Answer:
1 × 10² J
Step-by-step explanation:
Given data
[tex]\frac{129mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =57.7m/s[/tex]
2 oz × (1 kg/35.274 oz) = 0.06 kg
The work exerted by the player is equal to the kinetic energy (K) adquired by the ball.
K = 1/2 × m × v²
K = 1/2 × 0.06 kg × (57.7 m/s)²
K = 1 × 10² J
The work exerted by the player is 100 Joule.
Given that, speed [tex]=129mi/h[/tex]
[tex]v=129mi/h=\frac{129*1609.34}{3600}=57.7m/s[/tex]
Mass is 2 -oz.
2-oz = [tex]\frac{2*1kg}{35.274} =0.06kg[/tex]
So that, [tex]m=0.06kg[/tex]
The work by player to hit a serve in a racket sport is equal to kinetic energy.
Kinetic energy [tex]K.E=\frac{1}{2} *m*v^{2}[/tex]
[tex]KE=\frac{1}{2} *0.06*(57.7)^{2}\\\\KE=100[/tex]
Thus, the work exerted by the player is 100 Joule.
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