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podryga [215]
3 years ago
6

Why should a toolpath be verified on the screen of a CAM system prior to creating the program code?

Engineering
1 answer:
NISA [10]3 years ago
5 0

Answer:

The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons

1. analyze the machining strategy and identify which one is better for each piece.

2.Avoid the collision of the tool holder with the work piece.

3.Avoid the shock of the tool with the piece.

4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.

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A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
4 0
3 years ago
Read 2 more answers
You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from
Aleksandr [31]

Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

e) The gain K increases above % error as the  steady state speed increases

Explanation:

Given data:

Motor voltage  = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > \frac{g}{200}

7 0
3 years ago
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th
ddd [48]

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

4 0
3 years ago
Another name for your computer, running the web browser program is: Web user The client The mainframe Browsing agent
timofeeve [1]

Answer:

Browsing agent

Explanation:

Hope this helps!

6 0
3 years ago
Technician A says that excessive end play causes rapid wear of the sealing rings and possibly the sealing ring groove. Technicia
DerKrebs [107]

Answer:

Technician A only

Explanation:

Rapid wear of the sealing ring and the seal ring groove are result of excessive end play

7 0
3 years ago
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