All of the following safety tips are true EXCEPT Select one: a. It is not acceptable to handle broken glass with your bare hands b. It is acceptable to grasp the electrical cord when removing an electrical plug from its socket c. It is not acceptable to immerse hot glassware in cold water d. It is not acceptable to reuse dirty glassware
Answer:
Explanation:
B. you would grab the plug closest to the outlet
If you don't have enough experience, it's always best to leave socket changing to the experts. If you make a mistake, you might inflict harm and potentially endanger yourself and other people. Read on if you're interested in learning how to change a socket safely. Thus, option D is correct.
What, removing an electrical plug from its socket?Grip the plug, not the electrical cable, when taking an electrical plug out of its socket. Before handling an electrical switch, socket, or outlet, hands must be fully dry.
Reduce the extra so that it rests only on top of the existing plasterboard. If necessary, push it back a little by using your finger. Fill the dent with ready-mixed filler or powdered filler, whichever you want, and bring it flush with the surrounding wall. Allow to dry, then sand off any excess.
Therefore, It is acceptable to grasp the electrical cord when removing an electrical plug from its socket
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A micromechanical resonator is to be designed to have a Q factor of 1000 and a natural frequency of 2 kHz. Determine the system-damping factor and the system bandwidth.
Answer:
Explanation:
Given:
Q factor, =1000
natural frequency, [tex]f_n=2000~Hz[/tex]
Damping factor, [tex]\zeta=?[/tex]
Bandwidth, BW=?
We have the relation:
[tex]Q=\frac{1}{2\zeta}[/tex]
[tex]\zeta=\frac{1}{2Q}[/tex]
[tex]\zeta=\frac{1}{2\times 1000}[/tex]
[tex]\zeta=5\times 10^{-4}[/tex]
Bandwidth:
[tex]BW=\frac{f_n}{Q}[/tex]
[tex]BW=\frac{2000}{1000}[/tex]
[tex]BW=2~Hz[/tex]
a) Complete the following methods description using the correct tense for the verb in brackets. (This student is using passive voice rather than any human agents at the request of the instructor.) Student Lab Report Identical tensile test procedures were performed on all test specimens. Each of the metal specimens ____1____ [have] an indentation near the center to ensure that the fracture point would occur in this region. Tension tests ____2____ [conduct] as follows. Two pieces of reflective tape ____3____ [place] approximately 1 inch apart in the center of the specimen where the indentation 4 [locate]. The width and the thickness of the specimen at this location _____5_____ [measure] using a Vernier caliper. Then the specimen _____6____ [secure] in the MTS Load Frame. A laser extensometer _____7_____ [place] into position to measure the deformation of the specimen. The laser extensometer ______8_ __ [use] to measure the original distance between the pieces of reflective tape. The MTS ________9____ [set] to elongate the specimen one tenth of an inch every minute.
Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?
Solution :
B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.
C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.
D. The given solution provides integrity. Because it provides authentication and have not been changed.
E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.
Do you know who Candice is
Answer: Can these nuts fit in your mouth?
Explanation:
im just here for the points >:)
CO2 enters an adiabatic nozzle, operating at steady state, at 200 kPa, 1500 K, 5 m/s and exits at 100 kPa, 1400 K. The exit area of the nozzle is 10 cm2. Using the PG model, determine the exit velocity
Answer:
[tex]v_2=549.2 m/s\\[/tex]
Explanation:
Given:
[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]
Solution:
For [tex]Co_2[/tex] y=1.4
Since Nozzle is adiababic
So,
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]
Now,
[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]
in this expression the densities are constant
v = A √h
A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h = [tex]\frac{1}{454.55^2} \ v^2[/tex]
Δh = [tex]\frac{dh}{dv}[/tex] Δv
[tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
[tex]\frac{\delta v}{v} = 0.05[/tex]
[tex]\frac{\Delta h}{h}[/tex] = 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
[tex]\frac{\Delta h}{h}[/tex] = 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?
