Eureka is a telephone and Internet-based concierge services that specializes in obtaining things that are hard to find (e.g. Super Bowl tickets, first edition books from the 1500s, Faberge eggs). It currently employs 60 staff members who collectively provide 24-hour coverage (over three shifts). They answer the phones and respond to requests entered on the Eureka! Web site. Much of their work is spent on the phone and on computers searching on the Internet. The company has just leased a new office building and is about to wire it. They have bids from different companies to install (1) 100-Base T network (2) Wi-Fi network What would you recommend
Answers
attached below is the answer
I would recommend installing both a 100-Base T network and a Wi-Fi network in the new office building for Eureka.
Explanation:Based on the information provided, I would recommend installing both a 100-Base T network as well as a Wi-Fi network in the new office building for Eureka.
A 100-Base T network, also known as Ethernet, provides reliable and high-speed wired connections, perfect for the staff members who spend a significant amount of time on their computers searching the Internet. It ensures a stable and fast connection for their work.
On the other hand, a Wi-Fi network allows for flexible and convenient wireless connectivity, which is beneficial for staff members who frequently use their phones or move around the office space. It provides them with the freedom to connect wirelessly wherever they are in the building.
Learn more about Network installation here:https://brainly.com/question/33827192
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Related Questions
A coal-fired power plant is burning bituminous coal that has an energy content of 12,000 Btu/lb. The power plant is burning the coal at a rate of 110 lb/s. The efficiency of the powerplant is 33%, which means that 67% of the total energy from the coal is lost as waste heat.
a. A new heat exchanger is installed that is capable of using 0.12% of the waste heat to heat an air stream for use in another process in the plant. The air enters the heat exchanger at a volumetric flow rate of 2,000 acfm at 298 K and 1 atm. Assuming the heat exchanger is 90% efficient (i.e., 90% of the heat supplied to the heat exchanger is transferred to the air stream), at what temperature will the air leave the heat exchanger, assuming an exit pressure of 1 atm? Use an average specific heat of 0.244 Btu/lb-°R.
b. What will be the volumetric flow rate of the gas leaving the heat exchanger (in units of acfm)?
c. If the air exiting the heat exchanger has a moisture content of 12% (v/v), what is the moisture-corrected dry flow rate (in units of dry standard cubic meters per minute) at STP (1 atm, 298 K)?
Answers
Answer:
a) [tex]T_{out} = 2190.455 ^{\textdegree}R[/tex] b) [tex]\dot V_{air,out} = 8352.941 acfm[/tex] c) [tex]\dot V_{air,out,corr} = 7350.588 afcm[/tex]
Explanation:
a) The heat lost by the power plant is:
[tex]\dot Q_{loss} = (0.67) \cdot (110 \frac{lbm}{s} ) \cdot (12000 \frac{BTU}{lbm} )\\\dot Q_{loss} = 884400 BTU[/tex]
The waste heat used by heat exchanger is:
[tex]\dot Q_{used} = (0.0012) \cdot \dot Q_{loss}\\\dot Q_{used} = 1061.28 BTU[/tex]
Assuming that air behaves as an ideal gas, density is given by following expression:
[tex]\rho_{air} = \frac{P \cdot M_{air}}{R_{u} \cdot T}[/tex]
[tex]\rho_{air} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(298 K)}\\\rho_{air} = 1.145 \frac{kg}{m^{3}}[/tex]
The density unit is converted to pounds per cubic feet:
[tex]\rho_{air} = 1.145 \frac{kg}{m^{3}} \cdot (\frac{1 lb}{0.453 kg}) \cdot (\frac{0.304 m}{1 ft} )^{3} \\\rho_{air} = 0.071 \frac{lb}{ft^{3}}[/tex]
The heat received by air flow through heat exchanger is:
[tex]\dot Q_{air} = (0.90) \cdot \dot Q_{used}\\\dot Q_{air} = 955.152 BTU[/tex]
Outlet temperature can isolated from the following formula:
[tex]\dot Q_{air} = \rho_{air} \cdot \dot V_{air} \cdot c_{p,air} \cdot (T_{out} - T_{in})[/tex]
[tex]T_{out} =T_{in} + \frac{\dot Q_{air}}{\rho_{air} \cdot \dot V_{air} \cdot c_{p,air}}[/tex]
[tex]T_{out} = 536.4 ^{\textdegree}R + \frac{955.152 BTU}{(0.071\frac{lb}{ft^{3}})\cdot (33.333 \frac{ft}{s} )\cdot(0.244 \frac{BTU}{lb ^{\textdegree}R})}\\T_{out} = 2190.455 ^{\textdegree}R[/tex]
b) Due to the compressibility of air, density has to be calculated by using the approach from section a).
