(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²
<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>
<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always
Here mass number on the product side is given to us
so sum of mass number is given as
![A_1 + A_2 = 265 + 1 = 266](https://tex.z-dn.net/?f=A_1%20%2B%20A_2%20%3D%20265%20%2B%201%20%3D%20266%20)
now on the reactant side also the number must be same
![A_1' + A_2' = 58 + x](https://tex.z-dn.net/?f=A_1%27%20%2B%20A_2%27%20%3D%2058%20%2B%20x)
now we will have
![58 + x = 266](https://tex.z-dn.net/?f=58%20%2B%20x%20%3D%20266)
![x = 208](https://tex.z-dn.net/?f=x%20%3D%20208)
Now number of protons on product side is given as
![P_1 + P_2 = 108 + 0](https://tex.z-dn.net/?f=P_1%20%2B%20P_2%20%3D%20108%20%2B%200)
now we also know that atomic number of Fe is 26
so now we will have
![P_1' + P_2' = 108](https://tex.z-dn.net/?f=P_1%27%20%2B%20P_2%27%20%3D%20108)
![26 + P_2' = 108](https://tex.z-dn.net/?f=26%20%2B%20P_2%27%20%3D%20108)
![P_2' = 82](https://tex.z-dn.net/?f=P_2%27%20%3D%2082)
now the equation is given as
![_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X](https://tex.z-dn.net/?f=_%7B26%7D%5E%7B58%7DFe%20%2B%20_%7B82%7D%5E%7B208%7DPb%20%3D%20_%7B108%7D%5E%7B266%7DHs%20%2B%20_0%5E1X)
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