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2 years ago

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

Engineering

1 answer:

NISA [10]2 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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Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa

DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l ) = ∝ * ΔT

( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )

∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0

3 years ago

Find the velocity and acceleration of box B when point A moves vertically 1 m/s and it is 5 m

Goshia [24]

Answer:

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7 0

3 years ago

What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF

exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

C=capacitance

V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C= $\frac{Q}{V}$

=500× $10^{-12}$ × $\frac{1}{250}$

=2× $10^{-12}$

=2 pF

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3 years ago

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kow [346]

Answer:eyesight

Explanation:

7 0

3 years ago

Read 2 more answers

• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma

elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

$T(n) = 10\times 2n$

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

$\frac{T(k)}{T(n)} = 64$

$\frac{20k}{20n} = 64$

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0

3 years ago