A 1200-kg car is being raised at a constant speed if 25 cm/s by a crane,using a 20-m long steel cable that is 1.5 cm in diameter.Young's modulus for steel is 2.0 × 10¹¹ Pa.What is the change in length of the cable caused by the car? A)6.7 mm B)6.7 cm C)3.3 mm D)3.3 cm E)3.3 m

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The change in length ($\Delta L$) of the cable can be calculated using the formula derived from Hooke's Law for materials, which is $\Delta L = \frac{FL}{A E}$, where $F$ is the force applied (weight of the car in this case), $L$ is the original length of the cable, $A$ is the cross-sectional area of the cable, and $E$ is Young's modulus for the material of the cable.Given:- Mass of the car ($m$) = 1200 kg- Acceleration due to gravity ($g$) = 9.8 m/s$^2$- Length of the cable ($L$) = 20 m- Diameter of the cable ($d$) = 1.5 cm = 0.015 m (converted to meters)- Young's modulus for steel ($E$) = $2.0 \times 10^{11}$ PaFirst, calculate the force ($F$) due to the weight of the car: $F = mg = 1200 \times 9.8 = 11760$ N.Next, calculate the cross-sectional area ($A$) of the cable: $A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.015}{2}\right)^2 \approx 1.767 \times 10^{-4}$ m$^2$.Now, plug these values into the formula for $\Delta L$:$\Delta L = \frac{11760 \times 20}{1.767 \times 10^{-4} \times 2.0 \times 10^{11}} \approx 0.0067$ m, or 6.7 mm.

A 1200-Kg Car Is Being Raised at a Constant Speed