If you replace the text value in an associative dimension, the text value will not change when the

What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances

gavmur [86]

Explanation:

Air fuel ratio:

Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

$Air\ fuel\ ratio=\dfrac{mass\ of\ air}{mass\ of\ fuel}$

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.

When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.

When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.

5 0

3 years ago

Always refill your gas tank well before

Scorpion4ik [409]

I believe it’s c because you don’t want your gas to run real low, so I think it’s best to do it when your fuel.

8 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".