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zaharov [31]
3 years ago
7

A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr

oss section 1cm find the energy stored in rod .Y=3.5 ×10^10 Nm2​
Physics
1 answer:
Inga [223]3 years ago
3 0

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

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Explanation:

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Let the distance is r from both the wires on the same side.

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The work that is done when twice the load is lifted twice the distance is

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when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.

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