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Natasha_Volkova [10]
2 years ago
13

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

Engineering
1 answer:
gladu [14]2 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

#SPJ1

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Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

where;

[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

3 0
3 years ago
As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can
andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

2) 1.744 x 10 raised to power 9 firings

3) 2.18 x 10 raised to power 8  intake strokes

Explanation:

The step by step explanation is as shown in the attachment

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3 years ago
How can we love our country? Not by words but by deeds. - Jose P. Laurel
Vadim26 [7]

Answer:

1. You have the courage to help without expecting a reward.

2. Because actions are more eloquent than words. Actions are far more valuable and counted than  words, and that's how she inspired me.

3. Doing simple things that can make someone grateful and happy  without knowing that someone is inspired and motivated by your good deeds, and also doing some interesting things By.

Explanation:

6 0
2 years ago
Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0
3 years ago
Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
artcher [175]

Answer:

no it has to be removed

Explanation:

8 0
3 years ago
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