[tex]\rho_{air,out} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(1217 K)}\\\rho_{air,out} = 0.280 \frac{kg}{m^{3}}[/tex]
[tex]\rho_{air,out} = 0.017 \frac{lb}{ft^{3}}[/tex]
Volumetric flow can be found by Principle of Mass Conservation:
[tex]\dot V_{air, out} = \frac{\rho_{air}}{\rho_{air,out}}\cdot \dot V_{air}[/tex]
[tex]\dot V_{air,out} = 8352.941 acfm[/tex]
c) The moisture component indicates that 88 percent of volume is occupied by dry-air. The moisture-corrected dry flow rate is:
[tex]\dot V_{air,out,corr} = 0.88 \cdot (8352.941 afcm)\\\dot V_{air,out,corr} = 7350.588 afcm[/tex]
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high school and in first-year college physics courses, the formula is usually given asQ ΔH m Q mCp ΔT mCpT2 ␣T1 (2)T1(a) What assumption about Cp is required to go from Equation 1 to Equation 2? (b) The heat capacity (Cp) of liquid n-hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well-insulated vessel containing 2.00L of liquid n–C6H14 at T 300 K. A combustion reaction known to release 16.73 kJ of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be 3.10 K. In a separate experiment, it is found that 6.14kJ of heat is required to raise the temperature of everything in the system except the hexane by 3.10 K. Use these data to estimate Cp[kJ/(mol K)] for liquid n-hexane at T 300 K, assuming that the condition required for the validity ofEquation 2 is satisfied. Compare your result with a tabulated value.
Answers
Answer:
(a)
dQ = mdq
dq = [tex]C_p[/tex]dT
[tex]q = \int\limits^{T_2}_{T_1} {C_p} \, dT[/tex] = [tex]C_p[/tex] (T₂ - T₁)
From the above equations, the underlying assumption is that [tex]C_p[/tex] remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let [tex]C_{cal}[/tex] be heat constant of calorimeter
Q₂ = [tex]C_{cal}[/tex] ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m [tex]C_p[/tex] ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m [tex]C_p[/tex]' ΔT
[tex]C_p[/tex] = (16.73 - 6.14) / (15.085 x 3.10)
[tex]C_p[/tex] = 0.22646 KJ mol⁻¹ k⁻¹
At 500oC, the diffusion coefficient of Cu in Ni is 1.3 x 10-22 m2/s, and the activation energy for diffusion is 256,000 J/mol. From this information, determine the diffusion coefficient of Cu in Ni at 900 oC . (R = 8.314 J/(mol K)) 0oC = 273 K
Answers
Answer:
the diffusion coefficient at 900°C is D₂=8.9*10⁻¹⁸ m²/s
Explanation:
The dependence of the diffusion coefficient is similar to the dependence of chemical reaction rate with respect to temperature , where
D=D₀*e^(-Q/RT)
where
D₀= diffusion energy at T=∞
Q= activation energy
T= absolute temperature
D= diffusion coefficient at temperature T
R=ideal gas constant
for temperatures T₁ and T₂
D₁=D₀*e^(-Q/RT₁)
D₂=D₀*e^(-Q/RT₂)
dividing both equations and rearranging terms we get
D₂=D₁*e^[-Q/R(1/T₂-1/T₁)]
replacing values
D₂=D₁*e^[-Q/R(1/T₂-1/T₁)] = 1.3*10⁻²² m²/s*e^(-256,000 J/mol/8.314 J/(mol K)*(1/1073K-1/773K) = 8.9*10⁻¹⁸ m²/s
An insulated piston-cylinder device initially contains 300 L of air at 120 kPa and 17 oC. Air is now heated for 15 min by a 200-w resistance heater placed inside the cylinder. The pressure of air is maintained constant during this process. Determine the entropy change of air assuming (a) ) constant specific heats and (b) ) variable specific heats.
