The Rankine passive earth pressure on a 12-ft high retaining wall with backfill of granular soil having an internal angle of friction of 30° and a unit weight of 125 pcf is calculated to be 500 psf.
Explanation:The question relates to calculating the Rankine passive earth pressure on a retaining wall that is 12-ft high with backfill of granular soil. The internal angle of friction (φ) provided is 30°, and the unit weight (γ) of soil is 125 pcf (pounds per cubic foot). First, to calculate the passive earth pressure (Ψp), we use the Rankine theory formula: Ψp = γh [1 - sin(φ)]/[1 + sin(φ)], where h is the height of the wall. Substituting the given values, Ψp = 125 * 12 * [1 - sin(30°)]/[1 + sin(30°)]. Since sin(30°) = 0.5, the calculation simplifies to: Ψp = 125 * 12 * [1 - 0.5]/[1 + 0.5], which further simplifies to Ψp = 125 * 12 * 0.5/1.5. Therefore, the Rankine passive earth pressure on the wall amounts to Ψp = 500 psf (pounds per square foot).
Write a short poem that express your feelings towards one or more terminologies/jargons in electromagnetics (such as divergence, curl, gradient, Maxwell’s equations, electric flux, magnetic flux, boundary conditions, wave equation, time-harmonic form, phasor form, instantaneous form, complex exponential, standing wave, intrinsic impedance, propagation constant, phase shift, polarization, etc.), using the format of rhyming couplets. You can assume that a poetic license has been granted. State any other assumptions. Please write at least four sentences.
Answer / Explanation:
The divergence in my complex undulating movement assumed a sinusoidal wave movement that it began to form harmonic melodies even though the standing wave form seem static.
Never knew my exponential increment were causing intrinsic impedance causing a constant phase shift allowing me cross my boundary conditions.
I now therefore concur that the curl in every angle, gradient and wave pattern i assume are as a result of constant propagation of or the bid to exceed my boundary condition.
Now i feel caged in this electromagnetic field force while i dance i phasor form. Alas let it be known that every boundary met was polarized by me.
If you wanted to create a vehicle that requires a very low tractive effort in order to overcome resistance forces, what parameter would you change and how would you change it?
What are the challenges of designing a roadway based on a single design vehicle?
Answer:
The problems faced in designing a roadway based on a single design vary. The challenges are described in the explanation section below. The first part of the question "If you wanted to create a vehicle that requires a very low tractive effort in order to overcome resistance forces, what parameter would you change and how would you change it" is also answered in italics below
Explanation:
When you are designing the roadway geometry you commonly use a design vehicle that gives you the worst scenario viable in that road, which is translated in terms of speed, meaning that after all of the topography problems, localisation of essential regions that can´t be modified (urban regions or national parks for example), ambiental effect and economic issues are handled, the horizontal and vertical geometry goes to depend of the speed value that is going to regulate camber, radius of curvature, acceleration and deceleration longitudes and others.
Now, in terms of infrastructure or pavement engineering, after handling climate challenges, the worst scenario is about weight, that has to do with the total equivalent load that is going to resist that pavement till first failure.So you see, that the challenges of using a single design vehicle in the geometry and infrastructure design of a roadway has to do with the selection of the worst scenario possible.
On the other hand, knowing the LOS of the roadway, which is crucial for knowing how is going to be transit management, depends on evaluating the level of service in all kinds of scenarios, that means different types of vehicles speed and acceleration configurations. So using a single vehicle, in this case, is not going to give you the best real knowledge of the road.
For the first question, designing a vehicle with a low tractive effort means that, from the civil engineering point of view, in order to enhance that vehicle, vertical geometry is key for solving resistance forces, because you can design a road that brings much more speed and power to the vehicle. Friction is also a parameter that reduces speed in vehicles but is necessary for a safe travel so you have see if it is logic to reduce this parameter in order to enhance the design vehicle behavior.
If you want to bring more tractive effort to the vehicle; handling resistance forces caused by rolling resistance or air resistance, drawbar pull and other should be key for enhancing the vehicle behavior. If you want a more detailed answer in this topic you should ask a mechanical engineering expert
Consider each statement below and determine which are correct concerning dietary fiber. Select all that apply. View Available Hint(s) Select all that apply. Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. Consuming a high-fiber diet most likely promotes the health of the digestive system. Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day.
Answer:
Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. This is correct if you are consuming less than 25-30gms of fiber per day. exceding this limit won't be beneficial.
Consuming a high-fiber diet most likely promotes the health of the digestive system. This is correct. Fibers are important for the digestive system´s health, especially for intestines and colon.
Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. This is Incorrect. Fiber is absorbed and digested at a much slower rate than sugar or starch.
Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. This is incorrect. Consuming a diet high in dietary fiber would decrease the LDL.
Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day. This is incorrect. The data obtained by the University of California San Francisco said that currently the amount of fiber intake by Americans adults is about 15g a day, which is half the recommended amount.
Answer:
Consuming a high-fiber diet most likely promotes the health of the digestive system.
Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight.
Explanation:
True/False
An anemometer displays wind direction, wind speed, altitude and type of precipitation
Answer:FALSE
Explanation:
Soap is a very interesting chemical. We even discussed it on the discussion board. How does it work, exactly?
Answer:
Saponification is a process in which soap is formed from mixtures of sodium or potassium salts of fatty acids. These fatty acids are reacted high temperature of At 80°C-100°C with alkali to extract salt. These alkali can be sodium hydroxide or potassium hydroxide.
Soap has both polar (ionic) and non polar molecules due to which it has characteristics of both hydrophilic substance (having tendency to mix with water) and hydrophobic substance (have tendency to mix with oils) and due to this nature it can act as an emulsifier.
An emulsifier has tendency to diffuse one liquid into another liquid which is incapable of mixing with homogeneous liquid like water.
Cleansing action takes place due to presence of ionic and non-polar properties at same time, in combination with solubility principles. The ionic end of soap molecule is the salt end. It is hydrophilic (water soluble) in nature. The non-polar end cotains long hydrocarbon chains and is hydrophobic (water repellent).
When immiscible liquids like grease or oil mixed with soap water, non polar end (hydrophobic end) absorbs the dirt which means the soap will form the micelles and trap the dirt in it. As micelles is soluble in water it will remove the dirt with it.
Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum fill for that conduit is 40%?
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Conduit fill is a term used in electrical engineering to describe the maximum amount of space that can be occupied by wires in a conduit. In this case, the percentage of fill exceeds the allowed limit, indicating that the wires cannot be accommodated within the given conduit diameter.
Explanation:Conduit fill is a term used in electrical engineering to describe the maximum amount of space that can be occupied by wires in a conduit. It is important to ensure that the conduit is not overfilled, as this can cause overheating and other safety issues.
In this case, the conduit has a total area of 2.04 square inches and the wires occupy 0.93 square inches. To determine if the maximum fill is exceeded, we can calculate the percentage of fill:
Percentage of fill = (occupied area / total area) * 100
Plugging in the values given:
Calculating the percentage of fill:
Percentage of fill = (0.93 / 2.04) * 100 = 45.59%
Since the calculated percentage of fill (45.59%) is greater than the maximum fill allowed (40%), the 1½" conduit cannot be filled with the given wires within the allowed limit.
Learn more about Conduit fill here:https://brainly.com/question/32553721
#SPJ3
As a newly hired professional engineer your an executive with the firm strongly suggests that you attend a monthly dinner party that is held by an important client and major contractor. These parties are extravagant, paid for by the client and contractor, and attended by the executives from your company, the client, and the contractor. You are already working on one of the projects for this client. You want to please management and feel it would be a good way to learn more about the client and contractor in a casual setting, but have some reservations. Which of the following is allowed by the NCEES Model Rules of Professional Conduct?
A. Accept the invitation, but come up with creative excuses every month to not attend.
B. Attend the parties every 3 months, but not every month.
C. Decline the invitation and explain to your manager that to do otherwise is inappropriate for a registered professional engineer.
D. Attend the monthly parties to demonstrate to management that he understands the importance of pelasing the client
Answer:
C. Decline the invitation and explain to your manager that to do otherwise is inappropriate for a registered professional engineer.
Explanation:NCEES has 3 major rules with some sub sections which helps to uphold the professional conducts of it's members.
The rule that supports this professional conduct is rule II. LICENSEE’S OBLIGATION TO EMPLOYER AND CLIENTS subsection(d)
Subsection d states that a licensee shall not reveal any information about a client,contractor or his employer to a another party except it is required by Law.
Why is the experiment started with an initial load rather than starting at zero load?
Answer: The answer would depend on the experiment but in most cases zero load is not used to ensure that there is no overflow of current.
