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cluponka [151]
1 year ago
12

when discussing valve train components, technician a says stamped rocker arms are very strong and may be used in high-horsepower

engines. technician b says forged rocker arms are used in high-torque, high-horsepower applications. who is correct?
Engineering
1 answer:
mojhsa [17]1 year ago
5 0

From the discussion we know that tehcnician A is the correct statement abaout valve train components. Because of the stamped rocker arms arms are more resistant to more aggressive engine conditions.

<h3>Why stamped rocker arms are very strong for hig-horsepower engines? </h3>

Whether you're dealing mostly with small blocks from a junkyard or a heavily modified race engine, rocker arms play a big role in unleashing horsepower and reducing friction in the valve train. Automotive rocker arms are typically made of stamped steel or aluminum for high speed applications. Some rocker arms (called roller rocker arms) incorporate bearings at the contact points to reduce wear and friction at the contact points. So we know that technician A has correct statement.

Learn more about valve high component: brainly.com/question/28251926

#SPJ4

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The condensed Q formula may be used for operations in which the friction loss can be determined for:
yan [13]

The condensed Q formula may be used for operations in which the friction loss can be determined for a: 3, 4, or 5 inch hose.

<h3>What is a firehose friction loss?</h3>

A firehose friction loss can be defined as a measure of the effect of the resistance of water against the inner side of a firehose, which typically results in a pressure drop at the terminal end.

Generally, some of the factors that affect the resistance or friction in a firehose include:

  • Length of hose.
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Mathematically, the firehose friction loss can be calculated by using this formula:

FL = C × (Q/100)² × L/100.

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3 0
2 years ago
If the same amount of force were applied to all four balls in the picture, which would experience the greatest change in motion?
nydimaria [60]
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3 years ago
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25 points and brainliest is it A, B, C, D
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3 years ago
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The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the
Alina [70]

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

<u>Calculate the Principal stress, maximum in-plane shear stress and average normal stress</u>

Using Mohr's circle ( attached below )

<u>i) principal stresses:</u>

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

<u>ii) maximum in-plane shear stress</u>

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

<u>iii) average normal stress</u>

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

8 0
3 years ago
The current drawn by fluorescent lighting has a high total harmonic distortion. For this case, THD is calculated to be 88%. The
Alona [7]

Answer:

displacement power factor is 0.959087

Explanation:

given data

THD = 88%

true power factor = 0.72

solution

we get here total harmonic distribution THD is express as here

THD = \sqrt{\frac{1}{g^2}-1}       ..............1

her g is distortion factor

so put here value and we will get g that is

0.88² =   \frac{1}{g^2} -1    

solve it we get

g = 0.750714

and

displacement power factor is express as

DPF = \frac{PF}{g}   .................2  

put here value and we will get

DPF = \frac{0.72}{0.750714}    

DPF  = 0.959087

3 0
3 years ago
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