4.7 Answers to the Problem Set

1. 5.964 $\times$ 10¹⁸ kg
   
2. 2.306 $\times$ 10^-22 nt
   
3. 5.532 $\times$ 10^-65 nt
   
4. $\frac{\mu_0i}{2\pi R}$
   
5. $\frac{1}{2}kx^2$
   
6. Using the relations $F = ma$ and $F = G\frac{mM}{R^2}$, we get $a=\frac{GM}{R^2}$ where we define $a = g$. We substitute the relation $\frac{GM}{R}=gR$ into $v=\sqrt\frac{2GM}{R}$ to obtain $v=\sqrt{2gR}$. Substituting, we obtain $v=\sqrt{2*9.81*6368*10^3}$ m/s. We divide by 0.00062 to convert to miles/s. Obtaining v=6.93 miles/s.
7. 220.7 J
   
8. $\frac{1}{2}gt^2 = 10$ so it took $\sqrt{\frac{20}{g}}$ seconds to reach the ground. By that much time the vertical speed reaches $v = g\cdot t^2 = 14$ m/s. Since the initial horizontal speed was 3 m/s and it is unchanged, the speed at the ground is $\sqrt{14^2 + 3^2} = 14.3$ m/s.
9. 0.530 $\times$ 10^-10 m. Hence the diameter of a hydrogen atom is 1.06 $\times$ 10^-10 m (or approximately 1 angstrom unit in an earlier terminology).