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Savatey [412]
2 years ago
14

Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.

The carriage and half-nuts D. A1l of the above
Engineering
1 answer:
Maksim231197 [3]2 years ago
8 0

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts  

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2.11 Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface
alexandr402 [8]

Answer:

The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.

The cost associated with the heat loss through the windows for an 8-hour flight is:

For aerogel windows: $17.472 (most efficient)

For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:

\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

In this case:

\dot Q =k\frac{S}{L} \Delta T

If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).

To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):

Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

6 0
2 years ago
Nec ________ covers selection of time-delay fuses for motor- overload protection.
Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in  National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Learn more about motor- overload from

brainly.com/question/20738481

#SPJ1

6 0
1 year ago
What is the name for a program based on the way your brain works?
xxMikexx [17]

Answer:

KAT

Explanation:

I believe this is what ur looking for

8 0
3 years ago
Read 2 more answers
When should u check ur review mirror
vaieri [72.5K]

Answer:

All the time and to see if someone is behind you.

Explanation:

After you check your side mirrors you should check your review mirror to see who is behind you.

6 0
3 years ago
Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.
bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

<u>Given the following data:</u>

  • Emf = 520 Volts
  • Speed = 660 r.p.m
  • Number of armature conductors = 144 slots
  • Number of poles = 4 poles
  • Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

E = \frac{\theta ZN}{60} × \frac{P}{A}

<u>Where:</u>

  • E is the electromotive force in the DC generator.
  • Z is the total number of armature conductors.
  • N is the speed or armature rotation in r.p.m.
  • P is the number of poles.
  • A is the number of parallel paths in armature.
  • Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

Z = 144 × 2 × 3

Z = 864

Substituting the given parameters into the formula, we have;

520 = \frac{\theta (864)(660)}{60} × \frac{4}{2}

520 = \theta (864)(11) × 2

520 = 19008 \theta \\\\\Theta = \frac{520}{19008}

<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>

Therefore, the magnetic flux per pole is 0.0274 Weber.

Read more: brainly.com/question/15449812?referrer=searchResults

5 0
2 years ago
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