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3 years ago

15

Which option identifies what engineers will do to help in the following scenario?

2 answers:

Tasya [4]3 years ago

4 0

Answer:

a

Explanation:

MariettaO [177]3 years ago

3 0

Answer:

D.

Explanation:

"Engineers and scientists immediately worked diligently to design a system that would cap off the pipeline and stop the flow of the petroleum."

Trust me I know this is accurate because I have a reference.

Give brainliest

Thank you,

- massmaster34

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Resistance to impact is an example of a(n)

Anika [276]

Answer:

Mechanical property

Explanation:

MECHANICAL PROPERTIES can be defined as the ability of a metal or material to remain undamaged after different type of forces has been applied or used on them because forces or loads are often applied to metal, material or physical properties which is why MECHANICAL PROPERTIES enables us to know the strength , toughness as well as the hardness of metal and the way this metal perform or react when different forces are applied on them.

Lastly any metal, material or physical properties that has the strength , hardness and resistance to withstand or remain unaffected despite the loads or forces use on them is an example of MECHANICAL PROPERTIES.

Therefore Resistance to impact is an example of a(n) MECHANICAL PROPERTIES.

8 0

3 years ago

A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,

disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0

4 years ago

Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for

lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0

3 years ago

When you first start a car after it has been sitting for more than an hour, it pollutes up to ......times more than when the eng

lesya [120]

20 time more then when it’s warm

6 0

4 years ago

Read 2 more answers

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

6 0

3 years ago