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ivann1987 [24]
3 years ago
13

Parting tool purpose

Engineering
1 answer:
TiliK225 [7]3 years ago
6 0

Answer: Parting uses a blade-like cutting tool plunged directly into the workpiece to cut off the workpiece at a specific length. It is normally used to remove the finished end of a workpiece from the bar stock that is clamped in the chuck. Other uses include things such as cutting the head off a bolt.

Explanation:

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Identify and describe the three stages of a life cycle analysis.
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The LCA process is a systematic, phased approach and consists of four components: goal definition and scoping, inventory analysis, impact assessment, and interpretation. The standards are provided by the International Organisation for Standardisation (ISO) in ISO 14040 and 14044, and describe the four main phases of an LCA: Goal and scope definition. Inventory analysis. Impact assessment.

Hope this is helpful
5 0
3 years ago
Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh
Nutka1998 [239]

Answer:

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Explanation:

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3 years ago
Since the engineering design process may take the engineer back to its beginning, the process is considered ________
Valentin [98]

Answer:

Cyclical

Explanation:

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7 0
3 years ago
A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.783}{1}*100

DeltaV percent=21.7%

5 0
3 years ago
Antilaser eyewear should be worn when a laser​ level's output is greater than​ ____
creativ13 [48]

Answer: 5mW

Explanation:

8 0
3 years ago
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