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4 years ago

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

2 answers:

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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Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

3 years ago

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

Read 2 more answers

3 examples when friction is helpful?

Svetllana [295]

Some examples of when friction is helpful are: to help the movement of tires. When you walk, and also, when you erase. :)

4 0

3 years ago

The SI system uses three base units. Question 6 options: True False

GalinKa [24]

Answer:

The answer is false

Explanation:

Though the mostly used SI unit of measurement or the most popular units are the

Length,

Time and

Mass

i.e meter (m), seconds (s), kilogram (kg)

Aside all the above stated units for measurements there are other four basic units which are itemized bellow.

they are

1. Amount of substance - mole (mole)

2. Electric current - ampere (A)

3. Temperature - kelvin (K)

4. Luminous intensity - candela (cd)

6 0

3 years ago

Is the relationship between the length of a pendulum and its period is valid at all times?

Stels [109]

Yes it is valid all the times under the consideration of acceleration due to gravity .it is not valid on space where there is no influence of gravity

4 0

3 years ago