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Furkat [3]
3 years ago
14

What is the difference between backstopping and braking in terms of the applications in which they are used?

Engineering
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

Explained.

Explanation:

Backstopping and Braking both are used for stopping a rotating or translating object. Both are used in braking system of a machine.

Backstopping is generally used to halt the rotation of a shaft in one direction only whereas braking is used in halting a shaft in all direction. Backstopping is used in clutches in backstop clutches.

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On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned
Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

  • As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
  • As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

Learn more about project teamwork here:

brainly.com/question/14279121

7 0
3 years ago
The team needs to choose a primary view for the part drawing. Three team members make suggestions:
dexar [7]

Answer:

<u>Option 1</u>

Explanation:

As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :

  • <u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>
6 0
2 years ago
A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps,
masha68 [24]

Answer:

(a). max possible efficiency = 55.62%

(b). max power output = = 133.5 MW

Explanation:

From the question we were given the Maximum temperature in the system as

Tmax = 500°C

Minimum temperature in the system Tmin = 70°C

the Heat supplied to the boiler Qb = 240000 KJ/s

we use the temperature conversion factor from °C to K

given T(K) = T (°C) + 273

⇒ Tmax = 500 + 273 = 773 K

⇒ Tmin = 70 + 273 = 343 K

(a). we are to determine the maximum possible thermal efficiency;

(Πth)max = 1 - Tmin/Tmax

(Πth)max = 1 - 343/773  = 0.5562

(Πth)max = 55.62%

(b). to determine the maximum possible power output for the plant we have;

(Πth)max = Wmax/Qb

where Wmax rep the maximum power output

(Πth)max = 0.5562

Qb = 240000

∴ Wmax = 0.5562 × 240000 = 133505.6 Kw

Wmax = 133.5 MW

cheers i hope this helps

3 0
3 years ago
The current though a capacitor cannot change instantaneously.
malfutka [58]

Answer: The answer is (b) False

Explanation:

5 0
3 years ago
Read 2 more answers
Air pollutants produced by mining activities,and their characteristics​
marishachu [46]

Answer:

All activities during ore extraction, processing, handling, and transport depend on equipment, generators, processes, and materials that generate hazardous air pollutants such as particulate matter, heavy metals, carbon monoxide, sulfur dioxide, and nitrogen oxides.

Explanation:

5 0
4 years ago
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