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4 years ago

Does anyone know what I’m doing wrong here?

Engineering

1 answer:

Bingel [31]4 years ago

6 0

Answer:

It looks fine

Explanation:

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Which option identifies the goal in the user story in the following scenario?

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Answer:<u> to purchase organic jams</u>

Explanation:

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3 years ago

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For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an

VashaNatasha [74]

Answer:

Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

3 0

3 years ago

determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004

Afina-wow [57]

Answer:

hello your question is incomplete attached below is the complete question

A) optimum compressor ratio = 9.144

B) specific thrust = 2.155 N.s /kg

C) Thrust specific fuel consumption = 1670.4 kg/N.h

Explanation:

Given data :

Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k

γ = 1.4

attached below is the detailed solution

6 0

3 years ago

Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

DedPeter [7]

Answer:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

Explanation:

The internal energy is equal to:

$dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV$

The entropy is equal to:

$dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV$

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

$\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }$

Replacing:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

4 0

3 years ago

A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus

iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}\\\\$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5 $\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

7 0

3 years ago