l e t ′ s a s s u m e t h a t t h e l e n g t h o f t h e p o l e r e s t i n g r e l a t i v e t o t h e s h e d i s e q u a l t o L = 16 u n i t t h e n a c c o r d i n g t o t h e L o r e n t z t r a n s f o r m a t i o n s ( b y t h e t e r m s ) t h e l e n g t h o f a p o l e m o v i n g r e l a t i v e t o a s h e d a t t h e s p e e d o f v i s e q u a l t o L 2 = 8 u n i t . S o L 2 = L 1 − v 2 c 2 ; 1 − v 2 c 2 = 1 2 ; 1 − v 2 c 2 = 1 4 ; v = c 2 3 ; v = 3 × 1 0 8 2 × 1.73 ; v = 2.6 × 1 0 8 m c . A n s w e r : v = 2.6 × 1 0 8 m c o r v = 0.87 c let's\ assume\ that \;the\ length\ of \;the\ pole\\ resting \;relative\; to\\ the \;shed\; is\; equal \;to\ L=16\ unit\\ then\ according\ to\; the \;Lorentz\ transformations\\(by\ the\ terms) \;the\ length\ of\ a\ pole\\ moving\ relative \;to\ a\ shed\\ at\ the\ speed\ of\ v\ is\ equal\ to\;\frac{L}{2}=8\;unit .\;So\\\frac{L}{2}=L\sqrt{1-\frac{v^2}{c^2}}; \sqrt{1-\frac{v^2}{c^2}}=\frac{1}{2};\\1-\frac{v^2}{c^2}=\frac{1}{4}; v=\frac{c}{2}\sqrt{3}; v=\frac{3\times10^8}{2}\times1.73;\\v=2.6\times10^8\frac{m}{c}.\\Answer: v=2.6\times10^8\frac{m}{c}\;or\; v=0.87c l e t ′ s a ss u m e t ha t t h e l e n g t h o f t h e p o l e res t in g re l a t i v e t o t h e s h e d i s e q u a l t o L = 16 u ni t t h e n a ccor d in g t o t h e L ore n t z t r an s f or ma t i o n s ( b y t h e t er m s ) t h e l e n g t h o f a p o l e m o v in g re l a t i v e t o a s h e d a t t h e s p ee d o f v i s e q u a l t o 2 L = 8 u ni t . S o 2 L = L 1 − c 2 v 2 ; 1 − c 2 v 2 = 2 1 ; 1 − c 2 v 2 = 4 1 ; v = 2 c 3 ; v = 2 3 × 1 0 8 × 1.73 ; v = 2.6 × 1 0 8 c m . A n s w er : v = 2.6 × 1 0 8 c m or v = 0.87 c
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