Consider a frame S\' (x\', y\', z\', t\') moving with a velocity 2.87801 times 1

Question

Consider a frame S' (x', y', z', t') moving with a velocity 2.87801 times 10^8 m/s in the positive z direction relative to the frame S (x, y, z, t). For convenience, assume that the origins of the two reference frames overlap at time t = t' = 0. The relativistic coordinate transformation is observed length L = Z_2 - z_1 in the stationary reference frame given L' = z'_2 - z'_1 in the moving reference frame? The speed of light is 2.99792 times 10^8 m/s and the length in the primed frame is 56 m. Answer in units of m.

Explanation / Answer

length in the stationary frame L 0= L(length in prime frame)*gama

= 56/(1 - (v/c)^2)^(1/2)

= 56/(1 -( 2.87801/2.99792)^2)^(1/2)

= 56/(0.0784)^(1/2)

= 56/0.28

= 200 m

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