Answer:
given to you by others
Explanation:
Since it is other people who are giving you the feedback they may notice things that you are not aware you are doing. they can also give /help you to alter that thing and help you achieve you goal
To answer the question the antenna would probably burn up in the atmosphere. I have no idea how to get that thing on Earth if it's by itself. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
Answer:
Explanation:
From, the given information: we are not given any value for the mass, the proportionality constant and the distance
Assuming that:
the mass = 5 kg and the proportionality constant = 50 kg
the distance of the mass above the ground x(t) = 1000 m
Let's recall that:
![v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%5Cdfrac%7Bmg%7D%7Bb%7D%2B%20%28v_o%20-%20%5Cdfrac%7Bmg%7D%7Bb%7D%29%5Ee%5E%7B-bt%2Fm%7D)
Similarly, The equation of mption:
![x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cdfrac%7Bmg%7D%7Bb%7Dt%2B%5Cdfrac%7Bm%7D%7Bb%7D%20%28v_o%20-%20%5Cdfrac%7Bmg%7D%7Bb%7D%29%20%281-e%5E%7B-bt%2Fm%7D%29)
replacing our assumed values:
where ![v_=0 \ and \ g= 9.81](https://tex.z-dn.net/?f=v_%3D0%20%5C%20and%20%5C%20g%3D%209.81)
![x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%5Cdfrac%7B5%20%5Ctimes%209.81%7D%7B50%7Dt%2B%5Cdfrac%7B5%7D%7B50%7D%20%280%20-%20%5Cdfrac%7B%285%29%289.81%29%7D%7B50%7D%29%20%281-e%5E%7B-%2850%29t%2F5%7D%29)
![x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.981t%2B0.1%20%280%20-%200.981%29%20%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m)
![\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m%7D)
So, when the object hits the ground when x(t) = 1000
Then from above derived equation:
![\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%28t%29%20%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m%7D)
![1000= 0.981t-0.981(1-e^{-(10)t}) \ m](https://tex.z-dn.net/?f=1000%3D%200.981t-0.981%281-e%5E%7B-%2810%29t%7D%29%20%5C%20m)
By diregarding ![e^{-(10)t} \ m](https://tex.z-dn.net/?f=e%5E%7B-%2810%29t%7D%20%5C%20m)
![1000= 0.981t-0.981](https://tex.z-dn.net/?f=1000%3D%200.981t-0.981)
1000 + 0.981 = 0.981 t
1000.981 = 0.981 t
t = 1000.981/0.981
t = 1020.36 sec
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