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3 years ago

A Class A fire extingisher is for use on general combustibles such as:

Engineering

1 answer:

Tatiana [17]3 years ago

7 0

Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

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A 12-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

Ira Lisetskai [31]

Answer:

The smallest wire diameter that can be used is 1 cm

Explanation:

First, we find the smallest diameter using the criterion of maximum normal stress:

Max. Stress = 150 x 10^6 Pa = F/A

150 x 10^6 Pa = 12000 N/(πd²/4)

d² = (12000 N)(4)/(150 x 10^6 Pa)(π)

d = √1.0185 x 10^-4 m²

d = 0.010 m = 1 cm

Now, we find the smallest diameter using the criterion of maximum strain:

Max. Strain = Max. Change in Length/Original Length = 0.025 m/50 m

Max. Strain = 5 x 10^-4 mm/mm

Now,

Max. Strain = Stress/E = (F/A)/E = F/AE

using values:

5 x 10^-4 mm/mm = (12000 N)/(200 x 10^9 Pa)(πd²/4)

d =√(12000 N)(4)/(5 x 0^-4)(200 x 10^9 Pa)(π)

d = 0.012 m = 1.2 cm

Now, by comparison in both cases it can be noted that the smallest value of the diameter is <u>1 cm</u>, which is limited by maximum stress.

7 0

3 years ago

The energy flux associated with solar radiation incident on the outer surface of the earth’s atmosphere has been accurately meas

Degger [83]

Answer:

A) 3.868x10^26 W

B) 8176.678 K

C) 3.544x10^-7 m

D) 393.187 K

Explanation:

Detailed explanation and calculation is shown in the image below.

8 0

3 years ago

Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el

exis [7]

Answer:

a) V = 20 ∠30⁰ , I = 4 ∠-210⁰ Z inductive L = 0,0125 H

b) V = 9∠-60⁰ , I = 4 ∠ 190⁰ Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰ , I = 7 ∠ 150⁰ Z Inductive L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as function cosine

we know that sin ( x + 90 ) = cos x

Then sin ( 400*t -120 ) = cos ( 400*t - 120 -90 ) = cos ( 400t - 120 - 90)

Phasor form I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I ⇒ Z = 20 ∠30⁰ / 4 ∠-210⁰ Z = 5 ∠-180⁰

We notice that voltage advances the current then we are in presence of an inductive impedance

5 = wl ⇒ 5 = 400 *L ⇒ L = 0,0125 H

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t) = 4 sin ( 900t + 280 ) ⇒ i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 ) ⇒ I = 4 ∠ 190⁰

Z = V/I ⇒ Z = 9∠-60⁰ / 4 ∠ 190⁰ Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc = 1/ 900*C 1/wc = Z ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C ⇒ 2025 =1/C ⇒ C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 ) ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90) ⇒ i(t) = 7 sin (250t + 150)

Phasor Form I = 7 ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰ ⇒ Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L 250 L = 1,86 L = 1,86/250 L = 0,0074 H

7 0

3 years ago

According to the rules of dimension how should written notes appear

Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with horizontal but may be made at any convenient angle except vertical or horizontal.

7 0

3 years ago

Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and

Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒ P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0

4 years ago

A Class A fire extingisher is for use on general combustibles such as:​

A Class A fire extingisher is for use on general combustibles such as: