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3 years ago

Describe the are of mechanical engineering

Engineering

2 answers:

Alja [10]3 years ago

8 0

Bcc. Chilling. Vvcfgjkl

posledela3 years ago

5 0

Answer:

Mechanical engineering is an engineering discipline that combines engineering physics and mathematics principles with materials science to design, analyze, manufacture, and maintain mechanical systems.

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What is the difference Plastic vs elastic deformation.

Reika [66]

Answer:

What is the difference Plastic vs elastic deformation

Explanation:

The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.

The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.

8 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago

1. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined i

siniylev [52]

Answer:

don't know

Explanation:

huhuh

8 0

3 years ago

A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2

Furkat [3]

Answer:

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

$\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}$

The initial pressure is:

$P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}$

The speed at outlet is:

$v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}$

$v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }$

$v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )$

The initial pressure is:

$P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}$

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

7 0

3 years ago

Read 2 more answers

A cone-shaped part is to be fabricated using stereolithography. The radius of the cone at its base = 35 mm, and its height = 50

iren [92.7K]

Explanation:

volume = πR²h/3

= πx35²x50/3

= 192325/3

= 64166.725

number of layers n = 50mm/0.05

n = 1000layers

average volume = 64166.725/1000

= 64.167mm³

average area = 64.167/0.05

= 1283.34mm²

average time = 1283.34/900mm x 0.22

= 6.48

6.48 + 15 seconds

= 21.48 seconds

time required = 1000x21.48

= 21480 seconds

convert to minutes

21480/60

= 358 minutes

21480/3600

= 5.967 hours

8 0

3 years ago