Answers
Answer:
a) [tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]
b) [tex]\Delta S =0.395 \ \frac{kJ}{K}[/tex]
Explanation:
a)
Specific heat formula, we have:
[tex]mc(\theta_1-\theta_2)=Wt[/tex]
Where
m is mass
c is specific heat capacity of air at specific temp
[tex]\theta[/tex] is the temperature change
W is the power
t is the time
The ideal gas law is:
[tex]PV=nRT[/tex]
Where
P is the pressure
V is the volume
n is number of moles
T is temperature
R is the ideal gas constant
First, lets solve for [tex]\theta_2[/tex] in the 1st equation, remembering to use ideal gas law in it to have the variables that are given in the problem. Shown below:
[tex]mc(\theta_2-\theta_1)=Wt\\mc\theta_2-mc\theta_1=Wt\\mc\theta_2=mc\theta_1 + Wt\\\theta_2=\frac{mc\theta_1 + Wt}{mc}\\\theta_2=\theta_1+\frac{Wt}{mc}\\\theta_2=\theta_1+\frac{WtR\theta_1}{PVc}\\\theta_2=\theta_1(1+\frac{WtR}{PVc})[/tex]
Now, we know
Theta_1 is 17 celsius, which in Kelvin is 17 + 270 = 290K
Power is 200
Time is 15 mins = 15 * 60 = 900 seconds
R is 287 J/kg K
P is 120
V is 300 L
Specific heat of air at 290K is about 1005
Substituting we get:
[tex]\theta_2=\theta_1(1+\frac{WtR}{PVc})\\\theta_2=290(1+\frac{200*900*287}{120*300*1005})\\\theta_2=704 \ K[/tex]
Now, Entropy Change is:
[tex]\Delta S =mc Ln(\frac{\theta_2}{\theta_1})[/tex]
Again using the substitution equations, we have:
[tex]\Delta S = \frac{PV}{R\theta_1}cLn(1+\frac{WRt}{PVc})[/tex]
We know what the variables mean, we substitute the respective values, to get:
[tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]
b)
For variable specific heats, we need entropy value from entropy table:
s_2 = 2.58044
s_1 = 1.66802 [1.05 * 290 K]
The formula is:
[tex]\Delta S = m(s_2-s_1)\\\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)[/tex]
Substituting the values, we find the answer:
[tex]\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)\\\Delta S = \frac{120*300}{287*290}(2.58044-1.66802)\\\Delta S =0.395 \ \frac{kJ}{K}[/tex]
The maximum compressive load experienced by the same truss member would occur when the live load is __________. The magnitude of the maximum compressive load would be __________ the magnitude of the maximum tensile load.
Answers
Answer:
Maximum
Greater than
Explanation:
The maximum compressive load experienced by the same truss member would occur when the live load is maximum. The magnitude of the maximum compressive load would be greater than the magnitude of the maximum tensile load.
Set up the differential equation that describes the motion under the assumptions of this section. Solve the differential equation. State whether the motion of the spring-mass system is harmonic, damped oscillation, critically damped, oroverdamped. If the motion is a damped oscillation, rewrite in the form (22).33. The spring-mass system has an attached mass of 10 g. The spring constant is30 g/s 2 . A dashpot mechanism is attached, which has a damping coefficient of 40 g/s. The mass is pulled down and released. At time t = 0, the mass is 3 cmbelow the rest position and moving upward at 5 cm/s.34. A long spring has a mass of 1 slug attached to it. The spring stretches 16/13 ftand comes to rest. The damping coefficient is 2 slug/s. The mass is subjected toan impulsive force at time t = 0, which imparts a velocity of 5 ft/s downward.
