I have 12 * 10^-6 for the thermal expansion coef of steel
meters in --> meters out
18.3 * 12 * 10^-6 * (39) = 8564*10^-6 m
= .008564 m = 8.564 mm
Steel railroad tracks of length L = 18.30 m are laid at 10.0° C. How much space should be left between the track sections if they are to just touch when the temperature is T = 49.0° C? mm
Please help. I have tried this
L° α Δt = 18.3 * 13 10^-6 * (49 -10) = .92781cm which equals to 9.2681 mm I think but its wrong. please help.
2 answers
Thank you so much Damon :)!