Explanation:
there are three stages of heat treatment
1.hit the metal slowly to ensure that the metal maintains a uniform temperature
2.soak or hold the metal at a specific temperature for a alloted period of time
3.cool the metal to room temperature
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
A security engineer deploys a certificate from a commercial CA to the RADIUS server for use with the EAP-TLS wireless network. Authentication is failing, so the engineer examines the certificate's properties:
Issuer: (A commercial CA)
Valid from: (yesterday’s date)
Valid to: (one year from yesterday’s date)
Subject: CN=smithco.com
Public key: RSA (2040 bits)
Enhanced key usage: Client authentication (1.3.6.1.5.7.3.2)
Key usage: Digital signature, key encipherment (a0)
Which of the following is the MOST likely cause of the failure?
A. The certificate is missing the proper OID.
B. The certificate is missing wireless authentication in key usage.
C. The certificate is self-signed.
D. The certificate has expired.
Answer:
The MOST likely cause of the failure is:
A. The certificate is missing the proper OID.
Explanation:
OID (Object Identifier) is a globally unique group of characters, alphanumeric or numeric identifier, registered under the ISO registration standard to reference a specific object or object class (an entity). In computing, OID allows a server or end-user to retrieve an object without identifying the specific physical data location. Since OID is system-generated, it is immutable and can only be assigned to one object. It cannot be shared. Its presence in the certificate would have enabled the security engineer to authenticate the certificate.
Increasing following distance to
when encountering other motorists who follow too closely
is an example of appropriate implementation of the IPDE defensive driving strategy for the
maintenance of an appropriate Safety Cushion.
Two-seconds
Enree-seconds
Four-seconds
Twenty-seconds
Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
17- The cathodic polarization is ..... *
O
a- activation.
O
b- concentration
O c- both.
You toss two coins. If you get heads with the first coin, you stop. If you get tails, you toss it again. The second coin is tossed regardless. What is the ratio of heads to tails?
Answer:
1:1
Explanation:
A coin has either a head or tail.
Thus, probability of head = ½
Probability of tail = ½
The first coin is tossed twice and probability of head or tail on both tosses is still ½. Thus, they will have a ratio 1:1.
The second coin is still tossed regardless and thus continuously and so should have same ratio of 1:1.
Therefore, the ratio of the both of them must, also be 1:1.
determine if the fluid is satisfied
What is protection scheme?
Answer:
The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.
Explanation:
The devices that are used to protect the power systems from faults are called protection devices.
Calculate the pressure of dry O2 if the total pressure of O2 generated over water is measured to be 698 Torr and the temperature is 30.1 oC. P(H2O) = 19.8 torr.
If the volume of the O2 sample in the question above was 56.3 ml, what volume would the dry O2 occupy at 755 torr (assume the temp was unchanged).
Answer:
[tex]V_2=46mL[/tex]
Explanation:
From the question we are told that:
Pressure over Water [tex]P=698 Torr[/tex]
Temperature [tex]T= 30.1 \textdegree C[/tex]
Pressure of Water [tex]P(H2O) = 19.8 torr.[/tex]
Volume of O2 [tex]O_2=56.3[/tex]
Pressure of Dry O2 [tex]P_(0)=755torr[/tex]
Generally the equation for Total Pressure is mathematically given by
[tex]P_t = P_O + P_H[/tex]
Therefore
[tex]P_O=P_t-P_H[/tex]
[tex]P_O=638-19.8[/tex]
[tex]P_O=618.2torr[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]P_1*V_1 = P_2*V_2[/tex]
[tex]V_2=\frac{P_1*V_1}{P_2}[/tex]
Therefore
[tex]V_2=\frac{ 618.2*56.3}{755}[/tex]
[tex]V_2=46mL[/tex]
Hence,The volume would the dry O2 occupy at 755 torr
[tex]V_2=46mL[/tex]
deep invasion lead to???
Answer:
2
Explanation:
usjshshshjeoeieieiiekeiekeiekekekekekkske
Answer:
2
Explanation:
High separation between deep...........
You are responsible for notifying the DMV within 5 days of the sale using a Notice of Release of Liability form if you sell or transfer a vehicle to someone else.
Answer:
True
Explanation:
After you have sold or have made the transfer of ownership of a motor vehicle to another person, you are required to fill a notice of transfer and release of liability. You do this as a notification to let DMV be aware of a change of ownership of this vehicle. And it also serves to protect the previous owner from liabilities such as parking violations or traffic violations. This notification should be done within 5 Days of the transfer of ownership.
Thank you.
can you guys please introduce yourself
Answer: why?
Explanation:
State the factor that influence the frequency of the induced emf of an alternating quantity
Explanation:
conductor, flux, and movement of conductor in magnetic field are some of the factors that induce emf.
Which statement describes the relay between minerals and rocks ?
Answer:
•○●□■hey hi!■□●○•Explanation:
Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.
