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kupik [55]
3 years ago
9

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 mpa1m (75.0 ksi1in.). if the p

late is exposed to a tensile stress of 345 mpa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. assume a value of 1.0 for y.
Physics
1 answer:
Pie3 years ago
4 0
Using the equation σ =K/(Y √π a) ; a=1/π  (K/Yσ)^2. The critical stress required for initiating crack propagation is σ, plain strain fracture toughness is K, surface length of the crack is a, and dimensionless parameter us Y.Substituting the given parameters to the equation. 82.4 MPa √m for K, 345 MPa for σ, and 1 for Y in the equation of surface length of the crack.a = 1/π (K / Yσ )^2 = 1/π (82.4 / 1*345 )^2 = 0.01815 m= 18.15mm
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Hey can someone please help me and can u show your work plz plz plz plz
Tju [1.3M]

Answer:

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Explanation:

4 0
3 years ago
A 3.00-kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determ
aniked [119]

Answer:

a) k= 3594,7 N/m

b) v= 0.55 m/s

Explanation:

  • a)
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  • As we know that the surface is frictionless also, this change in kinetic energy must be equal to the change in the elastic potential energy of the spring.
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       \Delta K = \Delta U

       where \Delta K = \frac{1}{2}*m*v^{2}

       and \Delta U = \frac{1}{2} * k* \Delta x^{2}

  • Simplifying and replacing by the values, we get:

        3.00 kg* (1.8 m/s)^{2} = k* (0.052 m) ^{2}        

  • Solving for k:

k = \frac{3.00kg*(1.8m/s)^{2} }{(0.052m)^{2}} = 3594.7 N/m

  • k = 3594.7 N/m
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        v = \sqrt{\frac{k*\Delta x^{2}}{m} } = \sqrt{\frac{3594.7N/m*(0.016m)^{2} }{3.00kg}} = 0.55 m/s

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6 0
3 years ago
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Answer:

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Wb is the weight of the boy

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dg is the distance of the girl from the pivot

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Answer:

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Explanation:

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