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kotykmax [81]
2 years ago
12

Source of energy in wind power ​

Engineering
1 answer:
ohaa [14]2 years ago
3 0

The energy in which wind is used to generate electricity is known as wind energy. Wind turbines convert the kinetic energy in the wind into mechanical power. This mechanical power is then used for various purposes. Wind energy is a free and renewable resource. It is a clean and non-polluting source of energy.

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A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind
Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

N_F=n_F\times \frac{N_S}{n_S} here,

n_F = the number of unseeded faults = 6

N_S = number of seeded faults = 30

n_s = number of seeded faults found = 15

So NF will be calculated as,

N_F=6\times \frac{30}{15}=12

And the estimate of faults remaining is  N_F-n_F = 12 - 6 = 6

8 0
2 years ago
Identify one of the advantages of 3d modeling?
Naddik [55]

Answer:

Hello Adam here! (UWU)

Explanation:

The advantages of 3D modeling for designers is not limited to productivity and coordination, it is an excellent communication tool for both the designer and end user. 3D models can help spark important conversations during the design phase and potentially avoid costly construction mishaps.

Happy to Help! (>.O)

6 0
3 years ago
The angle formed between the SAI and the camber line is called the
hichkok12 [17]

Answer:C

Explanation:

8 0
3 years ago
How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?
nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

\frac{5400cu-yd}{1-0.25} =7200cu-yd

The volume at loose is:

V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd

If the Herrywampus has a capacity of 30 cubic yard:

\frac{8640cu-yd}{30cu-yd/trip} =288trip

b) By weight

The swell factor in terms of percent swell is equal to:

pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }

pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd

The weight of backfill is:

8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton

The Herrywampus has a capacity of 40 ton:

\frac{10800}{40ton/trip} =270trip

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0
3 years ago
: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the
Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}

V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s

\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}

taking f = 0.0185

at Z₁ = Z₂

\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P= 1.13\ psi

c) When flow is downhill  z₂-z₁ = -2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P=-0.601\ psi

7 0
3 years ago
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