To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:
![Q = \frac{\Delta P \pi r^4}{8\eta l}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5CDelta%20P%20%5Cpi%20r%5E4%7D%7B8%5Ceta%20l%7D)
Where:
are the viscosities of the concrete before and after the increase
l = Length of the vessel
= Radio of the vessel before and after the increase
= Change in the pressure
The rates of flow before and after he increase
Our values are given as:
10 times her resting rate
95% of its normal value
Increase of 50%
Plugging known information to get
![Q_1 = \frac{\Delta P \pi r^4}{8\eta l}](https://tex.z-dn.net/?f=Q_1%20%3D%20%5Cfrac%7B%5CDelta%20P%20%5Cpi%20r%5E4%7D%7B8%5Ceta%20l%7D)
![Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4](https://tex.z-dn.net/?f=Q_1%208%5Ceta_1%20l%20%3D%20%5CDelta%20P_1%20%5Cpi%20r_1%5E4)
![r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}](https://tex.z-dn.net/?f=r_1%5E4%20%3D%20%5Cfrac%7BQ_1%208%5Ceta_1%20l%7D%7B%5CDelta%20P_1%20%5Cpi%7D)
![r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}](https://tex.z-dn.net/?f=r_1%20%3D%20%28%5Cfrac%7BQ_1%208%5Ceta_1%20l%7D%7B%5CDelta%20P_1%20%5Cpi%7D%29%5E%7B1%2F4%7D)
![r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}](https://tex.z-dn.net/?f=r_2%20%3D%20%28%5Cfrac%7BQ_2%208%5Ceta_2%20l%7D%7B%5CDelta%20P_2%20%5Cpi%7D%29%5E%7B1%2F4%7D)
![r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}](https://tex.z-dn.net/?f=r_2%20%3D%20%28%5Cfrac%7B10Q_18%20%5Ctimes%200.95%5Ceta_1%20l%7D%7B1.5%5CDelta%20P_1%20%5Cpi%7D%29%5E%7B1%2F4%7D)
![r_2 = 1.586r_1](https://tex.z-dn.net/?f=r_2%20%3D%201.586r_1)
Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.
D
Giddy UP!!!!!!!!!!!!!!!!!!!!!
Answer:
2649600 Joules
Explanation:
Efficiency = 40%
m = Mass of air = 92000 kg
v = Velocity of wind = 12 m/s
Kinetic energy is given by
![K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20K%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2092000%5Ctimes%2012%5E2%5C%5C%5CRightarrow%20K%3D6624000%5C%20J)
The kinetic energy of the wind is 6624000 Joules
The wind turbine extracts 40% of the kinetic energy of the wind
![E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J](https://tex.z-dn.net/?f=E%3D0.4%5Ctimes%20K%5C%5C%5CRightarrow%20E%3D0.4%5Ctimes%206624000%5C%5C%5CRightarrow%20E%3D2649600%5C%20J)
The energy extracted by the turbine every second is 2649600 Joules
m=23.8kg a=8.97m/s^2 Fnet=? Fnet=ma=(23.8kg)(8.97m/s^2)=213.486N
Answer:
true can i get brainliest :)
Explanation: