The option that is the correct one concerning the uncontrolled burn phase is:
- The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned.
<h3>What is uncontrolled combustion?</h3>
Uncontrolled Combustion is known to be the the time and place in which a kind of an ignition will stop and it is said to be never fixed by anything in regards to the compression ignition engine as seen in SI engines.
Note that the four Stages of combustion are:
1. Pre-flame combustion
2. Uncontrolled combustion
3. Controlled combustion and
4. After burning
Hence, The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned as all the fuel need to burn out.
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We have that Zero signifies a perfect circle shape and 1 shows it maximum out of order shape.
From the question we are told
What does it mean when the orbital eccentricity of a planet is close to 1
Generally
Eccentricity
This in its simplest definition means to be eccentric which means to be a bit out of order or for the given subject at hand means to be a bit out of shape
Naturally the Eccentricity that an object possess is defined by two number 0(zero) to 1(one)
Where
Zero signifies a perfect circle shape and 1 shows it maximum out of order shape
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calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.
Answer:
2.2kWh
Explanation:
Given parameters:
Potential difference = 110v
Current = 20A
Unknown:
Power = ?
Solution:
To solve this problem, we use the expression below:
Power = IV
Power = 110 x 20 = 2200W
This is therefore 2.2kW
Power per hour = 2.2kWh
Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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Answer:
heat pressure, electron degeneracy, neutron degeneracy, and nothing
Explanation:
Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.
White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.
Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.
Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.