Answer:
(a) 1.66 x 10^-3 cm
(b) 17.66 N
Explanation:
diameter, d = 0.15 mm
radius, r = 0.075 mm = 0.075 x 10^-3 m
Young's modulus, Y = 0.2 x 10^10 N/m^2
Tensile strength = 1000 x 10^6 N/m^2
mass of spider, m = 0.50 g
length of the strand, L = 12 cm
(a) Let the silk strand be stretch by ΔL.
Use the formula for Young's modulus
Y = stress /strain
strain = Stress / Y
stress = force / area
Area, A = πr² = 3.14 x 0.075 x 10^-3 x 0.075 x 10^-3 = 1.766 x 10^-8 m^2
stress = mg / A
stress = 0.5 x 10^-3 x 9.8 / (1.766 x 10^-8) = 277423.92 N/m^2
strain = 277423.92 / (0.2 x 10^10) = 1.387 x 10^-4
ΔL = 1.387 x 10^-4 x L
ΔL = 1.387 x 10^-4 x 12 = 1.66 x 10^-3 cm
Thus, the stretch in length is 1.66 x 10^-3 cm .
(b) Tensile strength = maximum force / area
maximum weight = 1000 x 10^6 x 1.766 x 10^-8
maximum weight = 17.66 N
Thus, the maximum weight is 17.66 N.