Roboguru

Pada suatu alat pemercepat proton yang bernama SSC sebuah proton direncanakan akan mempunyai energi kinetik sampai 3 x 10-6 J. Hitunglah berapa kecepatan proton ini! Massa proton 1,67 x 10-27 kg.

Pertanyaan

Pada suatu alat pemercepat proton yang bernama SSC sebuah proton direncanakan akan mempunyai energi kinetik sampai 3 x 10-6 J. Hitunglah berapa kecepatan proton ini! Massa proton 1,67 x 10-27 kg.undefined 

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Pembahasan Soal:

Dari persamaan energi kinetik relatiivistik akan diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E subscript k end cell equals cell E minus E subscript o end cell row cell E subscript k end cell equals cell fraction numerator m subscript o c squared over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction minus m subscript o c squared end cell row cell E subscript k end cell equals cell m subscript o c squared left parenthesis fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction minus 1 right parenthesis end cell row cell fraction numerator 1 comma 67 cross times 10 to the power of negative 6 end exponent over denominator 1 comma 67 cross times 10 to the power of negative 27 end exponent times left parenthesis 3 cross times 10 to the power of 8 right parenthesis squared end fraction end cell equals cell left parenthesis fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction minus 1 right parenthesis end cell row cell 0 comma 111 cross times 10 to the power of 5 end cell equals cell left parenthesis fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction minus 1 right parenthesis end cell row cell 11110 plus 1 end cell equals cell fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction end cell row 11111 equals cell fraction numerator 1 over denominator square root of 1 minus begin display style v squared over c squared end style end root end fraction end cell row cell 1 comma 23 cross times 10 to the power of 6 end cell equals cell fraction numerator 1 over denominator 1 minus begin display style v squared over c squared end style end fraction end cell row cell 1 comma 23 cross times 10 to the power of 6 minus 1 end cell equals cell 1 comma 23 cross times 10 to the power of 6 v squared over c squared end cell row 1229999 equals cell 1230000 v squared over c squared end cell row cell v squared end cell equals cell fraction numerator 1229999 c squared over denominator 1230000 end fraction end cell row cell v squared end cell equals cell 0 comma 999999187 c squared end cell row v equals cell 0 comma 999999594 c end cell end table end style     

Jadi, kecepatan partikel tersebut adalah 0,999999594c atau hampir sama dengan c.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Maret 2021

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