A force of 62 Newtons is applied to a wire with a diameter of0.5 mm. What is the

Question

A force of 62 Newtons is applied to a wire with a diameter of0.5 mm. What is the tensile stress (in SI units) in the wire? A wire is 2.5 m long. After applying a force to one end of thewire, it is now 2.501 m long. Calculate the tensile strain in thewire.

The use of the micrometer caliper is described in Lab AA. Accordingto Lab AA, the smallest marked division on the micrometer caliperis Look up some typical values for Young's (or the elastic)modulus in your lecture textbook. Typical values (in SI units)are A force of 62 Newtons is applied to a wire with a diameter of0.5 mm. What is the tensile stress (in SI units) in the wire? A wire is 2.5 m long. After applying a force to one end of thewire, it is now 2.501 m long. Calculate the tensile strain in thewire.

The use of the micrometer caliper is described in Lab AA. Accordingto Lab AA, the smallest marked division on the micrometer caliperis Look up some typical values for Young's (or the elastic)modulus in your lecture textbook. Typical values (in SI units)are

Explanation / Answer

1.) Tensile stress = F/A A = r2 Tstress = 62/(*.252) 2.) Tensile strain = L/Lo Tstrain = (2.501-2.5)/2.5 3.) Lowest marked division is 1mm 4.) Just google/wiki "Young's modulus"

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