Final answer:
To find how the 1.00mm diameter nylon string is lengthened under 265 N of tension, we use the formula ΔL = (FL) / (AENylon) along with Young's modulus, converting diameters to meters for the cross-sectional area calculation.
Step-by-step explanation:
To determine how much the nylon string is lengthened from its untensioned length of 29.0 cm when under a tension of 265 N, we use Hooke's Law in conjunction with Young's modulus. The extension (ΔL) can be calculated using the formula ΔL = (FL) / (AENylon), where F is the force applied (265 N), L is the original length (29.0 cm), A is the cross-sectional area of the string, and ENylon is the Young's modulus of nylon (5.00×109 N/m2). First, we convert the diameter of the string from millimeters to meters and calculate the cross-sectional area (A = πd2/4). Then we can insert these values into our formula to find the change in length.
The cross-sectional area A is given by A = π (Ø/2)2 where Ø is the diameter. Converting the diameter to meters, we have Ø = 1.00 mm = 1.00×10-3 m. So, A = π (1.00×10-3 m /2)2.
Now, we can calculate the elongation ΔL: ΔL = (265 N)(0.29 m) / (π (1.00×10-3 m /2)2 (5.00×109 N/m2)).
Finally, we solve for ΔL to find how much the nylon string is lengthened under the applied tension.