3 years ago

Discuss with your neighbor your brain as a computer.

1 answer:

6 0

Senors are a type of device that produce a amount of change to the output to a known input stimulus.
Input signals are signals that receive data by the system and outputs the ones who are sent from it. Hope this helps ;)

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A stream of ethylene gas at 250°C and 3800 kPa expands isentropically in a turbine to 120 kPa. Determine the temperature of the

11Alexandr11 [23.1K]

Answer: for ideal situation

T2 = 16.5158K

Work = - 11.4J

For appropriate generalization correlation

T2 = 308.57K

Work = 177.797MJ

Explanation:detailed calculation and explanation is shown in the image below.

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3 years ago

Engineers need to be open-ended when dealing with their designs. Why?

melomori [17]

I think the answer would be A if its wrong I’m sorry

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3 years ago

In software engineering how do you apply design for change?

Pavlova-9 [17]

Answer:

it is reducely very iloretable chance for a software engineer to give an end to this question

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3 years ago

The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi

vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

$R=1+\frac{\rho * K_{d} }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65$

The time will take will be:

$t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days$

b) For toluene:

$R=1+\frac{2.6*3.3 }{0.37} = 24.19$

$t=\frac{12*24.19}{0.18} =1612.67days$

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3 years ago

Read 2 more answers

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in

Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:

σ = FL / π $R^{3}$

= 3000 $\times$ (40 $\times$ 10^-3) / π (5 $\times$ 10^-3)^3

σ = 305 $\times$ 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

= 2(305 $\times$ 10^6) (15 $\times$ 10^-3)^3 / 3(40 $\times$ 10^-3)

F = 17156 N.

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3 years ago