I think this is the solution:
1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4
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Explanation:
Formula to determine the critical crack is as follows.

= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, 
= 427.5
Hence, we will calculate the critical crack length as follows.
a = 
= 
= 
Therefore, largest size is as follows.
Largest size = 2a
= 
= 
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is
.
During the period of constant acceleration, the car's average speed is (1/2) (16 + 32) = 24 m/s.
At that average speed, it covers 240 meters in 10 seconds.
Answer: Because the fuse can disconnect the circuit only when the excess current flows completely through the neutral. ... Since, neutral is not a live conductor coming from the source, disconnecting a neutral line can only open the current path through neutral. But, the live phase still carries the charge.
Explanation: