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kvv77 [185]
2 years ago
6

Question Completion Status:

Engineering
1 answer:
Mashutka [201]2 years ago
6 0

Answer: c. Centre of pressure​

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

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A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kP
Semmy [17]

Answer:

Isothermal  expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ =  -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n  where P₃ = 2 MPa

Therefore V₃ = (\frac{1}{4} )^{\frac{1}{1.2} }*V_2  = 1.6904×10⁻² m³

Work = W₂ = \frac{P_2V_2-P_3V_3}{n-1} =  -34872.82 J

\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}     or T₃ = T_2*(\frac{P_3}{P_2})^\frac{n-1}{n} = 785.12 K

Isobaric compression we have  thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J

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2 years ago
Two basic types of mechanical fuel injector systems?​
REY [17]

Answer:

Single-point or throttle body injection. Port or multipoint fuel injection.

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2 years ago
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Who was the American founder and leader of the Shakers in the 1770’s who advocated equality, individual responsibility, and peac
DaniilM [7]

Answer: Ann Lee (1736-1784)

Explanation:

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3 years ago
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12. What procedure should you follow when taking measurements?
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I think the answer is b
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A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w
Alexxandr [17]

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P = \frac{230}{6}

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor = \frac{300}{V}

V = \frac{300}{38.33}

V = 7.826 ml

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3 years ago
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