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VikaD [51]
3 years ago
15

1. Discuss how products incorporate aesthetic design and why this appeals to target markets 2. Discuss how the universal design

process has impacted engineering design and the impact these expectations will have on the future of product design.
Engineering
1 answer:
Nikitich [7]3 years ago
3 0

<u>Explanation:</u>

<em>Remember, </em>to say a product is incorporated with aesthetic design implies that its overall appearance is designed to look beautiful to the eyes of the user/buyer. For example, a clothing company whose target market is mainly focused on women's clothing<u> would need to take into consideration that certain colors like pink, blue, etc are attractive to women more than men.</u> So they'll have to ensure the colors of their clothing are suitable to the needs of their target market.

The Universal Design process involves building products that can be used by a wide range of users at ease. For example, you may ask yourself: Is my product/service easily accesible to those with disabilities?

Other processes include;

  • Defining who the users (or universe) are of the products.
  • Involve consumers in the design.
  • Follow the existing standards of product design
  • Evaluate and review your universal design methods
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A PMDC machine is measured to va120Vdc, ia 5.5A at the electrical terminals and the shaft is measured to have Tmech 5.5Nm, ns120
san4es73 [151]

Answer:

(c) Generator, 95.5 %

Explanation:

given data

voltage va = 120V

current Ia = - 5.5 A

Tmech =  5.5Nm

ns = 1200 RPM

solution

first we get here electric input power that is express as

electric input power = va × Ia    ......1

put here value and we get

electric input power = 120 × -5-5

electric input power -660 W

here negative mean it generate power

and here

Pin ( mech) will be

Pin ( mech) = Tl × ω   .........2

Pin ( mech) = 5.5 × \frac{2\pi N}{60}    

Pin ( mech) =  5.5 × \frac{2\pi 1200}{60}  

Pin ( mech) = 691.150 W

and

efficiency will be here as

efficiency = \frac{660}{691.150}    

efficiency = 95.5 %

so correct option is (c) Generator, 95.5 %

5 0
3 years ago
Someone help me to do this?
Lera25 [3.4K]

Answer:

okay it is faster and easier

4 0
2 years ago
At a high school science fair, Connor won first place for his replica of the Golden Gate Bridge. Connor liked the project so muc
AnnZ [28]

Answer:

a

Explanation:

6 0
4 years ago
Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9
dybincka [34]

Answer:

The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

V = V_{in}+V_{p}-V_{e}-V_{out} (Eq. 1)

Where:

V - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

V_{in} - Inflow amount of water, measured in cubic feet per year.

V_{out} - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

V_{p} - Amount of water due to precipitation, measured in cubic feet per year.

V_{e} - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t

V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l} (Eq. 2)

Where:

f_{in} - Average watershed inflow, measured in cubic feet per second.

f_{out} - Average flow to be released, measured in cubic feet per second.

\Delta t - Yearly time, measured in seconds per year.

h_{p} - Change in lake height due to precipitation, measured in feet per year.

h_{e} - Change in lake height due to evaporation, measured in feet per year.

A_{l} - Surface area of the lake, measured in square feet.

If we know that f_{in} = 900\,\frac{ft^{3}}{s}, f_{out} = 300\,\frac{ft^{3}}{s}, \Delta t = 31,536,000\,\frac{second}{yr}, h_{p} = 32\,\frac{in}{yr}, h_{e} = 55\,\frac{in}{yr} and A_{l} = 47,000\,acres, the available amount of water for supply purposes in the Triangle area is:

V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}

The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

5 0
3 years ago
Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t
Natasha2012 [34]

Answer:

\dot W_{in} = 273.69\,kW

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

P = 20\,kPa

T = 60.06\,^{\textdegree}C

h = 251.42\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

Outlet (Subcooled Liquid)

P = 6000\,kPa

T = 60.06\,^{\textdegree}C

h = 257.502\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

The power input to the pump is computed hereafter:

\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)

\dot W_{in} = 273.69\,kW

8 0
4 years ago
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