Answer:
A) The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C[/tex]
B) The total charge that accumulates at the upper terminal is 50μC
C) If the current is stopped at t = 0.5 ms then total charge stored on upper terminal is 35.67μC
Explanation:
Given that:
[tex]I(t)=0 \quad \quad \quad \quad \quad t<0\\\\I(t)= 125e^{-2500t} \quad t\geq 0[/tex]
A) The charge that accumulates at the upper terminal for t > 0:
As we know
[tex]q(t)=\int {I(t)} \, dt[/tex]
for t > 0
[tex]q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]\\\\[/tex]
The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C---(1)[/tex]
B) The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C[/tex]
C) If the current is stopped at t = 0.5 ms then
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C[/tex]
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.
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.
Neon gas enters an insulated mixing chamber at 300 K, 1 bar with a mass flow rate of 1 kg/s. A second steam of carbon monoxide enters at 575 K, 1 bar at a mass flow rate of 0.5 kg/s. Assuming the mixture exits at 1 bar and the specific heat ratios for Neon and CO are constant, determine:
(a) The molar composition of the exiting mixture.
(b) The temperature of the exiting mixture, in K.
(c) The rate of entropy production, in kW/K.
Answer:
a) the molar fraction of neon at the exit is
xₙ= 0.735
and carbon monoxide
xₓ = 0.265
b) the final temperature is
T = 410.55 K
c) the rate of entropy production is
ΔS = 1.83 KW/K
Explanation:
denoting n for neon and x for carbon monoxide:
a) from a mass balance, the molar fraction of neon at the exit is:
outflow mass neon=inflow mass neon
xₙ = outflow mass neon/ (total outflow of mass) = inflow mass neon/ (total outflow of mass) = (1 kg/seg / 20.18 kg/kmol) / (1 kg/seg / 20.18 kg/kmol + 0.5 kg/seg / 28.01 kg/kmol) = 0.735
and the one of carbon monoxide is
xₓ = 1-xₙ = 1-0.735 = 0.265
b) from the first law of thermodynamics applied to an open system, then
Q - Wo = ΔH + ΔK + ΔV
where
Q= heat flow to the chamber = 0 ( insulated)
Wo= external work to the chamber = 0 ( there is no propeller to mix)
ΔH = variation of enthalpy
ΔK = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)
ΔV = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)
therefore
ΔH = 0 → H₂ - H₁ = 0 → H₂=H₁
if we assume ideal has behaviour of neon and carbon monoxide, then
H₁ = H ₙ₁ + H ₓ₁ = mₙ₁*cpₙ*Tn + mₓ₁*cpₓ*Tc
H₂ = (m ₙ+mₓ)*cp*T
for an ideal gas mixture
cp = ∑ cpi xi
therefore
mₙ*cpₙ*Tₙ + mₓ*cpₓ*Tₙ₁ = (m ₙ+mₓ)*∑ cpi xi*T
mₙ/(m ₙ+mₓ)*cpₙ*Tₙ + mₓ/(m ₙ+mₓ)*cpₓ*Tₓ = T ∑ cpi xi
xₙ* cpₙ*Tₙ +xₓ*cpₓ*Tₓ = T*( xₙ* cpₙ+xₓ*cpₓ)
T= [xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₙ₁ +[xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₓ₁
denoting
rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)
and
rₓ= xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)
T= rₙ *Tₙ +rₓ*Tₓ
for an neon , we can approximate its cv through the cv for an monoatomic ideal gas
cvₙ= 3/2 R , R= ideal gas constant=8.314 J/mol K=
since also for an ideal gas: cpₙ - cvₙ = R → cpₙ = 5/2 R
for the carbon monoxide , we can approximate its cv through the cv for an diatomic ideal gas
cvₓ= 7/2 R → cpₓ = 9/2 R
replacing values
rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ) = xₙ₁*5/2 R/ ( xₙ₁*5/2 R+xₓ*9/2 R) =
xₙ₁*5/(xₙ₁*5 + xₓ*9) = 5xₙ₁/(5 + 4*xₓ) = 5*0.735/(5+ 4*0.265) =0.598
since
rₙ + rₓ =1 → rₓ = 1-rₙ = 1- 0.598 = 0.402
then
T = rₙ *Tₙ +rₓ*Tₓ = 0.598 * 300 K + 0.402 * 575 K = 410.55 K
c) since there is no entropy changes due to heat transfer , the only change in entropy is due to the mixing process
since for a pure gas mixing process
ΔS = n*Cp* ln T₂/T₁ -n*R ln (P₂/P₁)
but P₂=P₁ (P=pressure)
ΔS = n*Cp* ln T₂/T₁ = n*Cp*ln T₂ - n*Cp*ln T₁ = S₂-S₁
for a gas mixture as end product
ΔS = (nₓ+nₙ)*Cp*ln T - (nₓ*Cpₓ*ln Tₓ + nₙ*Cpₙ*ln Tₙ)
ΔS = nₙ*Cpₙ ( (nₓ+nₙ)*Cp/[nₙ*Cpₙ]* ln T - ( nₓ*Cpₓ/ (nₙ*Cpₙ) *ln Tₓ + ln Tₙ)
ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)]
replacing values ,
ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)] = (1 kg/s/ 20.18*10 kg/kmol)* 5/2* 8.314 kJ/kmol K *[ 1/0.598 * ln 410.55 K-( 0.402/0.598 *ln 575 K + ln 300K)]
= 1.83 KW/K
"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]
