Answer:
Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation
, Cooling fans
, Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat
, Insulation of hot surfaces Elimination of steam leaks
, Cooled seats or benches for rest breaks
, Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.
Hope this helped you!
Explanation:
Answer:
im sorry but i cant find any studies about this and im 3 days late
Answer:
The difference in weight and size?
Explanation:
It explains itself :P
def calculate_pay(total_worked_hours, rate_per_hour):
if total_hours_worked > 40:
# anything over 40 hours earns the overtime rate
overtimeRate = 2 * rate
_per_hour
return (40 * rate_per_hour) + ((total_worked_hours - 40) * overtimeRate
else:
# if you didn't work over 40 hours, there is no overtime
overtime = 0
return total_worked_hours * rate_per_hour
Explanation:
- First we create the calculate_pay function which will takes 2 parameters.
- Secondly ,inside the function we check if the total_worked_hours is greater than 40 and then return the pay by calculating with the help of formula for work over 40 hours.
- If the total work hour is less than 40 then we return the pay by multiplying the total_worked_hours with rate_per_hour.
Answer:
The surface temperature of the ground is = 296.946K
Explanation:
Solution
Given
r₁= 0.05m
r₂= 0.08m
Tn =Ti = 77K
Ki = 0.0035 Wm-1K-1
Kg = 1 Wm-1K-1
Z= 2m
Now,
The outer type temperature (Skin temperature pipe)
Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)
Thus,
10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05
⇒ T₀ -77 = 231.72
T₀= 290.72K
The shape factor between the cylinder and he ground
S = 2πL/ln 4z/D
where L = length of pipe
D = outer layer of pipe
S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m
The heat gained in the pipe is = S * Kg * (Tg- T₀)
(10* 1) = 1.606 * 1* (Tg- 290.72)
Tg - 290.72 = 6.2266
Tg = 296.946K
Therefore the surface temperature to the ground is 296.946K
