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svp [43]
2 years ago
12

Which of the following is correct concerning the uncontrolled burn phase?

Physics
1 answer:
beks73 [17]2 years ago
7 0

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.

Learn more about burn phase from

brainly.com/question/14414049

#SPJ1

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What does flowing electrical charge produce?
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Useable energy, electricity

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3 years ago
F= (9.3 x 105 )(4.2 x 10-15)
kvv77 [185]

Answer:

3.906E-9 or 3.906 x 10-9

Explanation:

8 0
3 years ago
A bird flies 3.7 meters in 46 seconds, what is its speed?
victus00 [196]

Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds  

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.  

Below is the formula to find the speed.

Speed = Distance / Time

Now insert the given value in the formula.

Speed = 3.7 / 46 = 0.08 m/s

8 0
4 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

5 0
3 years ago
Read 2 more answers
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