Answer:
- the weight of the gravel is 1183.56 kg/m³
- the weight of sand is 588.98 kg/m³
- the weight of water is 165.784 kg/m³
Explanation:
Given the data in the question;
mixture content of the gravel = 1173 kg/m³
sand content = 582 kg/m³
free water content = 157 kg/m³
moist of gravel = 0.9%
absorption of gravel = 1.5%
moist of sand = 1.2%
absorption of sand = 1.5%
masses of gravel, sand, and water per cubic meter that should be used at the job site ?
To get the weight of moist gravel, we use the following equation;
Weight of moist gravel = content of gravel × ( 1 + moist of gravel )
we substitute
Weight of moist gravel = 1173 × ( 1 + 0.9%)
Weight of moist gravel = 1173 × ( 1 + 0.009)
Weight of moist gravel = 1173 × 1.009
Weight of moist gravel = 1183.56 kg/m³
Therefore, the weight of the gravel is 1183.56 kg/m³
To get weight of the sand, we use the following formula;
Weight of sand = sand content × ( 1 + moist of sand )
we substitute
Weight of sand = 582 × ( 1 + 1.2% )
Weight of sand = 582 × ( 1 + 0.012 )
Weight of sand = 582 × 1.012
Weight of sand = 588.98 kg/m³
Therefore, the weight of sand is 588.98 kg/m³
To get the weight of mix water ; we use the following equation;
[tex]Mix-Water_{weight[/tex] = water content - gravel content( moist of gravel - absorption of gravel) - sand content( moist of sand - absorption of sand )
so we substitute
[tex]Mix-Water_{weight[/tex] = 157 - 1173( 0.9% - 1.5%) - 582( 1.2% - 1.5% )
[tex]Mix-Water_{weight[/tex] = 157 - 1173( 0.009 - 0.015) - 582( 0.012 - 0.015 )
[tex]Mix-Water_{weight[/tex] = 157 - 1173( -0.006 ) - 582( -0.003 )
[tex]Mix-Water_{weight[/tex] = 157 + 7.038 + 1.746
[tex]Mix-Water_{weight[/tex] = 165.784 kg/m³
Therefore, the weight of water is 165.784 kg/m³
The angle grinder disc is turned at speeds that range from 5000 to rpm.
The angle grinder disc is turned at speeds that range from 5000 to 12,000 rpm.
What is an angle grinder?
An angle grinder is a tool for rough grinding and cutting. An angle grinder tool with the correct disc can replace a variety of instruments, making even the most time-consuming and labor-intensive tasks faster and easier.
The term rpm refers to the number of times per minute that something will go around in a circle. The term rpm stands for revolutions per minute.' Both engines were set at 2,500 rpm.
Therefore, the angle grinder disc is spun at speeds ranging from 5000 to 12,000 revolutions per minute.
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Which best describes the similarities and differences between the Engineering and Technology pathway and the Science and Math pathway?
a) Both pathways require workers to use teamwork, leadership, and teaching skills while the Engineering and Technology pathway requires clerical skills.
b) Both pathways require workers to have observational skills while the Engineering and Technology pathway requires CAD knowledge.
c) Both pathways require workers to be creative, while the Science and Math pathway requires specialized knowledge in subject areas like physics or chemistry.
d) Both pathways require workers to have patience and persistence while the Science and Math pathway requires physical stamina, dexterity, and use of technical equipment.
Answer:
Both pathways require workers to be creative, while the Science and Math pathway requires specialized knowledge in subject areas like physics or chemistry.
Answer:
Right B is right?
Explanation:
Which action is effective in preventing the transmission of food borne viruses and bacteria
Freezing prevents bacteria from growing. If food is heated sufficiently, parasites, viruses, and most bacteria are killed.
What is geotechical investigation?
A geotechnical investigation is primarily concerned with testing the physical properties of the soil, to provide recommendations for foundation requirements, excavation stability, drainage and buried concrete design.
Answer:
Geotechnical investigation is: A surface exploration and subsurface exploration of a site.
Explanation:
What does Geotechnical investigation mean?
It's a primarily concerned with testing the physical properties of the soil, to provide recommendations for foundation requirements, excavation stability, drainage and buried concrete design.
Why Geotechnical is important?
It allow's Engineers to evaluate the stability and strength of the ground, including slopes and soil deposits, assess risks such as soil aggressivity to buried concrete, and help to determine what type of foundations and earthworks would be required within a (project) for example.
A catapult has two rubber bands, each with a square cross-section with a width 4 mm and length 300 mm. In use its arms are stretched to three times their original length before release. Assume the the modulus of rubber is 10-3 GPa and that it does not change when the rubber is stretched. How much energy is stored in the catapult just before release?
The energy stored in the two rubber bands of the catapult just before release can be calculated as follows: Area of Rubber Bands = (4 mm) x (300 mm) = 1200 mm2, Modulus of Rubber = 10-3 GPa = 10,000 N/mm2 Strain = (3 x Original Length) / (Original Length) = Energy = (1200 mm2) x (10,000 N/mm2) x (3) = 36,000,000 Nmm or 36,000 J
What is energy?Energy is the capacity of a physical system to do work or produce heat. It is the fundamental source of virtually all movement and change on Earth. In physics, energy is a property of objects which can be transferred to other objects or converted into different forms. Examples of energy include light, electricity, heat, sound, and kinetic energy. Energy can be generated from a variety of sources including fossil fuels, nuclear power, and renewable sources such as wind, solar, and hydropower. Energy plays a crucial role in the economic and social development of societies, enabling the production of goods and services and the delivery of basic services such as heating, cooling, and lighting. The efficient and responsible use of energy is essential to sustainable development.
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The moist weight of 0.2 ft3 of a soil is 23 lb. The moisture content and the specific gravity of the soil solids are determined in the laboratory to be 11% and 2.7, respectively. Calculate the following:
a. Moist unit weight (lb/ft3)
b. Dry unit weight (lb/ft3)
c. Void ratio
d. Porosity
e. Degree of saturation (%)
f. Volume occupied by water (ft3)
In the given question, Moist unit weight = 115lbs/ft³, Dry unit weight = 103.6 lbs/ft³, Void ratio = 0.626, Porosity = 0.385 and Degree of saturation = 47.4%
What is Porosity?Porosity, also known as void fraction, is a measure of the void (i.e., "empty") spaces in a material. It ranges from 0 to 1, or as a percentage, between 0% and 100%, and represents the volume of voids over the total volume. Some tests, strictly speaking, measure the "accessible void," or the total volume of void space that is reachable from the surface (cf. closed-cell foam).
Porosity in a substance or part can be tested in a variety of ways, including through industrial CT scanning. The word "porosity" is used in a variety of disciplines, including pharmaceutics, ceramics, metallurgy, materials, manufacturing, petrophysics, hydrology, earth sciences, soil mechanics, and engineering.
a. Moist unit weight
W/V = 23/0.2 = 115lbs/ft³.
b. Dry unit weight
89℅ of 115 = 103.6 lbs/ft³
c. Void ratio = volume of void/ volume of total sample
= (2.7 × 62.4)/103.6
= 0.626
d. Porosity = 0.626/1+0.626
= 0.385
e. Degree of saturation = 47.4%
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A stall occurs when the smooth airflow over the unmanned airplane`s wing is disrupted, and the lift degenerates rapidly. This is caused when the wing
A stall happens when the smooth airflow over the unmanned airplane`s wing is disrupted, and the lift degenerates rapidly. This is caused when the wing "exceeds its critical angle of attack."
The angle of attack refers to the angle at which the airplane's wing meets the air that is flowing over it. When an airplane actually is taking off, it is lifting the nose up into the air. And, if that nose continues to rise ultimately it reaches a point where the air is not able to smoothly flow over the wing, causing the airplane to drop. And, this is the point when the airplane exceeds its critical angle of attack.
Exceeding the critical angle of attack is known to be a stall. This has no concern with the engine stalling, it just concerns with the wings not producing enough lift to keep the airplane in the air.
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b)
State the essential difference between a plain carbon steel
and an alloy steel
Answer:
Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.
Explanation:
Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.
The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.
For the systems described by the following equations, with the input x (t) and output y(t), determine which of the systems are linear and which are nonlinear. Dy / dt + 2y (t) = x2 (t) dy / dt + 3ty (t) = t2 x (t) 3y (t) + 2 = x (t) dy / dt + y2 (t) = x (t) (dy / dt )2 + 2y (t) = x (t) dy / dt + (sin t) y (t) = dx / dt + 2x (i) dy / dt + 2y (t) = x (t) dx / dt y (t) = f 1 -infinity x (tau) d tau
The first equation is linear, the second equation is nonlinear, the third equation is nonlinear, the fourth equation is nonlinear, the fifth equation is nonlinear, and the sixth equation is nonlinear.
