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Liono4ka [1.6K]
3 years ago
6

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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10. Identify one material we Mine and what we make with that material
Katen [24]

Answer:

We Mine Limestone. Its used to make cement, toothpaste or paints, soil conditioner, and rip rap stone.

Explanation:

7 0
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

5 0
3 years ago
What is the mode of operation of a ramp digital voltimeter​
liberstina [14]

Answer:

The operating principle of a ramp type digital voltmeter is to measure the time that a linear ramp voltage takes to change from level of input voltage to zero voltage (or vice versa).

7 0
1 year ago
A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a
Dafna1 [17]

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

5 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
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