The stiffness matrix can be defined as the matrix of material stiffness constants, which is a crucial mechanical material property for calculating mechanical structures' rigidity, elasticity, and strength.
The stiffness matrix for a [0/60/-60] glass/epoxy laminate will be discussed in this article.In structural mechanics, the stiffness matrix of a structure describes how much force is required to deform the structure under a given load. It is a critical property in the mechanics of materials, and it is used to calculate the strength, rigidity, and elasticity of a material.
The stiffness matrix for a [0/60/-60] glass/epoxy laminate is calculated using the properties of glass/epoxy unidirectional lamina from Table 2.2, assuming the lamina thickness is 0.005 m. The reduced stiffness matrix is first determined for the lamina, and it is then rotated to the global coordinate system to obtain the stiffness matrix for the lamina. Finally, the A, B, and D stiffness matrices are obtained using the stiffness matrix for the lamina.
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Soru No 3
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B) I am sorry for interrupting
C) You have a point there
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four subjects civil engineers need to study
Answer:
A civil engineering degree program applies mathematics and physical science to solve specific, real-world problems in commerce and industry. A strong civil engineering program typically emphasizes the practical use of geometry, trigonometry, and calculus in conjunction with physics, material science, and chemistry.
Explanation:
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Producer gas is fuel gas that is manufactured by blowing a coke or coal with air and steam simultaneously. it mainly consists of carbon monoxide, hydrogen, as well as substantial amounts of nitrogen. the caloric value of the producer gas is low. such a producer gas contains 28% co, 3.5% co2, 0.5% o2 and 68% n2. 100 kg of this gas is burned with 20% excess air. if the combustion is only 90% complete, determine the following: (a) the composition of the flue gas (b) the weight of the gaseous products (c) if the excess air increased to 40%, whether the combustion reaction in completed? justify your anaswer.
The flue gas resulting from burning producer gas with 20% excess air will contain carbon dioxide, nitrogen, oxygen, hydrogen, and water vapor.
If the combustion is only 90% complete, then the flue gas will contain more carbon dioxide and oxygen than if the reaction was 100% complete.
The general equation for burning producer gas is:
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O
]
Where C represents carbon, H represents hydrogen and O represents oxygen.
To calculate the composition of the flue gas, we need to know the amount of oxygen and fuel burned and the percentage of combustion completion.
In this case, we know that 100 kg of producer gas and 20% excess air were burned and the combustion was only 90% complete. We can then calculate the total amount of oxygen and fuel burned.
Amount of fuel = 100 kg
Amount of O2 = 100 kg x 20% excess air = 20 kg
Total mass of fuel and O2 = 120 kg
Now we need to calculate the amount of oxygen and fuel that is not combusted. This can be done by multiplying the total mass of fuel and O2 by the percentage of combustion completion.
Amount of oxygen and fuel not combusted = 120 kg x 10% = 12 kg
Now we can calculate the amount of oxygen and fuel combusted.
Amount of oxygen and fuel combusted = 120 kg - 12 kg = 108 kg
We can then calculate the amount of each gas produced by the combustion of the producer gas.
Amount of CO2 = 108 kg x (x/100) = 108 x 0.28 = 30.24 kg
Amount of H2O = 108 kg x (y/200) = 108 x 0.035 = 3.78 kg
Amount of O2 = 108 kg x (1/4) = 27 kg
Amount of N2 = 108 kg x (68/100) = 73.44 kg
Therefore, the composition of the flue gas is 30.24 kg of CO2, 3.78 kg of H2O, 27 kg of O2, and 73.44 kg of N2.
If the excess air is increased to 40%, then the total mass of fuel and O2 will be 140 kg and the amount of oxygen and fuel not combusted will be 14 kg. This means that the combustion reaction will not be complete and there will still be some unburned fuel and oxygen in the flue gas.
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A closed thermodynamic cycle consists of two isobaric and two isochoric processes. The processes operate between 0.43 MPa and 4.3 MPa and 0.28 m3 and 2.9 m3. Determine the total work (with appropriate sign) for one complete cycle, when the cycle operates in the clockwise direction. Provide your result in MJ.
