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nirvana33 [79]
2 years ago
13

Imagine yourself as an Engineer. How would you design your own Wind Turbine to generate electricity (at least 40 watts) at your

home or in some town area. Consider all the elements of the Engineering Design Process.
*Please write detailed answer*
Engineering
1 answer:
topjm [15]2 years ago
6 0

Answer: You need metal and other stuff

Explanation:

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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>

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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.

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5 0
1 year ago
The results of a _________ test will determine if there is suitable drainage and the size of the drain field that will be requir
topjm [15]

Answer:

The results of a percolation test will determine if there is suitable drainage and the size of the drain field that will be required for a septic system.

7 0
2 years ago
Sinks must be used for the correct intended purpose to prevent
irakobra [83]

Answer:

... spilling water or getting anything cascading onto the floor

8 0
2 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
2 years ago
For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta
pav-90 [236]

Answer:

khgy

Explanation:

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8 0
3 years ago
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