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Pertanyaan

Dimensi modulus Young adalah...

Dimensi modulus Young adalah...space 

  1. begin mathsize 14px style open square brackets straight M close square brackets open square brackets straight T to the power of negative 2 end exponent close square brackets end style 

  2. begin mathsize 14px style open square brackets straight M close square brackets open square brackets straight L close square brackets to the power of negative 1 end exponent open square brackets straight T close square brackets to the power of negative 2 end exponent end style 

  3. begin mathsize 14px style open square brackets straight M close square brackets open square brackets straight L close square brackets to the power of negative 1 end exponent open square brackets straight T close square brackets to the power of negative 1 end exponent end style 

  4. begin mathsize 14px style open square brackets straight M close square brackets open square brackets straight L close square brackets open square brackets straight T close square brackets end style 

  5. begin mathsize 14px style open square brackets straight L close square brackets open square brackets straight T close square brackets to the power of negative 2 end exponent end style 

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N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

jawaban yang tepat adalah B.

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Pembahasan

Pembahasan
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Diketahui Ditanyakan Dimensi Modulus Young ( E ) Jawab Dimensi Modulus Young dapat dicari menggunakan satuan dari persamaan berikut. Jadi, jawaban yang tepat adalah B.

Diketahui
begin mathsize 14px style Massa equals open square brackets straight M close square brackets Panjang equals open square brackets straight L close square brackets Waktu equals open square brackets straight T close square brackets end style 

Ditanyakan
Dimensi Modulus Young (E)

Jawab
Dimensi Modulus Young dapat dicari menggunakan satuan dari persamaan berikut.

begin mathsize 14px style E equals fraction numerator F L over denominator A capital delta L end fraction E equals fraction numerator k g times begin display style m over s squared end style times m over denominator m squared times m end fraction E equals fraction numerator k g times begin display style fraction numerator up diagonal strike m over denominator s squared end fraction end style times up diagonal strike m over denominator up diagonal strike m squared end strike times m end fraction E equals fraction numerator k g times s to the power of negative 2 end exponent over denominator m end fraction E equals k g times s to the power of negative 2 end exponent times m to the power of negative 1 end exponent E equals open square brackets straight M close square brackets open square brackets T close square brackets to the power of negative 2 end exponent open square brackets L close square brackets to the power of negative 1 end exponent E equals left square bracket straight M right square bracket open square brackets L close square brackets to the power of negative 1 end exponent open square brackets T close square brackets to the power of negative 2 end exponent  end style  

Jadi, jawaban yang tepat adalah B.

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