All categories

2 years ago

Which is a common cause of incomplete combustion

2 answers:

5 0

During complete combustion carbon and hydrogen combine with oxygen (O2) to produce carbon dioxide (CO2) and water (HO2) during that time it is not completely oxidized producing soot or carbon monoxide

valentina_108 [34]2 years ago

4 0

Answer: incomplete supply of oxygen

Explanation:

Combustion is a chemical reaction in which a fuel is reacted with excess supply of oxygen to result into complete oxidation of fuel to give carbon dioxide and water.

Example of complete combustion :

$C+O_2 \rightarrow CO_2$

Incomplete combustion happens when the supply of oxygen is limited and thus fuel can not be completely oxidized , thus forming carbon monoxide.

Example of incomplete combustion :

$C+\frac{1}{2}O_2\rightarrow CO$

You might be interested in

State coulomb's law mathematically

Vsevolod [243]

Coulomb's law is express as:

$\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ \text{ where} \\ k\text{ is Coulomb's constant} \\ q_1\text{ and }q_2\text{ are the charges} \\ r\text{ is the distance between charges} \end{gathered}$

7 0

4 months ago

Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger

Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

4 0

2 years ago

A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

Oksi-84 [34.3K]

Explanation:

The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

Total displacement = $135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East$

Total time travelled = $1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr$

$\text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}$