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1 year ago

Which is not one of the primary characteristic of unit testing:

Engineering

1 answer:

Neporo4naja [7]1 year ago

6 0

Answer:

Unit testing is a software development process in which the smallest testable parts of an application, called units, are individually and independently scrutinized for proper operation. This testing methodology is done during the development process by the software developers and sometimes QA staff. The main objective of unit testing is to isolate written code to test and determine if it works as intended.

Unit testing is an important step in the development process, because if done correctly, it can help detect early flaws in code which may be more difficult to find in later testing stages.

Unit testing is a component of test-driven development (TDD), a pragmatic methodology that takes a meticulous approach to building a product by means of continual testing and revision. This testing method is also the first level of software testing, which is performed before other testing methods such as integration testing. Unit tests are typically isolated to ensure a unit does not rely on any external code or functions. Testing can be done manually but is often automated. It might be helpful

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The most important element of green construction is that it is a(n) __________ approach to building. *

madam [21]

Environmentally friendly

Since it focuses on are sustainable and efficient with and are made with the future in mind.

5 0

3 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago

Whats the boolean expression of this circuit?

Natalija [7]

Answer:

G8 = x0'x2' +x0'x3' +x1x2

Explanation:

The expression can be written different ways, depending on the need to avoid hazards. One of them is ...

$G_8=\overline{X_0}\,\overline{X_2}+\overline{X_0}\,\overline{X_3}+X_1X_2$

A truth table and Karnaugh map are shown for the circuit. The terms used in the Boolean expression come from the corners, the upper half of the left- and right-columns, and the right half of the middle two rows. If a static hazard is to be avoided, a term x1x0' could be added representing the right column.

8 0

2 years ago

134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.

vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = $- 10^{\circ}C$ = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = $60^{\circ}C$ = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T = $60^{\circ}C$ :

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at P = 140 kPa, T = $- 10^{\circ}C$ :

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = $s'_{s}$ and $h'_{s} = 278.06 kJ/kg.K$ at 700kPa

(a) Isentropic efficiency of the compressor, $\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}$

$\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%$

(b) The temperature of the environment, $T_{e} = 27^{\circ}C$ = 273 + 27 = 300 K

Availability at state 1, $\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg$

Similarly for state 2, $\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg$

Now, the efficiency of the compressor as per the second law;

$\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757$ = 67.57%

4 0

3 years ago

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m

inn [45]

Answer:

$2.9\times 10^{-6}$

Explanation:

$Q_s$ = Energy for defect formation = 1.86 eV

T = Temperature = $573^{\circ}\text{C}=573+273.15=846.15\ \text{K}$

k = Boltzmann constant = $8.62\times 10^{-5}\ \text{eV/K}$

The fraction of lattice sites that are Schottky defects is given by

$\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}$

The required ratio is $2.9\times 10^{-6}$ .

6 0

2 years ago