Answer:
kinematic viscosity is 0.0149 ft²/s
Explanation:
given data
specific gravity S = 0.94
density ρ = 0.94 × 1000
viscosity μ = 13 Poise = 1.3 Pa-sec
we know 1 poise = 0.1 pas
to find out
kinematic viscosity
solution
we will apply here Kinematic viscosity formula that is
kinematic viscosity = 
..................1
put here value in equation 1
and here ρ is density and μ is viscosity
kinematic viscosity = 
kinematic viscosity = 1.382978 × 
m³/s
so kinematic viscosity is 0.0149 ft²/s
Answer: You need to lift a load of 15 tons (30,000 pounds) a distance of 25 feet. The distance is measured from the center pin of the crane to the center of the load. Once you determine the distance, look on that line for the largest capacity; that will indicate how many feet of boom must be extended
Explanation:
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Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s
We need to first find the parameter k from the equation below.
taking the natural logarithm of both sides:
Substituting values:
Also
![t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}](https://tex.z-dn.net/?f=t%5En%3D-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%5C%5Ct%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D)
Substituting values and y = 99% = 0.99
![t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s](https://tex.z-dn.net/?f=t%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D%3D%5Csqrt%5B1.7%5D%7B-%5Cfrac%7Bln%281-0.99%29%7D%7B2.76%2A10%5E%7B-4%7D%7D%7D%3D304.6s)
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion
