Answer:
3.59
4.32
Explanation:
We find the time factors and engine factor to solve for this.
Engine factor = (100%*15/60) + (60%*30/60) + (20%*15/60)
= 1x0.25 + 0.6x0.5 + 0.2x0.25
= 0.25 + 0.30 + 0.05
= 0.6
A. We find time factor
= 50/60 = 0.83 minutes
We then get consumption
= Time factor x engine x hp x .04
= 0.83 x 0.6 x 180 x 0.04
= 3.585
3.59 gallons in 1 hr
B. Time factor = 60/60 = 1
Consumption =
1x0.6x180x0.04
= 4.32 gallons in 1 hour
Answer:
874 psi
Explanation:
Given a sample mean (x') = 900,
and a standard error (SE) = 10
At 99% confidence, Z(critical) = 2.58
That gives 99% confidence interval as,
x' ± Z(critical) x SE = 900 ± 2.58 x 10
The value of the lower limit is,
900 - 25.8 = 874.2
≈ 874 psi
Answer:
ananswer my question please
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if then you have a thin walled cylinder
or using the diameter:
- if then you have a thin walled cylinder