Perhatikan kedua vektor kecepatan berikut!   Vektor resultan pada sumbu X dan sumbu Y adalah...

Pertanyaan

Perhatikan kedua vektor kecepatan berikut!

 

Vektor resultan pada sumbu X dan sumbu Y adalah...

  1. begin mathsize 14px style stack v subscript x with rightwards arrow on top equals left parenthesis 1 plus 2 square root of 3 right parenthesis space straight m divided by straight s semicolon space stack v subscript y with rightwards arrow on top equals left parenthesis square root of 3 plus 2 right parenthesis space straight m divided by straight s end style 

  2. begin mathsize 14px style stack v subscript x with rightwards arrow on top equals left parenthesis 2 square root of 3 minus 1 right parenthesis space straight m divided by straight s semicolon space stack v subscript y with rightwards arrow on top equals left parenthesis 2 minus square root of 3 right parenthesis space straight m divided by straight s end style 

  3. begin mathsize 14px style stack v subscript x with rightwards arrow on top equals left parenthesis 2 square root of 3 plus 1 right parenthesis space straight m divided by straight s semicolon space stack v subscript y with rightwards arrow on top equals left parenthesis square root of 3 minus 2 right parenthesis space straight m divided by straight s end style 

  4. begin mathsize 14px style stack v subscript x with rightwards arrow on top equals left parenthesis 1 minus 2 square root of 3 right parenthesis space straight m divided by straight s semicolon space stack v subscript y with rightwards arrow on top equals left parenthesis 2 minus square root of 3 right parenthesis space straight m divided by straight s end style 

  5. begin mathsize 14px style stack v subscript x with rightwards arrow on top equals left parenthesis negative 1 minus 2 square root of 3 right parenthesis space straight m divided by straight s semicolon space stack v subscript y with rightwards arrow on top equals left parenthesis negative 2 minus square root of 3 right parenthesis space straight m divided by straight s end style 

U. Dwi

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Diketahui
stack v subscript 1 with rightwards harpoon with barb upwards on top equals 4 space straight m divided by straight s stack v subscript 2 with rightwards harpoon with barb upwards on top equals 2 space straight m divided by straight s theta subscript 1 equals 30 degree theta subscript 2 equals 60 degree  

Ditanyakan
Vektor resultan pada sumbu X dan sumbu Y

Jawab
Tinjau resultan vektor di masing-masing sumbu.

resultan vektor di sumbu x

begin mathsize 14px style v with rightwards arrow on top subscript x equals v with rightwards arrow on top subscript 1 space cos space theta space minus space v with rightwards arrow on top subscript 2 space cos space theta v with rightwards arrow on top subscript x equals 4 space cos space 30 to the power of straight omicron minus 2 space cos space 60 to the power of straight omicron v with rightwards arrow on top subscript x equals left parenthesis 2 square root of 3 minus 1 right parenthesis space bevelled straight m over straight s end style 

resultan vektor di sumbu y

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Dengan demikian, vektor resultan pada sumbu X dan sumbu Y adalah stack bold v subscript bold x with bold rightwards arrow on top bold equals bold left parenthesis bold 2 square root of bold 3 bold minus bold 1 bold right parenthesis bold space bold m bold divided by bold s bold semicolon   stack bold v subscript bold y with bold rightwards arrow on top bold equals bold left parenthesis bold 2 bold minus square root of bold 3 bold right parenthesis bold space bold m bold divided by bold s.

Jadi, jawaban yang tepat adalah B.

1rb+

5.0 (1 rating)

Pertanyaan serupa

Tiga buah vektor  F1​​=(−15 i + 10 j​ ) satuan,  F2​​=(10 i + 5 j​ ) satuan, dan  F3​​=(5 i − 4 j​ ) satuan. Tentukan : a.   F1​​ + F2​​ − F3​​  b.   F1​​ − F2​​ − F3​​

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