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jekas [21]
2 years ago
8

Which type of work is a textile engineer most likely to do?

Engineering
1 answer:
pickupchik [31]2 years ago
7 0

Answer:

B because as a textile engineer, your job is to help design and create fabric, including the equipment and materials needed for fabrication.

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Saferty precautions of drill press​
OverLord2011 [107]
  1. Run drill at correct RPM for diameter of drill bit and material. ...
  2. Always hold work in a vise or clamp to the drill table.
  3. Use a correctly ground drill bit for the material being drilled. ...
  4. Use the proper cutting fluid for the material being drilled. ...
  5. Remove chips with a brush, never by hand.

3 0
2 years ago
Read 2 more answers
Imagine you work for the public housing agency of a city, and you have been charged with keeping track of who is living in the a
SOVA2 [1]

In this question, we are missing some of the information that is necessary in order to answer this question properly. However, we can look at what a relational database is in order to help you answer the question on your own.

A relational database is a set of tables from which data can be accessed. This can take place even without the need to reorganize the database tables. The programming interface of a relational database is the Structured Query Language (SQL). This approach was invented by E. F. Codd, who came up with it in 1970 while he was a programmer at IBM.

5 0
2 years ago
Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
motikmotik

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

8 0
3 years ago
In your opinion...
ch4aika [34]

Answer:no

TTHANLS FOR FREE POINTS

Explanation:

8 0
2 years ago
When would working with machinery be a common type of caught-in and caught-between<br> hazard?
tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between  hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

6 0
2 years ago
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