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Troyanec [42]
2 years ago
9

Several short trips taken from a cold start can use ....... as much fuel as a longer multi-purpose trip covering the same distan

ce when the engine is warm.
Physics
1 answer:
Lerok [7]2 years ago
4 0

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

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How many neutrons does element X have if it’s atomic number is 28 and it’s mass number is 81?
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Answer:

53

Explanation:

M-A=N

Mass-Atomic #= Neutrons

81-28= 53

3 0
3 years ago
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30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
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Answer:

I think it's the most important part in this

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3 years ago
Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
exis [7]

Answer:

it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
Rina8888 [55]
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r 
 v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
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5 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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