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Greeley [361]
2 years ago
11

B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers s

pecify this property before the system is actually implemented, how?
Engineering
1 answer:
Oksi-84 [34.3K]2 years ago
4 0

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

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Which permission do you need to shoot on the owner’s property?
Elena L [17]

Answer:

filming permit,

( MARK ME BRAINLIEST!!)

4 0
3 years ago
What happens to the duty cycle for a GMAW Gun when 75Ar/25COzgas
skad [1K]

So what happens is the host will not kill the y no se que hacer para no one can see it in

6 0
2 years ago
Is annealed polycrystalline copper a material that work hardens significantly or a material that exhibits a low work-hardening r
Olegator [25]

Those that harden under strain, such as the aluminum-magnesium alloys used in beverage cans and the copper-zinc alloy, brass, used for cartridges, which show more strain hardening than pure copper or aluminum, respectively.

When a material is deformed under a substantial amount of strain, strain hardening is seen as a strengthening process. Lamellar crystals and chain molecule orientation on a vast scale are the culprits. When plastic materials are stretched past their yield point, this phenomena is frequently seen. When a metal is stretched past its yield point, strain hardening occurs. The metal appears to get stronger and harder to deform as more stress is needed to cause additional plastic deformation. Strain hardening is directly related to fatigue.

Learn more about strain hardening here-

brainly.com/question/15058191

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6 0
1 year ago
The volume fraction of particles in a WC-particle Cu-matrix CERMET is 0.84 Calculate the minimum expected elastic modulus, in GP
Lynna [10]

Answer:

the minimum expected elastic modulus is 372.27 Gpa

Explanation:

First we put down the data in the given question;

Volume fraction V_f = 0.84

Volume fraction of matrix material V_m = 1 - 0.84 = 0.16

Elastic module of particle E_f = 682 GPa

Elastic module of matrix material E_m = 110 GPa

Now, the minimum expected elastic modulus will be;

E_{CT = (E_f × E_m ) / ( E_fV_m  + E_m V_f  )

so we substitute in our values

E_{CT = (682 × 110 ) / ( [ 682 × 0.16 ]  + [ 110 × 0.84] )

E_{CT = ( 75,020 ) / ( 109.12 + 92.4 )

E_{CT = 75,020 / 201.52

E_{CT = 372.27 Gpa

Therefore, the minimum expected elastic modulus is 372.27 Gpa

7 0
2 years ago
A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm a
kodGreya [7K]

Answer:0.3166N

Explanation:

Given data

Area \left ( A\right )=500 cm^2

Gap below top plate\left ( y_1\right )=20 mm

Gap above bottom plate\left ( y_2\right )=30 mm

SAE 30 oil viscosity =0.38 N-s/m^2

Velocity of middle plate\left ( v\right )=1 m/s

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force\left ( F\right )=\mu \frac{Av}{y}

Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ]

F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ]

F=31.66\times 10^{-2}=0.3166 N

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

8 0
3 years ago
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