2 years ago

Jnjn freeeeeeeeeeeeeeeeeeeeeeeeeeeeeee pointtttttttttt

Engineering

2 answers:

denis-greek [22]2 years ago

6 0

Answer:

thx for points :)

Explanation:

kvasek [131]2 years ago

5 0

Answer:

thx

Explanation:

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

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Answer:

elongation of the brass rod is 0.01956 mm

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diameter = 4.50 mm

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to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

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