What is equation?
An equation is an expression that contains an equal sign and relates two or more mathematical quantities. It typically consists of constants, variables, and symbols, and can be written in different forms such as standard form, slope-intercept form, and factored form. Equations are used to describe relationships between variables, and they can be used to solve a wide range of problems, from basic arithmetic to advanced calculus and physics. Equations are also used to model real-world phenomena, such as electricity, kinematics, and quantum mechanics. By studying equations and their solutions, we can gain a deeper understanding of the physical world around us.
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Need help with Part 1 – Additional Operations I MATLAB question
gove it to me. I'll do my best
Raw sugar cane is taken into a process to create sugar, which is essentially sucrose. the raw cane is approximately 16% sucrose, 63% water, and the rest fiber by mass. juice from the cane is extracted by passing the cane through a series of crushers. about 5% extra mass of water is added to the sugar cane prior to this step to help in the extraction process. the crushed cane and liquid juice is sent to a filter press that creates a cake that contains 4% of the weight of the cane juice, which has a composition similar to the overall non-fiber content of the raw cane. the filtrate is sent to an evaporator where enough water is evaporated to obtain a pale yellow juice that is 41% water. A series of vacuum processes removes enough water without damaging the sugars until you obtain a solution that is 91% sucrose. At this point, the mixture is fed to a crystallizer to produce a final product of sucrose that is 97.8% crystal. You control the process by measuring the flowrate of the solution into the crystallizer using a manometer that has mercury as the working fluid. The open-ended manometer shows a mercury height difference of 6.3 inches, , while the height of the sucrose solution in the manometer between the mercury and the pipe is 15.3 in.
The flowrate of the mixture into the crystallizer is related to the pressure with the following equation:
q(m^3/s) = 0.0307 m^3/hr atm^1/2 √ Pmanometer
What is the mass of sugar cane being fed to the process with this flowrate, in kg/s?
The mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.
What is flowrate?Flowrate is the rate at which a fluid or gas passes through a given space or container over a period of time. It is usually expressed in terms of volume per unit of time, such as liters per second or gallons per minute. Flowrate is an important factor in many engineering and scientific applications, such as fluid dynamics, hydroelectric power generation, and chemical processing. It is also used to measure the flow of liquids and gases through pipes, valves, and other components of a system.
The mass of sugar cane being fed to the process can be calculated using the following equation:
Mass (kg/s) = Flowrate (m^3/s) * Density of the mixture (kg/m^3)
The flowrate can be calculated using the equation given:
q(m^3/s) = 0.0307 m^3/hr atm^1/2 √ Pmanometer
Therefore, the flowrate is:
q(m^3/s) = 0.0307 m^3/hr * (atm^1/2 * (6.3 in/12 in/ft)^1/2)
q(m^3/s) = 0.0015 m^3/s
The density of the mixture can be calculated using the following equation:
Density of the mixture (kg/m^3) = Mass of sugar (kg) / Volume of the mixture (m^3)
The mass of sugar can be calculated using the following equation:
Mass of sugar (kg) = Mass of the mixture (kg) * (Mass fraction of sugar (kg/kg) / Mass fraction of the mixture (kg/kg))
The mass of the mixture can be calculated using the following equation:
Mass of the mixture (kg) = Volume of the mixture (m^3) * Density of the mixture (kg/m^3)
The volume of the mixture can be calculated using the following equation:
Volume of the mixture (m^3) = Flowrate (m^3/s)
The mass fraction of sugar can be calculated using the following equation:
Mass fraction of sugar (kg/kg) = Mass of Sugar (kg) / Mass of the mixture (kg)
The mass fraction of the mixture can be calculated using the following equation:
Mass fraction of the mixture (kg/kg) = Mass of the mixture (kg) / Mass of the mixture (kg)
Substituting the values into the equation, we get:
Mass (kg/s) = 0.0015 m^3/s * (0.91 kg/m^3)
Mass (kg/s) = 0.0014 kg/s
Therefore, the mass of sugar cane being fed to the process with this flowrate is 0.0014 kg/s.
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Problem 9.10. Which of the following sets of values is closest to iL(0), vC(0).
