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mamaluj [8]
4 years ago
6

Which stage occurs just before ignition in an internal combustion engine?

Physics
2 answers:
dangina [55]4 years ago
6 0

usually, the fuel pump turns on, along with all of the vehicle's electric systems, but on a large diesel engine, like in a semi, or agricultural equipment, the intake, and carburetor are activated first to allow diesel (fuel) into the engine to combustion can be produced once the engine turns over

ddd [48]4 years ago
5 0
I think it is this because compression stroke it needs to be compressed then open up when started.
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An object travels 20 m in 4 s heading south. What is it’s velocity?
vodka [1.7K]

Answer:

5 miles a second

Explanation:

20 divided by 4

hope it helps and for brainliest :)

6 0
3 years ago
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What is/are the energy transformation(s) that take place when using a wind turbine to generate usable energy?
Cerrena [4.2K]

Answer:

the answer is C

Explanation:

C) friction - mechanical - electrical

4 0
3 years ago
The value 10.00 has____ significant figures
natka813 [3]

Answer:

4

Explanation:

You assume that when they were writing two zeroes after the decimal point, they wanted to show that these are significant in determining the precision of this number.

8 0
3 years ago
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The cannon on a battleship can fire a shell a maximum distance of 26.0 km.
Alla [95]
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is \theta = 45^{\circ}.
In fact, the laws of motions on both x- and y- directions are
S_x(t)= v_0 cos \theta t
S_y(t)= v_0 \sin \theta t -  \frac{1}{2} gt^2
From the second equation, we get the time t at which the projectile hits the ground, by requiring S_y(t)=0, and we get:
t= \frac{2 v_0 \sin \theta}{g}
And inserting this value into Sx(t), we find
S_x(t) = 2  \frac{v_0^2}{g}  \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)
And this value is maximum when \theta=45^{\circ}, so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
S_x(t)=  \frac{v_0^2}{g} \sin (2\theta)
And by using S_x = 26 km=26000 m, we find the value of the initial velocity v0:
v_0 =  \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s
8 0
3 years ago
If a person is pushing a cart with a force of 40 Newtons and it
tia_tia [17]

Answer:

80kg

Explanation:

A = Fm

0.5 = 40m

m = 40/0.5

m = 80kg

5 0
3 years ago
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