Question

A basketball goal has a metal pole that is 3.50 m high at 21°C. On a cool day, at -11.5°C, it contracts by 0.00174 m.?

Answer 1

Answer:

1.53×10^-5 ( ⁰C)^-1

Answer 2

Answer:

1.53

Explanation:

α=Δl/(l Δt) here α is coefficient of linear expansion Δl is change in length l is original length andΔt is change in temperature

Δl=0.00174m , l=3.50m ,Δt=21-(-11-5)= 21+11.5 =32.5 ⁰C

α = 0.00174 / (3.50×32.5)

   = 1.53×10^-5 ( ⁰C)^-1