Problem 9.10. Which of the following sets of values is closest to the energy stored in the capacitor and inductor (in J) at t=0-
Problem 9.10. What is the value of Vmax (integer answer required) in Volts
Problem 9.10. Which of the following is closest to the expression for vc(t) in Volts
In the cicuit of Figure P9.10, suppose ui"(1) = 10 V, R= 10 Ω, C = 0.4 mF, L = 0.25 (a) Compure 0), vo) i(0), and i1(0*)
(b) Compute the energy stored in the inductor and the capacitor at t = 0
(c) Using only energy considerations, compute the maximum value of ), t> 0 (d) Find the analytical expression for vt), and verify the maximum value of v) computed in part (c)
The energy of the magnetic and electric fields that have been stored can differ. A completely charged capacitor is linked in series with an inductor.
A passive-terminal electric device called an inductor, often referred to as a coil, choke, or reactor, stores electricity in a magnetic field when electric current passes through it. An insulated string twisted into a coil is typically included in an inductor. A passive digital component called an inductor can store electrical energy in the form of magnetic electricity. In essence, it uses a conductor that has been twisted into a coil; as electricity enters the coil from left to right, this will produce a magnetic field that rotates counter clockwise. The energy of the magnetic and electric fields that have been stored can differ.
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5 . Which of the following is true about vehicles displaying a diamond-shaped sign that indicates a hazardous load
The following is true about vehicles displaying a diamond-shaped sign that indicates a hazardous load:
They must stop before a cross railroad track.
What is hazardous load?A hazardous load is any load determined by the Minister to be capable of posing a risk to health, safety, or property when transported on a public road. A hazardous load is defined as any load that has been prescribed by the Council in the Gazette as being capable of posing a risk to health, safety, and property when transported along the Regional Trunk Road Network.
A hazardous substance is any substance that is I listed, classified, or regulated in accordance with any environmental law; (ii) any petroleum product or by-product, asbestos-containing material, lead-based paint or plumbing, polychlorinated biphenyls, radioactive materials, or radon; or (iii) any other substance that is the subject of regulatory action by any governmental entity in accordance with any environmental law.
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Why is it so important that you need to know when calculate a bevel gear two (2) different numbers of gear teeth are to be know?
It is important to know the number of teeth on the two gears when calculating a bevel gear because the number of teeth determines the gear ratio, which in turn determines the torque and speed at which the gears will operate.
How to calculate the gear ratio?The gear ratio is calculated by dividing the number of teeth on the driving gear by the number of teeth on the driven gear.
If the number of teeth is not known, it would be impossible to accurately calculate the gear ratio and determine the performance of the gear set.
Additionally, the number of teeth also affects the meshing pattern and the contact ratio between gear teeth, which is also important to the gear's performance.
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Estimate properties and pipe diameter Determine the diameter of a steel pipe that is to carry 2000 gal/min of gasoline with a pressure drop of 5 psi per 100 ft of horizontal pipe. Pressure drop is a function of flow rate, length, diameter, and roughness. Either iterative methods OR equation solvers are necessary to solve implicit problems. Total head is the sum of the pressure, velocity, and elevation. What is the density of gasoline
Answer:
Diameter of pipe is 0.535 ft
Explanation:
see attachment, its works out 1st half
There are initially 500 rabbits (x) and 200 foxes (y) on Farmer Oat’s property near Riça, Jofostan. Use Polymath or MATLAB to plot the concentration of foxes and rabbits as a function of time for a period of up to 500 days. The predator–prey relationships are given by the following set of coupled ordinary differential equations:
dx/dt= k1x-k2x. Y
dy/dt= k3x. Y-k4y
Constant for growth of rabbits k1 = 0. 02
Constant for death of rabbits k2 = 0. 00004/(day × no. Of foxes)
Constant for growth of foxes after eating rabbits k3 = 0. 0004/(day × no. Of rabbits)
Constant for death of foxes k4 = 0. 04 What do your results look like for the case of k3 = 0. 00004/(day × no. Of rabbits) and tfinal = 800 days?
Also, plot the number of foxes versus the number of rabbits. Explain why the curves look
The curves look like a sine wave, with the number of rabbits and foxes both oscillating around a steady-state equilibrium. The presence of the constants for growth and death of both rabbits and foxes.
What is curves?
Curves is an international fitness franchise targeting women. It offers a 30-minute workout that combines strength training, cardio, and stretching. The circuit-style program is designed to help members reach their fitness goals while providing a supportive, non-threatening environment. The program also includes nutrition counseling, wellness education, and support from certified coaches.
