Question

1 pts A tensile stress of 50 kPa (1 kPa 1000 N/m) acting on 20-cm-long muscle stretches the muscle by 1 mm. what is the Young's modulus for this muscle? O10,000 kPa We must know cross-sectional area (or diameter) of muscle to find this. 250 kPa O 1,000 kPa 1 pts Question 3 The quadriceps the biceps muscle in the upper arm. Assuming that Young's elastic modulus is the same of both muscles, if the same tensile force were applied to the quadriceps as to the biceps, femoris muscle in the upper leg is about twice the diameter and twice the length of the quadriceps would stretch twice the length the biceps stretch. O the quadriceps would stretch a quarter the length the biceps stretch. the quadriceps would stretch half the length the biceps stretch. the quadriceps would stretch the same length the biceps stretch.

Answer 1

l = 20 cm Delta l = 1m stress = F/A = 50 kpa strain = Delta l/l = 1/200 young modulus = stress/strain y = 50 kpa /1/200 = 50 times 200 k pa = 10000 k pa diameter of quadriceps muscles = D length of quadriceps muscles of leg = L diameter of biceps an upper Arm = D/2 if tensile force applied is same then stress for quadriceps = F/pi d^2/4 strain = (Delta L)_leg/L strain = (Delta L)/L/2 y = stress/strain (Delta L)_leg = 2 (Delta L)_Arm The quadriceps would stretch twice the length the biceps stretch