Explanation: The internal resistance of any typical wire is minute compared to the resistive components that you are likely to utilize in any given experiment. Using I=V/R (Ohms law) , if the resistance is zero then the current that will flow through the wire would be much too high, causing the wire to burn out. The level of currents that the wire can tolerate are provided by the manufacturers. If you exceed those limits, than the wire will surely get permanently damaged. If you have no load at all than this will be known as a short circuit and might damage the source as well. If you have other components in the circuit than the overflow of current might in effect damage them too. This could also damage the bread board if you are using one.
Answer:
Q1. Why is the percentage elongation in 2 inches greater than in 8 inches?
Q2. Describe the Type of Information which may obtained from a character of fracture
Q3. Explain the difference, on the basis of test result between the ultimate strength and The true stress at fracture, cite your specific value to enhance your explanation.
Q4. What is the purpose of using the spherically seated compression plate?
Q5. Why is the experiment started with an initial load rather than starting at zero load?
Following are the answers to each of the questions
Answer 1.
The larger the radius of an elastic material the lower the elongation because the radius is inversely proportional to the percentage of elongation. A wire with a lower radius has a higher elongation than a material with higher radius
∈=PL/AE
Where P is the force applied, L is the original length of the material, A is the cross-sectional area which is πr^2, E is the Elastic Modulus.
Answer 2.
A fracture is the separation of material into two or more pieces usually caused by ultimate load or stress. Fracture strength is the stress required for a material to fail.
TYPES OF FRACTURE
Brittle fracture: brittle materials are materials that are liable to break easily under load for example cast iron. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.
In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated
Ductile fracture: they are material that are capable of being molded in to any shape easily. In ductile fracture a plastic deformation occurs like cracks or rough surface sign will be visible before it attains it ultimate failure state.
In ductile material, the cracks move slowly and in a gradual process together with plastic deformation. Cracks in ductile fracture will not grow unless there is increase in stress applied on the material
Answer 3.
The divergence in the values of true stress and engineering stress occurs only at large loads and displacements; or typically when the specimen is undergoing plastic deformation. That is because most materials have a elastic strain limit close to 0.2%. Note that the values of true stresses and strains are similar to their engineering counterparts below 0.2%.
If a material that strain hardens is tested in tension, the true stress-true strain curve will keep increasing till the specimen fails as the cross section area keeps decreasing continuously (even during necking). However, the slope of engineering stress-strain curve becomes positive and negative before and after necking respectively . This is because necking reduces the material's ability to harden and hence take larger loads. Hence the load starts dropping but remember that one still measures the stress as the load divided by the original cross section area. If the instantaneous cross section area were taken into account, as was done for true stress, the load drop is compensated by the acute reduction in cross section area.
In compression, the necking instability does not occur and this difference is not stark. the true stress-strain curve and engineering stress strain curve under compression more-or-less converge (unless specimen 'barrelling' is very prominent).
Answer 4
They are designed to be placed at the center at a point where loading is exerted in an electromechanical or hydraulic universal test machine. It provides a hardened surface when performing complex compression test in which uniform stress distribution is critical. They can be used to test the variety of materials like concrete, metals, wood and composite.
Answer 5
From ohms law, if the resistance is zero then the current that will flow through the experiment will be too high causing the wire to burn, since the level of current that the wire can tolerate has been produced by the manufacturer and if it exceed this limit, the wire will damage permanently. If there is no load at all this will be known as short circuit and might damage the source as well.
Explanation:
1. Elongation is the increase in length of a material after it has been stressed within the gauge length e.g. elastic materials like wires, spring and other materials that obeys Hooke's law.
2. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.
In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated .
4. Spherically seated compression plate are mounted to the cross head electromechanical or hydraulic universal test machine.
5. The internal resistance of any typical wire is low compared to the resistance of the component that are likely to be used for the experiment.
What do you think might be the explanation they are trying to test with the colored ice cubes?
Answer:
The layering and mixing effect of ice cubes in salt water and fresh water respectively.
Explanation:
This test helps to explain the melting of ice cubes which is more obvious when colored ice cubes are used, as it reveals the layering and mixing effect of ice cubes in salt water and fresh water respectively.
The mixing effect shows that ice cubes melt faster in fresh water (low density) than in salt water(high density).
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 201 MPa in the bolts and 142 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.
Answer:
24.87 mm
Explanation:
The area of the bolt is given by
[tex]A_b=\pi r^{2}[/tex]
Since diameter is 16mm, the radius is 16/2= 8 mm= 0.008 m
Area, [tex]A_b=\pi\times 0.008^{2}=0.000201062 m^{2}[/tex]
For safe design of the bolt, we use stress of 201 Mpa
[tex]\sigma_b=\frac {P}{A_b}[/tex] where P is the load and [tex]\sigma_b[/tex] is normal stress.
Making P the subject then
[tex]P=A_b \sigma_b[/tex]
Substituting the figures given and already calculated area of bolt
[tex]P=201\times 10^{6}\times 0.000201062 m^{2}=40413.4479 N[/tex]
The area of spacer is given by
[tex]\pi (r_o^{2}- r_i^{2})[/tex] where r is radius and the subscripts o and I denote inner and outer respectively
The value of 142 Mpa by default becomes the stress on spacer hence
[tex]\sigma_s=\frac {P}{A_s}[/tex] and making [tex]A_s[/tex] the subject then
[tex]A_s=\frac {P}{\sigma_s}[/tex]
[tex]\pi (r_o^{2}- r_i^{2})=\frac {P}{\sigma_s}[/tex]
Since we already calculated the value of P as 40413.4479 N and inner radius is 8mm= 0.008 m then
[tex]\pi (r_o^{2}- 0.008^{2})=\frac {40413.4479}{142\times 10^{6}}[/tex]
[tex]r_o^{2}=0.000154592[/tex]
[tex]r=\sqrt{0.000154592}=0.012433 m[/tex]
Therefore, [tex]d=2r=2*0.012433=0.024867 m=24.86697 mm\approx 24.87 mm[/tex]
Assuming 800 W / m 2 solar irradiance and a 35 % efficient solar panel, how much roof area should be covered to supply 10 A at 120 V ? A = 4.286 m 2 (within three decimal places) Given an average of 7 h o u r s of sunshine per day and a utility cost of $ 0.22 k W h , how much of the utility cost can such a solar panel save? Ignore the initial cost or any maintenance cost of the solar panels
Answer:
a) 4.286 m² b) $ 55.44/mo
Explanation:
If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.
If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:
A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²
As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:
Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²
b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:
E= 1200 W* 7 hs = 8.4 Kwh
If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:
Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44
For laminar flow over a flat plate the local heat transfer coefficient varies as hx 5 Cx20.5, where x is measured from the leading edge of the plate and C is a constant. Determine the ratio of the average convection heat transfer coefficient over the entire plate of length L to the local convection heat transfer coefficient at the end of the plate (x 5 L).
Answer:
ratio of the average convection heat transfer = 2
Explanation:
local heat transfer expression can be written as
[tex]H_{x}=Cx^{0.5}[/tex] (1)
C= Constant
x=measured from the leading edge of the plate
We need to find local heat transfer coefficient at
x=5L
so for equation 1 it can be written as
[tex]H_{x=5L}=C5L^{0.5}[/tex]
we can find the average heat transfer over entire lenght of plate as
[tex]H=\frac{1}{5L}\int\limits^5_0 {h_{x} } \, dx[/tex] (2)
subsitute [tex]H_{x=5L}=C5L^{0.5}[/tex] in equation 2
[tex]H=\frac{C}{5L}\int\limits^5_0 {x^{-0.5} } \, dx[/tex]
[tex]H=\frac{10C}{L}L^{-0.5}[/tex]
[tex]h=10CL^{-0.5}[/tex]
for ratio of the average convection heat transfer coefficient over the entire plate of length L to the local convection heat transfer coefficient at the end of the plate is given as
[tex]ratio = \frac{H}{H_{x} }[/tex]
Now putting values for H ,Hx and 5L for x
[tex]r=\frac{10CL^{-0.5}}{5CL^{-0.5} }[/tex]
[tex]r=2[/tex]
What is the the force available at the roadway surface to perform work?
Answer:
Tractive force or traction
Explanation:
The main purpose of the tractive forces is to improve the ability to transform the engine's energy into the vehicle's movement. There are several systems that have different qualities and uses. Here we explain how they work and what they are for.
In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:
- At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.
- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.
Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.
Find vC(t) for t ≥ 0 (in V), assuming the switch was open for a long time when closed at t = 0.
Assume the capacitor and inductor are initially uncharged.
(1) 3 – 4e–2t + e–8t
(2) 2 – e–2t + 2e–8t
(3) 6 – 6e–4t + 2e–6t
(4) 6 – 6(1+ t)e–8t
(5) 3 + (1+ 4t)e–2t
(6) 3 + [–cos(4t) + 4sin(2t)]e–2t
(7) 3 + [–4cos(4t) + sin(t)]e–t
(8) 3 + [–2cos(4t) + 4sin(2t)]e–4t
(9) None of the above
Answer:
So, we have to analyze the whole circuit system provided, then we will get to know that its an RL-Circuit along with that we can easily pick the best option from the given, which is mentioned below:
Explanation:
Option (1) 3- 4e-2t+e-8t is the best option to chose from the given options.
