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Stella [2.4K]
3 years ago
5

2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan

e- strain fracture toughness of 98.9 MPa √ m and a yield strength of 860 MPa . The flaw size resolution limit of the flaw de tection appa ratus is 3.0 mm . If the design stress is one- half the yield strength and the value of Y is 1.0, determine whether a critical flaw for this plate is subject to detection.
Physics
1 answer:
ikadub [295]3 years ago
5 0

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

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Explanation:

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
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Answer:

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E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

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\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

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Part (C) In this part we have to find the cutoff frequency

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5 0
3 years ago
NGSS Physics
vovangra [49]
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EXPLANATION:

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