Answer: because they take a lot of tie to form
The largest risks while designing a model to withstand a village include that the model does not mitigate the effects of the tsunami or only mitigates the effects partially, which would cause damages to the homes.
Designing a model to withstand the effect of any natural phenomenon such as an earthquake, fire or tsunami is not an easy task and will require the following cycle:
- Designing the model.
- Testing the model.
- Making changes or designing a new model.
In the case of a model for tsunamis, it is likely the following problems occur:
- The model does not protect the houses from tsunamis.
- The model does not protect the houses completely.
This would lead to negative effects such as:
- Damages in the houses.
- Dead or injured people.
- Destruction of infrastrcture.
Note: This question is incomplete because the context is missing; here is the missing part.
Protecting Your Model Village from Tsunamis this task, you will design a model village to withstand the effects of a tsunami.
Learn more about tsunami in: brainly.com/question/1126317
(a) Pushing the spring down gives it stored mechanical energy that turns into motion
Explanation:
Pushing on the spring causes the mechanical energy, of pushing on the spring, to be stored in the spring through potential elastic energy. Due to the elasticity of the spring, when the spring is released and resumes its initial shape the stored energy is released and can be used to do work such as motion.
Answer:
0.702 /s
Explanation:
Rate constant at 
Rate constant at 
Activation energy, 
Use the following equation to calculate 
Use the following equation to calculate
![\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cfrac%7B%5Cmathrm%7BK%7D_%7B2%7D%7D%7B%5Cmathrm%7B~K%7D_%7B1%7D%7D%3D%5Cfrac%7B%5Cmathrm%7BEa%7D%7D%7B%5Cmathrm%7BR%7D%7D%5Cleft%5B%5Cfrac%7B1%7D%7B%5Cmathrm%7B~T%7D_%7B1%7D%7D-%5Cfrac%7B1%7D%7B%5Cmathrm%7B~T%7D_%7B2%7D%7D%5Cright%5D)
Therefore,
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B1%7D%7B298%20%5Cmathrm%7B~K%7D%7D-%5Cfrac%7B1%7D%7B350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)
![\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%28%5Cfrac%7BK_%7B2%7D%7D%7B3.46%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7B~s%7D%5E%7B-1%7D%7D%5Cright%29%20%26%3D%5Cfrac%7B50.2%20%5Ctimes%2010%5E%7B3%7D%20%5Cmathrm%7B~J%7D%20%2F%20%5Cmathrm%7Bmol%7D%7D%7B8.314%20%5Cmathrm%7BJK%7D%5E%7B-1%7D%20%5Cmathrm%7B~mole%7D%5E%7B-1%7D%7D%5Cleft%5B%5Cfrac%7B52%20%5Cmathrm%7B~K%7D%7D%7B298%20%5Cmathrm%7B~K%7D%20%5Ctimes%20350%20%5Cmathrm%7B~K%7D%7D%5Cright%5D)
hence, the rate constant at 
is 0.702
Percent yield is calculated using:
Your actual yield is how much you actually had at the end of the experiment = 79g.
Your theoretical yield is how much you calculated you should have gotten in an ideal, perfect experiment = 104g.
So put those values into the equation:
