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3 years ago

Find the cost of fencing a rectangular park of length 10 m and breadth 5 m at the rate of? 10 per metre.

Engineering

1 answer:

ser-zykov [4K]3 years ago

3 0

It costs 300
Perimeter = 2(L+B)
2(10+5)
2(15) = 30
10 — 1metre
X — 30metres
30metres = 300

Hope that helps :)

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3 years ago

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$

$P_{real} = 2.3MW$

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$P_{real} = 3V_{ph}I_{ph} Cos\theta$

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$V_{ph} = \frac{V}{\sqrt{3}}$

$V_{ph} = \frac{690}{\sqrt{3}}$

$V_{ph} = 398.37V$

The current phase would be,

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Rearranging,

$I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}$

Replacing,

$I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}$

$I_{ph}= 2.26kA/phase$

Therefore the current per phase is 2.26kA

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3 years ago

primitive transportation and storage systems that make local distribution ineffective if not impossible, the lack of clean water

Vera_Pavlovna [14]

Primitive transportation and storage systems that make local distribution ineffective if not impossible, and the lack of clean water and the lack of effective sewer systems are all examples of the barrier called physical & environmental barriers.

<h3>What are the problems with having physical & environmental barriers?</h3>

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1 year ago

Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a g

brilliants [131]

Answer:

The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.

Explanation:

Concentration of Ge ( $C_{Ge}$ ) = 15%

Concentration of Si (C $_{Si}$ ) = 85%

Density of Germanium (ρ $_{Ge}$ ) = 5.32 g/cm³

Density of Silicon (ρ $_{Si}$ ) = 2.33 g/cm³

Atomic mass of Ge (A $_{Ge}$ )= 72.64 g/mol

To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:

No of Ge atoms/cm³=[Avogadro's Number* $C_{Ge}$ ]/([ $C_{Ge}$ *A $_{Ge}$ /ρ $_{Ge}$ )+(C $_{Si}$ *A $_{Ge}$ /ρ $_{Si}$ )]

= (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]

= (9.03x10²²)/(2.048+26.499)

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3 years ago

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Zolol [24]

hi here is your answer hope it helps or then sry

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