Answer:
Explanation:
For first isobaric process (expansion with constant pressure):
\(W_{1}=P_{1}(V_{2}-V_{1})=(0.43\times 10^{6})(2.9-0.28)=1.1266 \times 10^{6} J=1.1266 MJ(+) (Expansion)\)
Now, for the first isochoric: \(W_{2}=0\), and \(Q_{2}=\Delta U=U_{3}-U_{2}\)
For the second isobaric process (compression with constant pressure):
\(W_{3}=P_{2}(V_{1}-V_{2})=(4.3\times 10^{6})(0.28-2.9) =-11.266\times 10^{6}J=-11.266 MJ (Compression)\)
For the last isochoric: \(W_{4}=0, Q_{4}=\Delta U=U_{4}-U_{3}\)
So, the total work per cycle: \(W=W_{1}+W_{2}+W_{3}+W_{4}=1.1266+0-11.266+0=-10.1394 MJ\)
A cylindrical rod of copper originally 16.0 mm in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked yield strength in excess of 250 MPa and a ductility of at least 12% EL are desired. Furthermore, the final diameter must be 11.3 mm. Explain how this may be accomplished:
(1) show why it cannotbe realized by single drawing and (2) suggest proper procedures to fulfill the requirements.You will need to use following figuresto solve the problem.
Solution :
Given data:
Diameter of the copper cylindrical rod = 16 mm
Yield strength = 250 MPa
Calculating the percent cold work
\($\text{Percentage Cold Work} = \frac{\pi\left(\frac{d_0}{2}\right)^2-\pi\left(\frac{d_d}{2}\right)^2}{\pi\left(\frac{d_0}{2}}\right)^2} \times 100$\)
\($ = \frac{\pi\left(\frac{16}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{16}{2}}\right)^2} \times 100$\)
= 50% CW
Therefore, at \(50\% \ CW\), the yield strength of copper will be of the order of 330 MPa.
The ductility will be 4% elongation (EL).
Rather than performing drawing in single operation, we draw some of the fraction of total deformation, and then the anneal them to recrystallize and also finally w do cold work on the material for the second time to achieve its final diameter, ductility and yield strength.
\(21\% \ CW\)is required for a yield strength of \(250 \ MPa\). Similarly, a maximum of \(23\% \ CW\) is required for\(12\% \ EL\).
The average of the two values is \(22\% \ CW\). To achieve both the \(\text{specified yield strength and ductility,}\) the copper should be deformed to\(22\% \ CW\). The\(\text{ final diameter}\) after the first drawing and the initial diameter for the second drawing is \($d_0'$\) , then
\($22\% \ CW = \frac{\pi\left(\frac{d_0'}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{d_0'}{2}}\right)^2} \times 100$\)
\($d_0'=\frac{11.3}{\sqrt{1-\frac{22\% \ CW}{100}}}$\)
\($d_0'=12.8 \ mm$\)
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? (c) In bullet format compare and contrast the expected mechanical behavior of hypoeutectoid and hypereutectoid steels in terms of: (i) Yield strength (ii) Ductility (iii) Hardness (iv) Tensile strength (d) If you want to choose an alloy to make a knife or ax blade would you recommend a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your recommendation in 1-3 bullet points. (e) If you wanted a steel that was easy to machine to make a die to press powders or stamp a softer metal, would you choose a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your choice in 1-3 bullets.
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
(a)
Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.
Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.
(b)
The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.
Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite. The carbon concentration for both ferrites is 0.022 wt% C.
(c)
(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.
(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.
(iii) Hardness: hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.
(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.
(d)
I would recommend hypereutectoid steel alloy to make a knife or ax blade
1- Hardness is required at the surface of the blades.
2- Ductility is not needed for such application.
3- Due to constant impact, the material will not easily yield to stress.
(e)
I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.
1- hypoeutectoid steel alloys have high machinability, hence better productivity
2- It will be used on softer metals, hence its fitness for the application
3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.
Explanation:
See all together above
The primary energy source for the controller in a typical control system is either brainlythe primary energy source for the controller in a typical control system is either
Answer:
a pneumatic or electric power
Explanation:
The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."
This is because a typical control system has majorly four elements which include the following:
1. Sensor: this calculates the controlled variable
2. Controller: this receives and process inputs from the sensor to the controlled device as output
3. Controlled device: this tweak the controlled variable
4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power
(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are held constant at 300oC and 100oC, respectively.(b) Determine the heat loss per hour if the cross-sectional area of the sheet is 0.25 m2.(c) What will be the heat loss per hour if a sheet of soda-lime glass is used instead
Answer:
do the wam wam
Explanation:
The heat flux is =1038kW/m² , the heat lost per hour is =259.5 kW, the heat lost per hour using a sheet of soda- lime glass.
