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scoundrel [369]
2 years ago
15

) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p

ulses per revolution, what will be the number of pulses produced by a linear displacement of 200 mm?
Engineering
1 answer:
andrey2020 [161]2 years ago
5 0

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

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Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

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Temperature at saturation = 120°C

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Find attached the full solution to the question.

3 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

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<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

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cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm
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Answer:

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Explanation:

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Velocity (v):

= 40 mm/s

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= 38\times 38 \ mm^2

= 1444 \ mm^2

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= d\times v

= 1.25\times 40

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= \frac{1444}{50}

= 28.88 \ sec

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= 15 \ sec

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= 15+28.88

= 43.88 \ sec

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6 0
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