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Tiga buah massa berada di titik-titik sudut segitiga seperti pada gambar di samping. Jika m A ​ = 20 kg , m B ​ = 27 kg , m C ​ = 64 kg , dan G = 6 , 67 × 1 0 − 11 Nm 2 / kg 2 , tentukan gaya yang dirasakan benda A!

Tiga buah massa berada di titik-titik sudut segitiga seperti pada gambar di samping.

 

Jika , dan , tentukan gaya yang dirasakan benda A!

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Jawaban terverifikasi

Jawaban

gaya yang dirasakan benda A sebesar .

gaya yang dirasakan benda A sebesar begin mathsize 14px style bold 6 bold comma bold 67 bold cross times bold 10 to the power of bold minus bold 5 end exponent bold space bold N end style.

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Pembahasan

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Diketahui : Ditanya : benda A ? Jawab : Resultan gaya yang dirasakan benda A Jadi, gaya yang dirasakan benda A sebesar .

Diketahui :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript A end cell equals cell 20 space kg end cell row cell m subscript B end cell equals cell 27 space kg end cell row cell m subscript C end cell equals cell 64 space kg end cell row G equals cell 6 comma 67 cross times 10 to the power of negative 11 end exponent space Nm squared divided by kg squared end cell end table end style 

Ditanya : begin mathsize 14px style F subscript t o t a l end subscript end style benda A ?

Jawab :

 

Resultan gaya yang dirasakan benda A

 begin mathsize 14px style F subscript t o t a l end subscript equals square root of F subscript A B end subscript squared plus F subscript A C end subscript squared end root F subscript t o t a l end subscript equals square root of open parentheses fraction numerator G. m subscript A. m subscript B over denominator r subscript A B end subscript squared end fraction close parentheses squared plus open parentheses fraction numerator G. m subscript A. m subscript C over denominator r subscript A C end subscript squared end fraction close parentheses squared end root F subscript t o t a l end subscript equals square root of open parentheses fraction numerator G.20.27 over denominator 0 comma 03 squared end fraction close parentheses squared plus open parentheses fraction numerator G.20.64 over denominator 0 comma 04 squared end fraction close parentheses squared end root F subscript t o t a l end subscript equals square root of open parentheses fraction numerator G.20.27 over denominator 9 cross times 10 to the power of 4 end fraction close parentheses squared plus open parentheses fraction numerator G.20.64 over denominator 16 cross times 10 to the power of 4 end fraction close parentheses squared end root F subscript t o t a l end subscript equals square root of left parenthesis 6 G cross times 10 to the power of 5 right parenthesis squared plus left parenthesis 8 G cross times 10 to the power of 5 right parenthesis squared end root F subscript t o t a l end subscript equals square root of 36 G squared cross times 10 to the power of 10 plus 64 G squared cross times 10 to the power of 10 end root F subscript t o t a l end subscript equals square root of 100 G squared cross times 10 to the power of 10 end root F subscript t o t a l end subscript equals 10 G cross times 10 to the power of 5 F subscript t o t a l end subscript equals 10 cross times 10 to the power of 5.6 comma 67 cross times 10 to the power of negative 11 end exponent F subscript t o t a l end subscript equals 6 comma 67 cross times 10 to the power of negative 5 end exponent space straight N end style

Jadi, gaya yang dirasakan benda A sebesar begin mathsize 14px style bold 6 bold comma bold 67 bold cross times bold 10 to the power of bold minus bold 5 end exponent bold space bold N end style.

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