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Naily [24]
2 years ago
10

Difference between theory and practice?​

Engineering
1 answer:
Ronch [10]2 years ago
5 0

Answer:

There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.

Theory and Practice Explained

Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.

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g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],
yKpoI14uk [10]

Explanation:

\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}

Therefore,

-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}

Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}

5 0
3 years ago
Which of the following is a possible unit of ultimate tensile strength?
levacccp [35]

Answer:

Newton per square meter (N/m2)

Explanation:

Required

Unit of ultimate tensile strength

Ultimate tensile strength (U) is calculated using:

U = \frac{Ultimate\ Force}{Area}

The units of force is N (Newton) and the unit of Area is m^2

So, we have:

U = \frac{N}{m^2}

or

U = N/m^2

<em>Hence: (c) is correct</em>

4 0
2 years ago
Explain 3 ways that people in sports use engineering to increase their performance?
LenKa [72]
Designing systems for manufacturing, motion analysis or impact testing;
building and testing prototypes;
analyzing the human body to prevent injury;
developing or designing new light weight materials that will be more comfortable and withstand greater impacts or forces;
7 0
2 years ago
Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the
zysi [14]

<u>Explanation:</u>

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = \frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600m^{2}

let sunlight hours be 8 hours

Hence, solar power input on lawn,

=5.62×3600 = 20232 kJ/m^{2}/day

energy input in lawn = (600) (20232) (7)

                                  = 84974.4 mJ/week

Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

                                            = 3398.97 mJ

Mass of the clippers  \(=(30)(20)(1 \cdot 096)^{2}(667)\)

                                  \(=478632 \cdot 33\) pounds

Removing water content,

dried grass clippings \(=95726.46\) pound

                                    = 11533.25 gallons

Trash cans repaired  

                                     =\frac{11533}{50} =230.66\\=231 cans

By burning the gas, total energy input = 3398.97 MJ × 0.2

                                                                = 679.794 MJ

Efficiency of steeling engine  =  20%

Energy output by engine = 679.794 ×0.2

                                          = 135.96 mJ

Energy required by mover = 33.5565 mJ

Hence, Energy (output) ⇒ energy required

5 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
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