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Katyanochek1 [597]
3 years ago
7

Which of the following is an example of a computer simulation?

Engineering
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

b)prototype

Explanation:

_÷/#^'tixgixurykgdtusurdfjczut uxdkdktxkgxkt. it

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If you get a flat in the front of your car, your car will:
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stop and might even crash

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3 years ago
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wi
Eduardwww [97]

Answer: 255

255 turns are required to create 25 ohms of  secondary impedance.

Explanation:

Given that,

Number of turns in primary wire N₁ = 900

impedance on Primary wire Z₁ = 400 ohms

Number of turns in Secondary wire N₂ = ?

impedance on Secondary wire Z₂ = 25 ohms

we know that, the relationship between turn and impedance is

Zp / Zs = ( Np / Ns )²

(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²

there fore

Z₁ / Z₂ = ( N₁ / N₂ )²

Now we substitute

( 400 / 25 ) = ( 900 / N₂ )²

400 / 25 = 900² / N₂²

we cross multiple to get our N₂

400 × N₂² =  900² × 25

N₂² = ( 900² × 25 ) / 400

N₂² = ( 810000 × 25 ) / 400

N₂² = 20250000 / 400

N₂² = 50625

N₂ = √50625

N₂ = 225

Therefore 255 turns are required to create 25 ohms of  secondary impedance.

4 0
3 years ago
How did studying pagodas help engineers create earthquake-proof structures in modern society?A. Engineers learned that the desig
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D cuz i took the test

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3 years ago
Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the
Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

or

V = 0.0203 m/s

also,

Reynold's number, Re = \frac{VD}{\nu}

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = \frac{0.0203\times0.025}{0.833\times10^{-6}}

or

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity =  2 × average velocity = 2 × 0.0203 m/s

or

maximum velocity = 0.0406 m/s

5 0
3 years ago
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