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kow [346]
1 year ago
11

There are signs of oil spray on the compressor clutch hub and nearby underhood areas. Technician A says that a faulty compressor

clutch bearing could be the cause. Technician B says that a leaking compressor shaft seal could be the cause. Who is correct
Engineering
1 answer:
Nina [5.8K]1 year ago
8 0

The statements regarding the compressor as expressed by the technicians are correct. Hence, Both the technicians are correct.

<h3>What is a compressor?</h3>

A compressor is a mechanical tool that will increase the pressure of a gas with the aid of using decreasing its volume. An air compressor is a selected form of gas compressor.

Hence, The statements regarding the compressor as expressed by the technicians are correct. Hence, Both the technicians are correct.

learn more about compressors:

brainly.com/question/10174214

#SPJ1

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The art of manipulating, influencing, or deceiving you into taking some action that isn’t in your best interest or in the best i
Mrac [35]
Social engineering is the answer
7 0
2 years ago
Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downward
andrew-mc [135]

Answer:

We choose PTFE

Explanation:

Attached are the modulus density and modulus strength chart.

Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.

8 0
3 years ago
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea
vovangra [49]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

3 0
2 years ago
11. As __and___ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.
zvonat [6]

Answer:

Explanation:

whats the answeres two the question do they have a choise for you

5 0
10 months ago
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%

Where:

V_{gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{fe} - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%

Where:

m_{fe}, m_{gr} - Masses of the ferrite and graphite phases, measured in grams.

\rho_{fe}, \rho_{gr} - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%

\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%

If \rho_{gr} = 2.3\,\frac{g}{cm^{3}}, \rho_{fe} = 7.9\,\frac{g}{cm^{3}}, m_{gr} = 3.2\,g and m_{fe} = 96.8\,g, the volume percentage of graphite is:

\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%

\%V_{gr} = 10.197\,\%V

The volume percentage of graphite is 10.197 per cent.

5 0
3 years ago
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