The answer would be B, an electron because the proton is positive, neutron is neutral, and the nucleus is the center of the atom.
ωєℓℓ тнє ρнσѕρнσяι¢ α¢ι∂ мσℓє¢υℓєѕ αттα¢н тσ тнє мσℓє¢υℓєѕ σf тнє мιℓк, αи∂ тнαт ιи¢яєαѕєѕ тнє ∂єиѕιту αи∂ тнєи ѕєρєяαтєѕ тнєм fяσм тнє яєѕт σf тнє ℓιqυι∂ ιи ιт. тнє яємαιи∂єя σf тнє ℓιqυι∂ѕ,иσω нανιиg ℓєѕѕ ∂єиѕιту тнαи тнє ρнσѕρнσяι¢ α¢ι∂ѕ & тнє мιℓк мσℓє¢υℓєѕ, ѕσ ιт ιт иσω fℓσαтѕ σи тσρ.
нσρє ι ¢συℓ∂ нєℓρ уσυ.
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 =
Now, putting the given values into the above formula as follows.
HB =
=
=
= 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})}](https://tex.z-dn.net/?f=%5Csqrt%7B%28D%5E%7B2%7D%20-%20%5BD%20-%20%5Cfrac%7B2P%7D%7BHB%7D%20%5Cpi%20D%5D%5E%7B2%7D%29%7D)
= ![\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%2810%20mm%29%5E%7B2%7D%20-%20%5B10%20mm%20-%20%5Cfrac%7B2%20%5Ctimes%20500%20kg%7D%7B300%20%5Ctimes%203.14%20%5Ctimes%2010%20mm%7D%5D%5E%7B2%7D%7D)
= 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.
Answer: Ag^+ (aq) + NO3^-(aq) + Na^+ (aq) +Cl^- (aq) ⇒ AgCl(s) + NaNO3(aq)