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RoseWind [281]
3 years ago
6

-Why is it said that using faulty PPE could be just as dangerous as using no PPE at all?

Engineering
2 answers:
barxatty [35]3 years ago
6 0

Answer:

Because if you think you're safe, you're more likely to take less caution when in an environment where PPE is necessary. By using faulty PPE, you are putting yourself and others more at risk.

QveST [7]3 years ago
5 0

Answer:

Explanation:

"Safety helmet" redirects here. It is not to be confused with hard hat.

Drug Enforcement Administration (DEA) agents wearing Level B hazmat suits

Personal protective equipment (PPE) is protective clothing, helmets, goggles, or other garments or equipment designed to protect the wearer's body from injury or infection. The hazards addressed by protective equipment include physical, electrical, heat, chemicals, biohazards, and airborne particulate matter. Protective equipment may be worn for job-related occupational safety and health purposes, as well as for sports and other recreational activities. "Protective clothing" is applied to traditional categories of clothing, and "protective gear" applies to items such as pads, guards, shields, or masks, and others. PPE suits can be similar in appearance to a cleanroom suit.

The purpose of personal protective equipment is to reduce employee exposure to hazards when engineering controls and administrative controls are not feasible or effective to reduce these risks to acceptable levels. PPE is needed when there are hazards present. PPE has the serious limitation that it does not eliminate the hazard at the source and may result in employees being exposed to the hazard if the equipment fails.[1]

Any item of PPE imposes a barrier between the wearer/user and the working environment. This can create additional strains on the wearer; impair their ability to carry out their work and create significant levels of discomfort. Any of these can discourage wearers from using PPE correctly, therefore placing them at risk of injury, ill-health or, under extreme circumstances, death. Good ergonomic design can help to minimise these barriers and can therefore help to ensure safe and healthy working conditions through the correct use of PPE.

Practices of occupational safety and health can use hazard controls and interventions to mitigate workplace hazards, which pose a threat to the safety and quality of life of workers. The hierarchy of hazard controls provides a policy framework which ranks the types of hazard controls in terms of absolute risk reduction. At the top of the hierarchy are elimination and substitution, which remove the hazard entirely or replace the hazard with a safer alternative. If elimination or substitution measures cannot apply, engineering controls and administrative controls, which seek to design safer mechanisms and coach safer human behavior, are implemented. Personal protective equipment ranks last on the hierarchy of controls, as the workers are regularly exposed to the hazard, with a barrier of protection. The hierarchy of controls is important in acknowledging that, while personal protective equipment has tremendous utility, it is not the desired mechanism of control in terms of worker safety.rly PPE such as body armor, boots and gloves focused on protecting the wearer's body from physical injury. The plague doctors of sixteenth-century Europe also wore protective uniforms consisting of a full-length gown, helmet, glass eye coverings, gloves and boots (see Plague doctor costume) to prevent contagion when dealing with plague victims. These were made of thick material which was then covered in wax to make it water-resistant. A mask with a beak-like structure which was filled with pleasant-smelling flowers, herbs and spices to prevent the spread of miasma, the prescientific belief of bad smells which spread disease through the air.[2] In more recent years, scientific personal protective equipment is generally believed to have begun with the cloth facemasks promoted by Wu Lien-teh in the 1910–11 Manchurian pneumonic plague outbreak, although many Western medics doubted the efficacy of facemasks in preventing the spread of disease.[3]

Types

Personal protective equipment can be categorized by the area of the body protected, by the types of hazard, and by the type of garment or accessory. A single item, for example boots, may provide multiple forms of protection: a steel toe cap and steel insoles for protection of the feet from crushing or puncture injuries, impervious rubber and lining for protection from water and chemicals, high reflectivity and heat resistance for protection from radiant heat, and high electrical resistivity for protection from electric shock. The protective attributes of each piece of equipment must be compared with the hazards expected to be found in the workplace. More breathable types of personal protective equipment may not lead to more contamination but do result in greater user satisfaction.[4]

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3 years ago
The Cv factor for a valve is 48. Compute the head loss when 30 GPM of water passes through the valve.
dlinn [17]

Answer:

The head loss in Psi is 0.390625 psi.

Explanation:

Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.

Given:

Factor cv is 48.

Flow rate of water is 30 GPM.

GPM means gallon per minute.

Calculation:

Step1

Expression for head loss for the water is given as follows:

c_{v}=\frac{Q}{\sqrt{h}}

Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.

Step2

Substitute 48 for cv and 30 for Q in above equation as follows:

48=\frac{30}{\sqrt{h}}

{\sqrt{h}}=0.625

h = 0.390625 psi.

Thus, the head loss in Psi is 0.390625 psi.

 

5 0
3 years ago
An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator
MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit i_m=385mA=385\times 10^{-3}A=0.385A

Inductance of the inductor L=400mH=400\times 10^{-3}h=0.4H

Capacitance C=4.43\mu F=4.43\times 10^{-3}F

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm

Capacitive reactance will be equal to X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm

Impedance of the circuit will be Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm

So maximum voltage will be \Delta V_{max}=0.385\times 865.44=333.194volt

(B) Phase difference will be given as \Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70

So current will be leading by an angle 54.70

5 0
3 years ago
For the SR-latch below high levels of Set and Reset result in Q= 1 and 0, respectively. The next state is unknown when both inpu
dusya [7]

Answer:

hello your question lacks the required image attached to this answer is the image required

answer :  NOR1(q_) wave is complementary to NOR2(q)

Explanation:

Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_

Initial state is unknown i.e q = 0 and q_= 1

from the diagram the waveform reset and set

= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while  

from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )

From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.

From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table

also  from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.

since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)

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3 years ago
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