[د.امير]

Q1
A tow truck is pulling a car out of a ditch by means of a steel cable that has the length ( Lo) = 9.1 m
The radius ( r) = 0.50 cm.
= 0.005m
Then the area of the cross section is (A) = πr2
= (3.14)(0.005m)2
= 7.85*10-5m2
When the car just begins to move , the tension in the cable is (F) = 890 N
The young's modulus of the Y steel = 2.0 x 10¹¹ N/m²
We know that magnitude of F is given by
F = Y (ΔL/Lo)A
Then the stretched length in the cable is give by
Delta L = FL0/YA
= ( 890 N)(9.1m)/( 2.0 x 10¹¹ N/m²)(7.85*10-5m2)
= 8099 /15.7*106
= 515.859*10-6m
= 0.516mm