Answer:
The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output
I hope it is helpful
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
1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?
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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 Rankine power generation cycle operates with steam as the working fluid. The turbine produces 100 MW of power using 89 kg/s of steam entering at 700C and 5MPa. The steam exits the turbine at 0.10135 MPa. Saturated liquid water exits the condenser and is pumped back to 5Mpa before it is fed to an isobaric boiler. a. Draw a schematic of the cycle. Number the streams and label any constraints across the units. b. The turbine operates adiabatically but not reversibly. What is the temperature of the steam exiting the turbine
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.
Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
how he would plan and manage projects in the future?
Select an answer:
project highlights
major changes and risks
summary of schedule and cost performance
summary of project management effectiveness
Answer:
Dalton
The part of the close-out report that would describe how he would plan and manage projects in the future is:
summary of project management effectiveness
Explanation:
The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.
A steel component with ultimate tensile strength of 800 MPa and plane strain fracture toughness of 20 MPam is known to contain a tunnel (internal) crack of length 1.4 mm. This alloy is being considered for use in a cyclic loading application for which the design stresses vary from 0 to 410 MPa. Would you recommend this alloy for this application
Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.
A recessed luminaire bears no marking indicating that it is ""Identified for Through- Wiring."" Is it permitted to run branch-circuit conductors other than the conductors that supply the luminaire through the integral junction box on the luminaire?
Answer:
No it is not permitted
Explanation:
It is not permitted because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.
The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.
how to solve circuit theory using mesh analysis
Explanation:
Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.
This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.
__
Example
Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...
mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)
mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs
mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0
You will note that the matrix of equation coefficients is symmetric.
__
In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.
answer this qustion plz
Answer:
click here and I'm so happy I have been doing this weekend I can see the same way you could come back in a bit and then we are all of you to know you do for a bit of time with me or
Question 1: Determine the maximum load P the steel bracket can withstand if the steel bracket has a circular cross section with a diameter of 1.2 in, and has an allowable normal stress of allow
Complete Question
Complete Question is attached below
Answer:
[tex]P=1124.2ibf[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=1.2in[/tex]
Allowable Normal stress [tex]\sigma=27.5ksi=27.5 * 10^3 psi[/tex]
Generally the equation for Bending Stress is mathematically given by
[tex]\phi= \frac{32M}{ \pi d^3}[/tex]
[tex]\phi= {32 * 4 P}{\pi * 1.2^3}[/tex]
[tex]\phi=23.58 psi[/tex]
Generally the equation for Direct Normal Stress is mathematically given by
[tex]\sigma'=\frac{4P}{ \phi * 1.2^2}[/tex]
[tex]\sigma'= 0.88P psi[/tex]
Therefore
Total Normal stress
[tex]\sigma_T=23.58 + 0.88[/tex]
[tex]\sigma_T=24.46P[/tex]
Generally the equation for Allowable Stress is mathematically given by
[tex]\sigma=\sigma_T P[/tex]
[tex]P=\frac{\sigma}{\sigma_T}[/tex]
[tex]P=\frac{27.5 * 10^3}{24.46P}[/tex]
[tex]P=1124.2ibf[/tex]
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.
A 40-mm-diameter solid steel shaft, used as a torque transmitter, is replaced with a hollow shaft having a 40-mm outer diameter and a 36-mm inner diameter. If both materials have the same strength, what is the percentage reduction in torque transmission
Answer:
65.61%
Explanation:
we have the following information to answer this question
diameter of the solid steel shaft = 40 mm
outer diametr of the hollow shaft = 40mm
inner diametr pf the hollow shaft = 36mm
[tex]percentage reduction in torque transmission = \frac{Tsolid-Thollow}{Tsolid} *100[/tex]
= (40³ - (40⁴-36⁴)/40)/40³ * 100
= (40³ - 22009.6)/40³ * 100
= 41990.4/64000 * 100
= 0.6561 x 100
= 65.61%
percentage reduction in torque transmission = 65.61%
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]
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.
If the constant is added to every observation of data then arithmatic mean obtained is
Answer:
Explanation:
Increased by the constant. Take a very simple case.
4 + 5 + 6 = 15
The mean is 5 (obtained by dividing the total (15) by the number of terms (3).
Now add a constant say 6
4 + 6 = 10
5 + 6 = 11
6 + 6 = 12
Total = 33/3 = 11
So the mean 5 is increased by the constant 6.
Now do the same thing more symbolically.
4 + c
5 + c
6 + c
Total = 15 + 3c
Divide by 3 you get 5 + c
If you want a more formal proof involving n terms, leave a note.
. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
Answer:
not understand language
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
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
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
lists at least 6 units of measuring atmospheric pressure
Answer:
On my console displays for the ISS visiting vehicles, three units are used. The Americans use pounds per square inch (psi). The Russians use kilopascals (kPa). The Japanese use Torr - millimeters of mercury (mmHg). A fourth unit is simply the atmosphere, or multiples of it. So, for example, sea level air pressure (which is what we use onboard ISS) is defined as 1 atmosphere. That is equivalent to 14.7 psi, 101.3 kPa, or 760 mmHg.Here N represents newton which is SI unit of Force which is same as Kg.m/s2." role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2.2.
m represent metre which is SI unit of length.
Kg represent Kilogram which is SI unit of Mass.
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Hope it helps.
Thanks.
Answer:
Pounds per square inch (psi)
Kilopascals (kPa)
Millimeters of mercury (mmHg)
Pascal (Pa)
Megapascal (MPa)
Atmospheric pressure (atm)
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 %
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duct tape and wraps it around the pipe to stop the water from leaking. The duct tape in this situation is similar to _____ in that it/they _____.
Answer:
The answer is "Option a".
Explanation:
Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.
The Myelination was the myelinization mechanism of the neuron axon. Thus the option A is correct.
What is a Myelination?It is the process by which the brain's oligodendrocytes make layers of myelin and is wound around the neural axons and is seen as a layer of insulation for the transmission electrical potential down to the neuronal axon.
Find out more information about the sentence.
brainly.com/question/5636726
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
định khoản nghiệp vụ sau : tạm ứng cho nhân viên A đi công tác bằng tiền mặt 50.000
Answer:
report on a fight you have witnessed
Side milling cutter is an example of ______ milling cutter.
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.
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.
______ is not a type of digital signaling technique
Answer:
Data Rate Signaling
