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2 years ago

If you could help it would mean alot.

Engineering

2 answers:

scoundrel [369]2 years ago

8 0

Answer:

D is the correct choice.

Explanation:

I'm assuming that this is probably a phase in the textbook or progarm you are studying, and this is just a matter of reading thoroughly.

Engineers usually benefit from catching a mistake, and would also benfit from keeping record of a misstep in order to remain clear of that mistake in the future.

Have a great day, and mark me brainliest if I am most helpful!

gulaghasi [49]2 years ago

3 0

Answer:d

Explanation:

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A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter is = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Then

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which is =50 watt

Power =∅ΔP

50 = 1273885.3 ∅

5 0

2 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

2 years ago

Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?

lianna [129]

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents a crystal system which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have three axes which have only two of its axes equal in length.

5 0

2 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$