diketahui :
Lo = 10 cm
α = 1,9 x 10⁻⁵ /° c
ΔL = 0,50 mm = 0,05 cm
ditanya :
Δt
jawab :
ΔL = Lo . α . Δt
Δt = ΔL / ( Lo . α )
= 0,05 / ( 10 x 1,9 x 10⁻⁵ )
= 263,2° C
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diketahui :
Lo = 10 cm
α = 1,9 x 10⁻⁵ /° c
ΔL = 0,50 mm = 0,05 cm
ditanya :
Δt
jawab :
ΔL = Lo . α . Δt
Δt = ΔL / ( Lo . α )
= 0,05 / ( 10 x 1,9 x 10⁻⁵ )
= 263,2° C