Answer:
The answer would be B if im wrong SORRY ;)
Step-by-step explanation:
Answer:
the answer is "d" because the x region is negative 2 and the y region is a positive1
Answer:
Step-by-step explanation:
The model fo the shell is given by the following equation of equilibrium:

This first-order differential equation has separable variables, which are cleared herein:

The solution of this integral is:
![t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7Bm%7D%7B2b%7D%5Ccdot%20%5Cleft%5B%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7Bv%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%5Cright%29%20-%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%5Cright%29%5Cright%5D)

![\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%3D%5Ctan%20%5Cleft%5B-%5Cfrac%7B2%5Ccdot%20b%5Ccdot%20t%7D%7Bm%7D%20%2B%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%5Cright%5D)
![v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%5Cleft%20%5B%5Cfrac%7B%5Ctan%20%5Cleft%28-%5Cfrac%7B2%5Ccdot%20b%20%5Ccdot%20t%7D%7Bm%7D%20%20%5Cright%29%2B%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%7D%7B1%20-%20%5Cleft%28%5Cfrac%7B600%7D%7B%5Csqrt%7B%5Cfrac%7Bm%5Ccdot%20g%7D%7Bb%7D%20%7D%20%7D%20%20%5Cright%29%5Ccdot%20%5Ctan%20%5Cleft%28-%5Cfrac%7B2%5Ccdot%20b%20%5Ccdot%20t%7D%7Bm%7D%20%20%5Cright%29%20%7D%5Cright%5D)
Answer:
16
Step-by-step explanation:
good luck
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Answer:
18.66 ft/s
Step-by-step explanation:
The distance between you and the elevator is given by:

The rate of change for the distance between you and the elevator is given by:



Applying the chain rule:

Therefore, at x=300 and y = 500, dy/dt is:

The elevator is descending at 18.66 ft/s.