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A copper power line that is 50.0 m long will be shorter after a temperature drop of 30.0 [tex]^\circ C[/tex]. What is the coefficient of linear expansion for copper?

A) [tex]1.70 \times 10^{-5} \, ^\circ C^{-1}[/tex]
B) [tex]1.70 \times 10^{-6} \, ^\circ C^{-1}[/tex]
C) [tex]1.70 \times 10^{-4} \, ^\circ C^{-1}[/tex]
D) [tex]1.70 \times 10^{-3} \, ^\circ C^{-1}[/tex]

Answer :

Final answer:

The coefficient of linear expansion for copper can be calculated using the formula ΔL = α × L0 × ΔT. Given the length change, original length, and temperature change, the coefficient for copper approximates to 1.67×10^-5 K^-1, which is closest to option A) 1.70×10^-5°C^-1. So, the best option is , 1.70×10⁻⁵5 °C⁻¹.

Explanation:

To find the coefficient of linear expansion for copper, we can use the formula for linear expansion, which is:

ΔL = α × L0 × ΔT

Where:

ΔL is the change in length (m meters in this question)

α is the coefficient of linear expansion

L0 is the original length (50.0 meters)

ΔT is the change in temperature (30.0 degrees Celsius)

Since we have ΔL as m, L0 as 50.0 meters, and ΔT as -30.0 degrees Celsius (since the temperature drops, it is negative), we can rearrange the formula to solve for α:

α = ΔL / (L0 × ΔT)

α = m / (50.0 m × -30.0°C)

We can use the given coefficient of linear expansion for copper, which is 1.67 × 10^-5 K^-1 (or °C^-1 since the problem gives temperatures in °C), to find the length change m:

m = α × L0 × ΔT

m = 1.67 × 10^-5 K^-1 × 50.0 m × -30.0°C

m = -0.02505 m (since a negative change in length signifies a contraction)

Therefore, the correct answer from the given options would be A) 1.70×10^-5°C^-1, which is approximately equal to the given coefficient, 1.67×10^-5 K^-1.

So, the best option is , 1.70×10⁻⁵5 °C⁻¹.

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Rewritten by : Jeany