This is because the predator-prey relationships in this system are cyclical, with the fox population increasing as the rabbit population increases, and vice versa.
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Tech A says that both caution and danger signs indicate a potentially hazardous situation. Tech B says that an exhaust extraction hose is not needed if the vehicle is only going to run for a few minutes. Who is correct
According to the question of vehicle which is only going to run for a few minutes, tech A is correct.
What is vehicle?
Vehicles are any mode of transportation that can be used on land, sea, or air. They are an important part of modern transportation and include cars, trucks, buses, motorcycles, boats, trains, and airplanes. They provide an efficient way to travel, while also allowing people to transport goods and materials. Vehicles are powered by a variety of sources, including gas and electricity. Advanced technology has allowed for the development of self-driving vehicles, which are becoming increasingly popular.
Caution and danger signs indicate a potentially hazardous situation, and an exhaust extraction hose should be used whenever a vehicle is running, regardless of how long it is running for.
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round your answers to the nearest hundredths if needed -25+5s=8
Answer:
s = 6.6
-25 + 5 × 6.6 = 8
Just let me know if you need any more help.
The section is experiencing a positive bending moment about the z-axis, where Mz = 1.13 kN-m. What is the resulting stress at the top and bottom surfaces? And are they in tension or compression?
The resulting stress at the top and bottom surfaces is 837.5 N/m^2 and -837.5 N/m^2, respectively. The top surface is in compression and the bottom surface is in tension.
What is stress?
Stress in engineering is the amount of force per unit area placed on a material or structure. It can be physical, or it can be psychological. Stress can be positive or negative. Positive stress can cause a material to become stronger, while negative stress can cause it to become weaker or even fail. Stress can also refer to the amount of strain a material must endure before it yields or breaks. Stress is measured in terms of pounds per square inch (psi) or newtons per square meter (N/m2). Stress analysis is important in engineering as it helps to identify potential failure points in a design, or to determine how much a material can handle before it breaks. Stress analysis is used to ensure the safety and integrity of products, structures, and machines.
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A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. On day 5 the DO in the bottle measured 4 mg/L. After 30 days, the DO in the bottle measured 2 mg/L, and after 50 days, the DO in the bottle still measured 2 mg/L. At the beginning of the test, we added a nitrification inhibitor so we can assume that nitrification is not occurring, so only the carbonaceous BOD is being measured.
a) What is the BODs of the wastewater?
b) What is the ultimate carbonaceous BOD?
c) How much BOD remains after 5 days?
d) Based on the data above, estimate the reaction rate constant k (1/day).
Answer:
Explanation:
From the given information:
The BOD of the wastewater can be determined by using the formula:
[tex]BOD _5 = \dfrac{DO_i-DO_f}{\dfrac{V_s}{V_b}}[/tex]
where;
[tex]BOD _5 =[/tex]biochemical oxygen demand = ???
[tex]DO_i=[/tex] initial dissolved oxygen = 9 mg/L
[tex]DO_f=[/tex] final dissolved oxygen = 4 mg/L
[tex]V_b =[/tex] sample of the bottle, which is normally 300 mL
[tex]V_s[/tex] = sample volume = 30 mL
[tex]BOD _5 = \dfrac{9-4}{\dfrac{30}{300}}[/tex]
[tex]BOD _5 = \dfrac{5}{\dfrac{1}{10}}[/tex]
[tex]BOD _5 = 5\times{\dfrac{10}{1}}[/tex]
[tex]\mathbf{BOD _5 = 50 \ mg/L}[/tex]
The ultimate Carbonaceous BOD is estimated from the formula:
[tex]y_t = L_o-L_t \\ \\ y_t = L_o-L_oe^{-kt} \\ \\ y_t = L_o (1-e^{-kt})[/tex]
Making [tex]L_o[/tex] the subject, we have:
[tex]L_o = \dfrac{y_t}{(1-e^{-kt}} \\ \\ L_o = \dfrac{50}{1 - e^{-0.25*5}}[/tex]
[tex]L_o[/tex] = 70 mg/L
c) The BOD left over after five days = [tex]L_oe^{-kt}[/tex]
= [tex]70 \times e^{-0.25 *5}[/tex]
= 20 mg/L
d) The reaction constant rate is estimated as follows:
Recall that:
[tex]\mathbf{BOD _5 = 50 \ mg/L}[/tex]
Since DO measure 2 mg/L;
After 30 days, [tex]BOD_{30} = (9-2) \times 10 = 70 \ mg/L[/tex]
Therefore, the reaction rate constant is:
[tex]\dfrac{50 \ mg/L}{70 \ mg/L} =\dfrac{1-e^{-k*5}}{1-e^{-k*30}} \\ \\ 50 (1-e^{-k*30}) = 70 (1-e^{-k*5}) \\ \\ K = 0.25 /day[/tex]
Tech A says that all hazards can be removed from a shop. Tech B says that you should disconnect an air gun before inspecting it. Who is correct
According to the question of disconnecting an air gun before inspection, tech B is correct.
