Answer:
hello your question is incomplete here is the complete question
. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?
Answer : elements of the agreement
offeracceptancecapacity certaintyconsiderationintention to create legal relationExplanation:
Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid
Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.
capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .
certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions
Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement
Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement
an adiabatic compressor receives 1.5 meter cube per second of air at 30 degrees celsius and 101 kpa. The discharge pressure is 505 kpa and the power supplied is 325 kW, what is the discharge temperature
Answer:
The discharge temperature is 259.82 K
Explanation:
In this question, we are concerned with calculating the discharge temperature
Please check attachment for complete solution
Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes provide slip resistance. Who is correct
Given:
We have given two statements.
Statement 1: Proper footwear may include both leather and steel-toed shoes.
Statement 2: Leather-soled shoes provide slip resistance.
Find:
Which statement is true.
Solution:
A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.
Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.
Leather-soled shoes don't provide slop resistance.
Therefore, both the Technicians are wrong.
From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.
We are given the statements made by both technicians;
Technician A: Proper footwear may include both leather and steel-toed shoes.
Technician B: Leather-soled shoes provide slip resistance.
Now, they are talking about safety shoes to be worn in workshops.
A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.
This is the type of shoe that should be worn by technicians in the workshop.
Thus, Technician A is wrong because proper footwear does not include leather shoes.
Similarly, technician B is also wrong because leather shoes are not safety shoes.
Read more about slip resistant shoes at; https://brainly.com/question/17411739
One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 120 in., the height of the first bounce of the ball must be in the range 55 in. <= h <= 60 in. Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement. Any ideas on this?
Answer:
At temperature is and relative humidity is 86% therefore, the humidity ratio is 0.0223 and the specific volume is 14.289
At temperature is and Relative humidity is 40% therefore, the humidity ratio is 0.0066 and the specific volume is 13.535.
To calculate the mass of air can be calculated as follows:
Now , we going to calculate the volume,
The time which is required to fill the cistern can be calculated as follows:
Now, putting the value in above formula we get,
Therefore, the hours required to fill the cistern is 4.65 hours.
Explanation:
True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity
Answer:
An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration
Explanation:
While having a discussion, Technician A says that you should never install undersized tires on a vehicle. The vehicle will be lower, and the speedometer will no longer be accurate. Technician B says that the increase in engine rpm for a given speed will result in a decrease in fuel economy. Who is correct
Answer:
Both technician A and technician B are correct.
Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.
It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where it is traveling is 60°F. Determine the drag on the fin when the submarine is traveling at 2.5 ft/s.
Answer:
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
Explanation:
Given that:
The height of a triangular stabilizing fin on its stern is 1 ft tall
and it length is 2 ft long.
Temperature = 60 °F
The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.
From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.
The diagram can be found in the attached file below.
If we recall ,we know that;
Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]
the density of water ρ = 62.36 lb /ft³
[tex]Re_{max} = \dfrac{Ux}{v}[/tex]
[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]
[tex]Re_{max} = 414078.6749[/tex]
[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵
Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:
[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]
where;
[tex](2-x) dy[/tex] = strip area
[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]
Therefore;
[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]
[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]
Let note that y = 0.5x from what we have in the diagram,
so , x = y/0.5
By applying the rule of integration on both sides, we have:
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]
Let U = (2-2y)
-2dy = du
dy = -du/2
[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]
[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]
[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]
[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]
[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]
[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]
[tex]F_D = 1.071031071 \ lbf[/tex]
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
what are the conditions for sheet generator to build up its voltage?
Answer:
There are six conditions
1. Poles should contain some residual flux.
2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.
3. Resistance of field winding must be less than critical resistance.
4. Speed of prime mover of generator must be above critical speed.
5. Generator must be on load.
6. Brushes must have proper contact with commutators.
Explanation:
An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pressure ratio (rp) for constant volume heat addition, determine the efficiency and other values listed below. The gas constant for air is R = 0.287 kJ/ kg.K
T1= 310K
P1(kpa)= 100
r=11.5
rp =1.95
Required:
a. Determine the specific internal energy (kJ/kg) at state 1.
b. Determine the relative specific volume at state 1.
c. Determine the relative specific volume at state 2.
d. Determine the temperature (K) at state 2.
e. Determine the specific internal energy (kJ/kg) at state 2.
