Answer:
a. Rockwell 3. hardness
b. Instron 2. stress vs strain
c. Charpy 1. impact strength
d. Fatigue 4. Endurance Limit
e. Brinell 3. hardness
f. Izod 1. impact strength
Explanation:
Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.
Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.
Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.
Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.
The answer is true because if the effect is neglected, the saturation id region is considered true
Answer:
Explanation:
4140-40 I’d pick wood
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Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity


Temperature at 28 mm distance after t time = = 50 degree C
we know that

![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)

from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)

solving fot t we get
t = 2244.3 sec