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SVETLANKA909090 [29]
3 years ago
7

Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te

rminals can be damaged by over tightening the terminal bolt. Who is correctA. Technician AB. Technician BC. Both A and BD. Neither A and B.
Physics
1 answer:
zlopas [31]3 years ago
4 0

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

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A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s
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Answer:

The axial force is  P =  15.93 k N

Explanation:

From the question we are told that

     The diameter of the shaft steel is  d =  50mm

      The length of the cylindrical bushing  L =100mm

     The outer diameter of the cylindrical bushing  is  D =  70 \ mm

       The diametral interference is \delta _d = 0.005 mm

       The coefficient of friction is  \mu = 0.2

       The Young modulus of  steel is  207 *10^{3} MPa (N/mm^2)

The diametral interference is mathematically represented as

           \delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}

Where P_B is the pressure (stress) on the two object held together  

     So making P_B the subject

            P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}

Substituting values

                P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }

                 P_B = 5.069 MPa

Now he axial force required is

             P =  \mu * P_B * A

Where A is the area which is mathematically evaluated as

               \pi d l

So   P  =  \mu P_B \pi d l

Substituting values

      P =  0.2 * 5.069 * 3.142 * 50 * 100

       P =  15.93 k N

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