Answer:
A) 222.58 kJ / kg
B) 0.8897 M^3/ kg
c) 0.7737 m^3/kg
D) 746.542 k
E) 536.017 kj/kg
efficiency = 58% ( approximately )
Explanation:
Given Data :
Gas constant (R) = 0.287 kJ/ kg.K
T1 = 310 k
P1 ( Kpa ) = 100
r = 11.5 ( compression ratio )
rp = 1.95 ( pressure ratio )
A ) specific internal energy at state 1
= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg
B) Relative specific volume at state 1
= P1*V1 = R*T1 ( ideal gas equation )
V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3
V1 = 88.97 / 100 = 0.8897 M^3/ kg
C ) relative specific volume at state 2
Applying r ( compression ratio) = V1 / V2
11.5 = 0.8897 / V2
V2 = 0.8897 / 11.5 = 0.7737 m^3/kg
D) The temperature (k) at state 2
since the process is an Isentropic process we will apply the p-v-t relation
[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]
hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k
e) specific internal energy at state 2
= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg
efficiency = output /input = 390.3511 / 667.5448 ≈ 58%
attached is a free hand diagram of an Otto cycle is attached below
how can we prevent chemical hazards in labotary
Answer:
We can prevent it by:
a) By wearing GOOGLES.
b) By wearing our Lab coat.
c) Fire extinguisher should always be present in the lab.
d) Hand Gloves must be worn.
e) No playing in the lab.
f) No touching of things/equipment's e.g bottles, in the lab.
g) No eating/snacking in the lab.
h) Always pay attention, no gisting.
i) Adult/qualified person must be present in the lab with pupils/students.
Explanation:
Hope it helps.
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]
= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s
A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the maximum shear stress should not exceed 100 Mpa. b) Using the diameter, determine the maximum normal stress.
Answer:
a) 42 mm
b) 144.4 MPa
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]
[tex]d^{3}[/tex] = 7.638 x 10^-5
d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm
b) Normal stress = P/A
where A is the area
A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa
Under conditions for which the same roojm temperature is mainteined bt a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) y by considering a room whose air temperature is maintained at 20 Dagree C throughout the year, while the walls of the room are nominally at 27 Dagree C and 14 Dagree C in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of 32 Dagree C throughout the year and to have an emissivity of 0.90. The coefficient associated with heat transfer by natural convection between the person and the room air is approximately 2 W/m2 Middot K.
Answer:
Net heat transfers: during summer = 52.253 W/m^2 during winter = 119.375 w/m^2
Explanation:
Given data :
Room temperature throughout the year = 20⁰c = 293 k
Room temperature during summer = 27⁰c = 300 k
Room temperature during winter = 14⁰c = 287 k
surface temperature of a person throughout the year = 32⁰c = 305 k
coefficient of heat transfer by natural convection (h)= 2 w /m^2 k
emissivity = 0.9
An explanation to the condition of feeling chilled during the winter and comfortable during summer can be explained with the calculation below
The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as
q = hΔt = 2 * ( 305 - 293) = 2 * 12 = 24 w /m^2
also calculate heat transfer through radiation using this formula
[tex]q_{rad} =[/tex] εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]
ε = 0.90,(emissivity) σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)
during summer :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2
therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2
During winter :
[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3
therefore the net heat transfer during the winter
= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2
The re truck is moving forward at a speed of 35 mi /hr and is decelerating at the rate of 10 ft/sec2. Simultaneously, the ladder is being raised and extended. At the instant considered the angle is 30° and is increasing at the constant rate of 10 deg /sec. Also at this instant the extension b of the ladder is 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2. For this instant determine the acceleration of the end A of the ladder (a) with respect to the truck and (b) with respect to the ground.
Answer:
(a). -1.76i + 0.76j ft/s^2.
(b). -10.42i + 5.7j ft/s^2.
Explanation:
So, we are given the following data or parameters in the question/problem above;
=> "The fire truck is moving forward at a speed = 35 mi /hr and is decelerating at the rate = 10 ft/sec2."
=> "At the instant considered the angle = 30° and is increasing at the constant rate = 10 deg /sec. Also at this instant the extension b of the ladder = 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2"
STEP ONE: Convert the angular speed to rad/s and speed to ft/s.
Angular speed => π/18 rad/s
Speed => 35 × 1.46667 = 52.33 ft/s
STEP TWO: determine the coordinate form of the acceleration of the truck.
= - 10(cos 30°)i - sin(30°)j [π/18 × K] × 25i[π/18 × K] × 2i × [π/9 × K] - 1i.
= -10.42i + 5.7j ft/s^2.
PLEASE NOTE: The step TWO above is the solution to the second part (b) of this question which is the with respect to the ground.
STEP THREE:
This, from step two above;
( -10.42i + 5.7j ) - ( acceleration of the truck).
= - 10.42i + 5.7j - - 10(cos 30°)i - sin(30°)j = -1.76i + 0.76j ft/s^2.
