Answer:
sugar cube is at great pressure
Explanation:
this is because great pressure means its more solid
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})}]}](https://tex.z-dn.net/?f=%5Cfrac%7B2P%7D%7B%5Cpi%20D%20%5BD%20-%20%5Csqrt%7B%28D%5E%7B2%7D%20-%20d%5E%7B2%7D%29%7D%5D%7D)
Now, putting the given values into the above formula as follows.
HB = = ![\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Ctimes%201000%20kg%7D%7B3.14%20%5Ctimes%2010%20mm%20%5BD%20-%20%5Csqrt%7B%28%2810%20mm%29%5E%7B2%7D%20-%20%282.50%29%5E%7B2%7D%29%7D%5D%7D)
= 
= 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:
C). The Bohr-Rutherford model
Explanation:
The 'Bohr-Rutherford model' of the atom failed to elaborate on the attraction between some substances. It essentially targeted hydrogen atoms and failed to explain its stability across multi-electrons. The nature and processes of the chemical reactions remained unillustrated and thus, this is the key drawback of this model. Thus, <u>option C</u> is the correct answer.
The chemical reaction is expressed as:
2H2 + O2 = 2H2O
To determine the amount of oxygen used in the reaction, we use the amount of water produced and the relation of the substances in the reaction we do as follows:
209 g H2O ( 1 mol / 18.02 g ) ( 1 mol O2 / 2 mol H2O ) ( 32 g / 1 mol ) = 185.57 g O2
1. Group: Groups shows the number of valence electrons and are shown in columns on the periodic table.
2. Period: Periods show the number of energy levels each element has and are shown in rows on the periodic table (up and down).
3. Blocks are show the amount of orbitals each element has. There is an s block, d block, p block, and f block.
