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irina [24]
3 years ago
9

9. Technician A says to examine each bearing carefully for chips, pits, and scratches during servicing. Technician B says that d

iscoloration of a bearing
Indicates normal, acceptable wear. Who is correct?
A. Technician A only
B. Technician B only
©C. Both Technician A and B
D. Neither Technician A nor B
Engineering
2 answers:
Vanyuwa [196]3 years ago
8 0

Answer:

a. is correct.

Explanation:

During servicing you should examine each bearing carefully for chips, pits, and scratches, so Technician A is correct.

Discoloration of a bearingbis not acceptable wear, so Technician B is false.

Combine the answers: Only Technician A is correct therefore answer a. is the right answer.

Damm [24]3 years ago
4 0

Answer:

a

Explanation:

because we should see everything carefully

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2 years ago
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

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\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

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A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

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Answer:

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