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Dovator [93]
3 years ago
15

Vitellium (Vi) has the following composition:

Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

C It will be closer to the mass of Vi–193.

Explanation:

The average atomic mass of vitellium will be closer to the mass of Vi–193.

From the geonormal abundance we can derive the average atomic mass or relative atomic mass of the compound.

 Average atomic mass = sum of all the product of the mass of isotope x abundance of isotope

Average atomic mass = (10.861% x 187.9122) + (12.428% x 190.9047) + (76.11% x 192.8938)

 Average atomic mass  = 20.41 + 23.73 + 146.81  = 190.95amu

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
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Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

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\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

7 0
3 years ago
Read 2 more answers
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