All categories

3 years ago

Question

Engineering

1 answer:

Leto [7]3 years ago

4 0

Answer:

True

Explanation:

The CNC is the primary interface between the machine operator and the machine.

You might be interested in

Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L

Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power $Q1 = P tan(\phi)$

$= 14.91\times tan(cos^{-}0.8) = 11.18 kvar
$

Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power $P_3$ = 0 due to purely capacitiveload

Reactive power $Q_3$ = -20 Var

a) since all three loads are connected in parallel therefore

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

$I = \frac{P}{\sqrt{3}V_{L}}$

$= \frac{34910}{\sqrt{3}\times 480} = 41.99 A$

5 0

3 years ago

Five Kilograms of continuous boron fibers are introduced in a unidirectional orientation into of an 8kg aluminum matrix. Calcula

Lunna [17]

Answer:

Explanation:

Given that,

Mass of boron fiber in unidirectional orientation

Mb = 5kg = 5000g

Mass of aluminum fiber in unidirectional orientation

Ma = 8kg = 8000g

A. Density of the composite

Applying rule of mixing

ρc = 1•ρ1 + 2•ρ2

Where

ρc = density of composite

1 = Volume fraction of Boron

ρ1 = density composite of Boron

2 = Volume fraction of Aluminum

ρ2 = density composite of Aluminum

ρ1 = 2.36 g/cm³ constant

ρ2 = 2.7 g/cm³ constant

To Calculate fractional volume of Boron

1 = Vb / ( Vb + Va)

Vb = Volume of boron

Va = Volume of aluminium

Also

To Calculate fraction volume of aluminum

2= Va / ( Vb + Va)

So, we need to get Va and Vb

From density formula

density = mass / Volume

ρ1 = Mb / Vb

Vb = Mb / ρ1

Vb = 5000 / 2.36

Vb = 2118.64 cm³

Also ρ2 = Ma / Va

Va = Ma / ρ2

Va = 8000 / 2.7

Va = 2962.96 cm³

So,

1 = Vb / ( Vb + Va)

1 = 2118.64 / ( 2118.64 + 2962.96)

1 = 0.417

Also,

2= Va / ( Vb + Va)

2 = 2962.96 / ( 2118.64 + 2962.96)

2 = 0.583

Then, we have all the data needed

ρc = 1•ρ1 + 2•ρ2

ρc = 0.417 × 2.36 + 0.583 × 2.7

ρc = 2.56 g/cm³

The density of the composite is 2.56g/cm³

B. Modulus of elasticity parallel to the fibers

Modulus of elasticity is defined at the ratio of shear stress to shear strain

The relation for modulus of elasticity is given as

Ec = = 1•Eb+ 2•Ea

Ea = Elasticity of aluminium

Eb = Elasticity of Boron

Ec = Modulus of elasticity parallel to the fiber

Where modulus of elastic of aluminum is

Ea = 69 × 10³ MPa

Modulus of elastic of boron is

Eb = 450 × 10³ Mpa

Then,

Ec = = 1•Eb+ 2•Ea

Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³

Ec = 227.877 × 10³ MPa

Ec ≈ 228 × 10³ MPa

The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa

OR Ec = 227.877 GPa

Ec ≈ 228GPa

C. modulus of elasticity perpendicular to the fibers?

The relation of modulus of elasticity perpendicular to the fibers is

1 / Ec = 1 / Eb+ 2 / Ea

1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³

1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6

1 / Ec = 9.376 × 10^-6

Taking reciprocal

Ec = 106.66 × 10^3 Mpa

Ec ≈ 107 × 10^3 MPa

Note that the unit of Modulus has been in MPa,

7 0

3 years ago

Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default

mina [271]

Answer:

Explanation:

As a result of the codes used to making this solved, i was unable to upload my answer, therefore i'm sending a screenshot of the codes.

thank you for your understanding,

cheers i hope this helps

6 0

4 years ago

Two technicians are discussing scan-tool use. Technician A says that a special diagnostic interface device is needed to connect

Anna71 [15]

Answer:

both technicians are correct, I haven't taken the test

7 0

3 years ago

Read 2 more answers

La liberacion de energia se facilita por

lakkis [162]

Answer:

Enzimas

Explanation:

6 0

3 years ago