A piece of metal 0.12 m² in cross sectional area is stretched by a force of 74 * 10⁵ N.The fractional change in length is 5.11 * 10^-4 ($\alpha$ = 19* 10^-6 C^-1,Y = 9.0 * 10¹⁰ N/m²).The piece of metal was:

Question

Quizplus · Accepted Answer

We can use the formula ΔL/L = Y(ΔT)/α to find the change in temperature.

Plugging in the given values, we get:

5.11 × 10^-4 = (9.0 × 10^10 N/m^2)(ΔT)/(19 × 10^10 N/m^2)(0.12m)

Simplifying, we get:

ΔT = 9.2°C

Since the fractional change in length is positive, we know that the metal experienced tensile stress. From the formula for thermal expansion, we know that as temperature increases, length increases. Therefore, to counteract the tensile stress and bring the fractional change in length back down to zero, the metal must be cooled.

Thus, the answer is C) cooled by 9.2°C.

A Piece of Metal 0 α\alphaα = 19* 10-6 C-1,Y = 9

A Piece of Metal 0 $\alpha$ = 19* 10^-6 C^-1,Y = 9