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svlad2 [7]
3 years ago
10

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

dth b 10 mm (0.4 in.); the distance between support points is 45 mm (1.75 in.). Compute the flexural strength if the load at fracture is 290 N (65 lb
Engineering
1 answer:
Anon25 [30]3 years ago
4 0

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

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By using the second equation of motion to find the distance S;

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D_{police} = 0.98t^2 + 141.12 + 23.52t

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(187.8+15.65 \ t-1.525 t^2)=  0.98t^2 + 141.12 + 23.52t

(187.8 - 141.12)  + (15.65 \ t  -  23.52t)  -( 1.525 t^2    - 0.98t^2)  =   0

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Total time  required for the police car  to over take the automobile = 15.02 sec

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