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3 years ago

In the four-stroke internal combustion engine, fuel is ignited during the ____ stroke.

2 answers:

7 0

The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.

Leona [35]3 years ago

3 0

The power stroke, I believe.
Sorry if this does not help.

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Which considerations are used to calculate a windchill factor? Select two options.

vlada-n [284]

Wind speed and air temperature are used to calculate a windchill factor.

<u>Explanation:</u>

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Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.

The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.

There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.

7 0

3 years ago

Read 2 more answers

What is the magnitude of the gravitational force acting on the earth due to the sun?

expeople1 [14]

Answer: $3.524(10)^{22}N$

Explanation:

According to Newton's law of Gravitation, the force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m1=1.99(10)^{30}kg$ is the mass of the Sun

$m2=5.972(10)^{24}kg$ is the mass of the Earth

$r=1.50(10)^{11}m$ is the distance between the Sun and the Earth

Substituting the values in (1):

$F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}$ (2)

Finally:

$F=3.524(10)^{22}N$ This is the gravitational force acting on the earth due to the sun

3 0

3 years ago

Read 2 more answers

What is the displacement of a car in 50 s if it is travelling with a velocity of 25 m/s?

Schach [20]

Answer:

v =25 m/s

time= 50 s

Velocity =Displacement/Time

Displacement = Velocity × Time

S = 25×50

s=1250m

Explanation:

v =25 m/s

time= 50 s

Velocity =Displacement/Time

Displacement = Velocity × Time

7 0

3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$