Answer:
The road will narrow
P.S. please add more information when asking a question
To solve the problem it is necessary to take into account the concepts related to the Reynolds Number and the Force of drag on the bodies subjected to a Fluid.
The Reynolds number for the Prototype and the Model must therefore be preserved,
![Re_p = Re_m](https://tex.z-dn.net/?f=Re_p%20%3D%20Re_m)
![\frac{V_mL_m}{\upsilon_m} = \frac{V_pL_p}{\upsilon_p}](https://tex.z-dn.net/?f=%5Cfrac%7BV_mL_m%7D%7B%5Cupsilon_m%7D%20%3D%20%5Cfrac%7BV_pL_p%7D%7B%5Cupsilon_p%7D)
Re-arrange for the speed of the model we have,
![V_m = \frac{L_p}{L_m}\frac{\upsilon_m}{\upsilon_p}V_p](https://tex.z-dn.net/?f=V_m%20%3D%20%5Cfrac%7BL_p%7D%7BL_m%7D%5Cfrac%7B%5Cupsilon_m%7D%7B%5Cupsilon_p%7DV_p)
Our values at 20°C would be given of the table of Physical Properties of water where
![\upsilon_m=1*10^{-6}m^2/s](https://tex.z-dn.net/?f=%5Cupsilon_m%3D1%2A10%5E%7B-6%7Dm%5E2%2Fs)
![\rho_m = 998kg/m^3](https://tex.z-dn.net/?f=%5Crho_m%20%3D%20998kg%2Fm%5E3)
While for the values previous given we have
![V_p = 2m/s](https://tex.z-dn.net/?f=V_p%20%3D%202m%2Fs)
![\upsilon_m=1*10^{-6}m^2/s](https://tex.z-dn.net/?f=%5Cupsilon_m%3D1%2A10%5E%7B-6%7Dm%5E2%2Fs)
![\upsilon_p=1.4*10^{-6}](https://tex.z-dn.net/?f=%5Cupsilon_p%3D1.4%2A10%5E%7B-6%7D)
And we have a Ratio between the prototype and the model of 16:1, then
![V_m = \frac{L_p}{L_m}\frac{\upsilon_m}{\upsilon_p}V_p](https://tex.z-dn.net/?f=V_m%20%3D%20%5Cfrac%7BL_p%7D%7BL_m%7D%5Cfrac%7B%5Cupsilon_m%7D%7B%5Cupsilon_p%7DV_p)
![V_m = \frac{16}{1}\frac{1*10^{-6}}{1.4*10^{-6}}*2](https://tex.z-dn.net/?f=V_m%20%3D%20%5Cfrac%7B16%7D%7B1%7D%5Cfrac%7B1%2A10%5E%7B-6%7D%7D%7B1.4%2A10%5E%7B-6%7D%7D%2A2)
![V_m = 22.857m/s](https://tex.z-dn.net/?f=V_m%20%3D%2022.857m%2Fs)
PART B) To calculate the ratio of the drag force now we have to,
![\frac{F_{DM}}{F_{DP}} = \frac{L_m}{L_p}^2\frac{V_m}{V_p}^2\frac{\rho_m}{\rho_p}](https://tex.z-dn.net/?f=%5Cfrac%7BF_%7BDM%7D%7D%7BF_%7BDP%7D%7D%20%3D%20%5Cfrac%7BL_m%7D%7BL_p%7D%5E2%5Cfrac%7BV_m%7D%7BV_p%7D%5E2%5Cfrac%7B%5Crho_m%7D%7B%5Crho_p%7D)
Replacing with our values we have,
![\frac{F_{DM}}{F_{DP}} = \frac{1}{16}^2\frac{22.857}{2}^2\frac{998}{1015}](https://tex.z-dn.net/?f=%5Cfrac%7BF_%7BDM%7D%7D%7BF_%7BDP%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B16%7D%5E2%5Cfrac%7B22.857%7D%7B2%7D%5E2%5Cfrac%7B998%7D%7B1015%7D)
![\frac{F_{DM}}{F_{DP}} = 0.5016](https://tex.z-dn.net/?f=%5Cfrac%7BF_%7BDM%7D%7D%7BF_%7BDP%7D%7D%20%3D%200.5016)
Therefore the ratio of drag force for prototype and model is 0.5016
Answer:
- B. Precipitation require the diffusional activation energy plus an additional energy to form the precipitate.
Explanation:
Precipitation is the creation of a solid from a solution. When the reaction occurs in a liquid solution, the solid formed is called the precipitate.
The formation of a precipitate indicates the occurrence of a chemical reaction.
Precipitation of carbide requires alot of energy which the diffusion activational energy alone cannot achieve and this was calculated to be 225.6 kJ/mol.