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 2100 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. Determine
(a) the net power developed, in MW
(b) the rate of heat addition in the combustor, in MW
(c) the thermal efficiency of the cycle
To determine the net power developed, the rate of heat addition, and the thermal efficiency for an air-standard Brayton cycle with specified inlet conditions, compressor pressure ratio, maximum cycle temperature, and efficiencies, detailed thermodynamic calculations considering the given parameters are required.
Explanation:The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, commonly used in jet engines and gas turbine engines. The cycle involves air entering a compressor, being compressed, then heated in a combustor before expanding through a turbine, and finally being released to the environment. This question asks to calculate the net power developed, the rate of heat addition in the combustor, and the thermal efficiency of the cycle given specific conditions and efficiencies of the compressor and turbine.
Unfortunately, without the necessary thermodynamic equations and properties of air provided, along with the specific steps and intermediate states of the cycle, it is impossible to perform the detailed calculations needed to answer this question precisely. Typically, solving this problem would involve using the thermodynamic relations for an ideal gas, the isentropic process equations for the compressor and turbine, and the definitions of isentropic efficiency. The conservation of energy principle would be applied to find the net work output and heat added, and subsequently, the thermal efficiency would be calculated.
As such, this question requires a detailed understanding of thermodynamics, specifically the principles governing the Brayton cycle and the equations for calculating work done, heat transfer, and efficiency in thermodynamic cycles.
Letm(t) = sinc2(t) andmh(t) be its Hilbert transform. Denote the upper side band modulatedwaveform asxUSB(t) =Am(t) cos(2000πt)−Amh(t) sin(2000πt)
(a) Sketch the Fourier transforms ofm(t) andmh(t).
(b) Sketch the Fourier transformX(f) of the complex baseband ofxUSB(t) with respectto carrier frequency 1000.
Answer:
The Hilbert transform we use in mathematics it usually apply on a function consisting of a real variable and produces another function of a real variable.
Explanation:
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.
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In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. (a) Using the data provided, determine the volume fraction of fibers required.
Answer:
a) 0.26
b) 1077 MPa
Explanation:
a) The following equation can be used to determine the volume fraction:
[tex]\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}[/tex]
[tex]\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}[/tex]
[tex]32.3 = \frac{260V_f}{2.8-2.8V_f}[/tex]
[tex]V_f = 0.26[/tex]
b) Tensile strength can be found by using the following equation:
[tex]\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077[/tex] MPa
Chairs and bolsters are used to:
a. keep formwork from moving during concrete placement.
b. maintain proper spacing between members during installation of precast concrete members.
c. help during posttensioning of concrete slabs.
d. splice reinforcing bars that require greater length.
e. support reinforcing bars in beams and slabs, prior to concrete placement.
Answer:e. support reinforcing bars in beams and slabs, prior to concrete placement.
Explanation: Chairs are used for construction of foundations,large steel supports,deck constructions and for underground works. Chairs can be for rebar support ( rebar support chairs),post tension chairs.
Bolsters are usually long made of metallic materials mainly used as support for construction of different infrastructures like roads it helps to ensure the concretes and other construction materials stay firmly connected. Image 1 is a metal be chair
Image 2 is a metal bolster
A series RC circuit contains a 0.5 microfarad capacitor and a 500 ohm resistor. The circuit is operating at a frequency of 2,000 Hz. What is the impedance of the circuit?
