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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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How old are legos? Who created them? Why did they create them?
Sunny_sXe [5.5K]
“How old are the legos” 89 years old.


“Who created them?” Ole Kirk Christiansen.


“Why did they created them?” As a you man, Christiansen turned his love of whittling and playing with wood into a Business and, in 1916, he opened his own shop. Since times were so hard, Christiansen made the hard decision to use his wood to create inexpensive goods that might actually sell. Among them were Cheap toys.





Hope this helps•


Bye~



•Kate•
4 0
3 years ago
Primary Creep: slope (creep rate) decreases with time
Igoryamba

Answer:

true

Explanation:

Creep is known as the time dependent deformation of structure due to constant load acting on the body.

Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

5 0
3 years ago
As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
katen-ka-za [31]

yes it will

Explanation:

5 0
3 years ago
The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne
pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

5 0
3 years ago
6.3.9 A coin was tossed n = 1000 times, and the proportion of heads observed was
4vir4ik [10]

Answer:

No

Explanation:

51 / 100 = 510 / 1000

Chance of getting a head is 1 / 2 of total throws

= 1 / 2 × 1000

= 500 is the probability

and the number of heads was just 10 more the the probability...if the was a greater gap, there would be evidence to say the coin is unfair

4 0
4 years ago
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