Several short trips taken from a cold start can use ...twice... as much fuel as a longer multi-purpose trip covering the same distance when the engine is warm.
In cold weather, properly designed gasoline aids in engine starting, while in hot weather, it helps prevent vapor lock. In order to meet the requirements of a modern engine, the fuel must have the volatility for which the engine's fuel system was built and an antiknock quality strong enough to prevent knock during routine operation.
During the intake phase, the air and fuel are combined before being introduced into the cylinder. The spark ignites the fuel-air mixture after the piston compresses it, resulting in combustion. During the power stroke, the piston is propelled by the expansion of the combustion gases.
To learn more about engine and fuel please visit -
brainly.com/question/5181209
#SPJ4
Answer:

Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, 
Let the N number of electrons have charge -40 nC, such that,
Now, mass of one electron = 
Therefore, mass of N electrons = 
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is 
Then,

It is the required fraction of mass of the droplet.
Answer:
speed
Explanation:
its the speed
because the formula for deriving speed is distance/time
the unit for distance is metre
the unit for time is seconds
so speed is metre/seconds
so speed us m/s.
Well here in the sticks we call 'em fletchings haha