The Engineer of Record may exercise control over a project by means of electronic communication devices.a. Trueb. False
Answer:
False
Explanation:
The Engineer of Record for all projects shall be a registered Professional Engineer (P.E.)
Some Detailed Responsibilities of Engineer of Record
The Engineer of record shall attend
Pre-Construction Conferences,Pressure Tests,Lift Station Start-ups,Pre-Final Walk-through andFinal Inspections, etc.All these responsibilties require his physical presence which cannot be achieved through electronic communication.
A rigid tank that contains 4.0 kg of N2 at 25°C and 550 kPa is connected to another rigid tank that contains 6.0 kg of O2 at 25°C and 150 kPa. The valve connecting the two tanks is opened, and the two gases are allowed to mix. If the final mixture temperature is 25°C, determine the volume of each tank and the final mixture pressure.
Answer:
the volume of the oxygen tank is
V ox = 0.844 m³
the volume of the nitrogen tank is
V ni = 2.359 m³
the final pressure is
P = 255. 534 kPa
Explanation:
taking into account that
n= m/M
where
n= number of moles , m = mass, M = molecular weight
then
n oxigen = m ox / M ox = 6.0 kg/(32 gr/mol) *1000gr/kg = 187.5 moles
n nitrogen = m ni / M ni = 4.0 kg/(28 gr/mol) *1000gr/kg = 142.857 moles
from the ideal gas law
P*V=n*R*T
where P= absolute pressure, V= volume occupied by the gas , R= ideal gas constant= 8.314 J/(mol K) , T= absolute temperature
V=n*R*T/P
replacing values
for the oxygen tank , T ox= 25°C= 298 K , P = 550 kPa= 550000 Pa ,
V ox =n*R*T/P = 187.5 mol* 8.314 J/(mol K)* 298 K/ 550000 Pa = 0.844 m³
V ox = 0.844 m³
for the nitrogen tank, T ni= 298 K , P = 150000 Pa
V ni =n*R*T/P = 142.857 mol* 8.314 J/(mol K)* 298 K/ 150000 Pa = 2.359 m³
V ni = 2.359 m³
when the gases mix , they occupy a volume of
V = V ox + V ni = 0.844 m³ + 2.359 m³ = 3.203 m³
and total number of moles of gas of the mixture is
n = n oxigen + n nitrogen = 187.5 moles + 142.857 moles = 330.357 moles
therefore
P*V=n*R*T
P = n*R*T/V
replacing values
P = n*R*T/V = 330.357 mol*8.314 J/(mol K)* 298 K/ 3.203 m³ *1 kPa/1000Pa = 255. 534 kPa
P = 255. 534 kPa
Draw the hierarchy chart and then plan the logic for a program needed by Hometown Bank. The program determines a monthly checking account fee. Input includes an account balance and the number of times the account was overdrawn. The output is the fee, which is 1 percent of the balance minus 5 dollars for each time the account was overdrawn. Use three modules. The main program declares global variables and calls housekeeping, detail, and end-of-job modules. The housekeeping module prompts for and accepts a balances. The detail module prompts for and accepts the number of overdrafts, computes the fee, and displays the result. The end-of-job module displays the message Thanks for using this program.
Answer:
Explanation:
we have to make three functions namely
1 Housekeeping
2 Details
3 end of Job
for variable int
for character string
The hierarchy chart.
As per the chart of the hierarchy, the plan of logic is for making a program that is needed for doing hometown banks and the program determines the checking for an account fee. This includes the account balance and the number of accounts overdrawn. The fee is measured in percentage and each time the account was overdrawn.
Thus the answer is functions include housekeeping, details, and end of job module.
The chart follows the three main parts as housekeeping, details, and job module where the housekeeping is making several deletions, followed by the details and end at the stop.
The details have a outbalance prompt then input balance and then return. The job module has output overdraft then input fee and then output fee and return.
Find out more information about the hierarchy chart.
brainly.com/question/25720264.
Alcohol consumption tends to cause more ___________ behavior.
Answer:
Aggressive behavior
Explanation:
Alcohol consumption tends to cause more Aggressive behavior.
The consumption of alcohol plays a more role in our culture but drinking of too much alcohol can cause drowsiness, vomiting, Upset stomach, slurred speech, heart damage, infertile, numbness lung infections, and many more. Also too much alcohol can cause violence, anger and so on in the society.
