Roboguru

Suatu partikel memiliki energi diam sedang bergerak dengan energi kinetik dan kecepatan sedemikian rupa hingga . untuk partikel besarnya adalah ...

Suatu partikel memiliki energi diam begin mathsize 14px style E subscript 0 end style sedang bergerak dengan energi kinetik begin mathsize 14px style E K end style dan kecepatan begin mathsize 14px style v end style sedemikian rupa hingga begin mathsize 14px style v over c equals 0 comma 99 end stylebegin mathsize 14px style fraction numerator E K over denominator E subscript 0 end fraction end style untuk partikel besarnya adalah ...undefined

  1. 2

  2. 4

  3. 6,1

  4. 9

  5. 5

Jawaban:

begin mathsize 14px style v over c equals 0 comma 99 space rightwards double arrow v equals 0 comma 99 c  E k equals E subscript t o t end subscript minus E subscript 0 space space space space space equals m c squared minus m subscript 0 c squared space space space space space equals open parentheses m minus m subscript 0 close parentheses c squared space space space space space equals open parentheses fraction numerator m subscript 0 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction minus m subscript 0 close parentheses c squared space space space space space equals open parentheses fraction numerator 1 over denominator square root of 1 minus begin display style open parentheses 0 comma 99 c close parentheses squared over c squared end style end root end fraction minus 1 close parentheses m subscript 0 c squared space space space space space equals open parentheses fraction numerator 1 over denominator square root of 1 minus begin display style fraction numerator 0 comma 9801 c squared over denominator c squared end fraction end style end root end fraction minus 1 close parentheses m subscript 0 c squared space space space space space equals open parentheses fraction numerator 1 over denominator square root of 1 minus 0 comma 9801 end root end fraction minus 1 close parentheses m subscript 0 c squared space space space space space equals open parentheses fraction numerator 1 over denominator square root of 0 comma 0199 end root end fraction minus 1 close parentheses m subscript 0 c squared space space space space space equals open parentheses fraction numerator 1 over denominator 0 comma 14 end fraction minus 1 close parentheses m subscript 0 c squared space space space space space equals open parentheses 7 comma 1 minus 1 close parentheses m subscript 0 c squared space space space space space equals 6 comma 1 m subscript 0 c squared space space space space space equals 6 comma 1 E subscript 0  E k equals 6 comma 1 E subscript 0 fraction numerator E k over denominator E subscript 0 end fraction equals 6 comma 1 end style 

Jadi, jawaban yang benar adalah C.

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