The metal wire will experience a contraction of 19.76 mm when it is cooled from 39 to -1°C, with a coefficient of thermal expansion of 26 x 10^-6 °C^-1.
The question deals with the thermal expansion of a metal wire as its temperature changes. Using the linear expansion formula ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature, we can calculate the thermal contraction of the wire when it is cooled.
The original length of the wire is given as 19 m, the temperature change (ΔT) is from 39 to -1°C, thus the change in temperature is 39 - (-1) = 40°C. The coefficient of thermal expansion is given as 26 x 10-6 °C-1.
The change in length (ΔL) can be calculated as follows:
ΔL = αLΔT
ΔL = (26 x 10-6 °C-1) × (19 m) × (40°C)
ΔL = 0.01976 m or 19.76 mm
Because the wire is cooling, the change in length is negative, indicating that the wire contracts. Therefore, the metal wire will contract by 19.76 mm.
The complete question is- A metal wire 19 m long is cooled from 39 to -1°C. How much of a change in length will the wire experience if the coefficient of thermal expansion for this metal is 26 x 10-6 (°C) ?? If the change in length is negative (i.e., the wire shrinks), insert a minus sign before the numerical answer.