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svet-max [94.6K]
4 years ago
8

2. A Class B fire includes burning:

Engineering
1 answer:
vovangra [49]4 years ago
3 0

Answer: The answer is C. Gasoline and greases and here is some more info

In fire classes, a Class B fire is a fire in flammable liquids or flammable gases, petroleum greases, tars, oils, oil-based paints, solvents, lacquers, or alcohols. For example, propane, natural gas, gasoline and kerosene fires are types of Class B fires.

Explanation:

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A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature  = - 35 degree C

Maximum expansion can be determined as

\Delta L = L \alpha (T_{max} - T_{min})

where , \alpha is expansion coefficient = 12\times 10^{-6} degree C

SO, \Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))

\Delta L = 0.45 m = 450 mm

number of minimum expansion joints is calculated as

n = \frac{450}{20} = 22.5

b) length of each bridge

Length = \frac{500}{22.5} = 22.22 m

8 0
3 years ago
Calculate and plot the radial and circumferential stress distribution in the left ventricle at the end of systole (p 5 80 mmHg;
alexandr402 [8]

Answer:

62990.08 N/M

Explanation:

Circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a stress in directions coplanar with but perpendicular to the symmetry axis.

See attachment for the step by step solution of the given problem..

8 0
3 years ago
The tolerance for a geometric can be easily identified by reading the _________.
baherus [9]

Answer:

C

Explanation:

4 0
3 years ago
Read 2 more answers
What are the searching algorithms used by search engines?
Juliette [100K]
Search engines use specific algorithms based on their data size and structure to produce a return value.
Linear Search Algorithm. ...
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6 0
3 years ago
• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the
Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

\sigma a =  \sqrt{(12 \times 1)^2+3\times (0\times 1)^2}  

\sigma a = 12 kpsi

and

\sigma m = \sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}     ...........2

put here value and we get

\sigma m = \sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}  

\sigma m = 17.34 kpsi

now we apply here goodman line equation here that is

\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

\frac{17.34}{63.800} +  \frac{12}{31.9} = \frac{1}{FOS}  

solve it we get

FOS = 1.5432

6 0
3 years ago
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