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Sagot :
The second skater accelerates at 6 [tex]m/s^2[/tex] to the right.
- Let the first skater be F.
- Let the second skater be S.
Given the following data:
- Mass of F = 100 kg
- Mass of S = 50 kg
- Acceleration of F = 3 [tex]m/s^2[/tex]
To find the acceleration of the second ice skater, we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]Force = Mass[/tex] × [tex]acceleration[/tex]
Substituting the given values, we have;
[tex]Force = 100[/tex] × [tex]3[/tex]
Force = 300 Newton
Note: The same force is acting on the two ice skaters and the second ice skater would be pushed in the opposite direction (right).
Now, we would find the acceleration of the second ice skater:
[tex]Acceleration = \frac{300}{50}[/tex]
Acceleration of S = 6 [tex]m/s^2[/tex]
Therefore, the acceleration of the second ice skater is 6 [tex]m/s^2[/tex] to the right.
Read more here: https://brainly.com/question/24029674
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