1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of s
Anettt [7]

Answer:

33.4

Explanation:

Step 1:

\sumMo=0 (moment about the origin)

Fb(15)-Fc(15)=0

Fb=Fc

Step 2:

\sumFx=0

-Fb-Fccos\theta+Ncsin\theta=0

Fc=0.3Nc=Fb

-0.3Nc-0.3Nccos\theta+Ncsin\theta=0

(-0.3-cos\theta+sin\theta)Nc=0----(1)

\sumFy=0

Nccos\theta+Fcsin\theta-Nb=0

Nccos\theta+0.3Ncsin\theta-Nc=0

Nc[cos\theta+0.3sin\theta-1]=0--------(2)

Solving eq (1) and eq (2)

\theta=33.4

Step 3:

As the roller is a two force member

2(90-\phi)+\theta=180

\phi=\theta/2

\phi=Tan(\muN/N)-1

\phi=16.7

\theta=2x16.7=33.4

5 0
2 years ago
What are two reasons why Hunter edjucation is important?<br><br><br> 30 pts
sasho [114]

Answer:

Hunter Education is Important Because It Helps To: Prevent hunting and shooting accidents. Improve hunter behavior to maintain public acceptance of hunting.

Explanation:

6 0
3 years ago
The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
Anni [7]

Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

8 0
3 years ago
A spacecraft is fueled using hydrazine ​(N2H4​; molecular weight of 32 grams per mole​ [g/mol]) and carries 1 comma 630 kilogram
Varvara68 [4.7K]

Answer:

attached below

Explanation:

7 0
3 years ago
a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc
Elena L [17]

Answer:

Point force (Qp) = 704 kn/m²

Explanation:

Given:

length = 19 m

Width = 0.5 m

fs = 4

Vicinity of the pile = 25

Find:

Point force (Qp)

Computation:

Point force (Qp) = fs²(l+v)

Point force (Qp) = 4²(25+19)

Point force (Qp) = 16(44)

Point force (Qp) = 704 kn/m²

5 0
2 years ago
Other questions:
  • Why is low voltage advantageous in arc welding?
    5·1 answer
  • When in a flow do the streamlines, streak lines and timelines coincide?
    14·1 answer
  • A string of ASCII characters has been converted to hexadecimal resulting in the following message: 4A EF 62 73 73 F4 E5 76 E5 Of
    6·1 answer
  • A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w
    7·1 answer
  • Sam promises to pay Sandy $2,000 in four years and another $3,000 four years later for a loan of $2,000 from Sandy today. What i
    8·1 answer
  • Home safety and security is an _________<br><br> process. (7 Letters)<br><br> Answer
    10·1 answer
  • Name eight safety electrical devices including their functions and effects if not present.​
    15·1 answer
  • (Architecture) Sarah is an environmental activist. She frequently conducts various programs and activities in her community to p
    15·1 answer
  • Write a program that takes three numbers as input from the user, and prints the largest.
    12·1 answer
  • The sum of forces on node 2 (upper-left) is ______.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!