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Maurinko [17]
3 years ago
9

One employee was climbing a metal ladder to hand an electric drill to the journeyman installer on a scaffold about five feet abo

ve him.
When the victim reached the third rung from the bottom of the ladder he received an electric shock that killed him. The investigation
revealed that the extension cord had a missing grounding prong and that a conductor on the green grounding wire was making
intermittent contact with the energizing black wire thereby energizing the entire length of the grounding wire and the drill's frame.
The drill was not double insulated. What could have been done to prevent this accident?
Stathatay
U
d
f orce
to
ed equipment Grounding conductor program to protect employees on construction sites
Engineering
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

ACCIDENT PREVENTION RECOMMENDATIONS  

Fatal Fact made us to understand some of the prevention techniques as stated below.

1. Use approved ground fault circuit interrupters or an assured equipment grounding conductor program to protect employees on construction sites.  

2. Use equipment that provides a permanent and continuous path from circuits, equipment, structures, conduit or enclosures to ground.

3. Inspect electrical tools and equipment daily and remove damaged or defective equipment from use until it is repaired.

Explanation:

In order to gain a better understanding of the answer above let explain some terms

Ground Fault Circuit Interrupters :

    A ground fault circuit interrupter (GFCI), or Residual Current Device (RCD) is a type of circuit breaker which shuts off electric power when it senses an imbalance between the outgoing and incoming current. The main purpose is to protect people from an electric shock caused when some of the current travels through a person's body due to an electrical fault such as a short circuit, insulation failure, or equipment malfunction.

So the first statement is implying that in order to prevent this accident that  this device (GFCI) should have  been used in that  construction site, or as  an alternative before the  construction commenced  the company should have drafted a lay down conductor program(i.e. a step by step conductor program) that assured equipment grounding  in order to protect employees on construction sites

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A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0
3 years ago
Tires can be recycled instead of thrown out.<br> True<br> False
Arisa [49]

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

8 0
2 years ago
There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

5 0
2 years ago
Question 74
torisob [31]

Answer:

C). rearview mirror (at least of four inches by four inches).

Explanation:

<u>A 'rearview mirror' would be needed in case one is required to pull an individual on water skis after a PWC(personal watercraft) that is rated for carrying two persons only</u>. This wide-angle mirror for rearview would allow the operator to monitor the person who is being towed constantly along with riding the ski at the same time. It is considered illegal to tow a person on a vessel without having a rearview mirror and at the same time, the limited capacity must also be followed strictly. Hence, <u>option C</u> is the correct answer.

7 0
2 years ago
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