All categories

3 years ago

For which situation will structural firefighting protective clothing provide you with adequate protection

Engineering

1 answer:

jeka943 years ago

3 0

Missing part

The options are:

A. You will be exposed to splashes of the material.

B. You will have to handle the material.

C. There are high atmospheric concentrations of the material.

D. None of the above.

Answer:

None of the above

Explanation:

Structural firefighting protective clothing can't give adequate protection when one is exposed to splashes of the material. Moreover, it can't help in a situation where one is required to handle the material or when there are high atmospheric concentrations of the material.

SFC may be used when the following conditions are met:

• Contact with splashes of extremely hazardous materials is unlikely.

• Total atmospheric concentrations do not contain high levels of chemicals toxic to the skin.

• There are no adverse effects from chemical exposure to small areas of unprotected skin.

• Live victims who are in need of rescue, such as those found in a terrorist attack.

You might be interested in

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

kiruha [24]

Answer:

percentage change in volume = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × $10^{11}$ N/m²

bulk stress = $10^{7}$ N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = $\frac{bulk\ stress}{bulk\ strain}$ ...............1

put here value and we get

3.5 × $10^{11}$ = $\frac{10^7}{bulk\ strain}$

solve it we get

bulk strain = 2.8571 × $10^{-5}$

and

bulk strain = $\frac{change\ volume}{original\ volume}$

so that percentage change in volume is = 2.8571 × $10^{-5}$ × 100

percentage change in volume = 0.00285 %

6 0

3 years ago

Always refill your gas tank well before

Scorpion4ik [409]

I believe it’s c because you don’t want your gas to run real low, so I think it’s best to do it when your fuel.

8 0

3 years ago

The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j

Anna [14]

Answer:

the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

Explanation:

Given that;

volume of cut = 25,100 m³

Volume of dry soil fill = 23,300 m³

Weight of the soil will be;

⇒ 93% × 18.3 kN/m³ × 23,300 m³

= 0.93 × 426390 kN 3

= 396,542.7 kN

Optimum moisture content = 12.9 %

Required amount of moisture = (12.9 - 8.3)% = 4.6 %

So,

Weight of water required = 4.6% × 396,542.7 = 18241 kN

Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

6 0

3 years ago

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

5 0

2 years ago

Think about a packed lunch and the

Mazyrski [523]

Answer:

They are combination structures made up of solid, and shell structures.

Explanation:

Thinking of a lunch pack which is shaped as a solid bag made up of different compartments to hold food, what comes to mind are;

1. The solid structure: This is seen in the form of the bag which has solid makeup.

2. The shell structure: This is seen in the different compartments holding the different types of foods. Shell structures are usually in the form of containers made to hold something in.

4 0

3 years ago