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.
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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
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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
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
QUESTION 6
Which of the following is NOT a resume format?
01. Chronological
O2. Portfolio
3. Functional
04. Combination
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
Forces always act in equal and opposite pairs
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]
How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including the requirements needed?
Answer:
The application of regulations in locations are very important.
Explanation:
The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.
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.
What two units of measurement are used to classify engine sizes?
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:
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
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
Where do greywater pipes generally feed into?
-Vent stack
-Water heater
-Waste stack
-Main supply
Answer:
c Waste stack
Explanation:
The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform
Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
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.