A computer chip inside an orbiting satellite heats up during operation. Since there is no air in the satellite the only cooling will be by radiation. The chip is 1.50 cm wide by 2.50 cm long with a negligible thickness; it has an emissivity of 0.600; and dissipates 0.420 Watts during usage (assume that the chip radiates heat evenly from both the top and bottom surfaces). Assuming the inside of the satellite is held at a constant temperature of -100°C; calculate the surface temperature of the chip in °C
Answer:
89.967°C
Explanation:
According to Stefan-Boltzmann's law,
E = μΩA(Ts⁴ - Ti⁴)
Where,
Emissitivity, μ = 0.6,
Stefan-boltzmann's constant, Ω = 5.67 x 10^-8 W/m²K⁴
Surface area, A = 1.5 x 2.5 x 10-⁴ = 3.75 x 10-⁴ m²
Satellite temperature, Ti = -100°C = 173K
Power dissipated, E = 0.42 W
Since we're evaluating the temperature for both the top and bottom surfaces,
E = 2 x μΩA(Ts⁴ - Ti⁴)
0.42 = 2 x 0.6 x 3.75 x 10-⁴ x 5.67 x 10^-8 x (Ts⁴ - 173⁴)
0.0165 x 10^-12 = Ts⁴ - 8.957 x 10^8
Ts⁴ = 173.57 x 10^8
Taking 4th root of both sides,
Ts = 362.97 K
Ts = (362.97 - 273)°C
Ts = 89.967°C
What type of cable communicates binary data by changing the voltage between two ranges?
Copper cables are used to communicate binary data by altering the voltage between two ranges.
What type of cable communicates Binary dataA pair of Twisted copper cables send data through a network by transmitting pulses of electricity that represent binary data.
On this note, to make sure cables are transmitting information in a way that can be understood by the recipient, they follow the Ethernet standards. This twisted pair cables are commonly known as Ethernet cables.
Read more on cable communications:
https://brainly.com/question/7142635
Which one of the following activities is not exempt from licensure pursuant to Chapter 471, F.S.? A person practicing engineering on property owned by him or her. A full time electrical engineer of Progress Energy Corp. A civil engineer employed full time by the U.S. Army Corps of Engineers. An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp. Applicants for licensure with degrees from foreign institutions are required to document "substantial equivalency to ABET criteria to the FBPE. They can do this by: Providing a transcript from their institution to the Board. Providing a notarized certification that they have completed the requisite college credit hours set forth in Rules 61615-20.007(2)(a) thru (2)(d), F.A.C. Getting the evaluation of substantial equivalency from a provider of the service that is approved by the FBPE Passing the Principles & Practice examination In order to verify an applicant's experience, the FBPE: Follows guidelines set forth in Rule 61615 20.002, F.A.C. Relies on information obtained from personal references Requires evidence of employment from employers or supervisors who are employed in the engineering profession. • All of the above.
An Independent consultant working on the design of an electrical distribution system for a corporation is not exempt from licensure. Applicants with foreign degrees must prove ABET equivalency, and the FBPE verifies experience through multiple documentation and procedures.
Explanation:The activity that is not exempt from licensure pursuant to Chapter 471, F.S. is 'An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp.' This scenario indicates an individual offering engineering services to the public, which requires a license. On the other hand, individuals practicing engineering on their own properties, full-time employees of a utility company, and civil engineers employed by the U.S. Army Corps of Engineers typically fall under exemptions to licensure requirements in many states, including Florida.
For applicants with degrees from foreign institutions to document "substantial equivalency to ABET criteria to the FBPE," they can achieve this by providing a transcript from their institution, providing a notarized certification of completed credit hours as per the specified rules, getting the evaluation from a service provider approved by the FBPE, or passing the Principles & Practice examination. To verify an applicant's experience, the FBPE requires various forms of documentation and follows several procedures, including adherence to specific rules, consideration of personal references, and requiring evidence of employment in the engineering profession – essentially, they consider all of the above methods.
In RSA Digital Signature, Suppose Bob wants to send a signed message (x = 4) to Alice. The first steps are exactly t eps are exactly the same as it is done for an RSA encryption: Bob computes his RSA parameters and sends the public key to Alice. We know p = 3, q=11, and bob choose e=3. (Hint: We learn RSA algorithm and key generation in Week 7) (1) What is the public key pair Bob sends to Alice? (2) What is the value of signature s? (3) What is the value of verkpubA(x,s)? Show all intermediate steps clearly.
