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alexandr1967 [171]
3 years ago
9

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
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