l = 25 м; Т1 = –30°С; Т2 = 30°С; α = 1,1 ⋅ 10–5 К–1; Δl – ?
Δl = l(1 + α(T2 – T1)) – l = lα(T2 – T1) =
= 25 м ⋅ 1,1 ⋅ 10–5 К–1 ⋅ (30°С – (–30°С)) = 1,5 ⋅ 10–2 м = 1,5 см
l = 25 м; Т1 = –30°С; Т2 = 30°С; α = 1,1 ⋅ 10–5 К–1; Δl – ?
Δl = l(1 + α(T2 – T1)) – l = lα(T2 – T1) =
= 25 м ⋅ 1,1 ⋅ 10–5 К–1 ⋅ (30°С – (–30°С)) = 1,5 ⋅ 10–2 м = 1,5 см