Answer:
[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]
Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):
[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]
Now, the isentropic efficiency of the turbine can be given as follows:
[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
Answer:
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=480/0 \textdegree V[/tex]
Power [tex]P=120kW[/tex]
Initial Power factor [tex]p.f_1=0.77 lagging[/tex]
Final Power factor [tex]p.f_2=0.9 lagging[/tex]
Generally the equation for Reactive Power is mathematically given by
Q=P(tan \theta_2-tan \theta_1)
Since
[tex]p.f_1=0.77[/tex]
[tex]cos \theta_1 =0.77[/tex]
[tex]\theta_1=cos^{-1}0.77[/tex]
[tex]\theta_1=39.65 \textdegree[/tex]
And
[tex]p.f_2=0.9[/tex]
[tex]cos \theta_2 =0.9[/tex]
[tex]\theta_2=cos^{-1}0.9[/tex]
[tex]\theta_2=25.84 \textdegree[/tex]
Therefore
[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=-41.33VAR[/tex]
Therefore
The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
Calculate the Principal stress, maximum in-plane shear stress and average normal stress
Using Mohr's circle ( attached below )
i) principal stresses:
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
ii) maximum in-plane shear stress
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
iii) average normal stress
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl.
For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely.
Kb for CH3NH2 = 4.4 x 10-4.
What species are present based off what is being added?
200.00 mL HCl added
yes no H+
yes no H2O
yes no Cl-
yes no CH3NH2
yes no CH3NH3+
300.00 mL HCl added
yes no H+
yes no H2O
yes no Cl-
yes no CH3NH2
yes no CH3NH3+
Calculate the pH at the equivalence point for this titration?
Answer:
The answers are in the explanation. The pH is 5.91
Explanation:
The CH3NH2 reacts with HCl as follows:
CH3NH2 + HCl → CH3NH3⁺ + Cl⁻
When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:
no H+. All reacted
yes H2O. Because the water is present in the solutions of HCl and CH3NH2
yes Cl-. Is a product of the reaction
Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+
yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES
When 300.00mL of HCl are added, 100mL are in excess:
yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.
yes H2O. Is present because the reactants are diluted.
yes Cl-. Is a product of reaction and comes from HCl.
Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+
yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.
To find the pH:
At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:
0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+
The equilibrium of CH3NH3+ is:
Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]
As both [H+] [CH3NH2] comes from the same equilibrium:
[H+] = [CH3NH2] = X
2.273x10-11 = [X] [X] / [0.0667M]
1.5159x10-12 = X²
X = 1.23x10-6M = [H+]
As pH = -log [H+]
pH = 5.91
The pH at the equivalent point for this titration is "5.91".
pH Calculation:[tex]CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\[/tex]
We must now quantify the pH well at the equivalence point.
We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.
Calculating the Moles in [tex]CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles[/tex]
Calculating the Moles in [tex]HCl = 0.0200 \ moles[/tex]
Calculating the volume of [tex]HCl[/tex]:
[tex]\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\[/tex]
[tex]= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\[/tex]
Calculating the reaction among the acid and base:
[tex]CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-[/tex]
[tex]0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200[/tex]
Therefore the conjugate acid of the bases exists at the standard solution.
Then we must calculate the new molar mass of [tex]CH_3NH_3^+[/tex].
Total volume[tex]= 100 + 200 = 300\ mL = 0.300\ L[/tex]
[tex][CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M[/tex]
Using the ICE table
[tex]CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+[/tex]
[tex]I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}[/tex]
Calculating [tex]K_a[/tex] from [tex]K_b[/tex]
[tex]\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\[/tex]
[tex]= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}[/tex]
The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.
[tex]\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\[/tex]
We have the formula to calculate pH.
[tex]\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91[/tex]
The pH at the equivalent point for this titration is "5.91".
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A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.
Answer:
Tensile stress = 0.1855Kpsi
Total deformation = 0.0012243 in
Unit strain = 1.855 *10^-5 or 18.55μ
Change in the rod diameter = 5.02 * 10^ -6 in
Explanation:
Data given: D= 0.82 in
L = 5.5 ft * 12 = 66 in
load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm
Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²
∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²
Rt = 0.1855 Kpsi
∴ Total deformation = PL / AE = Rt * L/ Eal
= 0.1855 * 10³ * 66 / 10000 * 10³
= 0.0012243 in
∴the unit strains = total deformation / L = 0.0012243/ 66
=0.00001855 = 1.855 *10^-5
= 18.55μ
∴ Change in rod Δd/ d = μ ΔL/L
= (0.33) 1.855 *10^-5 * 0.82
= 5.02 * 10^ -6 in
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree
Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Answer:
a. 164 °F b. 91.11 °C c. 1439.54 kJ
Explanation:
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.
