Answer:
The powertrain system is one of the major systems on the vehicle.
Tech B states that this system usually contains the engine and transmission.
Technician B is correct.
Explanation:
The powertrain system contains three interconnected systems: the engine, the transmission, and the drivetrain. The engine part is the power source of the powertrain that ensures vehicular movement. The transmission system converts the engine's power into the normal movement of the vehicle. The drivetrain consists of the transmission, driveshafts, and the axles. These three parts of the powertrain system work seamlessly to ensure that the vehicle is able to move with some speed back and front, as may be desired.
Saturated refrigerant-134a vapor at 15 psia is compressed reversibly in an adiabatic compressor to 80 psia. Determine the work input to the compressor. Use the tables for R-134a.
The initial enthalpy and the entropy of the saturated water can be found out from the table of A-12E
i.e. [tex]$h_1= 101 \ \text{Btu/lbm}$[/tex]
[tex]$s_1 = 0.22717 \text{ Btu/lbmR}$[/tex]
Since the process mentioned above is an adiabatic compression process, the entropy will remain constant throughout the process. Therefore, we take the value of entropy and the final pressure using the table with few interpolations and also approximations to find the final enthalpy. It is given by :
[tex]$h_2= 116.09 \text{ Btu/lbm}$[/tex]
So the work input from the energy balance equation :
[tex]$\dot{W} + \dot{m}h_1 = \dot{m}h_2$[/tex]
[tex]$w=h_2 - h_1$[/tex]
= 116.09 - 101
= 15.09
Therefore, [tex]$w= 15.09 \text{ Btu/lbm}$[/tex]
Find the derivative of y = sin(ln(5x2 − 2x))
Answer:
[tex]y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)[/tex]
Explanation:
Let [tex]y = \sin[\ln(5\cdot x^{2}-2\cdot x)][/tex] and we proceed to find the derivative by the following steps:
1) [tex]y = \sin[\ln(5\cdot x^{2}-2\cdot x)][/tex] Given
2) [tex]y = \sin [\ln[x\cdot (5\cdot x - 2)]][/tex] Distributive property
3) [tex]y = \sin[\ln x + \ln (5\cdot x - 2 )][/tex] [tex]\ln (a\cdot b) = \ln a + \ln b[/tex]
4) [tex]y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)[/tex] [tex]\frac{d}{dx} (\sin x) = \cos x[/tex]/[tex]\frac{d}{dx}(\ln x) = \frac{1}{x}[/tex]/[tex]\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}[/tex]/Rule of chain/Result
Omg I just got 17/25 questions wrong using this on an Ag test , but got 100’s every time on health
Answer:
sorry im answering questions for the points cuz im built dfferent
Explanation:
Answer: That happens to me too sometimes. I hate when people give the wrong answers...like if u don't know the answer, then don't respond
please help me solve this
Answer:
Following are the solution to this question:
Explanation:
[tex]\to \Sigma F_x = ma_x \\\\\to 3200 t^2 -440 (9.81) (\frac{8}{17}) = 440 (a) \\\\\to 3200 t^2 -2031.24 = 440 (a)\\\\\to a= 7.27t^2 -4.616\\\\\to dv=a.dt \\\\\to \int^{v}_{2} dv = \int^{2}_{0} (7.27t^2 -4.616) dt\\\\\to v-2 = 7.27 \frac{t^3}{3} - 4.616 t^{2}_{0}\\\\\to v-2 = 7.27 \frac{2^3}{3} - 4.616 {2} \\\\\to v-2 = 10.154\\\\\to v= 12.154 \ \frac{m}{s}\\[/tex]
Most of the work that engineers do with fluids occurs in nature. True False
Answer:
It's False
Explanation:
I did the assignment
How did ancient machines pave the way for improvement
Answer:
----------
Explanation:
Ancient machines have paved the way for improvement by being the foundation for change. these machines made it possible to find flaws so in the next generations, they could fix, develop, and produce better quality machines. eventually this process has reached the present time but after today, the process will continue to produce even better quality machines than the time before. I hope this helps!
A set of civil construction plans shows a distance of 131 meters across the front of a piece of property. You need to convert this value to feet. You know that one foot equals approximately 0.305 meters. What is the distance in feet, to the nearest tenth of a foot?
The distance when the civil construction plans shows a distance of 131 meters across the front of a piece of property is 430 feet.
How to calculate the distance?From the information given, it was stated that the civil construction plans shows a distance of 131 meters across the front of a piece of property.
Therefore, since one foot equals approximately 0.305 meters, the distance will be:
= 131/0.305
= 429.5 feet
= 430 feet approximately
In conclusion, the distance is 430 feet.
Learn more about distance on:
https://brainly.com/question/17273444
A 280 km long pipeline connects two pumping stations. It is desired to pump 0.56 m3/s of oil through a 0.62 m diameter line, the discharge station is 250 m lower in elevation than the upstream station, and the discharge pressure is to be maintained at 300 kPa. The oil has a dynamic viscosity of 3.645 x 10-3 Pa·s and a density of 810 kg/m3. The pipe is constructed of commercial steel and the inlet pressure may be taken as 1 atmosphere. Considering the mechanical-energy friction loss Ff, determine the power required to pump the oil if the power efficiency is 70%. Develop a sketch of the piping system considered.
Ignore energy losses due to contraction, expansion & 90° bends!!!
what document is the primary reference document when making ethical decisions?
What is electrical energy being transformed into? How do you know?
For the beam loaded as shown in Fig Q2a. perform the following task:
Calculate the support reactions at A and E;
(i)
Draw the shear force diagram for the beam showing all important
values;
>
(iii)
Draw the bending moment diagram for the beam showing all
important values.
6 kN
10 KN
2 kN/m
B
D
1 m
1 m
1 m
1 m
Answer:
a
Explanation:
aaaaaa
When handling chemicals and solvents, technicians are recommended to
Answer:
To wear PPEHave prior knowledge of explosive levels and elemental propertiesKnow procedure to eliminate any threatAnswer
have PPE on wash hands after
Explanation:
1 british gallon = .....litres.
a:4.546
b:3.785
c.5.456
d.7.385
Answer:
c
Explanation:
I live in britian
are considered a form of personal protective equipment (PPE) for eyes.
Answer:
All of the above
Explanation:
A centrifugal pump is installed in a pipeline to raise water14.9 m into an elevated holding tank. The length of the ductileiron pipeline (f=0.019) connecting the reservoirs is 22.4 m. Thepipe is 5.0 cm in diameter and the performance curve of the pump isgiven by Hp =23.9?7.59Q2 where Hp is in meters, Q is in liters persecond, and the equation is valid for flows up to 1.5 L/s. Usingthis pump, what flow do you expect in the pipeline if minor lossesare ignored? What pump head is required for this flow?
Answer:
Hp = 23.86 m
Hence, 23.86 m head pump is required for this flow.
Explanation:
Solution:
Data Given:
Elevation = Z = 14.9 m
Friction of the ductile iron pipe = f = 0.019
Length of the ductile iron pipe connecting the reservoirs = 22.4 m
Diameter of the Pipe = D = 0.5 cm
Radius of the Pipe = r = D/2 = 0.25 cm
Performance Curve of the Pump = Hp = 23.9 - 7.59[tex]Q^{2}[/tex] , where Hp is in meters.
[tex]Q^{}[/tex] is in Liters/second and equation is valid for 1.5 L/s
We are asked to find out the pump head required for this flow.
We need to apply the energy equation for this problem:
Energy Equation:
[tex]\frac{P1}{\beta 1} + \frac{v1^{2} }{2g} + Z1 + Hp = \frac{P2}{\beta 2} + \frac{v2^{2} }{2g} + Z2 + H_{L}[/tex]
where, [tex]\beta[/tex] = specific weight of the fluid.
v1 = 0
P1 = P2 = P
Z1 = 0
Z2 = 14.9
Hence,
New equation is:
[tex]Hp = \frac{v2^{2} }{2g} + Z2 + H_{L}[/tex]
And
[tex]H_{L}[/tex] = [tex]\frac{fLv^{2} }{2gD}[/tex]
So,
[tex]Hp = \frac{v2^{2} }{2g} + Z2 + \frac{fLv^{2} }{2gD}[/tex]
Now, converting formula of v into terms of Q, we get
Hp = 23.9 - 7.59[tex]Q^{2}[/tex]
[tex]Hp = \frac{Q2^{2} }{2(9.81)(\frac{\pi }{4}(0.05^{2}))^{2} } + 14.9 + \frac{(0.09)(22.4)Q^{2} }{12.1(0.05)^{2} }[/tex]
23.9 - 7.59[tex]Q^{2}[/tex] = [tex]\frac{Q2^{2} }{2(9.81)(\frac{\pi }{4}(0.05^{2}))^{2} } + 14.9 + \frac{(0.09)(22.4)Q^{2} }{12.1(0.05)^{2} }[/tex]
Solving for Q, we will get
9 - 7.59[tex]Q^{2}[/tex] = 13233.71[tex]Q^{2}[/tex] + 112555.37[tex]Q^{2}[/tex]
[tex]Q^{}[/tex] = 7.154 x [tex]10^{-5}[/tex] [tex]m^{3}/s[/tex]
Converting it into L/s
Q = 0.0715 L/s
Putting this value of Q into the Hp equation to get the required answer:
Hp = 23.9 - 7.59[tex]Q^{2}[/tex]
Hp = 23.9 - 7.59 x [tex]0.0715^{2}[/tex]
Hp = Pump head
Hp = 23.86 m
Hence, 23.86 m head pump is required for this flow.
Two technicians are discussing hd-obd for trucks technician a says that the simplest most familiar definition is the diagnostic function of electronic control system to identify or self-diagnose system fault and report fault codes technician b says automotive and light duty truck versions of OBD is called OBD2 who is correct
Answer:
Two technicians are discussing HD-OBD for trucks
Technician a says that the simplest most familiar definition is the diagnostic function of electronic control system to identify or self-diagnose system fault and report fault codes.
Technician A is correct.
Explanation:
Onboard Diagnostics (OBD) is a vehicle's self-diagnostic tool which enables technicians to access the diagnostic information for the purpose of performance monitoring and analyzing repair needs of vehicles. OBD2 is the diagnostic system that is included in every vehicle made after 1996. The diagnostics can check the engine and other vehicle systems and also communicate possible vehicular glitches that happen.
Typical journal-to-bearing clearance is......
A) 0.00015 to 0.00018 inch
B)0.0005 to 0.0025 inch
C) 0.15 to 0.25 inch
D)0.02 to 0.035 inch
The typical journal-to-bearing clearance for engines is,
B) 0.0005 to 0.0025 inch
Given that,
To find the typical journal-to-bearing clearance for engines.
Now, The range 0.0005 to 0.0025 inch represents the recommended clearance between the rotating journal of the crankshaft and the bearing surface to ensure proper lubrication and minimal friction.
Hence, The correct option for typical journal-to-bearing clearance for engines is,
B) 0.0005 to 0.0025 inch.
To learn more about journal-to-bearing visit:
https://brainly.com/question/16971319
#SPJ4
Find and write the mathematical problem formulation of shortest path problem
Which one of the following would appear as a thin line on a drawing?
A. Centerline
B. Cutting plane line
C. Short break line
D. Outline
Answer:
try b
Explanation:
It is preferable not to drive a vehicle in reverse if you have an obstructed view, but if you do, make sure:
Answer:
While driving in reverse gear it necessary to have reverse audible alarm , a spotter and a rear view mirror.
Explanation:
It is prefer that to not drive a vehicle in reverse. But if we are driving in reverse gear it is necessary to have a reverse audible alarm , a spotter and a rear view mirror.
one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller number by 4
Answer:
a = 40
b = 29
Explanation:
Give a place holder for the numbers that we don't know.
Lets call the two numbers a and b.
From the given info, we can write an expression and solve it:
"one number is 11 more than another number"
a = 11 + b
from this, we know that a > b.
''three times the larger number exceeds four times the smaller number by 4"
3a = 4b + 4
Now we have 2 equations, we can use them to solve using whatever method you want.
a = 11 + b
3a = 4b + 4
I will be using matrices RREF to solve for this.
a - b = 11
3a - 4b = 4
[tex]\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}[/tex]
[tex]\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}[/tex]
a = 40
b = 29
A small probe P is gently forced against the circular surface with a vertical force F as shown . Determine the n- and t-components of this force as functions of the horizontal positions.
This question is incomplete, its missing an image which will be uploaded along this Answer.
Answer:
the normal component of force F_n is F((√(r²-s²)) / r)
the tangential component of force F_t is F(s/r)
Explanation:
Given the data in the image;
from the free body diagram, we write the expression for ∅
sin∅ = s/r
cos∅ = (√(r²-s²)) / r
now expression for normal component of force is;
F_n = Fcos∅
we substitute
F_n = F((√(r²-s²)) / r)
Therefore, the normal component of force F_n is F((√(r²-s²)) / r)
Also for force F_t
F_t = Fsin∅
we substitute
F_t = F(s/r)
Therefore, the tangential component of force F_t is F(s/r)
What should Kimi have done to ensure that she was acting legally and ethically?
Answer:
Kimi have done to ensure that she was acting legally and ethically by following the measure that are discussed below in detail.
Explanation:
Legal measures are those measures that are outlined in regulatory laws. Ethical measures are based on the human postulates of morality and wrongdoing. The distinctions between them are these: Legal measures are based on transcribed law, while ethical measures are based on human rights and violations.