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xenn [34]
3 years ago
14

A 10 cm long tensile specimen with a yield stress of 350 MPa is pull to a stress of 300 MPa and then the load is released. What

is the length of the specimen after the load is released?
Physics
1 answer:
Advocard [28]3 years ago
3 0

Answer:

The length of the specimen after the load is released is 11.67 cm

Explanation:

Given;

yield stress, Y = 350 MPa

ultimate tensile stress, T = 300 MPa

Elongation factor, e = yield stress, Y / ultimate tensile stress, T

Elongation factor, e = 350 Mpa / 300 Mpa

Elongation factor, e = 1.1667

New length of the specimen = 1.1667 x 10 cm = 11.67 cm

Therefore, when the load is released from 10 cm long tensile specimen, the length of the specimen becomes 11.67 cm

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In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t
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Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       Em_{f} = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

         h = ½ v² / g

let's calculate

         h = ½ 41² / 0.0636

         h = 13,205.4 m

b) For this part we are asked to use the law of universal gravitation, write the energy

starting point. Satellite surface

           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            Em_{f}= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

     

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

          r_f = 2.12 106 m

to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

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