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allsm [11]
2 years ago
9

Question 1 A design team completes their high-fidelity prototype of a responsive website. Before they hand off designs to the en

gineering team, they need to confirm that the designs are final. What questions should they answer before handing off the designs
Engineering
1 answer:
MissTica2 years ago
5 0

A question the design team should answer before handing off the designs is: are the designs a true representation of the intended end user experience?

<h3>What is a website?</h3>

A website can be defined as a collective name that is used to describe series of webpages that are interconnected or linked together with the same domain name.

In Computer technology, the main goal of a high-fidelity prototype is to understand how end users would interact with a website and areas to improve the design.

In conclusion, the design team should answer whether or not the designs are a true representation of the intended end user experience before handing off the designs.

Read more on website here: brainly.com/question/26324021

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Does anyone know obamas last name???? please help its for a friend I swear!!!111!!11!
ANTONII [103]

Answer:

Explanation:

Barack Hussein Obama

6 0
3 years ago
Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process
malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = 3\times 0.133\ kPa

At the end of pumping, temperature of air, T = 5^{\circ}C = 278 K

After the rise in the air temperature, T' = 20^{\circ}C = 293 K

Now, we know the ideal gas eqn:

PV = mRT

So

P = \frac{m}{V}RT

P = \rho_{a}RT          (1)

where

P = Pressure

V = Volume

\rho_{a} = air\ density

R = Rydberg's constant

T = Temperature

Using eqn (1):

P = \rho_{a}RT

\rho_{a} = \frac{P}{RT}

\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}

Now, at constant volume the final pressure, P' is given by:

\frac{P}{T} = \frac{P'}{T'}

P' = \frac{P}{T}\times T'

P' = \frac{3}{278}\times 293 = 3.16 torr

7 0
4 years ago
What steel type and ASTM designation is preferred for W-shapes?
Ulleksa [173]

Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

3 0
4 years ago
Calculate the diameter of pneumatic cylinder to drive a mechanism. The force exerted by the mechanism is 80 N. The return force
sashaice [31]

Answer:

14.27 mm

Explanation:

Force = 80 Newton

Pressure exerted = 5 bar = 5×10⁵ Pa

Pressure exerted = Force/Area

Area = Force/Pressure

Area=\frac{80}{500000}\\\Rightarrow Area=0.00016\ m^2

Area of the cylinder=πd²/4

\Rightarrow \frac{\pi}{4}d^2=0.00016\\\Rightarrow d^2=\frac{4\times 0.00016}{\pi}\\\Rightarrow d^2=0.0002037\\\Rightarrow d=0.01427\ m

Hence diameter of cylinder 14.27 mm

5 0
3 years ago
Describe the importance of ferrite and austenite stabilizing elements in steels
podryga [215]

Answer:

The importance of ferrite and austenite stabilizing elements in steels .

Explanation:

Alloying -

The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .

The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .

Stabilizing austenite -

The process by which temperature is increased , in which Austenite exists .

Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .

Hence, the elements are -

Cobalt , Nickel , Manganese , Copper.

The examples of the Austenitic steels are -

Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.

Stabilizing ferrite –

The process by which temperature is decreased , in which austenite exists .

Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.

Hence, the elements are -  

Aluminium , Silicon , Tungsten , Chromium , Molybdenum , Vanadium

The examples of the Ferritic steels are -

F-Cr alloys , transformer sheets steel ( 3% Si ).

3 0
4 years ago
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