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goblinko [34]
3 years ago
11

Ball joints on a vehicle equipped with MacPherson struts are being inspected for wear. Which of the following would be the corre

ct method to check the load bearing joint?
Question 2 options:

Leave the vehicle weight on the tires, turn the steering wheel and listen for a popping noise.


Raise the vehicle by the subframe and measure the joints vertical movement.


Raise the vehicle by the lower control arm and measure the joints vertical movement.


Raise the vehicle by the subframe and turn the steering wheel and listen for a popping noise.
Engineering
1 answer:
Aleks04 [339]3 years ago
5 0

he following would be the correct method to check the load bearing joint is Raise the vehicle by the subframe and turn the steering wheel and listen for a popping noise.

Explanation:

  • Ball joints on a vehicle equipped with MacPherson struts are being inspected for wear. It can be checked by raising the vehicle by the subframe and turn the steering wheel and listen for a popping noise.
  • When a joint of CV becomes rusted or damaged, the flexibility of axle becomes out of order and starts making a constant clicking noise whenever the wheels are turned around.
  • The struts are the most important part of the suspension system.
  • The metallic clunking noise is found to be one of the most noticeable and frequent symptoms that shows of a bad ball joint which gives a bad clunking or knocking noise whenever the suspension operates in an up and down motion.
  • The ball joint can be broken in the following ways:
  • the ball detaching from the socket and stud breakage, nonetheless of the type of breakage, but it is catastrophic.
  • When the joint of the ball completely breaks down, the wheel becomes free to move in all the directions.

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Answer:

Check  the 2nd, 3rd and 4th statements.

Explanation:

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3 years ago
A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and
Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

    = \sqrt{79837.128< 4.899^o}   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

   = \sqrt{1.5456*10^{-6} < 175.1^0} * 500

   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

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In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

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                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

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A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

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Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

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