Answer:
1) The mass of the continent is approximately [tex]1.608\times 10^{21}[/tex] kilograms.
2) The kinetic energy of the continent is approximately [tex]8.04\times 10^{16}[/tex] joules.
3) The speed of the 77 kg-jogger would be approximately [tex]45.698\times 10^{6}[/tex] meters per second.
Explanation:
1) The mass of the North American continent can be estimated by using the following formula under the assumption that rock has an uniform density:
[tex]m = \rho \cdot L^{2}\cdot h[/tex] (1)
Where:
[tex]m[/tex] - Mass of the continent, measured in kilograms.
[tex]\rho[/tex] - Average density of the rock, measured in kilograms per cubic meter.
[tex]L[/tex] - Side of the continent, measured in meters.
[tex]h[/tex] - Depth of the continent, measured in meters.
If we know that [tex]\rho = 2620\,\frac{kg}{m^{3}}[/tex], [tex]L = 4.450\times 10^{6}\,m[/tex] and [tex]h = 31\times 10^{3}\,m[/tex], then the mass of the continent is:
[tex]m = \left(2620\,\frac{kg}{m^{3}} \right)\cdot (4.450\times 10^{6}\,m)^{2}\cdot (31\times 10^{3}\,m)[/tex]
[tex]m = 1.608\times 10^{21}\,kg[/tex]
The mass of the continent is approximately [tex]1.608\times 10^{21}[/tex] kilograms.
2) By assuming that continent can be represented as a particle, we define its kinetic energy as:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]v[/tex] - Motion rate of the continent, measured in meters per second.
If we know that [tex]m = 1.608\times 10^{21}\,kg[/tex] and [tex]v = 1\times 10^{-2}\,\frac{m}{s}[/tex], then the kinetic energy of the continent is:
[tex]K = \frac{1}{2}\cdot (1.608\times 10^{21}\,kg)\cdot \left(1\times 10^{-2}\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 8.04\times 10^{16}\,J[/tex]
The kinetic energy of the continent is approximately [tex]8.04\times 10^{16}[/tex] joules.
3) The speed of the jogger is derived from the definition of translational kinetic energy:
[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]
If we know that [tex]K = 8.04\times 10^{16}\,J[/tex] and [tex]m = 77\,kg[/tex], then the expected speed of the jogger is:
[tex]v = \sqrt{\frac{2\cdot (8.04\times 10^{16}\,J)}{77\,kg} }[/tex]
[tex]v\approx 45.698\times 10^{6}\,\frac{m}{s}[/tex]
The speed of the 77 kg-jogger would be approximately [tex]45.698\times 10^{6}[/tex] meters per second.
