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3 years ago

(b) The application of user centred design can lead to innovative products such as the umbrella shown in Figure 9.

Engineering

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Answer:

the concept of user centred design is the idea of putting your clients best needs in mind. you need to be empathic and understanding of how to make their life easier. by doing research you can do this. for example the design in figure 9 shows an umbrella with a canopy. the canopy can deflect rain and still have the easy design of an umbrella. this makes it not only easier to use but functional.

Explanation:

i wrote this myself so copy it or just rewrite it, it doesnt matter

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3 years ago

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victus00 [196]

Answer: Engineering

Explanation:

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3 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

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solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

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3 years ago

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dezoksy [38]

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3 years ago

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Bars A and B have cross-sectional areas of 400 m2 and a modulus of elasticity of 200 GPa. A gap exists between bar A and the rig

NISA [10]

Answer:

axial stress in bar B = 25Mpa.

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Explanation:

PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.

So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;

The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.

STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,

Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.

STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,

The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.

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