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Andreas93 [3]
2 years ago
14

Tec A says that high octane fuel is harder to ignite than lower octane. Tec B says that using high octane fuel in an engine desi

gned for lower octane will produce better fuel economy and power. Who is correct?
Engineering
2 answers:
vlada-n [284]2 years ago
7 0

Answer:

To my understanding higher octane fuel has a higher detonation point, so you compress it more, which will result in more power. Also yes higher octane fuel in a car expecting lower should get better gas mileage and efficiency.

Explanation:

Kryger [21]2 years ago
3 0

Answer:

Tec A says that high octane fuel is harder to ignite than lower octane. Tec B says that using high octane fuel in an engine designed for lower octane will produce better fuel economy and power. Who is correct?

Explanation:

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A two-dimensional flow field described by
Oduvanchick [21]

Answer:

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Explanation:

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5 0
3 years ago
Create a flowchart that describes the following algorithm and then write Python code to implement the algorithm. Write the Pytho
Naya [18.7K]

Answer:

<em>Python code is as follows: </em>

********************************************************************************

#function to get number up to any number of decimal places

def toFixed(value, digits):

return "%.*f" % (digits, value)

print("Enter the price: ", end='', flush=True) #prompt for the input of price

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totalCost = price + 0.05 * price #calculating cost

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3 0
3 years ago
A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o
Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0
3 years ago
Movimiento
goblinko [34]

wth u should clear ur question

5 0
3 years ago
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