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.?
2 answers:
Answer:
<u>1.53</u>×10^-5 ( ⁰C)^-1
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
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