Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Answer:
It would break I think need to try it out
Explanation:
Answer:
<em>0.0386 hr</em>
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Explanation:
Area = 565 cm^2 = 0.0565 m^2 (1 cm^2 = 0.0001 m^2)
flux state rate = 220 mole/m^2-day
<em>There are 24 hrs in a day,</em> therefore rate in hrs will be
220/24 = 9.17 mole/m^2-hr
mass of water = 0.4 kg
molar mass of water = (1 x 2) + 16 = 18 kg/mole
therefore,
<em>mole of water = mass of water/molar mass of water</em>
mole of water = 0.4/18 = 0.02 mole
<em>mole flux = mole/area</em> = 0.02/0.0565 = 0.354 mol/m^2
<em>time that will be taken will be for water to pass = mole flux/mole flux rate</em>
time = 0.354/9.17 = <em>0.0386 hr</em>
Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw
Answer:
-2/√3 atan ((2t + 1)/√3) + C
Explanation:
∫ (t − 1) / (1 − t³) dt
Factor the difference of cubes:
∫ (t − 1) / ((1 − t)(1 + t + t²)) dt
Divide:
∫ -1 / (1 + t + t²) dt
-∫ 1 / (t² + t + 1) dt
Complete the square:
-∫ 1 / (t² + t + ¼ + ¾) dt
-∫ 4 / (4t² + 4t + 1 + 3) dt
-∫ 4 / ((2t + 1)² + 3) dt
If u = 2t + 1, du = 2 dt:
-∫ 2 / (u² + 3) du
Use an integral table, or use trigonometric substitution:
-2 (1/√3) atan (u/√3) + C
-2/√3 atan (u/√3) + C
Substitute back:
-2/√3 atan ((2t + 1)/√3) + C