Question

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. Oc

Answer 1

c < a = b

Step-by-step explanation:

to arrange tensile stress from smallest to largest

tensile stress =
`\sigma = (F)/(A)`

a) A rod 50.0 cm long with cross-sectional area 1.00 mm² and with a 200 N force applied on each end.

`\sigma = (F)/(A)`

`\sigma = (200)/(1)`

`\sigma = 200\ Mpa`

b) A rod 25.0 cm long with cross-sectional area 1.00 mm² and with a 200 N force applied on each end.

`\sigma = (F)/(A)`

`\sigma = (200)/(1)`

`\sigma = 200\ Mpa`

c) A rod 20 cm long with cross-sectional area 2.00 mm² and with a 100 N force applied on each end.

`\sigma = (F)/(A)`

`\sigma = (100)/(2)`

`\sigma = 50\ Mpa`

hence the order comes out to be

c < a = b