Answer:
what r u on
Explanation:
"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when network usage = 39.3% ? (Give answer is miliseconds, rounded to one decimal place, without units. So for an answer of 0.10423 seconds you would enter "104.2" without the qu"
Answer:
Explanation:
effective delay = delay when no traffic x [tex]\frac{100}{100- network\r usage}[/tex]
effective delay = [tex]84.1 \times \frac{100}{100-39.3}=138.55024711697ms[/tex]
Why is the percent elongation in 2 inches greater than in 8 inches?
Elongation in a bar is
[tex]\epsilon = \frac{PL}{\pi r^2 E}[/tex]
Where,
P = Applied force
L = Lengthof the specified rod
A = Cross-sectional area [tex](\pi r^2)[/tex]
E = Modulus of Elasticity
Performing a quick analysis we can realize that the larger the radius, the lower the elongation percentage. The radius is inversely proportional to the percentage of elongation. For this reason in a 2in bar the change will be GREATER than that of an 8in bar.
. How are cybersecurity issues similar to and different from privacy issues affecting cybertechnology?
Consider a 4.45-m × 4.45-m × 4.45-m cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of 547 K and 1100 K, respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace. (Given: The view factor from the ceiling to the floor of the furnace F12 = 0.2, σ = 5.67 × 10-8.)
Answer: -1543371.65837 W
= - 1543.372 kW.
Explanation:
Using the equation;
Q= EσA -------------------------------------------------------------------------------------------(1).
Where Q= net rate of radiation heat transfer between the floor and the ceiling of the furnace, σ = Boltzmann's constant, A= area of the cube, E = emissitivity.
Recall that the emissitivity of a black body is equals to one(1).
From the question, the parameters given are; The view factor from the ceiling to the floor of the furnace,F12 = 0.2, σ = 5.67 × 10-8., A= (4.45×4.45) m.
Slotting in the parameters into the equation;
Q= EσA[T(2)^4 - (T(1)^4] ---------------------------------------------------------------------(2).
Therefore, Q= (1)× (5.67×10^-8) × (4.45×4.45) m × [(547)^4 - (1100)^4]
= 0.0000011228 × (89526025681 - 1.4641×10^12).
= 0.0000011228×(-1.374574×10^12)
= -1543371.65837 W
= -1543.372 kW.
An airplane, of overall length 55 ft, is designed to fly at 680 m/s at 8000-m standard altitude. A one-thirtieth-scale model is to be tested in a pressurized helium wind tunnel at 20°C. What is the appropriate tunnel pressure in atm? Even at this (high) pressure, exact dynamic similarity is not achieved. Why?
Answer:
appropriate tunnel pressure is 384.64 ATM
Even at this (high) pressure, exact dynamic similarity is not achieved because the specific heat ratio of helium is approximately 1.66 and is not equal to that of air which is approximately 1.40
Explanation:
It is too long to type here.
Kindly check the attached file for the calculation.
How does a beam deform when it is subjected to an upward load in the middle and supported at both ends? a. Both the top and bottom of the beam compress. b. The top of the beam stretches and the bottom compresses. c. Both the top and bottom of the beam stretch. d. The top of the beam compresses and the bottom stretches.
Answer:b
Explanation:
A beam is subjected to an upward load in the middle and supported in both ends. The upper fibers i.e. a topmost layer of the beam is in tension while bottom fiber i.e. fiber of bottom layer is in compression because they are under compressive force.
Fibers above the neutral axis are in tension and fibers below the neutral axis are in compression.
So option b is correct
2. * Give a scatterplot of this data and comment on the direction, form and strength of this relationship. a. Determine the least-squares estimate equation for this data set. b. Give the r2, comment on what that means. c. Give the residual plot based on the least-squares estimate equation. d. Test if this least-squares estimate equation specify a useful relationship between commuting distance and commuting time.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of refrigerant-134a as the working fluid. Is it known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net work input to the cycle I s 22 kJ. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine he minimum pressure in the cycle.
