Answer:
Calculate the individual residential lifetime cancer risk.
Risk = Potential factor x CDI …… (1)
Calculate the value of C
C =
t = d / v
t = 150 miles / 2 mile per hrs.
t = 75 hrs
t = 75/24
t = 3.13 days
Substitute the obtained value in (2).
C =
C = 0.72 x e^(-0.1*3.13)
C = 0.72 x 0.7313
C = 0.526 mg/L
Substitute the obtained value in (1).
Risk = Potential factor x CDI
Risk = 0.30kg.d/mg x 0.526mg/L x 2L/d x 350day/365days
Risk = 0.3026
Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
<u>Calculate the Principal stress, maximum in-plane shear stress and average normal stress</u>
Using Mohr's circle ( attached below )
<u>i) principal stresses:</u>
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
<u>ii) maximum in-plane shear stress</u>
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
<u>iii) average normal stress</u>
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
