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timama [110]
3 years ago
9

If a m = 74.7 kg m=74.7 kg person were traveling at v = 0.800 c v=0.800c , where c c is the speed of light, what would be the ra

tio of the person's relativistic kinetic energy to the person's classical kinetic energy?
Physics
1 answer:
igomit [66]3 years ago
4 0

Answer:

\frac{E}{E_c} =3.125

Explanation:

The kinetic energy of a rigid body that travels at a speed v is given by the expression:

E_c=\frac{1}{2} mv^2

The equivalence between mass and energy established by the theory of relativity is given by:

E=mc^2

This formula states that the equivalent energy E can be calculated as the mass m multiplied by the speed of light c squared.

Where c is approximately 3\times 10^{8} m/s

Hence:

E_c=\frac{1}{2} (74.7)*(0.8*3\times 10^{8} )^2=2.15136\times 10^{18} J

E=(74.7)*(3\times 10^{8} )^2 =6.723\times 10^{18} J

Therefore,  the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy is:

\frac{E}{E_c} =\frac{6.723\times 10^{18}}{2.15136\times 10^{18}} =3.125

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Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

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        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

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