(A) 500 ohms
(B) 809 ohms
(C) 2,006 ohms
(D) 525 ohms
Answer:
[tex]\left | Z_{in}} \right | = 524.72 \approx 525[/tex]ohm
Explanation:
Given data:
capacitor = 0.5 micro farad
resistor = 500 ohm
Frequency = 2000 Hz
Impedance of any circuit is calculated by using following equation
[tex]Z_{in} = Resistor - \frac{j}{\omega_c}[/tex]
[tex] = 500 - \frac{j}{2\pi*2000*0.5 \mu}[/tex]
=500 - j 159.155
[tex]\left | Z_{in}} \right | = \sqrt{(500^2 + 159.155^2)}[/tex]
[tex]\left | Z_{in}} \right | = 524.72 \approx 525[/tex]ohm
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.
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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:
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.
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Depending on the environmental demands, there are different types of organizational structures, including __________.
Answer:
Functional (Centralized) Organization Divisional Organization Team-Based Organization Product-Based Organization Modular OrganizationMatrix OrganizationExplanation:
Organization structure:
refers to the idea of how people are supposed to work and coordinate in an organization to maintain a healthy and effective work environment.
There are various types of organizational structures which depends on several factors. There is no single best organization structure. Each structure has its own advantages and disadvantages. In order to select a structure the organization's vision, mission, culture values, goals are to be identified first.
A room in a building has the following characteristics: Sensible heat gain: 90,000 Btu/hr, Latent heat gain: 40,000 Btu/hr, Supply air: 3,600 cfm @ 55 deg. F db
Air ventilation: 700 cfm, The room is kept at 78 deg. F db and 45% rh. Outdoor air is 92 deg. F db 76 deg. F wb.
What is the wet bulb wb air temperature entering EAT and wb temperature of air leaving the cooling coils LAT in the room?
[LAT = 55 deg. F db, given, need to calculate EAT db, LAT wb. The EAT db of the air entering the cooling coils is the mixed temperature = [(room temp db)(2,900) cfm + (outside air temp db)(700)]/3,600 cfm. Find EAT wb and h of EAT from chart. Find h of outside air from chart. Then Q for outside air = (4.5)(?h)(cfm) where cfm is outside air ventilation = about 26,800 Btu/hr (show this), then total Q = (sensible heat gain) + (latent heat gain) + (outside air heat gain) = about 157,000 Btu/hr (show this), then total Q = (4.5)(?h)(total cfm), then find (?h) from total Q, then find h of leaving air from ?h, then find wb temp of leaving air LAT wb from chart. We get EAT = 80.7 deg. F db/66.2 deg. F wb, LAT = 51.7 deg. F wb.]
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.
Technician A says that load-carrying ball joints should always have some play in them. Technician B says that follower ball joints should never have some play in them. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
The right answer is d.
Explanation:
The amount of play that a ball joint can have (if any) depends on the application of the joint itself and the suspension system. For each vehicle, the manufacturer must warn wich is the maximum tolerance of play. For some load-carrying ball joints, the amount of play should be minimum or null (like in some heavy-duty vehicles). Therefore technician A is wrong. In most cases in urban vehicles, it is acceptable some level of play for follower ball joints (these joints are normally unloaded). Therefore the technician B is also wrong. That means neither technicians A nor B statements are correct.
There are different functions of Technician. The two Technician are wrong so Neither A nor B is the right answer.
In Ball Joint Inspection , using the old rule of thumb which state that ball joints if more than .050 inches of play are worn and it does will not hold true for all vehicles.
There are some ball joints which does not have visible play while others can hold up to . 250 inch or more of play and still be work well.
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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
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.
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.
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.
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horizontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe.
Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
[tex]Q = \dfrac{\Delta P \pi D^4}{128 \mu L}[/tex]
[tex]Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}[/tex]
Q = 5.06 x 10⁻⁸ m³/s
The nozzle bends the flow from vertically upward to 30 degrees with the horizontal as it discharges the water (at 20 degrees C) to the atmosphere at V = 125 ft/s.
The volume of water within the nozzle itself (above the flange) is 100 lb.
