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BartSMP [9]
3 years ago
14

A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable su

rface crack length if the surface energy of polystyrene is 0.50 J/m2 (2.86 103 in.-lbf/in.2 ). Assume a modulus of elasticity of 3.0 GPa (0.435 106 psi).

Physics
2 answers:
torisob [31]3 years ago
6 0

Answer:

Maximum allowable surface crack length =  l =  0.000611m

Explanation:

Tensile stress = σ = 1.25 MPa = 1.25 × 10^6 N/m2

Surface energy of polystyrene = Es = 0.50 J/m2

Elastic Mudolus = E = 3 GPa = 3 × 10^9 N/m2

Maximum allowable surface crack length = l = ?

We know that:

       l = 2EEs/πσ^2 ....... (i)

By putttin the values in equation (i)

       l = (2)(0.50)(3 × 10^9)/(3.14)(1.25 × 10^6)^2

       l =  0.000611m

Hence, the Maximum allowable surface crack length will be 0.000611m.

Scorpion4ik [409]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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Marie observed people at a store. Which is a qualifier observation she may have made ?
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Complete question is;

Marie observed people at a store. Which is a qualitative observation she may have made?

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6 0
3 years ago
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Answer:

a) Andrea's initial momentum, 200 kg m/s

b) Andrea's final momentum, 0

c) Impulse,  = - 200 Ns

d) The force that the seat belt exerts on Andrea, - 400 N

Explanation:

Given data,

The initial velocity of the car is, u = 40 m/s

The mass of Andrea, m = 50 kg

The time period of deceleration, a = 0.5 s

The final velocity of the car, v = 0

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The negative sign indicated that the momentum is decreased.

d) The force that the seat belt exerts on Andrea

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3 years ago
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