## 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).