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Serga [27]
2 years ago
15

What is a beta testing ?

Engineering
2 answers:
olganol [36]2 years ago
5 0
Beta testing is one of the final steps in your software development lifecycle (SDLC) before a product goes live. Also referred to as user testing or customer validation, beta testing aims to ensure that end users are satisfied with a software product before you make it generally available (GA).
cupoosta [38]2 years ago
4 0
Beta testing is a final step software development ( which means before a product is posted for other people to see) it’s supposed to make sure the user likes it or what they think about it.


Hope this helps :)
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Answer:

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2 years ago
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The simply supported beam in the Figure has a rectangular cross-section 150 mm wide and 240 mm high.
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Same question idea but different values... I hope I helped you... Don't forget to put a heart mark

4 0
3 years ago
It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?
vovangra [49]

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

5 0
3 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
3 years ago
Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.
Sholpan [36]

Answer:

Frequency = 3.5294\times 10^{14}s^{-1}

Wavenumber = 1.1765\times 10^6m^{-1}

Energy = 2.3365\times 10^{-19}J

Energy = 1.4579 eV

Energy = 2.3365\times 10^{-12}erg

Explanation:

As we are given the wavelength = 850 nm

conversion used : (1nm=10^{-9}m)

So, wavelength is  850\times 10^{-9}m

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where, c is the speed of light having value = 3\times 10^8m/s

So, Frequency is:

Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}

Frequency=3.5294\times 10^{14}s^{-1}

Wavenumber is the reciprocal of wavelength.  

So,  

Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}

Wavenumber=1.1765\times 10^6m^{-1}

Also,  

Energy=h\times frequency

where, h is Plank's constant having value as 6.62\times 10^{-34}J.s

So,  

Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})

Energy=2.3365\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=1.4579eV

Also,  

1J=10^7erg

So,  

Energy=(2.3365\times 10^{-19})\times 10^7erg

Energy=2.3365\times 10^{-12}erg

5 0
3 years ago
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