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3 years ago

Do you get a better performance using premier gasoline (Octane number 93) for your compact car?

Engineering

1 answer:

topjm [15]3 years ago

6 0

Answer:

Yes

Explanation:

As we know that octane number resist the engine from knocking.If knocking can prevent that automatically the performance of engine will increases.If octane number is 100 then it means that knocking tendency in the engine is zero.So higher the octane number better will the performance of the engine.

Generally octane number is 87 but for premier gasoline is 92 or 93.

So we can say that if octane number is 93 then car will give better performance

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Katen [24]

Answer:

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Explanation:

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3 years ago

A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for

Svet_ta [14]

Answer:

Time taken by the $1\mu m$ diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = $1\times 10^{- 6} m$

Diameter of the smaller droplet, R' = $0.5\times 10^{- 6} m$

Now,

Volume of the droplet, V = $\frac{4}{3}\pi R^{3}$

Volume of the smaller droplet, V' = $\frac{4}{3}\pi R'^{3}$

Volume of the droplet ∝ Time taken for complete evaporation

Thus

$\frac{V}{V'} = \frac{t}{t'}$

where

t' = taken taken by smaller droplet

$\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}$

$\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}$

t' = $60\times 10^{- 9} s = 60 ns$

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What is the mode of operation of a ramp digital voltimeter

liberstina [14]

Answer:

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1 year ago

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