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4 years ago

Two factors that can be modified to optimize the cutting processes are feed and depth, and cutting velocity. a) True b) False

Engineering

1 answer:

Kisachek [45]4 years ago

4 0

Answer:

True

Explanation:

The process of cutting is used to cut an object with the help of physical forces.

This process includes cutting like shearing, drilling, etc.

The cutting process makes use of the mechanical tools to maintain the contact of cutter with the object.

This process can be optimized by modification of the cutting depth, cutting velocity and feed.

The process optimization also depends on the cutting fluid as these are the deterministic factors for the cutting condition.

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Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

AURORKA [14]

Answer:

a) $w_{out} = 281.55\,\frac{kJ}{kg}$ , b) $s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

$-q_{out} - w_{out} + h_{in}-h_{out} = 0$

$w_{out} = h_{in} - h_{out}-q_{out}$

$w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}$

$w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}$

$w_{out} = 281.55\,\frac{kJ}{kg}$

b) The entropy production is determined after the Second Law of Thermodynamics:

$-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0$

$s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}$

$s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)$

$s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)$

$s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

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3 years ago

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Answer:

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3 years ago

The technique of smoothing out joint compound on either side of a joint is known as which of the following

Black_prince [1.1K]

Answer:

a is the answer to your questions

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3 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

cohesion = 25 kPa

internal friction angle = 5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

and we know formula for $N_{\gamma }$ is

$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

so here Ф is very less $N_{\gamma }$ should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0

4 years ago

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$

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3 years ago