B i believe BUT IF IT DONT WORK IM SORRY LOLZ
D. Electromagnetic
I hope this helps!
Answer: Net force=20N a=F/m=4
Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of
= 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:
![\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BN_%7Bo%7D%7D%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7Bt%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7D)
where
N = No. of nuclei left after time t
= No. of nuclei initially started with
![\frac{N}{N_{o}} = 1\times 10^{- 4}](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BN_%7Bo%7D%7D%20%3D%201%5Ctimes%2010%5E%7B-%204%7D)
(Since, 100% - 99.99% = 0.01%)
Thus
![1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-%204%7D%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7Bt%7D%7B28.5%7D%7D%7D)
Taking log on both the sides:
![- 4 = \frac{t}{28.5}log\frac{1}{2}](https://tex.z-dn.net/?f=-%204%20%3D%20%5Cfrac%7Bt%7D%7B28.5%7Dlog%5Cfrac%7B1%7D%7B2%7D)
![t = \frac{-4\times 28.5}{log\frac{1}{2}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-4%5Ctimes%2028.5%7D%7Blog%5Cfrac%7B1%7D%7B2%7D%7D)
t = 387.69 yrs
C - line segment im not sure but im guessing that's the answer if that's helps.