I believe it is D. Earth spinning on it's axis.
Explanation:
Nuclear reactions are the reactions in which nucleus of an atom changes either by splitting or joining with the nucleus of another atom.
There are two types of nuclear reactions.
- Nuclear fission - In this process, large atomic nuclei splits into smaller nuclei.
- Nuclear fusion - In this process, two small nuclei combine together to form a large nuclei.
Both nuclear fission and fusion processes involve nuclei of atoms.
For example, ![^{0}n + ^{235}_{92}U \rightarrow ^{236}_{92}U \rightarrow ^{144}_{56}Ba + ^{89}_{36}Kr + 3n + 177 MeV](https://tex.z-dn.net/?f=%5E%7B0%7Dn%20%2B%20%5E%7B235%7D_%7B92%7DU%20%5Crightarrow%20%5E%7B236%7D_%7B92%7DU%20%5Crightarrow%20%5E%7B144%7D_%7B56%7DBa%20%2B%20%5E%7B89%7D_%7B36%7DKr%20%2B%203n%20%2B%20177%20MeV)
Thus, we can conclude that statements which are true are as follows.
- Nuclear reactions involve the nuclei of atoms.
- The products of nuclear reactions are lighter than the reactants.
a. The disk starts at rest, so its angular displacement at time
is
![\theta=\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%5Calpha2t%5E2)
It rotates 44.5 rad in this time, so we have
![44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}](https://tex.z-dn.net/?f=44.5%5C%2C%5Cmathrm%7Brad%7D%3D%5Cdfrac%5Calpha2%286.00%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%5Calpha%3D2.47%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D)
b. Since acceleration is constant, the average angular velocity is
![\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Comega_f%2B%5Comega_i%7D2%3D%5Cdfrac%7B%5Comega_f%7D2)
where
is the angular velocity achieved after 6.00 s. The velocity of the disk at time
is
![\omega=\alpha t](https://tex.z-dn.net/?f=%5Comega%3D%5Calpha%20t)
so we have
![\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_f%3D%5Cleft%282.47%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%286.00%5C%2C%5Cmathrm%20s%29%3D14.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
making the average velocity
![\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B14.8%5Cfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%7D2%3D7.42%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
Another way to find the average velocity is to compute it directly via
![\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5CDelta%5Ctheta%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B44.5%5C%2C%5Crm%20rad%7D%7B6.00%5C%2C%5Crm%20s%7D%3D7.42%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
c. We already found this using the first method in part (b),
![\omega=14.8\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega%3D14.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
d. We already know
![\theta=\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%5Calpha2t%5E2)
so this is just a matter of plugging in
. We get
![\theta=179\,\mathrm{rad}](https://tex.z-dn.net/?f=%5Ctheta%3D179%5C%2C%5Cmathrm%7Brad%7D)
Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that
![\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2](https://tex.z-dn.net/?f=%5Ctheta%3D44.5%5C%2C%5Cmathrm%7Brad%7D%2B%5Cleft%2814.8%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%5Cright%29t%2B%5Cdfrac%5Calpha2t%5E2)
Then for
we would get the same
.
The correct answer is a, lunar eclipse