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Answer :
The force exerted by the skater is 575.2 N.
Force exerted by the skater
The force exerted by the skater is calculated by taking net force as follows;
[tex]Fcos(\theta) = F_c[/tex]
where;
- Fc is the centripetal force on the circular path
[tex]Fcos(\theta) = m\omega ^2 r[/tex]
The angular speed for each complete turn in 2 seconds is calculated as follows;
[tex]\omega = \frac{2\pi \ rad}{2 \ s} = \pi \ rad/s[/tex]
The force exerted by the skater is calculated as follows;
[tex]F = \frac{m \omega ^2 r cos(\theta)}{cos(\theta)} \\\\F = \frac{47 \times (\pi)^2 \times 1.24cos(45)}{cos(45)} \\\\F = 47 \times (\pi)^2 \times 1.24\\\\F = 575.2 \ N[/tex]
Learn more about centripetal force here: https://brainly.com/question/20905151
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Answer:
The magnitude of the force is 575.2 N
Explanation:
the force exerted by the male skater must be equal to the centripetal force:
F*cos45° = m*r*w^2, where
m = 47 kg
r = 1.24*cos45° m
w = (2*π)/2 = π radians/s
Replacing values:
F*cos45° = 47*(1.21*cos45°)*π^2
F = 47*1.24*π^2 = 575.2 N
