Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
Answer:
speed by mass attain is 55.86 m/s
Explanation:
given data
glucose = 10 g
mass = 100 kg
to find out
speed by mass attain
solution
we know glucose have 180 g molecular weight and
that 1 g glucose produce energy = 2816/180 × 10³ J
so here 10 g of glucose produce energy = 1.56 ×
J
so here energy release = 1/2 × mv²
1.56 ×
= 1/2 × (100)v²
v² = 3.12 × 10³
and v = 55.86 m/s
so speed by mass attain is 55.86 m/s
Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved
answer : Electric power generated = 216 * 10^6 kJ
money saved = $0.XY * 60000 kwh
Explanation:
<u>Calculating the amount of electric power generated by wind turbine</u>
power generated = ( 30 * 2000 ) kWh = 60000 kWh
Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ
<u>Calculate money saved by school per year </u>
$0.XY * 60000 kwh
Answer:
<em><u>The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'</u></em>
<em><u>Mark</u></em><em><u> </u></em><em><u>as</u></em><em><u> brilliant</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> </u></em>
Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1

For pipe 2

Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV




head loss (h)

Now putting the all values

So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.