Calculation of heat fluxThe thickness of steel( t) = 10mm = 10× 10^-³m
The temperature difference on both sides = 300-100
∆T = 200°C
But the formula for heat flux = q = k∆T/t
Where K = thermal conductivity for steel = 51.9W/mK.
Substitute the variables into the formula for heat flux;
q = 51.9 × 200/10 × 10-³
q = 10380 × 10³/10
q = 10380000/10
q = 1038000 W/m² = 1038kW/m²
To calculate the heat lost per hour if the cross sectional area is = 0.25 m2 use the formula q × A
= 1038kW/m² × 0.25 m2
= 259.5 kW.
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What is the first step in the decision-making process?
Answer: identify your decision. relize you need to make a decision.
Explanation:
Given a 250Ω strain gage with a gage factor of 1 which is mounted to a metal bar 0.6m long. The bar is stretched under a tension force and the resistance changes to 251.4 Ω. How much was the bar stretched? _ _mm (Answer in mm to 2 decimal places) What is the length of the bar after it is stretched?
Answer:
the bar was stretched by \(\mathbf{\Delta L = 3.36 \ mm}\)
the length of the after it was stretched is \(\mathbf{L_{new} = 603.36 \ mm}\)
Explanation:
From the information given:
The strain gauge resistance R = 250 Ω
The gauge factor = 1
The original length L = 0.6 m = 600 mm
After the bar is being stretched under tension force;
the new resistance \(R_{new} = 251.42\)
The gauge factor \(G = \dfrac{\Delta R/R}{\Delta L /L }\)
where;
\(\Delta R = R_{new} - R\) and \(\Delta L = L_{new} - L\)
ΔR = 251.4 - 250
ΔR = 1.4 Ω
\(\Delta L = L_{new} - L\)
\(L_{new} = L + L (\dfrac{\Delta R/R}{G})\)
\(L_{new} = 0.6 + 0.6 (\dfrac{\Delta 1.4/250}{1})\)
\(L_{new} = 0.60336 \ m\)
\(\mathbf{L_{new} = 603.36 \ mm}\)
Thus, the length of the after it was stretched is \(\mathbf{L_{new} = 603.36 \ mm}\)
Thus, the bar was stretched by \(\Delta L = L_{new} - L\)
\(\Delta L = (603.36 - 600) \ mm\)
\(\mathbf{\Delta L = 3.36 \ mm}\)
What is your creative solution for clean, sustainable, and accessible energy?
True or false?Inventory-risk costs are a form of carrying costs.
or...................
Answer:
i believe it is true
Explanation:
Listed following are a series of statements that each make a claim. Classify these as either testable by accepted methods of science or non- testable by accepted methods of science. Be sure to note that this question does not ask whether a statement would pass or fail a test; it only asks whether it is testable in principle. View Available Hint(s) Reset Help Hurricane Katrina was an act of God Mars once had liquid water on its surface People bom under the sign of Sagittarius are twice as likely to be teachers as anyone else Vince Young is the greatest quarterback of all time. Bacteria acquire resistance to antibiotics through changes in their DNA Earth orbits the Sun every 365.25 days. There will be a solar eclipse nextTuesday at 11 am testable by science not testable by science
The statements that are testable by accepted methods of science are:
- "Mars once had liquid water on its surface"
- "Bacteria acquire resistance to antibiotics through changes in their DNA"
- "Earth orbits the Sun every 365.25 days"
- "There will be a solar eclipse next Tuesday at 11 am"
The statements that are not testable by accepted methods of science are:
- "Hurricane Katrina was an act of God"
- "People born under the sign of Sagittarius are twice as likely to be teachers as anyone else"
- "Vince Young is the greatest quarterback of all time"
The testable statements can be investigated and proven or disproven through the use of accepted scientific methods, such as observation, experimentation, and data analysis. The non-testable statements, on the other hand, cannot be proven or disproven through these methods because they are based on subjective beliefs or opinions.
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What is the purpose of OSHA's typical minimal lockout procedure? The typical minimal lockout procedure:
Before employees perform any servicing or maintenance where the unexpected energization or start-up of the machine or equipment or release of stored energy could cause injury, it shall be used to ensure that the machine or equipment is stopped, isolated from all potentially hazardous energy sources, and locked out.
