Answer:
(a). The HB of this material is 217.8
(b). The diameter of an indentation is 1.45 mm.
Explanation:
Given that,
Diameter of Brinell hardness D= 10.0 m
Diameter of steel alloy d= 2.4 mm
Load = 1000 kg
(a). We need to calculate the HB of this material
Using formula of Brinell hardness
Put the value into the formula
(b). Given that,
Hardness = 300 HB
Load = 500 kg
We need to calculate the diameter of an indentation
Using formula of diameter
![d=\sqrt{D^2-[D-\dfrac{2P}{(HB)\pi D}]^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7BD%5E2-%5BD-%5Cdfrac%7B2P%7D%7B%28HB%29%5Cpi%20D%7D%5D%5E2%7D)
Put the value into the formula
![d=\sqrt{10^2-[10-\dfrac{2\times500}{300\pi\times10}]^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B10%5E2-%5B10-%5Cdfrac%7B2%5Ctimes500%7D%7B300%5Cpi%5Ctimes10%7D%5D%5E2%7D)
Hence,(a). The HB of this material is 217.8
(b). The diameter of an indentation is 1.45 mm.
Answer:
g'(10) = 
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x <em> </em><em>(Identity)</em>
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = 
--equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = 
+ 2
= 8
x = 
Since f(x) = + 2
f'(x) = -
f'() = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -
Close to 14 is considered a strong base.
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