Answer:
Explanation:
From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.
So, in light of that, to determine the electric charges values & signs on each of them, we have:
on a = +3 C
on b = -7 C
on c = -7 C
on d = +3 C
on e = -7 C
Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing.
<span>mass = density*volume </span>
<span>volume = (4/3)*pi*r^3 </span>
<span>volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 </span>
<span>mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere </span>
<span>Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 </span>
<span>Radius = cube root (volume / (4/3) / pi) = 2.87 cm. </span>
<span>I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.
hope this helped :)
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