Can anyone tell me all the corrects answers to these? I’m sorry if this is the wrong subject I’m not sure what to put it under but I really need help!
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Answer 1
Answer:

Answer:

Crankshaft position sensor - F     I can't quite make out the letter but it's the thing at the bottom almost touching the notched wheel.

Coil Module - B  

Knock Sensor - D

Coil Pack -E

Fuse Block - A

Powertrain Control Module - C


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The average flow speed in a constant-diameter section of the Alaskan pipeline is 2.5 m/s. At the inlet, the pressure is 8.25 MPa (gage) and the elevation is 45 m; at the outlet, the pressure is 350 kPa (gage) and the elevation is 115 m. Calculate the head loss in this section of pipeline.

Answers

Answer:

head loss = 805.327 m

Explanation:

given data

average flow speed v = 2.5 m/s

inlet pressure Pi = 8.25 MPa

elevation Zi =  45 m

outlet pressure Po = 350 kPa

elevation Zo = 115 m

we consider oil Specific Gravity = 0.92

to find out

head loss in this section of pipeline

solution

we find here head loss that is inlet and outlet  

Hi = (Pi)/(\rho g) +(Vi^2)/(2g) +Zi    ..............1

put here value  

Hi = (8.25*10^6)/(920*9.81) +(2.5^2)/(2*9.81) +45

Hi = 959.425 m

and

Hout = (Pout)/(\rho g) +(Vout^2)/(2g) +Zout    ..............2

put here value  

H out = (350*10^3)/(920*9.81) +(2.5^2)/(2*9.81) +115  

H out = 154.098 m

so  

head loss is = Hi - H out  

head loss is = 959.425 - 154.098  

head loss = 805.327 m

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?

Answers

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

Y = 1 - e^(-kt^n)

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^(-k^146^2.1)

0.50 = e^(-k306.6)

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26* 10^(-3)

for 86 % completion

0.86 = 1 - e^{-2.26* 10^(-3) * t^(2.1)}

e^{-2.26* 10^(-3) * t^(2.1)} = 0.14

-2.26* 10^(-3) * t^(2.1) = ln(0.14)

t^(2.1) = 869.96

t = 25.10 sec

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.​

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

  • Vertical slope, = 2
  • Horizontal slope = 3
  • Clockwise moment = 330 ft-lb
  • Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

  • the A coordinate = (0, 5)
  • the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2,  2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

What is a SAFETY CHECK

Answers

Answer:

Safety check is defined as rounding to make sure that the patients and the milieu (patients living quarters) is secured and free of harmful items that can be used to hurt someone.

Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?

Answers

Answer: Provided in the explanation section

Explanation:

Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.

In this style, leader lays emphasis on the relation with the followers.

It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.

It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.

It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.

A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.

 

cheers i hope this helped !!

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer:

v = 250[1 - {e^(-6000t)}] mV

Explanation:

The voltage across a capacitor at a time t, is given by:

v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)                 ----------------(i)

Where;

v(t) = voltage at time t

t_(0) = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3e^(-6000t) mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)    

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt            

v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt             [Solve the integral]

v = (3)/(2*10^(-6)*(-6000))  {e^(-6000t)}|_0^t

v = (-3000)/(12)  {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]

v = -250 {e^(-6000t)} - [-250]

v = -250 {e^(-6000t)} + 250

v = 250 -250 {e^(-6000t)}

v = 250[1 - {e^(-6000t)}]

Therefore, the voltage across the capacitor is v = 250[1 - {e^(-6000t)}] mV

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