Answers to the Problem Set
- 5.964 \(\times\) 1018 kg
- 2.306 \(\times\) 10-22 nt
- 5.532 \(\times\) 10-65 nt
- \(\frac{\mu_0i}{2\pi R}\)
- \(\frac{1}{2}kx^2\)
- 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.
- 220.7 J
- \(\frac{1}{2}gt^2 = 10\) 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.
- 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).