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:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
Answer: 35.3 °
Explanation:
Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.
The simplest crystal structures are those that have present only a single atom at each lattice point.
body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms
With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:
[1 1 0] [1 1 1]
λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))
Cos^-1 2/ sqrt 6
= 35.386°
Answer:
#WeirdestQuestionOfAllTime
Explanation:
False you have to do the work and perricapate