What is gun?
Gun is a type of weapon that is typically used to fire bullets at a target. It is typically composed of a barrel, a chamber, a firing mechanism, and ammunition. Guns have been used for centuries as a tool for hunting, self-defense, and warfare. Guns are typically categorized by the type of ammunition they use, such as pistol, shotgun, or rifle. Guns can be used to shoot at a variety of targets, such as animals, clay pigeons, and even human targets. Guns are also used in competitive shooting sports. Although guns can be used for a variety of purposes, they have the
Although it is not possible to completely remove all hazards from a shop, it is important to take precautions such as disconnecting an air gun before inspecting it to help reduce the risk of injury.
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It is desired to develop a strong Aluminum-based alloy with a shear strength on the order of G/100 by precipitation hardening, where G is the shear modulus. Calculate the necessary precipitate spacing, and estimate the required percentage of the precipitate phase. Assume that the precipitate phase consists of spherical particles of radius of R=40 nm with centers uniformly distributed in simple cubic lattice. Does it matter whether the precipitates strain the matrix locally around the interface or not?
Answer:
a) L = 50.01 nm
b) 99.98%
Explanation:
Determine the necessary precipitate spacing and estimate the required percentage of precipitate phase
Given that the shear strength = G/100 by precipitation hardening
G = shear modulus
Radius of particles = 40 nm
attached below is the detailed solution
a) precipitate spacing ( L ) = √( 2π / f ) * R₁
= √ ( 2π / 0.00719956 ) * 1.693 nm
hence L = 50.01 nm
b) Determine The percentage of precipitate phase that is required
( precipitation strength / peak strength ) * 100
= ( 18.4583 / 18.4604 ) * 100
= 99.98%
A bar of copper is subjected to a tensile load. The elastic modulus of the copper is 110 GPa, its yield strength is 240 MPa, its tensile strength is 400 MPa, and its Poisson's ratio is 0.340. If the original length of the rod is 250 mm and it has a round cross-section with an original diameter of 9.50 mm, what load (in N) would need to be applied in order for the rod's length to be 250.1 mm
Answer:
The answer is "3,118 N"
Explanation:
Using the hookes laws:
[tex]y=\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}[/tex]
So,
[tex]F=\frac{y \times \Delta L \times A}{L}\\\\[/tex]
[tex]=\frac{110 \times 10^{3} \times 0.1 \times \frac{\pi}{4} \times 9.5^2 }{250}\\\\=\frac{110 \times 10^{3} \times 0.1 \times \frac{3.14}{4} \times 90.25 }{250}\\\\=3,117.235 \ N\\\\=3,118 \ N\\\\[/tex]
In a single-family residence, a line-voltage thermostat is primarily used to control _____.
Answer:
In a single-family residence, a line-voltage thermostat is primarily used to control heating and cooling systems.
5 Systems Modeling
es / SPE(2201 / General / Business System Modelling CAT
2. Business Process Modelling is important to a business due to the following advantages except:
(2 marks)
O a. None of the above
h
O b. Enhances Customization of Business Processes
O c Enhances Competitive advantage
O d. Enhances Process Communication
age
Next pag
Answer: None of the above
Explanation:
Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.
Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.
It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.
Therefore, the answer to the question will be "None of the above".
please i want to paraphrase this paragraph please helppppppppp don't skip!!!!!!
