All categories

3 years ago

Why is gasoline so flammable

2 answers:

8 0

Flammable and combustible liquids themselves do not burn. It is the mixture of their vapours and air that burns. Gasoline, with a flashpoint of -40°C (-40°F), is a flammable liquid. Even at temperatures as low as -40°C (-40°F), it gives off enough vapour to form a burnable mixture in air.

Hope this helps

I looked this up

Don't forget to rate and thanks me

liq [111]3 years ago

7 0

It is flammable because of the low flashpoint. Gasoline's flashpoint is -40 degrees Fahrenheit, which essentially means that the vapors released into the oxygenated air can be lit ablaze in low temperatures.

You might be interested in

Cars A and B are racing each other along the same straight road in the following manner:

zysi [14]

Answer:

$\frac{x_{o}}{v_B-V_A} =t$

Explanation:

Represent the car's position as a function

$x_o=$ "head start"

$x_{A}(t) = x_{o} + v_{A}t\\x_B(t)=v_Bt \\$

$Remember: v_B>v_A$

"cathching up means" that $x_A(t)=x_B(t)$

$x_{o} + v_{A}t =v_Bt\\x_{o} = v_Bt -v_{A}t\\x_{o} = t(v_B-V_A)\\\frac{x_{o}}{v_B-V_A} =t, where \ v_B>v_A$

8 0

3 years ago

Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

$\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.$

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

5 0

2 years ago

Can someone help me on 2 science question,

gavmur [86]

1. I think you should compare diagrams of moon phases from the textbook to diagrams of moon phases online. Because if you pick D it will take to long and C will help you out whith 3 different things to look at.

2. The moon changes in appearances from the perspective of people on earth because it's revolving around the planet and the earth is revolving around the sun, so A. Hoped this helped.

8 0

2 years ago

Read 2 more answers

What is the pressure at 5000 km below the surface of the earth

attashe74 [19]

Answer:

The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.

Explanation:

7 0

2 years ago

How can u tell matched forces act on objects?

mamaluj [8]

Answer:If an object's speed changes, or if it changes the direction it's moving in,

then there must be forces acting on it. There is no other way for any of

these things to happen.

Once in a while, there may be a group of forces (two or more) acting on

an object, and the group of forces may turn out to be "balanced". When

that happens, the object's speed will remain constant, and ... if the speed

is not zero ... it will continue moving in a straight line. In that case, it's not

possible to tell by looking at it whether there are any forces acting on it

3 0

2 years ago