Answer: -
71
Explanation: -
From the diagram, we see that volume goes from 70 to 75 in 5 markings.
Each marking is for 1.
It is the most accuracy possible as there is no smaller marking.
Significant figures expresses the required amount of accuracy.
Thus the volume indicated from the diagram is 71 .
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
Nucleic Acids are polymers made of nucleotideds
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.
40.28 sorry if it is wrong:) but I tried