Answers
Answer:
Explanation:
The principle and application of differential equation is applied and the steps with detailed and appropriate substitution is as carefully shown in the attached files.
m = mass attached
k = spring constant
c = damping coefficient
def find_missing_letters(sentence): the input sentence is an array/list (created from normalize()) returns a sorted list of letters that are NOT in the sentence use the built in Python type set to solve (see previous lesson) def find_missing letters algorithm(sentence): the input sentence is an array/list (created from normalize()) returns a sorted array of letters that are NOT in the sentence you must NOT use the set type Hints for find_missing_letters algorithm: create your own algorithm and use only the data types (other than set) that we have discussed the set datatype is removed during the running of the tests (so don't use it) the output should be the same as that from find_missing_letters (easy to test) if you find yourself writing over 30 lines of code, you are proba
Answers
Answer:
Follow step by step explabation to get answer to the question and also see attachments for output.
Explanation:
code:
def normalize(input_string):
l=list();
for i in input_string:
if (i>='a' and i<='z') or (i>='A' and i<='Z'):
l.append(i.lower())
return l
def find_missing_letters_algorithm(sentence,prevsentece):
l=list();
for i in prevsentece:
i=i.lower();
if i not in sentence:
l.append(i)
return l
print(find_missing_letters_algorithm(normalize("O.K. #1 Python!"),"O.K. #1 Python!"))
An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 240 V. Assume a plate separation of d 1.77 cm and a plate area of A-25.0 cm. when the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
(a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC after Q pC
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
(c) Determine the change in energy (in nJ) of the capacitor nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 240 V potential difference Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q, - after Q Determine the capacitance (in F) and potential difference (in V) after immersion. pC pC Determine the change in energy (in nJ) of the capacitor. nJ
Answers
Answer:
a. 3.0 X 10^-10 C; 2.4 X 10 ^-8 C b. 1.0 X 10^-10 C; V= 240+3 =243 V c. 36 nJ d. 12.2 nJ
Explanation:
a. C=Q/V = ε0εrA/d => Q=εoεrAV/d; since dielectric constant for free space is 8.85 x 10^-12 and that of distilled water is 80, this shows that other dielectric material is free space with εr=1, Q=5.21 x 10^-15 C; εr for water is 80;
c. Energy, E =1/2(QV)
There is a large push in the united states currently to convert incandescent light bulbs more energy-efficient technologies, including compact fluorescent lights and light-emitting diodes. The lumen [lm] is the SI unit of luminous flux, a measure of the perceived power of light. To test the power usage, you run
an experiment and measure the following data. Create a proper plot of these data, with electrical consumption (EC) on the ordinate.
Electrical Consumption [W]
Incandescent Compact
Luminous Flux [lm] 120 Volt Fluorescent
80 16
200 4
400 38 8
600 55
750 68 13
1,250 18
1,400 105 19
Answers
Answer:
See the attached picture.
Explanation:
See the attached picture for explanation.
Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressure ratio is 11, the turbine inlet temperature is 1400 K, and air exits the nozzle 26 kPa. The diffuser and the nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of 85% and 90%, respectively. There is no pressure drop for flow through the combustor. Kinetic energy is negligible everywhere except at the diffuser inlet and the nozzle exit. On the basis of air-standard analysis, determine: (a)the pressures and temperatures at each state, (b)the rate of heat addition to the air passing through the combustor, (c) the velocity at the nozzle exit.
Answers
Answer:
Explanation:
Answer:
Explanation:
Answer:
Explanation:
This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.
(a). here we are asked to determine the Temperature and Pressure.