Title 29 Code of Federal Regulations (CFR) Part 1910.147, the OSHA standard for The Control of Hazardous Energy (Lockout/Tagout), covers the practices and procedures required to shut down machinery or equipment in order to prevent the release of hazardous energy while workers perform servicing and maintenance.
This procedure must be followed to guarantee that all energy sources that could be hazardous have been shut off and that the machinery, apparatus, or system that has to be maintained has been rendered inoperable by lockout or similar measures.
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On a freeway, drivers should look farther ahead than they normally would on a city street:
a. because it takes 500 feet to stop a vehicle on a freeway.
b. in order to spot potential hazards and dangers ahead.
c. because there are multiple lanes.
The looking farther ahead is a crucial practice for maintaining situational awareness and promoting safe driving on freeways.
Why should drivers look farther ahead on a freeway compared to a city street?Looking farther ahead on a freeway is important because it allows drivers to anticipate and react to potential hazards and dangers in a timely manner.
Freeways generally have higher speeds and multiple lanes, which increases the complexity of the driving environment compared to city streets.
By looking farther ahead, drivers can identify any upcoming obstacles, changes in traffic patterns, or potential hazards such as accidents, road debris, or merging vehicles.
This extended visibility gives drivers more time to adjust their speed, change lanes, or take other necessary actions to ensure their safety and the safety of others on the freeway.
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The most common type of pressure gauge is the 1. Piston 2. Linkage 3. Bourdon Tube 4. 5. Temperature
Answer:
Bourdon Tube gauge
Explanation:
The most popular type of pressure gauge in several countries is the Bourdon pressure tube gauge, that is used to determine medium and high loads. Bourdon tube will measure pressures ranging between 600 mbar - 4,000 bar. While the inner pressure is greater than the exterior pressure, the tube pushes forward, and vise versa.
Employees cannot be held legally responsible for an environmental violation.
Answer:
It does not take into consideration what the responsible party knew about the law or regulation they violated. Environmental criminal liability is triggered through some level of intent.
In regards to whether an employee can be held legally liable for environmental violations, this statement is False.
Environmental ViolationsThese are offences against existing environmental laws. Apply to those who commit it and those who willfully ignored the crime.This means that if an employee notices that a company is responsible for environmental violations, and does nothing about it, they could be held legally responsible for the violation as well.
In conclusion, this is true.
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A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à 2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à 3: Constant volume cooling to p3=5 bar Process 3à 4: Constant pressure heating Process 4à 1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.
Answer:
\(W_{net} = - 1223 kJ\)
Explanation:
State 1:
\(P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg\)
State 2:
\(P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg\)
State 3:
\(P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263\)
\(u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg\)
State 4:
\(P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg\)
Calculate the network done for the cycle
\(W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ\)
The least value of the extraterrestrial solar radiation is in ed 2.00 Select one: on O a. Autumn O b. Spring Oc Constant throughout the year O d. Summer O e. Winter Apyranometer with a shading ring to measure diffuse radiation at location with latitude 0° on Fe a reading of 100 W/m2 What is the adjusted reading of this pyranometer?
The least value of extraterrestrial solar radiation is constant throughout the year. Therefore, the answer is option (c) Constant throughout the year. The extraterrestrial solar radiation is the amount of solar radiation received from the sun at the top of the Earth's atmosphere, the solar constant (Isc). The adjusted reading of the pyranometer is 82 W/m2.
The least value of extraterrestrial solar radiation is constant throughout the year. A pyranometer is an instrument used for measuring the solar irradiance (total energy flux from the Sun) that falls upon a planar surface. A pyranometer with a shading ring is used to measure diffuse radiation at a location with a latitude of 0°.If a pyranometer with a shading ring is used to measure diffuse radiation at a location with a latitude of 0° and it has a reading of 100 W/m2, the adjusted reading of this pyranometer will be as follows:
Let’s assume the correction factor of a pyranometer with a shading ring is 0.18. Therefore the adjusted reading of pyranometer is calculated as: Adjusted reading = (1 - 0.18) × reading= 0.82 × 100= 82 (W/m2).
Therefore, the adjusted reading of the pyranometer is 82 W/m2.
