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Brilliant_brown [7]
3 years ago
8

List three reasons for surfacing metals.

Engineering
2 answers:
Dmitrij [34]3 years ago
8 0
Chemical,fatigue,and metal transfer or adhesion.
Solnce55 [7]3 years ago
4 0

Answer:

JUST BECUASE

WHY NOT

LOOKS BETTER

Explanation:

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Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support fo
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solution in the picture attached

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One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
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False i think it would be
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Air flows through a convergent-divergent duct with an inlet area of 5 cm² and an exit area of 3.8 cm². At the inlet section, the
Luda [366]

Answer:

The mass flow rate is 0.27 kg/s

The exit velocity is 76.1 m/s

The exit pressure is 695 KPa

Explanation:

Assuming the flow to be steady state and the behavior of air as an ideal gas.

The mass flow rate of the air is given as:

Mass Flow Rate = ρ x A1 x V1

where,

ρ = density of air

A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²

V1 = inlet velocity = 100 m/s

For density using general gas equation:

PV = nRT

PV = (m/M)RT

PM/RT = ρ

ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k

ρ = 7.11 kg/m³

Therefore,

Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)

<u>Mass Flow Rate = 0.27 kg/s = 270 g/s</u>

Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:

Mass Flow Rate = ρ x A2 x V2

where,

ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)

A2 = exit area = 5 cm² = 5 x 10^-4 m²

V2 = exit velocity = ?

Therefore:

0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2

<u>V2 = 76.1 m/s</u>

Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:

P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2

Since, both inlet and exit are at same temperature.

Therefore, h1 = h2, and those terms will cancel out.

P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²

P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²

P2 = P1 + (1/2) ρ (V1² - V2²)

P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]

P2 = 680000 Pa + 14962.25 Pa

<u>P2 = 694962.25 Pa = 695 KPa</u>

4 0
2 years ago
How deep does electrical conduit need to be buried?
GenaCL600 [577]
In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.
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What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should ne
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Answer:

What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value

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