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answer:
the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Explanation:
From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
The Critical Stress for a maximum internal crack can be expressed by the formula:
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
where;
[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation
[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa
Y = dimensionless parameter
a = length of the internal crack
given that ;
the maximum internal crack length is 8.6 mm
half length of the internal crack will be 8.6 mm/2 = 4.3mm
half length of the internal crack a = 4.3 × 10⁻³ m
From :
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]
[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]
[tex]Y= \dfrac{26}{13.01748802}[/tex]
[tex]Y=1.99731315[/tex]
[tex]Y \approx 1.997[/tex]
For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
when the length of the internal crack a = 3mm
half length of the internal crack will be 3.0 mm / 2 = 1.5 mm
half length of the internal crack a =1.5 × 10⁻³ m
From;
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]
[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]
[tex]\sigma_c =189.6595506[/tex]
[tex]\sigma_c =[/tex] 189.66 MPa
Thus; the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
You subjected a rod under the cyclic stress with the maximum stress of 200 MPa and minimum stress of 20 MPa. The fatigue limit was determined to be ~100 MPa. How many cycles can this materials sustain before failure?
Answer:
The material will not fail
Explanation:
A rod subjected under cyclic stress will fail if the cyclic stress it is subjected to is a constant maximum value that is above the fatigue limit of the rod. but in this problem the Rod is subjected to a cyclic stress that ranges from 200 MPa(maximum stress) and 20 MPa ( minimum stress). this simply means that at all times the Rod will not experience maximum stress of 200 MPa and its Fatigue limit is also set at ~100 MPa
attached is the diagram showing the cyclic stress the rod is subjected to
Which of these properties generally applies to transfer lines? A. It is difficult to introduce product modifications. B. It is difficult to store products between individual machines. C. The process can run even if one of the machines fails. D. Products can take different paths through the system.
Answer:
It is difficult to introduce products modification
1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
Answer:
Following are the answer to this question:
Explanation:
1)
Following the four operations in the airlines:
Landside operations:
In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.
Airside Operations:
It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.
Billing and invoicing Operations:
This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.
Information management:
In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.
2)
It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.
what's the maximum shear on a 3.0 m beam carrying 10 kN/m?
Answer:
max shear = R = V = 15 kN
Explanation:
given:
load = 10 kn/m
span = 3m
max shear = R = V = wL / 2
max shear = R = V = (10 * 3) / 2
max shear = R = V = 15 kN
A state of stress that occurs at a point on the free surface of the of a solid body is = 50 MPa σ x , =10 MPa σ y , and = −15 MPa xy τ .
a. Evaluate the two principal normal stresses and the one principal shear stress that can be found by coordinate system rotations in the x-y plane and give the coordinate system rotations.
b. Determine the maximum normal stress and the maximum shear stress at this point.
c. Sketch the rotations and stress magnitudes
Answer:
A) 5 MPa , 55 MPa
B) maximum stress = 55 MPa, maximum shear stress = 25 MPa
Explanation:
using the given Data
free surface of a solid body
α[tex]_{x}[/tex] = 50 MPa, α[tex]_{y}[/tex] = 10 MPa , t[tex]_{xy}[/tex] = -15 MPa
attached below is the detailed solution to the question
Answer:
I think its A) 5 MPa , 55 MPa
B) ms = 55 MPa, mss= 25 MPa
α = 50 MPa, α = 10 MPa , t = -15 MPa
g 940 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073
Describe the meaning of the different symbols and abbreviations found on the drawings/documents that they use (such as BS8888, surface finish to be achieved, linear and geometric tolerances, electronic components, weld symbols and profiles, pressure and flow characteristics, torque values, imperial and metric systems of measurement, tolerancing and fixed reference points)
Answer:
Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.
There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.
Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.
Explanation:
An AISI/SAE 4340-A steel rod with the yield strength of 450 MPa, 2.0 m long will be subjected to a tensile force, must have the minimum weight possible, and must behave elastically for this load. The elastic modulus of steel is 207 GPa. What is the engineering strain of the rod
Answer: 0.002174
Explanation:
Given that the
Yield strength rho = 450 MPa
Length = 2 m
Elastic modulus E= 207 GPa
According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə
E = rho/ə
Make ə the subject of formula
ə = rho/ E
ə = (450 × 10^6) / (207 × 10^9)
ə = 2.174 × 10^-3
Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3
Elastic Strain has no S.I Units.
"A three-phase, 60-Hz, 4-pole, 440-V (line-line, rms) induction-motor drive has a full-load (rated) speed of 1750 rpm. The rated torque is 40 Nm. Keeping the air gap flux-density peak constant at its rated value, (a) plot the torque-speed characteristics (the linear portion) for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz. (b) This motor is supplying a load whose torque demand increases linearly with speed, such that it equals the rated torque of the motor at the rated motor speed. Calculate the speeds of operation at the four values of frequency in part (a)."