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14.13 compare thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating and (b) according to possible molecular structures.\
Thermoplastics become softer when heated and firmer when they cool. however when they are heated, thermosetting polymers become harder.
a. The thermoset polymer can only be heated once; once heated, it cannot be reheated to allow for further molding. This polymer becomes stronger when heated. The chemical composition of thermoplastic polymers, on the other hand, is unaffected by repeated heating and remolding.
b. In contrast to thermoplastic polymers, which have a linear and flexible structure that allows for molding and reheating, thermoset polymers have strong covalent cross connections in their molecular structure.
The main distinction between the two is that thermoset, a material that becomes stronger when heated, cannot be remolded or heated once it has been formed, whereas thermoplastics may be warmed, remolded, and cooled as needed without resulting in any chemical changes.
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Where are drivers most likely to encounter distracted pedestrians?
Answer: Intersections.
Explanation:
A driver is most likely to encounter distracted pedestrians at playgrounds, school areas, and intersections.
Drivers are most likely to encounter distracted pedestrians in urban areas, particularly near busy intersections and shopping centers. Pedestrians may be chatting on their phones, listening to music, or engaged in other activities that take their attention away from the road. As pedestrians become more and more absorbed in their smartphones, they may become complacent in watching out for passing cars or crossed streets, leading to many cases of distracted pedestrians crossing roads at odd or dangerous times. Drivers must, therefore, always remain vigilant to the potential of hitting a distracted pedestrian in urban areas.
Hence, a driver is most likely to encounter distracted pedestrians at playgrounds, school areas, and intersections.
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A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in a pressure drop of 0.183 lb/in? for 1.73 mi. What is a pipe diameter in inches if the mass rate of flow is 7000 lb/h?
From the information, let us list the parameters given to solve for the diameter of the pipe.
Given that:
the specific gravity of the liquid (SG) = 2.6the density of the liquid = (S.G × density of water) = 2.6 × 1000 kg/m³= 2600 kg/m³
Using the standard conversion rates:
The viscosity = 2.0 cP = 0.002 kg/m.sThe pressure drop ΔP = 0.183 lbf/in²Since 1 lbf/in² = 6894.76 N/m²
∴
The pressure drop ΔP = 0.183 lbf/in² = 1261.74 N/m²The length of the pipe = 1.73 mi = 2784.165 mMass flowrate = 7000 lb/h = 0.882 kg/sFrom the given information, let's start by determining the volumetric flow rate of the liquid in the pipe:
\(\mathbf{The \ volumetric \ flow \ rate \ ( Q) = \dfrac{mass \ flow \ rate}{density \ of \ the \ liquid}}\)
\(\mathbf{Q = \dfrac{0.882 \ kg/s}{2600 \ kg/m^3}}\)
Q = 0.00034 m³/s
In a cylindrical flow pipe, using the formula for the pressure drop to estimated the pipe diameter, we have:
\(\mathsf{\dfrac{\Delta P}{\rho g}= \dfrac{8fLQ^2}{\pi^2gd^5}} --- (1)\)
where (f) can be computed as;
\(f = \dfrac{64}{\dfrac{\rho vd}{\mu}}\)
\(f = \dfrac{64}{\dfrac{\rho Qd}{A\mu}}\)
replacing the values from the above-listed parameters, we have:
\(f = \dfrac{64}{\dfrac{2600 \times 0.00034 \times d}{\dfrac{\pi}{4}(d)^2 \times 0.002}}\)
\(f = \dfrac{64}{2600 \times 0.00034 \times d} \times \dfrac{\dfrac{\pi}{4}(d)^2 \times 0.002}{1}\)
f = 0.1137d
From equation (1), Recall that:
\(\mathsf{\dfrac{\Delta P}{\rho g}= \dfrac{8fLQ^2}{\pi^2gd^5}}\)
\(\mathsf{\dfrac{\Delta P}{\rho }= \dfrac{8fLQ^2}{\pi^2d^5}}\)
Replacing the values, we have;
\(\mathsf{\dfrac{1261.74}{2600}= \dfrac{8\times 0.1173(d) \times (2784) \times (0.00034)^2}{\pi^2(d)^5}}\)
\(\mathsf{0.48528= \dfrac{2.966\times 10^{-5}}{(d)^4}}\)
\(\mathsf{d^4= \dfrac{2.966\times 10^{-5}}{0.48528}}\)
\(\mathsf{d^4= 6.11193538 \times 10^{-5}}\)
\(\mathbf{d = \sqrt[4]{ 6.11193538 \times 10^{-5}}}\)
d = 0.0884 m
d = 88.4 mm
since 1 mm = 0.0393701 inch
∴
88.4 mm will be = 3.48 inches
Therefore, we can conclude that the diameter of the pipe = 3.48 inches
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Two glass tubes diameter 2 and 4 mm respectively, are attached to the side of a water tank
to measure the level inside the tank,(0 = 0.074N/m). Use this information to express the
capillary rise in the tube in the form h= mr + c where m and c are constants and r is the
tube radius and hence determine the ideal tube diameter.
