Answer:
[tex]\frac{E}{E_c} =3.125[/tex]
Explanation:
The kinetic energy of a rigid body that travels at a speed v is given by the expression:
[tex]E_c=\frac{1}{2} mv^2[/tex]
The equivalence between mass and energy established by the theory of relativity is given by:
[tex]E=mc^2[/tex]
This formula states that the equivalent energy [tex]E[/tex] can be calculated as the mass [tex]m[/tex] multiplied by the speed of light [tex]c[/tex] squared.
Where [tex]c[/tex] is approximately [tex]3\times 10^{8} m/s[/tex]
Hence:
[tex]E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J[/tex]
[tex]E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J[/tex]
Therefore, the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:
[tex]\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125[/tex]
