MOS, we will use the formula;
Mass = Density * volume
we are given density = 2.20mg/m³
Thus;
Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³
Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g
From periodic table, the molar mass of SiO₂ = 60.08 g/mol
Then;
Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol
Number of moles of MOS = 5.86 × 10⁻¹⁸ moles
Now, 1 mol of SiO₂ is formed from 1 mol of Si and 2 mol of O. Thus;
5.86×10⁻¹⁸ mol of SiO₂ will have;
5.86×10⁻¹⁸ moles of Si
and (5.86×10⁻¹⁸ .2) = 11.72 × 10⁻¹⁸ moles of O₂
From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;
5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si
11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O
Read more about number of atoms at; https://brainly.com/question/3157958
Allura Red Moles
Question of the Day {QOD}:
If you decided to drink massive amounts of Cherry Kool-AidTM on a dare, would you die from Allura Red toxicity or water intoxication first?
5. Answer the Question of the Day. Support your answer with specific evidence from your experimentation including LD50 calculations. In your discussion, calculate both the amount of grams and moles of Allura red in Kool-Aid as well as the volume of Kool-Aid you would need to ingest to reach the median lethal dose of Allura red
The concentration of Allura red used was 1.89x10^-4 M
Kool aid absorbance (after dilution: 1mL kool aid, 9mL water): 0.453
Kool aid concentration: 1.97x10^-4 M
LD50 water: 90g/kg
LD50 (rats and mice): 6,000 - 10,000 mg/kg
Answer:
Die of intoxication by water first
Explanation:
We assume that the weight of the man is 154.35 pounds which is 70 kg
LD50 water = 90g per kg
Maximum concentration = 90x70
= 6300grams
Convert grams to liters
6300/100
= 6.3 litres
From here we get amount of kool aid
6.3 x 1.97x10^-4
= 1.24x10^-3
= 1.24grams
1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.
Discuss the nature of materials causing turbidity in
(a) River water during flash flood
(b) Polluted river water
(c) Domestic wastewater
Answer:
a
Explanation:
Which one of the following answer options are your employers responsibility
Where are your answer options?
Answer:
Implement a hazard communication program
Explanation: i took the quiz
the removed soil at an excavation site is also called spoil?
Answer:
True, That is correct. Soil removed from an excavation site is indeed called spoil.
Spoil definition: The waste material (such as soil) brought up during the course of an excavation.
Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D
Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Thus, The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.
It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.
Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.
Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Learn more about Soil, refer to the link:
https://brainly.com/question/31227835
#SPJ6
Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0) = 0 and derive y"
Answer:
Explanation:
Given that:
[tex]y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)[/tex]
where;
[tex]g'(0) = \dfrac{1}{m}[/tex] and [tex]mg"+kg = 0[/tex]
[tex]\text{Using Leibniz Formula to prove the above equation:}[/tex]
[tex]\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt[/tex]
So, [tex]y = \int ^t_0 g' (t-s) f(s) \ ds[/tex]
[tex]\text{By differentiation with respect to t;}[/tex]
[tex]y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds[/tex]
[tex]y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)[/tex]
[tex]Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\[/tex]
[tex]Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}[/tex]
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation:
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.
Answer: hello your question lacks the required diagram attached below is the diagram
answer : 29528.1 N/m^2
Explanation:
Given data :
dimensions of tank :
Length = 5-m
Width = 4-m
Depth = 2.5-m
acceleration of tank = 2m/s^2
Determine the maximum gage pressure in the tank
Pa ( pressure at point A ) = s*g*h1
= 10^3 * 9.81 * 3.01
= 29528.1 N/m^2
attached below is the remaining part of the solution
Consider the following chain-reaction mechanism for the high-temperatureformation of nitric oxide, i.e., the Zeldovich mechanism:
O+ N2 → NO + N
N + O2 → NO+ O
Write out expressions for d[NO] / dt and d[N] / dt.
Answer: hello attached below is the properly written chain reaction to your question
answer :
d[NO] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]
d[N] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]
Explanation:
write out expressions for d[NO] / dt and d[N] / dt
Given :
properly written chain reaction ( attached below)
Expression for d[NO] / dt can be written as
[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]
Expression for d[N] / dt can be written as
[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]
After earning a bachelor's degree, one must do which of the following before taking the PE examination to receive a Professional Engineering license?
A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit
Answer:
T = 438.87 N.m
Explanation:
The power required to raise the 4961 N load in 10 meters for 2 minutes is:
[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]
P = Torque × W
[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]
[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]
[tex]T = \dfrac{413.42}{0.942}[/tex]
T = 438.87 N.m
Where do greywater pipes generally feed into?
-Vent stack
-Water heater
-Waste stack
-Main supply
Answer:
c Waste stack
Explanation:
What two units of measurement are used to classify engine sizes?
