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3 years ago

Since the engineering design process may take the engineer back to its beginning, the process is considered ________

Engineering

1 answer:

Valentin [98]3 years ago

7 0

Answer:

Cyclical

Explanation:

I looked at the next question on edgenuity and it said it in the question.

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10. What does a profile of a river from its headwaters to its mouth typically show?

irina1246 [14]

Answer:

c. an abrupt increase followed by a gradual decrease

Explanation:

At the headwater, the flow gradient starts high but then slowly decreases as the river moves downstream to its mouth.

3 0

2 years ago

Which clauses in the Software Engineering Code of Ethics are upheld by a whistleblower (check all that apply). a. "Respect confi

garri49 [273]

Answer:

c. and d

Explanation:

As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,

In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.

6 0

3 years ago

How can I draw this image in 2D form

Ket [755]

Answer:

no it is not 2D

Explanation:

it is 3D

ok so follow these steps

- make hole

-make square

-make triangle

ok now your figure is ready

5 0

2 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-10

masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = $\int\limits^t_0 {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\$

$3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750$

5 0

3 years ago