An iron rod of length = 50 cm at temperature 12 celcius at what temperature will it become 50.12
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Answered by
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L' = L + L@(T2-T1)
50.12 = 50 + 50×12×10^(-6)[T2-12]
REARRANGING:
6×10^(-4)[T2-12] = 0.12
T2 -12 = 200
T2 = 212
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Answered by
0
L' = L + L@(T2-T1)
50.12 = 50 + 50×12×10^(-6)[T2-12]
REARRANGING:
6×10^(-4)[T2-12] = 0.12
T2 -12 = 200
T2 = 212
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