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Answer:

10,30,40,50 and above

The average volume of each of the coin is 2.5 × 10⁻⁶m³.

Given the data in the question;

  • Number of coin; [tex]n_c = 10[/tex]
  • Initial volume; [tex]v_1 = 75ml = 7.5*10^{-5}m^3[/tex]
  • Final volume; [tex]v_2 = 100ml = 1.0*10^{-4}m^3[/tex]

To determine the average volume of each coin, we calculate change in volume after the coins were dropped into the graduated cylinder.

[tex]\delta V = v_2 - v_1\\\\\delta V = 1.0*10^{-4}m^3 - 7.5*10^{-5}m^3\\\\\delta V = 2.5*10^{-5}m^3\\[/tex]

Hence, [tex]\delta V = volume\ of\ the\ 10 coins[/tex]

Now, volume of each of the coins will be;

[tex]V_{each\ coins} = \frac{\delta V}{n_c} \\\\V_{each\ coins} = \frac{2.5*10^{-5}m^3}{10} \\\\V_{each\ coins} = 2.5*10^{-6}m^3[/tex]

Therefore, the average volume of each of the coin is 2.5 × 10⁻⁶m³.

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