So, Δ°F = 212 °F - 48 °F = 164 °F
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
To find the degree change in Celsius, we convert the initial and final temperature to Celsius.
°C = 5(°F - 32)/9
So, 48 °F in Celsius is
°C₁ = 5(48 - 32)/9
°C₁ = 5(16)/9
°C₁ = 80/9
°C₁ = 8.89 °C
Also, 212 °F in Celsius is
°C₂ = 5(212 - 32)/9
°C₂ = 5(180)/9
°C₂ = 5(20)
°C₂ = 100 °C
So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.
So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ
An intersection with a four phase signal has a displayed red time of 35 seconds, a start-up lost time of 2 seconds, a yellow time of 4 seconds, and an all red time of 1 second per phase. The total lost time is typically calculated as ____ seconds per cycle.
Answer:
53 sec / cycle
Explanation:
Displayed red time = 35 seconds
Start up lost time = 2 seconds
Yellow time = 4 seconds
Red time = 1 second
Total lost time L = 2n + r
L = lost time
n = number of phase
R = red time
35+2+4+4*1
= 45
L = 2x4+45
= 53 sec/cycle
The total lost time is typically calculated as 53 seconds per cycle
A 5.74 kg rock is thrown upwards with a force of 317 N at a location where the local gravitational acceleration is 9.81 m/s^2. What is the net acceleration of the rock?
Answer:
[tex]a=45.31m/s^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=5.74[/tex]
Force [tex]F=317N[/tex]
Gravitational Acceleration [tex]g=9.81m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F-mg=ma[/tex]
[tex]317-5.74*9.81=5.74 a[/tex]
[tex]a=\frac{260.7}{5.74}[/tex]
[tex]a=45.31m/s^2[/tex]
Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?
Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]
Therefore, the current consumed is 3.3 A
Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2
Draw a sinusoidal signal and illustrate how quantization and sampling is handled by
using relevant grids.
Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct
Answer:
Both are correct.
Explanation:
Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.
plsssssss help me here
Explain why veracity, value, and visualization can also be said to apply to relational databases as well as Big Data.
Answer:
Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...
Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?
Answer:
jsow
hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh
Explanation:
j
grudbA signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ration of 0.9, calculate the minimum necessary cycle length.
Answer:
[tex]T_o=82.1sec[/tex]
Explanation:
From the question we are told that:
Lost Time [tex]t=12secs[/tex]
Sum of critical flow ratios [tex]X=0.72[/tex]
Generally the Webster Method's equation for Optimum cycle time is is mathematically given by
[tex]T_o=\frac{1.5t+5}{1-x}[/tex]
[tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]
[tex]T_o=82.1sec[/tex]
Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).
After finishing computation, the app sends some data to the cloud, regardless of the mode it’s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.
Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?
The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.
Why reducing leads to increasing wages?Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.
Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.
Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.
Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.
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If there is a discrepancy between Chick-fil-A food safety requirements and local Health Department
regulations, what should Team Member do?
The following should be done by the team member:
It is important to follow both Chick-fil-A food safety requirements and local Health Department regulations. In the case when there is a discrepancy between the two, always follow the more stringent requirement. Any other appearance or grooming issue not covered in these materials may be addressed at the discretion of the Operator.
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Add a pair of radio buttons to your form, each nested in its own label element.
One should have the option of car and the other should have the option of bike.
Both should share the name attribute of “vehicle” to create a radio group
Make sure the radio buttons are nested with the form
Make sure that the name attributes appear after the type
Answer:
The code is as follows:
<form name = "myForm">
<div>
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
</div>
<div>
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
</div>
</form>
Explanation:
This defines the first button
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
This defines the second button
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
The code is self-explanatory, as it follows all the required details in the question
In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is
Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
hence the correct option is :
N010 GO2 X7.0 Y2.0 15.0 J2.0
ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:
Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2