A truck is pulling a car out of a ditch by me | vedclass

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Question

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot l}{A \cdot \Delta l}$,where $F$ is the force,$l$ is the original length,$A$ i

Vedclass · Accepted Answer

$4.64 	imes 10^{-4} \, m$

Vedclass · Answer

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot l}{A \cdot \Delta l}$,where $F$ is the force,$l$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the change in length.Rearranging for $\Delta l$,we get $\Delta l = \frac{F \cdot l}{A \cdot Y}$.The cross-sectional area $A = \pi r^2$,where $r = 5 	imes 10^{-3} \, m$.Substituting the values: $A = 3.14 	imes (5 	imes 10^{-3})^2 = 3.14 	imes 25 	imes 10^{-6} = 7.85 	imes 10^{-5} \, m^2$.Now,$\Delta l = \frac{800 	imes 9.1}{7.85 	imes 10^{-5} 	imes 2 	imes 10^{11}}$.$\Delta l = \frac{7280}{15.7 	imes 10^6} = 463.69 	imes 10^{-6} \, m = 4.6369 	imes 10^{-4} \, m$.Rounding to two decimal places,the stretch is approximately $4.64 	imes 10^{-4} \, m$.

$A$ truck is pulling a car out of a ditch by means of a steel cable that is $9.1 \, m$ long and has a radius of $5 \, mm$. When the car just begins to move,the tension in the cable is $800 \, N$. How much has the cable stretched? (Young's modulus for steel is $2 \times 10^{11} \, N/m^2$)

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