☆♡hope this helps♡☆The impeller shaft of a fluid agitator transmits 20 kW at 430 rpm. If the allowable shear stress in the impeller shaft must be limited to 65 MPa, determine (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 36 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)
Given :
Power, P = 20 kW
Speed, N = 430 rpm
Allowable shear stress, τ = 65 MPa
Torque in the shaft is given by :
[tex]$P=\frac{2 \pi NT}{60}$[/tex]
[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]
T = 444.37 N.m
Diameter of the solid shaft is
[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]
[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]
[tex]$d=\sqrt[3]{34.83} $[/tex]
d = 3.265 m
d = 326.5 mm
Internal diameter of the hollow shaft is :
[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]
[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]
[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]
[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]
[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]
[tex]$d_i^4=300000$[/tex]
[tex]$d_i = 23.40$[/tex] mm
Percentage savings in the weight is given by :
Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]
[tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]
[tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]
[tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]
[tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]
[tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]
[tex]$=\frac{105553 }{106602} \times 100$[/tex]
= 99.01 %
A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
Final PDF to printer
13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
A 2-stage dcv that has an internal pilot does not work well (if at all) on
Answer:
i really font onow why tbh eot you
I want to explain what 2000 feet looks like to young children so that they can imagine it in class
Answer:
maybe take a really common toy kids play with or often see, find the average height for the toy and do the math to see how many of those toys stacked ontop of eachother would make up 2000 feet. For example (this isn't accurate btw just an idea of what it would sound like but) "Have you ever seen a barbie doll? well if you stack 400 barbie dolls ontop of their head it would be equal to 2000 feet."
Explanation:
sometimes taking common or beloved objects children have into your examples makes them have a better image of how small or how big something is.
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
The number-average molecular weight of a poly (styrene-butadiene) alternating copolymer is 1,350,000 g/mol. What is the average number of styrene and butadiene repeat units per molecule.
a) 6,806
b) 6,944
c) 4,801
d) 8,544
In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The artery bifurcates (splits) into two identical blood vessels that are each 3mm in diameter. What are the average velocity and the mass flow rate upstream and downstream of the bifurcation? The density of blood is 1.06g·cm−3
9514 1404 393
Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
__
The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diameter is 3 cm. Estimate (a) the mass flow rate through the nozzle and (b) the Mach number at the throat.
Answer:
a) [tex]m=0.17kg/s[/tex]
b) [tex]Ma=0.89[/tex]
Explanation:
From the question we are told that:
Pressure [tex]P=60kPa[/tex]
Diameter [tex]d=3cm[/tex]
Generally at sea level
[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]
Generally the Power series equation for Mach number is mathematically given by
[tex]\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}[/tex]
[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]Ma=0.89[/tex]
Therefore
Mass flow rate
[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\rho=0.848kg/m^3[/tex]
Generally the equation for Velocity at throat is mathematically given by
[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])
Where
[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]
[tex]T_e=248[/tex]
Therefore
[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]
Generally the equation for Mass flow rate is mathematically given by
[tex]m=\rho*A*V[/tex]
[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]
[tex]m=0.17kg/s[/tex]
2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.
Write the code using the do-while loop to force the user to enter a number in the range [20,50]
Answer:
Mark as brainlist pls hello
A 40kg steel casting (Cp=0.5kJkg-1K-1) at a temperature of 4500C is quenched in 150kg of oil (Cp=2.5kJkg-1K-1) at 250C. If there are no heat losses, what is the change in entropy of?
(i) The casing.
(ii) The oil.
(iii) Both considered.
Hãy trình bày các bộ phận chính trong một bộ điều khiển điện tử (ECU) dùng trên ô tô. Cho biết công dụng của từng thành phần.
Answer:
sorry but I can't understand this Language.
Explanation:
unable to answer sorry
hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer equations. For steady, incompressible, and two-dimensional flow, neglecting gravity, the result is delta u/ delta x + delta v/ delta y= 0; u delta u/ delta x +v delta u/ delta y= -1/p(delta u/ delta x)+ v delta^2 u/ delta y^2 Use L and V0 as characteristic length and velocity, respectively. Non-dimensionalize these equations and identify the similarity parameters that result.
Answer: Attached below is the well written question and solution
answer:
i) Attached below
ii) similar parameter = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Explanation:
Using ; L as characteristic length and Vo as reference velocity
i) Nondimensionalize the equations
ii) Identifying similarity parameters
the similar parameters are = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Attached below is the detailed solution
Different metabolic control systems have different characteristic time scales for a control response to be achieved. Match the time scale with the control system.
a. Covalent modification
b. Allosteric control
c. Gene expression
1. Seconds to minutes
2. Milliseconds
3. Hours
Answer:
a. Covalent modification = Seconds to minutes
b. Allosteric control = Milliseconds
c. Gene expression = Hours
Explanation:
Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.
8- Concentration polarization occurs on the surface of the.......
a- cathode.
b- anode.
C- both
d-ption 4
Explanation:
Concentration overpotential, ηc,
I hope it helps you
The input sin(20) is sampled at 20 ms intervals by using impulse train sampling: i. Construct the input and sampled signal spectra.
Solution :
Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]
[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]
[tex]$=\frac{1}{20}$[/tex]
= 0.05 kHz
[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex] rad/s
We know that,
FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]
The sampled signal is :
[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]
So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
[tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
