Answer:
a) 50μC
b) 37.45 m/s
Explanation:
a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.
Thus, you have:

Hence, each sphere has a charge of 50μC.
b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:
![\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J](https://tex.z-dn.net/?f=%5CDelta%20W%3D%5CDelta%20K%5C%5C%5C%5C%5Cint_%7B0.4%7D%5E%5Cinfty%20Fdr%3D%5Cfrac%7B1%7D%7B2%7Dm%5Bv%5E2-v_o%5E2%5D%5C%5C%5C%5CF%3Dk%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5C%5C%5C%5Cv_o%3D0m%2Fs%5C%5C%5C%5Cm%3D0.08kg%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3DkQ%5E2%5B-%5Cfrac%7B1%7D%7Br%7D%5D_%7B0.4%7D%5E%7B%5Cinfty%7D%3D%5Cfrac%7BkQ%5E2%7D%7B0.4m%7D%3D%5Cfrac%7B%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%2850%2A10%5E%7B-6%7DC%29%5E2%7D%7B0.4m%7D%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3D56.125J)
where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2
Next, you equal the total work to the change in K:

hence, the speed of the spheres is 37.45 m/s
Using a punnet square,
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The offspring will be 50% Heterozygous dominant and 50% homozygous recessive.
<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>