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Snowcat [4.5K]
3 years ago
15

Why must fire extinguishers be routinely maintained

Engineering
2 answers:
sukhopar [10]3 years ago
7 0

Answer:

Fire extinguishers controls the amount of pollution caused by smoke and burning. Although they have no expiring date, they won't last forever. Extinguishers should work for 5 to 15 years, but fire extinguishers be routinely maintained to make sure they remain effective.

Explanation:

BRAINLEST PLZ

vagabundo [1.1K]3 years ago
7 0

Answer:

Fire extinguishers control the amount of pollution caused by smoke and burning. Although they have no expiring date, they won't last forever. Extinguishers should work for 5 to 15 years, but fire extinguishers are routinely maintained to make sure they remain effective.

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Answer:

See attached images

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4 years ago
Liệt kê 10 quá trình sản xuất trong công nghiệp có sử dụng chất xúc tác
Ugo [173]

Answer: English please!

Explanation:

3 0
3 years ago
End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf
sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 -  5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y)    .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

3 0
4 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
4 years ago
Need help finding the pats of a transformer (Electronics Science) Any help would be appreciated.
Yakvenalex [24]

Answer:

21 coils

Explanation:

3 0
3 years ago
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