Find the horizontal and vertical forces that must be applied to the flange (by the pipe below it) to hold it in place.Area of flange = 1.0 ft^2Area of nozzle = 0.50 ft^2Volume of area above flange = 1.8 ft^3Vertical height from flange to nozzle = 2 ft
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.
Answer:
Fy = -11267.294 lbf
Explanation:
given data
nozzle flow = 30 degrees
discharges the water = 20 degrees C
volume of water = 100 lb
Area of flange = 1.0 ft²
Area of nozzle = 0.50 ft²
Volume of area flange = 1.8 ft³
Vertical height flange to nozzle = 2 ft
solution
we will apply here continuity equation that is
A1 × V1 = A2 × V2 .............1
put here value and we get volume V1 that is
V1 = [tex]\frac{0.5\times 125}{1}[/tex]
V1 = 62.5 ft/s
and
now we will apply here Bernoulli equation that is
[tex]\frac{p1}{\gamma 1} + \frac{V1^2}{2g} + z1 = \frac{p2}{\gamma 2} + \frac{V2^2}{2g} + z2[/tex] .............................2
put here value and we will get
p1 = 0 + [tex]\frac{62.4}{2\times 32.2}(125^2 - 62.5^2) + 62.4 (2)[/tex]
p1 = 11479.614 psf
so here moment in y will be
∑ Fy = m [ (Vo)y - (Vi)x ]
so here we get
p1 ×A1 + Fy - Wn - Ww = [tex]\rho[/tex] Q [ V2 × sin30 - V1 ]
put here value and we get Fy
1147.614 × 1 + Fy - 100 - (62.4 × 1.8) = (1.94) × (0.5 ×130) × (125sin30 - 62.5)
solve it we get
Fy = -11267.294 lbf
Water vapor at 6 MPa, 500°C enters a turbine operating at steady state and expands to 20 kPa. The mass flow rate is 3 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible.
Determine:
(a) the isentropic turbine efficiency and
(b) the rate of entropy production within the turbine, in kW/K.
Answer:
a)75.8%
b)2.517KW/K
Explanation:
Hello!
To solve this problem follow the steps below
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
1. use thermodynamic tables to find the following variables.
a.enthalpy and entropy at the turbine entrance
h1=Enthalpy(Water;T=500;P=6000)
=3422KJ/kg
s1=Entropy(Water;T=500;P=6000)
=6.881KJ/kgK
b. enthalpy and ideal entropy at the turbine outlet
h2i=Enthalpy(Water;s=6.881;P=20)
=2267KJ/kg
s2i=s1=6.881KJ/kgK
2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output
W=m(h1-h2)
h2=h1-W/m
h2r=3422-2626/3=2546.6KJ/kg
3.
find efficiency with the following equation
[tex]eficiency=\frac{h1-h2r}{h1-h2i}[/tex]
[tex]\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%[/tex]
4.
find the real entropy at the turbine exit
s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK
5.Finally find the entropy generated, using the following equation
ΔS=m(s2-s1)=(3kg/s)(7.72 KJ/kgK-6.881 KJ/kgK)=2.517KW/K
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.
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 engineer is considering time of convergence in a new Layer 3 environment design. Which two attributes must be considered? (Choose two)
A.Addition of a valid forwarding path
B.Loss of a valid forwarding path
C.SPT timers update
D.OSPF database updates
E.Forwarding table updates
Answer:
Options B and E
Explanation:
To give sustainable environmental design when considering time of convergence in layer 3, an engineer must consider the loss of a valid forwarding path and the table updates since these will determine whether the design becomes fit or not.
Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.
Answer / Explanation:
On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.
WHILE,
True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.
Lot ABCD between two parallel street lines is 350.00 ft deep and has a 220.00-ft frontage (AB) on one street and a 260.00-ft frontage (CD) on the other. Interior angles at A and B are equal, as are those at C and D. What distances AE and BF should be laid off by a surveyor to divide the lot into two equal areas by means of a line EF parallel to AB?
Answer:
240.83 ft
Explanation:
The distances AE and BF will be equal = 182.58 ft
The area of the lot will be the product of the depth and the average of the two frontages
= ( 350 * (220 + 260)/2) = 84000 ft
Half of the area becomes 42000 ft
<A = arctan (20/350) = 3. 3°
Hence 42000=h/2*(220 + 220 + 2h*tan3.3)
Solving, we obtain h= 182.28ft
EF = 220 + (2*182.28*tan 3.3)
= 240.83ft
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