Answer:
I = 1205.69 Lx
Explanation:
The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀
I = I₀ sin θ
in this case with the initial data we can calculate the initial irradiance
I₀ =
I₀ = 1600 /sin 53
I₀ = 2003.42 lx
for when the angle is θ = 37º
I = 2003.42 sin 37
I = 1205.69 Lx
Answer:
a) benzene = 910 days
b) toluene = 1612.67 days
Explanation:
Given:
Kd = 1.8 L/kg (benzene)
Kd = 3.3 L/kg (toluene)
psolid = solids density = 2.6 kg/L
K = 2.9x10⁻⁵m/s
pores = n = 0.37
water table = 0.4 m
ground water = 15 m
u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s
a) For benzene:
The time will take will be:
b) For toluene:
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
