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earnstyle [38]
2 years ago
12

Respiratory protection ____________________________ against fumes and gases.

Engineering
1 answer:
anygoal [31]2 years ago
3 0

Respiratory protection <u>provides fool proof protection</u> against fumes and gases during wielding.

<h3>What is PPE?</h3>

PPE is an acronym for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which is designed and developed to offer protection to different parts of the body while working in a potentially hazardous environment.

Some examples of personal protective equipment (PPE) that are used to protect the different parts of the body include the following:

  • Respirators
  • Face mask
  • Face shield
  • Gloves
  • Boots
  • Helmet

In conclusion, respiratory protection <u>provides fool proof protection</u> against fumes and gases during wielding.

Read more on PPE here: brainly.com/question/19131588

#SPJ1

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A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
A contractor is planning on including several skylights in each unit of a residential development. What type of worker would she
Sladkaya [172]

Answer:

Glazier

Explanation:

Glaziers are workers who specializes in cutting and installation of glass works.

They work with glass in various surfaces and settings, such as cutting and installing windows and doors, skylights, storefronts, display cases, mirrors, facades, interior walls, etc.

Thus, the type of worker the contractor will hire for this project is a Glazier

8 0
3 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
4 years ago
Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr
baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0
3 years ago
The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th
Alja [10]

Answer:

Explanation:

y'''+y=0---(i)

General solution

y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)

y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1

in equation (iii)

c_1=c_2=-1

Therefore,

\large\boxed{y=-1-\cos x-2\sin x}

5 0
3 years ago
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