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Gnom [1K]
3 years ago
5

Increasing rake angle will decrease the tendency for Built-Up-Edge to form. a) True b) False

Engineering
1 answer:
anygoal [31]3 years ago
7 0

Answer:

a)True

Explanation:

While machining of ductile material and high feed and low cutting speed welding action take place between tool material and chip material this welding action is called built up edge. Built up edge action takes place due to high temperature.

To decrease the built up edge action

1. Increase the rake angle.

2. Increase the cutting speed

3. Decrease the feed rate

4. Use cutting fluid

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Consider a hot working operation in which there is no work hardening, but the flow stress is sensitive to the strain rate. Calcu
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Explanation:

You can use Hollomon's equation to estimate this.

Y_f is the flow stress.

K is the strenght coefficient or constant.

e is the strain

n is the strain hardening exponent.

You should be able to solve for e given the data in the problem statement.

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What would the Select lines need to be to send data for the fifth bit in an 8-bit system (S0 being the MSB and S2 being the LSB)
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Answer:

A. S0 = 1, S1 = 0, S2 = 0

lines need to send data for the fifth bit in an 8 bit system

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Future solution for air pollution in new zealand
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Answer:

New Zealand may use some of these solutions to prevent air pollution

Explanation:

Using public transports.

Recycle and Reuse

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The two basic categories of electrical switches are manual and ??
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Read 2 more answers
A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i
Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

so

E₁ = \frac{1}{2}mv²

we substitute

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m =  5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = \frac{1}{2}mv₂²

we substitute

E₂ = \frac{1}{2} × 1082.2065 × ( 43.362 )²

E₂ = \frac{1}{2} × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

6 0
3 years ago
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