The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 6 when 16.0% of the crack length is removed.

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

([tex]p_t[/tex])₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2

where ([tex]p_t[/tex])₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]  

divide both sides by 2(σ₀)₁

[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex]  =  36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]

1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂

([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁

([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.