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
Answer:
B
Explanation:
The impulse experienced by an object is the force•time.
Answer:
Increase
Explanation:
The plane strain fracture toughness of a metal is expected to increase with rising temperature.
Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area![= \sigma = 1.99*10^{-7} C/m^2](https://tex.z-dn.net/?f=%3Cstrong%3E%20%3D%20%5Csigma%20%3D%201.99%2A10%5E%7B-7%7D%20C%2Fm%5E2%3C%2Fstrong%3E)
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is![E = \frac{\sigma}{\epsilon}](https://tex.z-dn.net/?f=%20E%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Cepsilon%7D)
force acting on charge is F = q E
Work done by charge q id![\Delta X =\frac{ q\sigma \Delta x}{\epsilon}](https://tex.z-dn.net/?f=%20%5CDelta%20X%20%3D%5Cfrac%7B%20q%5Csigma%20%5CDelta%20x%7D%7B%5Cepsilon%7D)
this work done is converted into kinectic enerrgy
![\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%5Cfrac%7B%20q%5Csigma%20%5CDelta%20x%7D%7B%5Cepsilon%7D)
solving for v
![v = \sqrt{\frac{2q\Delta x}{\epsilon m}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2q%5CDelta%20x%7D%7B%5Cepsilon%20m%7D)
![\epsilon = 8.85*10^{-12} Nm2/C2](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%208.85%2A10%5E%7B-12%7D%20Nm2%2FC2)
![v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%201.6%2A10%5E%7B-19%7D1.99%2A10%5E%7B-7%7D%2A1.69%2A10%5E%7B-2%7D%7D%7B8.85%2A10%5E%7B-12%7D%20%2A9.1%2A10%5E%7B-31%7D%7D)
v = 1.15*10^{7} m/s
None can.
A clinical thermometer only measures temperatures above +30°C.
Mercury and alcohol are both frozen solid at -50°C.