Of the cost reduction strategies for workers' compensation mentioned in the required readings, which one do you think would work
1 answer:
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Answer:
1.737 kJ
Explanation:
Thinking process:
Step 1
Data:
Area of the shaft = 0.8 cm²
Combined mass of shaft and piston = (24.5 + 0.5) kg
= 25 kg
Piston diameter = 0.1 m
External atmospheric pressure = 1 bar = 101.3 kPa
Pressure inside the gas cylinder = 3 bar = 3 × 101.3 kPa
g = 9.81 m/s²
Step 2
Draw a free body diagram
Step 3: calculations
area of the piston = 0.0314 m²
Change in the elevation of the piston,
z =
=
= 0.82 m
Next, we evaluate the work done by the shaft:
= (1668) ( 0.082)
= 1. 37 kJ
Net area for work done = A (piston) - Area of shaft
=
= 77.7 cm²
= 0.007774 m²
Work done in overcoming atmospheric pressure:
Wₐ = PAZ
=101.3 kPa * 0.007774 * 0.82
= 0.637 kJ
total work = work done by shaft + work to overcome atmospheric pressure = 0.367 + 1.37
= 1.737 kJ Ans
Answer:
(a) R₁ / (R₁ + )
(b) - / (R₁ +
)
Explanation:
(a)
i - Using voltage division rule =
v₀ = v₁ [ / (R₁+
)]
The sensitivity is
ii - = dv₀/d
*
/v₀
= R₁ / (R₁ + )
(b)
i - Using voltage divisor rule = v₀ = v₁ [ / (R₁+
)]
ii - = dv₀/d
*
/v₀
= - / (R₁ +
)