Given that the properties of Air;
ha = 230.02 KJ/Kg
Ta = 230 K
Pra = 0.5477
From the energy balance equation for a diffuser;
ha + Va²/2 = h₁ + V₁²/2
h₁ = ha + Va²/2 (where V₁²/2 = 0)
h₁ = 230.02 + 220²/2 ˣ 1/10³
h₁ = 254.22 KJ/Kg
⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg
from this we have;
Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]
Pr₁ = 0.77759
therefore T₁ = 254.15K
P₁ = (Pr₁/Pra)Pa
= 0.77759/0.5477 ˣ 26
P₁ = 36.91 kPa
now we calculate Pr₂
Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349
⇒ now we obtain properties of air at
Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg
calculating the enthalpy of air at state 2
ηc = h₁ - h₂ / h₁ - h₂
0.85 = 254.22 - 505.387 / 254.22 - h₂
h₂ = 549.71 KJ/Kg
to obtain the properties of air at h₂ = 549.71 KJ/Kg
T₂ = 545.15 K
⇒ to calculate the pressure of air at state 2
P₂/P₁ = 11
P₂ = 11 ˣ 36.913
p₂ = 406.043 kPa
but pressure of air at state 3 is the same,
i.e. P₂ = P₃ = 406.043 kPa
P₃ = 406.043 kPa
To obtain the properties of air at
T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5
for cases of turbojet engine,
we have that work output from turbine = work input to the compressor
Wt = Wr
(h₃ - h₄) = (h₂ - h₁)
h₄ = h₃ - h₂ + h₁
= 1515.42 - 549.71 + 254.22
h₄ = 1219.93 kJ/Kg
properties of air at h₄ = 1219.93 kJ/Kg
T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
T₄ = 1150.58 K
Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
Pr₄ = 200.5636
Calculating the ideal enthalpy of the air at state 4;
Лr = h₃ - h₄ / h₃ - h₄*
0.9 = 1515.42 - 1219.93 / 1515.42 - h₄
h₄* = 1187.09 kJ/Kg
now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg
P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]
P₄* = 181.316
P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process
P₄ = 181.316/450.5 * 406.043
P₄ = 163.42 kPa
For the 4-5 process;
Pr₅ = (P₅/P₄)Pr₄
Pr₅ = 26/163.42 * 200.56 = 31.9095
to obtain the properties of air at Pr₅ = 31.9095
h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]
h₅ = 734.09 KJ/Kg
T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]
T₅ = 719.32 K
(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;
QH = m(h₃-h₂)
QH = 25(1515.42 - 549.71)
QH = 24142.75 kW
(c). To calculate the velocity at the nozzle exit;
we apply steady energy equation of a flow to nozzle
h₄ + V₄²/2 = h₅ + V₅²/2
h₄ + 0 = h₅₅ + V₅²/2
1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2
therefore, V₅ = 985.74 m/s
cheers i hope this helps
g A plane stress element has components sigma x = 160 MPa, tau xy = 100 MPa (CW). Determine the two values pf sigma y for which the maximum shear stress is 73 MPa.
Answers
Answer:
The question is mentioned in the attachment.
Explanation:
An NMOS amplifier is to be designed to provide a 0.50-V peak output signal across a 50- kΩ load that can be used as a drain resistor. If a gain of at least 5 V/V is needed, what gm is required?
Answers
Answer:
The gm required is 0.1 [tex]\frac{mA}{V}[/tex]
Explanation:
Knowing
Vpeak = 0.50 V
Rd = 50 kΩ
Av = 5 V/V
To find the transconductance we should to use the equation
Av = gm * Rd
so,
gm = Av/Rd
gm = 5/50 = 0.1 [tex]\frac{mA}{V}[/tex]
Three large plates are separated bythin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. At what speed and in what direction must the bottom plate be moved to hold the center plate stationary?
Answers
To keep the center plate stationary, the bottom plate must move at 2 m/s to the left, counteracting the top plate's movement to the right at the same speed.
Explanation:Three large plates are separated by thin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. To hold the center plate stationary, the velocity profile of the fluid layers between the plates indicates that for every action, there is an equal and opposite reaction according to Newton's third law. Given that the top plate moves at 2 m/s, to keep the center plate stationary, the bottom plate must also move at 2 m/s but in the opposite direction, which is to the left.
This setup demonstrates the principles of fluid dynamics, where the velocity gradient across the thickness of the fluid layers causes a shear stress that is directly proportional to the velocity gradient according to Newton's law of viscosity. The equilibrium of forces ensures the central plate remains stationary if the bottom plate's movement perfectly counterbalances the top plate's movement.
The bottom plate should move left at 2 m/s to keep the center plate stationary relative to the top plate.
To hold the center plate stationary, the relative velocity between the top and bottom plates should be zero. Since the top plate moves to the right at 2 m/s, the bottom plate must move to the left at the same speed to cancel out this motion.
