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3 years ago

6

saan nag tungo si Aguinaldo at ilang pinuno ng kilusan pagkatapos mapairal ang kasunduan na pansamantalang nag dulot ng kapayapa

an

1 answer:

avanturin [10]3 years ago

6 0

Answer:

sa china po

Explanation:

sana makatulong ako

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Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the

OLga [1]

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

$Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3$

For A992 STREL,

$F_y= 50\ ks$

Check for complete section:

$\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}$

Design the strength of beam = $\phi_b Z_x F_y\\\\$

$=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\$

8 0

3 years ago

Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc

aksik [14]

Answer:

a) $\eta = 13.455\%$ , b) $E_{day} = 812.716\,kJ$ , c) $C_{month. total} = 19.505\, USD$ , d) $t = 40.588\,years$

Explanation:

a) The area of the solar panel is:

$A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}$

$A = 1.858\,m^{2}$

The energy potential is determined herein:

$\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})$

$\dot E_{o} = 1858\,W$

The efficiency of the solar panel is:

$\eta = \frac{\dot E}{\dot E_{o}}\times 100\%$

$\eta = \frac{250\,W}{1858\,W}\times 100\%$

$\eta = 13.455\%$

b) The energy generated by the solar panel is presented below:

$E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )$

$E_{day} = 812.716\,kJ$

c) The energy generated per month and per panel is:

$E_{month} = 30\cdot E_{day}$

$E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)$

$E_{month} = 6.773\,kWh$

Monthly energy savings due to the use of 18 panels are:

$C_{month, total} = 18\cdot E_{month}\cdot c$

$C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )$

$C_{month. total} = 19.505\, USD$

d) The payback of the solar energy system is:

$t = \frac{9500\,USD}{12\cdot (19.505\,USD)}$

$t = 40.588\,years$

6 0

3 years ago

A metal alloy has been tested in a tensile test with the following results for the flow curve

MA_775_DIABLO [31]

Answer:

dhjdjs

Explanation:

ehejsikajaudiieisisjsjsjsjsjjsjsjsjsjs

4 0

3 years ago

You are diving a car out in rural America and suddenly you get a flat tire. you pull off to the side of the road. you check your

julsineya [31]

Answer:

Walk to saftey!

Explanation:

4 0

3 years ago

(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a

Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 × $10^{-5}$ %

Explanation:

No. of phosphorus atoms = 5 × $10^{17} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100$

PCT = $10^{-3} = 0.001$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × $10^{15} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100$

PCT = 0.4 × $10^{-5}$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

7 0

3 years ago