All categories

3 years ago

Which method used to attack ground cover fires is generally used against fires that are very hot, very big, and very fast?

Chemistry

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

Indirect attack (ground cover)

Explanation:

Indirect attack (ground cover) is a method of controlling a groundcover fire whereby the control line is erected or situated at a remote from the main fire edge, and the fuel between the two points was burned.

The indirect attack is set up at some distance

from the raging fire. The indirect attack method is normally used to put out

fires that are too fast, too big or too hot for a direct attack method to work.

adjustments can be made when required

You might be interested in

Which best describes the difference between radiation and conduction? (1 point)

NNADVOKAT [17]

Answer:

B) Conduction requires two objects to be in physical contact.

Explanation:

Conduction is when two objects are directly touching each other. So, the objects must come in physical contact with each other.

6 0

3 years ago

I give brainliest

Tatiana [17]

Answer: During winter the Northern Hemisphere leans away from the sun, there are fewer daylight hours and the sun hits us at an angle. This makes it appear lower in the sky. In equatorial regions the length of days and the directness of sunlight don't change as much.

Hope this helps :)

8 0

3 years ago

Read 2 more answers

How tungsten conducts electricity

Elis [28]

Answer:

Yes

Explanation:

The virtue that tungsten is metal it means that it does conduct electricity. Tungsten does conduct electricity currents quite well, such as when compared to iron and nickel. It is, however, not a good conductor as compared to other metals though it does get the job done.

8 0

3 years ago

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

3 years ago

After fertilization, female cones become very: sticky light O hard

S_A_V [24]

Answer:

light

Explanation:

sorry don't now this one

8 0

3 years ago