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3 years ago

To put out a class D metal fire, you must _______ the fire.

Engineering

1 answer:

gladu [14]3 years ago

7 0

To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.

<h3>What is a Class D fire?</h3>

A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.

Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.

By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.

You can learn more about extinguishing fires here https://brainly.in/question/760550

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3 years ago

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a

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Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre = 183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

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183.5919 km/gallon = 114.08 miles/gallon (highway)

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3 years ago

Please calculate the current for the circuit below

Aleks04 [339]

Answer:

The answer is " $I = 0.0085106383 \ A$ "

Explanation:

Given:

$R= 470 \ \Omega \\\\V= 4 \ v$

Formula:

$\to V=IR\\\\\to I = \frac{V}{R}\\\\$

$= \frac{4}{470}\\\\ = 0.0085106383 \ A$

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3 years ago

Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is ap

telo118 [61]

Answer:

e = 3.97*10^-4

Explanation:

1.8 mm = 0.0018 m

2.6*10^4 mm = 26 m

Elongation is The ratio between the stretched length and the original length.

e = L/L0

This is calculated with Hooke's law:

e = σ/E

Where

σ: normal stress

E: elastic constant

σ = P/A

Where

P: normal load

A: cross section

A = π/4 * d^2

Therefore:

e = P / (A * E)

e = 4 * P / (π * d^2 * E)

e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4

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