contestada

An observer in frame S′ is moving to the right (+x-direction) at speed u = 0.600c away from a stationary observer in frame S. The observer in S′ measures the speed v′ of a particle moving to the right away from her. What speed v does the observer in S measure for the particle if (a) v′=0.400c; (b) v′=0.900c; (c) v′=0.990c?

Respuesta :

Hashirriaz830

Answer:

a) 0.806c

b) 0.97c

c) 0.997c

Explanation:

given:

[tex]v_{x}'=v'\\ v_{x} =v[/tex]

solution:

a) [tex]v=\frac{v'+u}{1+\frac{uv'}{c^2} }= \frac{0.400+0.600}{1+(0.400)(0.600) }=0.806c[/tex]

b)[tex]v=\frac{v'+u}{1+\frac{uv'}{c^2} }= \frac{0.900+0.600}{1+(0.900)(0.600) }=0.97c[/tex]

c) [tex]v=\frac{v'+u}{1+\frac{uv'}{c^2} }= \frac{0.99+0.600}{1+(0.99)(0.600) }=0.997c[/tex]

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