Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
Answer:
A,C, and D
Explanation:
Potible ladders have to configure with many designs in mind but the most evedent is that they are usally unstable
BRAINLIEST PLS
Answer:
The answer is "828.75"
Explanation:
Please find the correct question:
For W21x93 BEAM,
![Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3](https://tex.z-dn.net/?f=Z_x%20%3D%20221.00%20in%5E3%20%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%3D32.3)
For A992 STREL,
![F_y= 50\ ks](https://tex.z-dn.net/?f=F_y%3D%2050%5C%20%20ks)
Check for complete section:
![\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bb_t%7D%7B2t_f%7D%20%3D4.53%20%3C%20%5Cfrac%7B65%7D%7B%5Csqrt%7Bf_y%20%3D%209.19%7D%7D%5C%5C%5C%5C%5Cto%20%5Cfrac%7Bh%7D%7Bt_w%7D%20%3D32.3%20%3C%20%5Cfrac%7B640%7D%7B%5Csqrt%7Bf_y%20%3D%2090.5%7D%7D)
Design the strength of beam =![\phi_b Z_x F_y\\\\](https://tex.z-dn.net/?f=%5Cphi_b%20Z_x%20F_y%5C%5C%5C%5C)
![=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\](https://tex.z-dn.net/?f=%3D0.9%20%5Ctimes%20221%20%5Ctimes%2050%5C%5C%5C%5C%3D9945%20%5C%20in%20%5C%20%5C%20kips%5C%5C%5C%5C%3D%5Cfrac%7B9945%7D%7B12%7D%5C%5C%5C%5C%3D%20828.75%20%5Cft%20%5C%20kips%20%5C%5C)
Answer:
The answer is "
and 157.5 MPa".
Explanation:
In point A:
The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point
![y = yo + \frac{k}{\sqrt{x}}](https://tex.z-dn.net/?f=y%20%3D%20%20yo%20%2B%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Bx%7D%7D)
Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.
Replacing the equation for each condition:
![y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k](https://tex.z-dn.net/?f=y%20%3D%20y_o%20%2B%20%5Cfrac%7Bk%7D%7B%5Csqrt%7B%281%20%5Ctimes%2010%5E%7B-2%7D%29%7D%7D%5C%5C%5C%5C%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20230%20%3D%20yo%20%2B%2010k%5C%5C%5C%5C%20y%20%3D%20yo%20%2B%20%5Cfrac%7Bk%7D%7B%5Csqrt%7B%286%5Ctimes%2010%5E%7B-3%7D%29%7D%7D%5C%5C%5C%5C275%20%3D%20yo%20%2B%2012.90k)
People can get yo = 275 MPa with both equations and k= 15.5 Mpa
.
To substitute the answer,
![310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm](https://tex.z-dn.net/?f=310%20%3D%20275%20%2B%20%5Cfrac%7B%2815.5%29%7D%7B%5Csqrt%7Bx%7D%7D%5C%5C%5C%5Cx%20%3D%200.00435%20%5C%20mm%20%3D%204.35%20%5Ctimes%2010%5E%7B-3%7D%5C%20%20mm)
In point b:
The equation is ![\sigma y = \sigma 0 + k y d^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Csigma%20y%20%3D%20%5Csigma%200%20%2B%20k%20y%20d%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
equation is:
![75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\](https://tex.z-dn.net/?f=75%20%3D%20%5Csigma%20o%2B4%20ky%20%5C%5C%5C%5C175%20%3D%20%5Csigma%20o%2B12%20ky%5C%5C%5C%5Cky%20%3D%2012.5%20MPa%20%28mm%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%5C%5C%20%5Csigma%200%20%3D%2025%20MPa%5C%5C%5C%5Cd%3D%208.9%20%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5Cd%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%7D%20%3D10.6%20mm%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C)
by putting the above value in the formula we get the
value that is= 157.5 MPa