Answer:
Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. A rod 25.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. .00 mm2 and with a 100 N on each end. (c) A rod 20.0 cm long with cross-sectional area 2 force applied Order the rods according to the tensile stress on each rod, from smallest to largest.
Step-by-step explanation:
Tensile stress is given as =
σ = Force/Area
A. Given that
Length of rod = 50cm
Cross-sectional Area A = 1mm²
A = 1 × 10^-6m²
Force applied F = 200N
Then, σ = F/A
σ = 200/1×10^-6
σ = 200 × 10^6 N/m²
σ = 200 Mpa
B. Given that,
Length of rod = 25cm.
Cross-sectional Area A = 1mm²
A = 1 × 10^-6m²
Force applied F = 200N
Then, σ = F/A
σ = 200/1×10^-6
σ = 200 × 10^6 N/m²
σ = 200 Mpa
C. Given that,
Length of rod = 20cm.
Cross-sectional Area A = 2mm²
A = 2 × 10^-6m²
Force applied F = 100N
Then, σ = F/A
σ = 100/2×10^-6
σ = 50 × 10^6 N/m²
σ = 50 Mpa
It is noticed that
50Mpa < 20Mpa =200Mpa
Then, C < B = A
The tensile strength of rod A and B are equal and their tensile strength is greater than that of Rod C