Cellular network towers are configured in such a way so that they avoid what type of problem?
Answer:
They are configured this way to avoid overlap issues.
Explanation:
using data in Appendix A ,calculate the number of atoms in 1 tonne of iron
\(1.0783*10^{28}(atoms).\)
Explanation:Since I don't have access to "Appendix A", I'll solve the problem using data from the periodic table.
1. Determine the molar mass of iron.
According to the periodic table, the molar mass of iron is:
55.845g/mole.
2. Convert 1 tonne to grams.\(1(tonne)*1000=1000kg\\1000kg*1000=1000000g=10^6g\)
3. Apply rule of 3.\(55.845g\) ----------- \(1 mole\)
\(10^6g\) ----------- \(x\)
\(x=\frac{10^6*1}{55.845}=17906.7061(moles)\)
4. Determine the amount of atoms.Considering that there are, approximately, \(6.022*10^{23}\) atoms in a mole of any element, apply another rule of 3.
1 mole --------------------- \(6.022*10^{23}(atoms)\)
\(17906.7061(moles)\) --------------------- x
\(x=\frac{17906.7061*6.022*10^{23}}{1}=1.0783*10^{28}\).
PLEASE HELP give an example of how scientists, technologists, engineers, and mathematicians may work together when inventing or innovating a new product. Include the role of each profession.
An example of them working together would be something like building a robot. Technologists would work on this project bc they would code the robot. Engineers would work on this bc they would design the interior. A mathematician would work in this bc a mathematician would calculate the angels of when a robot picks something up or when it turns. A scientist would work on this bc the scientist know about physics. You need physics so u know how much force to apply, how much energy to use, how much mass u need, and how much charge u need.
Explanation:
A runner ran a 600 m race in 2 min 17 seconds. Calculate his average speed in m/sec.
you can think of the ____ in a method declaration as a funnel into the method.
You can think of the parameters in a method declaration as a funnel into the method.
You can think of the parameters in a method declaration as a funnel into the method. Parameters are the inputs that a method takes in, and they define the type and name of the data that the method needs to execute its logic. When a method is called with arguments, the arguments are passed as values to the corresponding parameters in the method declaration. The parameters act as a funnel that channels the input data into the method's logic. The number and type of parameters in a method declaration determine how the method can be used and what data it can handle. By defining parameters with specific data types, a method can ensure that it receives the correct data and can operate on it correctly. The parameter names can also provide information about the purpose of the data being passed in, making the method more readable and understandable to other developers. When a method is called with arguments, the values of the arguments are copied into the corresponding parameters in the method declaration. The method then uses these parameter values to perform its logic and compute a result, which is returned to the caller. By defining parameters in a method declaration, developers can create modular and reusable code that can be used in different contexts and with different input data.
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A 0.5 m rod of some material elongates 0.1 mm on heating from 20 to 112°C. Determine the value of the linear coefficient of thermal expansion [in (°C)-1] for this material.
Answer:linear coefficient of thermal expansion of the rod =2.17 x10^-6(°C)^-1
Explanation:
The Linear thermal expansion is calculated as
ΔL = αLΔT,
where ΔL = change in length L,
ΔT = the change in temperature,
α is the coefficient of linear expansion
Change in length ΔL= 0.1 mm =0.1/1000= 0.0001m
change in temperature,ΔT= Final - Initial temperature
= (112- 20)°C=92°C
Solving, we have that
ΔL = αLΔT
α=ΔL/LΔT
= 0.0001 / 0.5 x 92
=0.0001/46
=2.17 x10^-6 / °C
The linear coefficient of thermal expansion of the rod, α =2.17 x10^-6(°C)^-1
what is the dimensions of beta
Answer:
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