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scZoUnD [109]
3 years ago
6

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o

f 89,000 N (20,000 lbf), and experiences an elongation of 0.10 mm (4.0 x 10^-3 in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.
Engineering
1 answer:
grigory [225]3 years ago
7 0

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

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3 years ago
Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi
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Answer:

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Explanation:

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\vec F_{A} = \vec F_{AB} + \vec F_{AC}

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SpyIntel [72]

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