2 years ago

Periodic trends are examined by looking down the _ of elements or by looking across the _ of the elements

Chemistry

1 answer:

nikklg [1K]2 years ago

6 0

Answer: yo sorry this a hard one

Explanation:

bro

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A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat

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Explanation:

(a) The given data is as follows.

Load applied (P) = 1000 kg

Indentation produced (d) = 2.50 mm

BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$

Now, putting the given values into the above formula as follows.

HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$ = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$

= $\frac{2000}{9.98}$

= 200

Therefore, the Brinell HArdness is 200.

(b) The given data is as follows.

Brinell Hardness = 300

Load (P) = 500 kg

BHI diameter (D) = 10 mm

Indentation produced (d) = ?

d = $\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}$

= $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

= 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

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