A single direct acting controller used to control space temperature and set up for 72°F setpoint with an 8°F throttling range assigned. The system pressure range is from 3 to 13 psig. Predict the following: (a) Controller PB setting when remote sensors with spans of 50°F, 150°F, and 200°F are used. (b) Controller pressure range. (c) Controller setpoint. (d) Controller output pressure at T = 72°F using the same sensors as in (a). (e) Controller output pressure at T = 76°F. (f) Predict the output pressure of the controller at T = 80°F when th
The controller PB is 0.22, the range of the controller pressure is between 3 to 13 psig, the controller setpoint is at 72°F, and the output pressure at that temperature is 3 psig while the controller output pressure at 76°F is 6 psig
What is Controller SystemsA controller system is a type of feedback system used to regulate the performance of an output device by comparing the output to a desired set point. It uses the difference between the two values (error) to continuously adjust a control input to reduce the error. In simple terms, the controller system monitors, adjusts, and regulates the operations of the output device to achieve a desired result. Common examples of controllers include thermostats, cruise control systems, and PID controllers.
(a) Controller PB setting when remote sensors with spans of 50°F, 150°F, and 200°F are used:
This can be calculated as;
PB = (72°F - 50°F) / (200°F - 50°F) = 0.22
(b) Controller pressure range:
The range of the controller pressure is 3 - 13 psig
(c) Controller setpoint:
The controller set point is 72°F
(d)
The controller output pressure when T = 72°F while using the same sensors as in (a) is 3 psig
(e)
The controller output pressure at T = 76°F is 6.6 psig
(f) To predict the output pressure of the controller when T = 80°F with three remote sensors having spans of 50°F, 150°F, and 200°F are used will be 11.2 psig
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Which of the following is NOT a factor in the quality of digital video? a.- frame rate and resolution c. compression technique b. memory technology in your camera de color and bit depth
The choices that best describe the process of digitization are: Option A: binary-stored video. Option C converts analog video to digital video.
Information transformation into a digital format is referred to as digitization. Be aware that in this format, data is organized into discrete units of data called bits. As a result, the following options best represent the digitization process: Option A: binary-stored video. Option C converts analog video to digital video. This is due to the fact that digital video is a key technology for both video conferencing and video messaging in addition to being a key technology for digital television. This was demonstrated by its use in messaging apps for confidential conversations and for hosting virtual conferences for business meetings amongst personnel.
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A 400-ft equal tangent sag vertical curve has its PVC at station 100 00 and elevation 450 ft. The initial grade is -4.0% and the final grade is 2.5%. Determine the elevation of the lowest point of the curve g
The elevation of lowest point of the curve is 445.077 ft.
What is elevation?Height above or below the mean sea level is referred to as elevation. A map's elevation can be depicted either by labelling the precise elevations of specific points or by using contour lines, which link points of the same elevation. Topographic maps are depicted as having elevations.
Calculate rate of change of the curve as below:
[tex]$$\begin{aligned}r & =\frac{g_2-g_1}{L} \\& =\frac{2.5-(-4.0)}{\left(\frac{400}{100}\right)} \\& =1.625 \%\end{aligned}$$[/tex]
Calculate distance from PC to the lowest point as below:
[tex]$$\begin{aligned}X & =\frac{-g_1}{r} \\& =\frac{-(-4.0 \%)}{1.625 \%} \\& =2.462 \mathrm{ft}\end{aligned}$$[/tex]
Calculate the elevation of the lowest point of the curve as below:
[tex]$$\begin{aligned}Y & =Y_{P C}+g_1 X+\frac{r}{2} X^2 \\& =450\mathrm{ft}+(-4.0 \times 2.462)+\frac{1.625}{2}(2.462)^2 \\& =450\mathrm{ft}-9.848 \mathrm{ft}+4.925 \mathrm{ft} \\& =\mathbf{445.077} \mathrm{ft}\end{aligned}$$[/tex]
Thus, the elevation of lowest point of the curve is 445.077ft.
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Tech A says that one approved way to clean dust off brakes is with compressed air. Tech B says that some auto parts may contain asbestos. Who is correct
Tech B is correct it's a verified fact that some auto parts may contain asbestos.
Do auto parts contain asbestos?
Asbestos has been used in a wide range of automotive products, including brakes, clutches, hood liners, gaskets, heat shields, and many others.
Drum and disc brakes were traditionally made with 35% to 60% asbestos. It is still legal in the United States to sell asbestos-containing auto parts, and many brake and clutch parts contain up to 35% asbestos.
Contaminated parts have been found in vehicles of all types, including cars, trucks, motorcycles, buses, trains, and military vehicles. Elevators and other types of transportation machinery also used asbestos brakes.
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