Answer:
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
Explanation:
a) The mass flow rate through the nozzle can be calculated with the following equation:
![\dot{m_{i}} = \rho_{i} v_{i}A_{i}](https://tex.z-dn.net/?f=%20%5Cdot%7Bm_%7Bi%7D%7D%20%3D%20%5Crho_%7Bi%7D%20v_%7Bi%7DA_%7Bi%7D%20)
Where:
: is the initial velocity = 20 m/s
: is the inlet area of the nozzle = 60 cm²
: is the density of entrance = 2.21 kg/m³
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
![\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}](https://tex.z-dn.net/?f=%20%5Crho_%7Bi%7D%20v_%7Bi%7DA_%7Bi%7D%20%3D%20%5Crho_%7Bf%7D%20v_%7Bf%7DA_%7Bf%7D%20)
![0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}](https://tex.z-dn.net/?f=%200.27%20kg%2Fs%20%3D%200.762%20kg%2Fm%5E%7B3%7D%2A150%20m%2Fs%2AA_%7Bf%7D%20)
![A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}](https://tex.z-dn.net/?f=%20A_%7Bf%7D%20%3D%20%5Cfrac%7B0.27%20kg%2Fs%7D%7B0.762%20kg%2Fm%5E%7B3%7D%2A150%20m%2Fs%7D%20%3D%200.00236%20m%5E%7B2%7D%2A%5Cfrac%7B%28100%20cm%29%5E%7B2%7D%7D%7B1%20m%5E%7B2%7D%7D%20%3D%2023.6%20cm%5E%7B2%7D%20)
Therefore, the exit area of the nozzle is 23.6 cm².
I hope it helps you!
Your answer is ''Uniform''.
Hope this helps :)
Answer:
True
Explanation:
Electronegativity difference of less than 0.4 characterized covalent bonds. For two atoms with an electronegativity difference of between 0.4 and 2.0, a polar covalent bond is formed-one that is neither ionic nor totally covalent.
Answer: 10.36m/s
How? Just divide 200m by 19.3 and you will get how fast he ran per m/s