Question

Sepotong kawat luas penampangnya 4mm² diregangkan oleh gaya 3,2 N. bila panjang kawat mula mula adlh 8 cm dan modulus elastisitasnya 32x10pngkat9 N/m².brpa mm pertambahan panjang kawat tsb ?

Answer 1

Diket:
A = 4mm² = 4.10⁻⁶ m²
F = 3,2 N
lo = 8 cm = 8.10⁻² m
E = 32.10⁹ N/m²

Dit: Δl.....?

E = F/A x lo/Δl
32.10⁹ = 3,2/4.10⁻⁶ x 8.10⁻²/Δl
32.10⁹ = 25,6.10⁻²/4.10⁻⁶ Δl
128.10³ Δl = 25,6.10⁻²
Δl = 25,6.10⁻²/128.10³
Δl = 0,2.10⁻⁵ m
Δl = 0,2.10⁻² mm = 2.10⁻³ mm