The velocity of the bottom plate [tex](\(v_b\))[/tex] relative to the center plate is the sum of the velocities of the bottom plate relative to the top plate ( [tex]\(v_{b/t}\))[/tex] and the velocity of the top plate relative to the center plate
[tex](\(v_{t/c}\))[/tex] . Since [tex]\(v_{t/c}\)[/tex] is given as 2 m/s to the right, to make [tex]\(v_b\)[/tex] zero, [tex]\(v_{b/t}\)[/tex] should be 2 m/s to the left.
So, the bottom plate must be moved to the left at 2 m/s to hold the center plate stationary relative to the top plate. This ensures that the relative velocity between the top and bottom plates is zero, maintaining the center plate stationary.
4. Three routes connect an origin and a destination with performance functions tl = 8 + 0.5x1, t2 = 1 + 2x2, and t3 = 3 + 0.75x3, with the x's expressed in thousands of vehicles per hour and the t's expressed in minutes. If the peak-hour traffic demand is 3400 vehicles, determine user-equilibrium traffic flows. Note: there are 823 vehicles that must pass by a daycare in route 1.
Answers
Answer:
X1 = 2081.64
X2 = 523.91
X3 = 1394.45
Explanation:
See the attached pictures for detailed explanation.
For a bronze alloy, the stress at which plastic deformation begins is 275 MPa and the modulus of elasticity is 115 GPa. If the original specimen length is listed above, what is the maximum length to which it may be stretched without causing plastic deformation
Answers
Answer:
maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm
Explanation:
given data
stress = 275 MPa
modulus of elasticity = 115 GPa
solution
we will apply here modulus of elasticity formula that is
modulus of elasticity = stress ÷ strain ....................1
here we consider original length is 200 mm
so strain is = [tex]\frac{\triangle L}{original\ length }[/tex] ..........2
put here value in equation 1 we get
115 × [tex]10^{9}[/tex] = [tex]\frac{275 \times 10^{6}}{\frac{x}{200} }[/tex]
solve it we get
x = [tex]\frac{11}{23}[/tex]
x = 0.4782
so maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm
You are running a software company and have a series of n jobs that must be pre-processed first on a supercomputer before being moved to a smaller PC. You have only one super- computer, but you have n PCs so the second stage can be performed in in parallel. More specifically, your jobs are described as J- (s1f), J2 (22).., Jn-(Sn,n), where job J needs si units of time to be pre-processed on the super-computer and fi units of time on the PC You need to work out an order in which to give the jobs to the super-computer. As soon as the first job is done on the super-computer, it can be moved to the PC for finishing; at that point a second job can be given to the super-computer; when the second job is done it can go straight to a PC since the PCs can work in parallel, and so on. So if the jobs are processed in the order given, job J finishes at time (sk) fi. A schedule is an ordering of the jobs to be given to the super-computer. The completion time is the point at which all jobs have finished being processed on the PCs. We wish to minimize the completion time. (a) Give an efficient (greedy!) algorithm for computing the optimal order in which to proces the jobs so that the completion time is minimized. (b) Describe the greedy choice your algorithm makes and show that it is correct.
Answers
in this exercise we have to use the knowledge of software to explain its efficiency, as:
A) Relationship between the software performance level and the amount of resources used, under pre-defined usage conditions.
B) An algorithm is efficient if it doesn't waste time for nothing. In other words, an algorithm is efficient if it is faster than other algorithms for the same problem.
What is a software?Software can be defined as a set of instructions that allow the user to control an electronic device. In a computer, for example, the physical parts and peripherals make up the hardware, but you need software so that the components know how they are supposed to work.
A) Greedy algorithms are a class of algorithm that will choose the local optimum at any time to find the optimum solution for the problem. Since only one super computer is there, it can process the jobs sequentially and its processing time is constant.
B) The selfish choice secondhand above is selecting the jobs accompanying greater value of PC occasion at each time. Thus the accomplishment period can be weakened because the jobs accompanying bigger PC time will be done first. Thus the overhead period after perfecting the excellent calculating processing maybe removed.
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