Jawabanmu

2014-05-31T00:00:32+07:00
Momen gaya = F/r
40 =  \frac{F}{0,01}
F= 0,4 N

Tegangan = F/A
 \frac{0,4}{1/4 π d^2}
 \frac{0,4}{1/4 π 0,0001}
 \frac{0,4}{0,000025 π}
 \frac{16000}{π} N/m^2
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