Answer:
[tex]P_m_i_n= 356.9 KPa[/tex]
Explanation:
Coefficient of performance (COP) of refrigeration cycle is given by:
[tex]COP=\frac{1}{\frac{T_H}{T_L}-1 }[/tex]
We are given:
[tex]T_H=1.2T_L[/tex]
[tex]COP=\frac{1}{1.2-1 }[/tex]
COP= 5
We can also write Coefficient of performance (COP) of refrigeration cycle as:
[tex]COP_R=\frac{Q_L}{W_i_n}[/tex]
Amount of heat absorbed by low temperature reservoir can be found as:
[tex]Q_L=COP_R * W_i_n[/tex]
[tex]Q_L=5 * 22 KJ[/tex]
[tex]Q_L=110 KJ[/tex]
According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:
[tex]Q_H=Q_L + W_i_n[/tex]
[tex]Q_H=110 KJ + 22 KJ[/tex]
[tex]Q_H=132 KJ[/tex]
We are given the mass of 0.96 kg. So,
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]q_H=\frac{132 KJ}{0.96Kg}[/tex]
[tex]q_H=137.5 KJ/Kg[/tex]
Since it is a saturated liquid-vapour mixture [tex]q_H=h_f_g[/tex].
[tex]q_H=h_f_g=137.5 KJ/Kg[/tex]
From Refrigerant 134-a tables [tex]T_H[/tex] at [tex]h_f_g=137.5 KJ/Kg[/tex] is 61.3 C. (We calculated this by interpolation)
Converting [tex]T_H[/tex] from Celsius to Kelvin:
[tex]61.3^{o} C+273 = 334.3^{o} K[/tex]
[tex]T_H= 334.3^{o} K[/tex]
We are given:
[tex]T_H=1.2T_L[/tex]
[tex]T_L=\frac{T_H}{1.2}[/tex]
[tex]T_L=\frac{334.3}{1.2}[/tex]
[tex]T_L=278.58^{o} K[/tex]
Converting [tex]T_L[/tex] from Kelvin to Celsius:
[tex]278.58^{o} K-273 = 5.58^{o} C [/tex]
[tex]T_L= 5.58^{o} C [/tex]
From Refrigerant 134-a tables [tex]P_m_i_n[/tex] at [tex]T_L=5.58^{o} C[/tex] is 356.9 KPa. (We calculated this by interpolation).
[tex]P_m_i_n= 356.9 KPa[/tex]
The minimum pressure in the Carnot refrigeration cycle is 356.9 kPa.
Using,
Coefficient of performance (COP) of the refrigeration cycle. The COP is a measure of how efficient the refrigeration cycle is.
It is given by the formula:
COP = 1 / ([tex]T_H[/tex] /[tex]T_L[/tex] - 1),
where
[tex]T_H[/tex] is the maximum absolute temperature and [tex]T_L[/tex] is the minimum absolute temperature.
We are told that [tex]T_H[/tex] is 1.2 times [tex]T_L[/tex].
COP = 1 / (1.2 - 1).
Simplifying
COP = 5.
We can also express the COP as the ratio of the heat absorbed by the low-temperature reservoir to the net work input.
Using this relationship,
The heat absorbed ([tex]Q_L[/tex]) is equal to 5 times the net work input, which is 5 * 22 kJ
= 110 kJ.
According to the first law of thermodynamics, the heat rejected by the hot reservoir ([tex]Q_H[/tex]) is equal to the sum of the heat absorbed ([tex]Q_L[/tex]) and the net work input (22 kJ).
So,
[tex]Q_H[/tex] = 110 kJ + 22 kJ
= 132 kJ.
Since we are given the mass of the refrigerant (0.96 kg), we can find the heat transfer per unit mass ([tex]q_H[/tex]) using QH/m.
[tex]q_{H}[/tex]= 132 kJ / 0.96 kg
= 137.5 kJ/kg.
Since the refrigerant is in a saturated liquid-vapor mixture state, [tex]q_H[/tex] represents the enthalpy change from liquid to vapor.
By looking up the tables for refrigerant-134a, we can find the corresponding temperature ([tex]T_H[/tex]) at [tex]q_H[/tex] = 137.5 kJ/kg.
After conversion to Kelvin, [tex]T_H[/tex] is found to be 334.3 K.
Using the relationship [tex]T_H[/tex] = 1.2[tex]T_L[/tex],
Dividing [tex]T_H[/tex] by 1.2,
[tex]T_L[/tex] = 334.3 K / 1.2 =
278.58 K.
Converting this back to Celsius, we get [tex]T_L[/tex] = 5.58 °C.
Finally, by looking up the tables for refrigerant-134a, we can find the minimum pressure ([tex]P_{min}[/tex]) at [tex]T_L[/tex] = 5.58 °C.
The value is determined to be 356.9 kPa.
So, the minimum pressure in the cycle is 356.9 kPa.
Know more about Carnot refrigeration cycle,
https://brainly.com/question/29234662
#SPJ6