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3 years ago

In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.

Engineering

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.

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Answer: speaks Portuguese

Eu disse a todos a tradução para que possam te ajudar

Explanation: Y’all can help I have no idea

QUESTION 13. Explain the use of quotation marks in the excerpt "I will give each of you

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3 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng

Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

Decrease the length of the pipes: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
Reduce the surface roughness of the pipes: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
Increase the pipe diameter: By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

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1 year ago

Common car loan duration

chubhunter [2.5K]

Answer:

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Explanation:

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3 years ago

A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0

Aneli [31]

Answer:

Q=339.5W

T2=805.3K

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.

3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature

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4 years ago