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Alika [10]
2 years ago
13

In what ways did electrical switches have to change to progress from the

Engineering
1 answer:
frez [133]2 years ago
7 0

Answer:

The answer is D. Electrical switches had to become faster and much smaller.

Explanation:

I just took the test on ~ap.ex and got it right :)

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Explanation:

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Imagine yourself as an Engineer. How would you design your own Wind Turbine to generate electricity (at least 40 watts) at your
topjm [15]

Answer: You need metal and other stuff

Explanation:

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2 years ago
Is making design tradeoffs the same as, or different from, the practice of combining the best parts of different solutions
mezya [45]

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Explanation:

Using a process of informed decision-making, the designer or design team compares different solutions to the requirements of the problem and either chooses

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A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge
Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg =  1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi  T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

    T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S  * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0
3 years ago
Atmospheric air at a pressure of 1 atm and dry bulb temperature of 28 C has a wet-bulb temperature of 20 C.
IrinaK [193]

Answer:

a) \phi = 48\%, b) \omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}, c) h = 58\,\frac{kJ}{kg\,Air}, d) T = 17^{\textdegree}C, e) P_{v} = 1.831\,kPa

Explanation:

a) The relative humidity is given by the intersection of the dry bulb and wet bulb temperatures:

\phi = 48\%

b) The humidity ratio is:

\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}

c) The enthalpy is:

h = 58\,\frac{kJ}{kg\,Air}

d) The dew-point temperature is:

T = 17^{\textdegree}C

e) The water vapor pressure is the product of the relative humidity and the saturation pressure evaluated at dry bulb temperature:

P_{v} = \phi \cdot P_{sat}

P_{v} = 0.48\cdot (3.816\,kPa)

P_{v} = 1.831\,kPa

7 0
3 years ago
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