Answer:
B-as deadly storms that claim lives in th great pines
Explanation:
I hope it helped if it did pls make me brainliest
Answer:
474.59 mg/L
Explanation:
Given that
BOD = 30 mg/L
Original BOD = 30 mg/L × dilution factor
Original BOD = 30 mg/L × 10 = 300 mg/L

here
is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day

Answer:
Explanation:
Initial conditions


Final conditions


Steady flow energy equation
![\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%5Cleft%20%5B%20h_1%2B%5Cfrac%7Bv_1%5E2%7D%7B2%7D%2Bgz_1%5Cright%20%5D%2B%5Cdot%7BQ%7D%3D%5Cdot%7Bm%7D%5Cleft%20%5B%20h_2%2B%5Btex%5D%5Cfrac%7Bv_2%5E2%7D%7B2%7D%2Bgz_2%5Cright%20%5D%2B%5Cdot%7BW%7D)
![\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%5Cleft%20%5B%20c_pT_1%2B%5Cfrac%7B0%5E2%7D%7B2%7D%2Bg0%5Cright%20%5D%2B%5Cdot%7BQ%7D%3D%5Cdot%7Bm%7D%5Cleft%20%5B%20c_pT_2%2B%5Cfrac%7B0%5E2%7D%7B2%7D%2Bg0%5Cright%20%5D%2B%5Cdot%7BW%7D)
![\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121](https://tex.z-dn.net/?f=%5Cdot%7Bm%7Dc_p%5Cleft%20%5B%20T_1-T_2%5Cright%20%5D%2B%5Cleft%20%5B%20-5hp%5Cright%20%5D%3D%5Cdot%7BW%7D%20-5%5Ctimes%20746%5Ctimes%203.4121)
![-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]](https://tex.z-dn.net/?f=-4%5Cdot%7Bm%7D-%5Cdot%7Bm%7D%5Ctimes%200.24%5Ctimes%20%5Cleft%20%5B%20400-60%5Cright%20%5D)



Answer:
Explanation:
Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)
fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)
vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)
BFFS = 50+5, BFFS =55 (given) fLW= 6.6
TLC=6+3=9 fLC= 0.65
fM= 0.0
fA= 1.0
FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)
Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)
After: fA= 3.0
FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)
Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln
Answer:
Fatigue factor of safety is 2.0267
Explanation:
Solution is attached below.