Answer:
A) slip(s1) = (1800 - 1750) / 1800 = 0.0277
B) 1800 rpm, 1350 rpm, 900 rpm, 450 rpm
Explanation:
Given data
frequency = 60 Hz,
Line - line rms = 440 - V
3 phase induction-motor drive
number of poles = 4
Full-load rated speed = 1750
rated torque = 40 Nm
A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below
first we calculate the rated speed:
Ns = [tex]\frac{120f}{p}[/tex]
f = 60 Hz . p (number of poles) = 4
Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800
full loaded rated value = 1750.
slip(s1) = (1800 - 1750) / 1800 = 0.0277
considering a linearly condition the slip is low
[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]
S1 = 0.0277
f1 = 60 Hz
hence s2 = 0.018 therefore Ns2 = 1500
B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz
for 60 Hz :
Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm
for 45 Hz:
Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm
for 30 Hz:
Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm
for 15 Hz:
Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm
An inventor claims to have developed a heat pump that produces a 200-kW heating effect for a 293 K heatedzone while only using 75 kW of power and a heat source at 273 K. Justify the validity of this claim.
Answer:
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
Explanation:
Heat generated Q = 200 kW
power input W = 75 kW
Temperature of heated region [tex]T_{h}[/tex] = 293 K
Temperature of heat source [tex]T_{c}[/tex] = 273 K
For this engine,
coefficient of performance COP = Q/W = 200/75 = 2.67
The maximum theoretical COP obtainable for a heat pump is given as
COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] = [tex]\frac{293 }{293 - 273 }[/tex] = 14.65
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
Some characteristics of clay products such as (a) density, (b) firing distortion, (c) strength, (d) corrosion resistance, and (e) thermal conductivity are affected by the extent of vitrification. Will they increase or decrease with increasing degree of vitrification?
1. (a) increase (b) decrease (c) increase (d) decrease (e) increase
2. (a) decrease (b) increase (c) increase (d) increase (e) decrease
3. (a) decrease (b) decrease (c) increase (d) decrease (e) decrease
4. (a) increase (b) increase (c) increase (d) increase (e) increase
5. (a) increase (b) decrease (c) decrease (d) increase (e) decrease
Explanation:
1. increase This due to increase in the pore volume.
2.increase . This is due to the fact that more liquid phase will be present at the firing.
3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.
4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.
5. Increase , glass has higher thermal conductivity than the pores it fills.
When you shift your focus, everything you
see is still in perfect focus.
True or false
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
Describe the similarities and differences between circuits with resistors combined in series and circuits with resistors combined in parallel
Answer:
from the below explanation... we can say that, in the series circuit, flowing current remains the same at each part of the circuit. While in parallel circuits, the voltage across two endpoints of the branches is the same as the supplied voltage.
Explanation:
1.
The components in a series circuit are arranged in a single path from one end of supply to another end. However, the multiple components in a parallel circuit are arranged in multiple paths wrt the two end terminals of the battery.
2.
In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit.
3.
In the series circuit, different voltage exists across each component in the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.
4.
A fault in one of the components of the series circuit causes hindrance in the operation of a complete circuit. As against fault in a single component in a parallel network do not hinder the functioning of another part of the circuit.
5.
The detection of a fault in case of a series circuit is difficult, but it is quite easy in parallel circuits.
6.
The equivalent resistance in case of a series circuit is always more than the highest value of resistance in the series connection. While the equivalent resistance in the parallel circuit is always less than any of the individual resistances in parallel combination.
what is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?
Answer:
P=11 kW
Explanation:
Given that
Number of poles= 8
I.E.C. 180L motor frame
From data book , for 8 poles motor at 50 Hz
Speed = 730 rpm
Power factor = 0.75
Efficiency at 100 % load= 89.3 %
Efficiency at 50 % load= 89.1 %
Output power = 11 kW
Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.
P=11 kW
the partner who can lose only what he or she has invested in business is the
Answer:
partner itself
Explanation:
Answer:
limited partner
hope this answer correct :)
A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2. Deflection.
Answer:
1. Slope = 53.3 x 10⁻⁶
2. Deflection = -0.00016m
Explanation:
given:
let L = 4 m (span of cantilever beam)
let w = 300 N/m (distributed load)
let EI =60 MNm² (flexural stiffness)
dy w * L³ 300 x 4³
1. slope = ------- = --------- = ------------------- = 53.3 x 10⁻⁶
dx 6EI 6 x 60x10⁶
wL⁴ 300 x 4⁴
2. Deflection = y = - ----------- = - ------------------ = -0.00016m
8EI 8 x 60x10⁶
therefore the deflection is 0.16mm downwards.