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in your wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 570 turns. The diameter of the tube is 8.10 cm, and the length of the wire-wrapped portion is 35.0 cm. You pull out your calculator to determine the following.
a. the inductance of the coil (in mH)
b. the emf generated in the coil if the current in the wire increases at the rate of 3.00 A/s (Enter the magnitude in mV.)
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Determine (a) the maximum flow rate of water (5-point) and (b) the pressure difference across the pump (5-point). Assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible
Answer:
a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s
b) The pressure difference across the pump is approximately 293.118 kPa
Explanation:
The efficiency of the pump = 78%
The power of the pump = 5 -kW
The height of the pool above the underground water, h = 30 m
The diameter of the pipe on the intake side = 7 cm
The diameter of the pipe on the discharge side = 5 cm
a) The maximum flowrate of the pump is given as follows;
[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]
Where;
P = The power of the pump
Q = The flowrate of the pump
ρ = The density of the fluid = 997 kg/m³
h = The head of the pump = 30 m
g = The acceleration due to gravity ≈ 9.8 m/s²
[tex]\eta_t[/tex] = The efficiency of the pump = 78%
[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]
[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s
The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s
b) The pressure difference across the pump, ΔP = ρ·g·h
∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa
The pressure difference across the pump, ΔP ≈ 293.118 kPa
A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?
Answer:
10.007
Explanation:
Assuming we have to find out the compression ratio of the engine
Given information
Cubic capacity of the engine, V = 245 cc
Clearance volume, V_c = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]
plugging values we get
245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]
Solving we get D =7 cm
therefore, L= 7/1.1 =6.36 cm
Now,
the compression ratio is given as:
r =(V+V_c)/V_c
on substituting the values, we get
r = (245+27.2)/27.2 =10.007
Hence, Compression ratio = 10.007
The earth can be assumed flat when *
Az = 0 degree
El = 50 degree
Az = 50 degree
El = 0 degree
Answer:
El =50 degre
Explanation:
Yan ang answer
4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views
Answer:
C Electrical Connections
Explanation:
In reading says . However, electrical
connections aren’t shown in construction drawings.
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevation difference between the free surfaces upstream and downstream of the dam is 165 ft. Water is to be supplied at a rate of 7000 lbm/s. The electrical power generated is measured to be 1546 hp and the generator efficiency of 92%. (1 hp = 550 lbf.ft/s).
Determine:
a. the overall efficiency of the turbine-generator.
b. the mechanical efficiency of the turbine.
Answer:
a) 75%
b) 82%
Explanation:
Assumptions:
[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]
Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]
Conversions:
[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]
Analysis:
Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:
[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]
Then;
[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]
gh = 0.491 kJ/kg
[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]
= 1559 kW
Therefore; the overall efficiency is:
[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]
[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]
= 0.75
= 75%
b) mechanical efficiency of the turbine:
[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]
thus;
[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]
In Databrawl, what team is most powerful?
A) Virus
B) Malware
C) Firewall security
D) Programs
Answer:
I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.
Explanation:
Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temperature, although lightly-crosslinked NR can partially crystallize at room temperature when stretched. Apparently, Tm is elevated upon stretching which allows for crystallization above the Tm of the unstretched polymer. Explain.
Answer:
Explanation:
Crystalline melting temperature (Tm) is termed as the temperature required for a crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).
Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.
For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.
The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.
The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:
1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.
2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.
3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.
Required:
Determine the normal stress σ in the steel tube.
Answer:
Explanation:
From the information given:
[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]
The total load is distributed across both the rod and tube:
[tex]P = P_1+P_2 --- (1)[/tex]
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
[tex]\delta_1=\delta_2[/tex]
[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]
[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]
[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]
[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]
[tex]P_2 = 6.6212 \ P_1[/tex]
Replace [tex]P_2[/tex] into equation (1)
[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]
Finally, to determine the normal stress in aluminum rod:
[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]
[tex]\sigma_1 = - 23.523 \ MPa}[/tex]
Thus, the normal stress = 23.523 MPa in compression.
The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and usingit topreheat the water entering the boiler. In this process,liquid water at 50oC and 1000 kPa is mixed withsuperheated steam at 200oC and 1000 kPa. If the plantoperators want to produce a saturated liquid at 1000kPa, what ratio of mass flow rates of water andsuperheatedsteam are required
Answer:
Explanation:
This is Answer....
How does distribution add value to goods and services being sold,
including intellectual property?
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Explanation:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.
Explanation:
hope it helps <33
Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.
Answer:
Explanation:
The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.
From the given information:
The elastic modulus is:
[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]
Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.
The shear yield strength by using von Mises criteria is estimated as;
[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]
The shear yield strength by using Tresca criteria is:
[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]
Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Answer:
Attached below is the derived next state equations
Explanation:
Attached below is the derived next state equations
used for the solution of the given problem.
electrical engineering
Answer:
Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.
Explanation: