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ale4655 [162]
2 years ago
13

can anyone give me tips on adding HP to my 2014 dodge charger SE? it uses the 3.6L220 CI 24 valve v6 vvt (variable valve timing)

Pentastar engine. I'm looking to make 350-400 hp as it is 292 stock. If it helps to know, the motor has about 105k miles on it.
Engineering
1 answer:
marissa [1.9K]2 years ago
5 0
<h2>ANSWER</h2><h2></h2>

I had a couple of answers for this, but when I checked nothing

was right, so im not sure.

<h2></h2>

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Time complexity of merge sort
vovangra [49]

Answer:

The correct answer is "O (n\times Log n)". A further explanation is given below.

Explanation:

  • Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be O (n\times Log n) as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
  • As an unsorted array, it needs an equivalent amount of unnecessary capacity. Therefore, large unsorted arrays are not appropriate for having to search.
6 0
3 years ago
4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer h
brilliants [131]

Answer:

V_{p (load)} = 28,3 V - 0,7 V = 27,6 V

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

Explanation:

The peak voltage after the 6 to 1 step down is V_{p} = \frac{120}{6} \sqrt{2} =  28,3V. Then, the peak voltage of the rectified output is V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi}  = \frac{55,2 V}{\pi } = 17,6 VV_{d}[/tex] and according to the statement, the diodes can be modeled to be V_{d} = 0,7 V. Then, the peak voltage in the load is V_{p (load)} = 28,3 V - 0,7 V = 27,6 V.

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The average current in the load is calculated as:

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

4 0
3 years ago
Describe a simple process
Sunny_sXe [5.5K]

Answer:

Use linking words.

Explanation:

8 0
3 years ago
While discussing the diagnosis of an EI system in which the crankshaft and camshaft sensor tests are satisfactory but a spark te
Lubov Fominskaja [6]

Answer:

Both A and B  are correct.

Explanation:

The crankshaft sensor is attached to the crankshaft pulley on the

back side of the crankshaft. It is used to pass a signal to the PCM

which controls the ignition of an engine. This sensor helps in providing some basic timing for the signal. If there is a faulty sensor present, the signal will not be received by the PCM. So, it will affect the engine while starting. One reason can be a defective

assembly of the coil.

If there is a faulty \mathrm{PCM}, the signal will not be received properly. This will not allow the passing of signals to the ignition system. This will cause a misfire. One of the most common causes of

misfire is a faulty ignition coil. This will affect the working of a

sensor. So, prevent engine from starting.

5 0
2 years ago
Subject : SCIENCE
Leviafan [203]

Answer:

Increase the eletric current

slow down the resistance

Explanation:

7 0
2 years ago
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