Fcheerleader/ Areacheerpiston = Ffootball/ Areafootball piston
Fcheerleader= 57 kg
Ffootball= 4players(130 kg)=520 kg
Areacheer piston=πr2=π(.095m)2=.0283 (can add more decimal places if needed)
Areafootball piston= Ffootball x Areacheer piston/ Fcheerleader
= (520kg)(.0283m2)/(57kg)
= .258 m2
Area=πr2
radius=√(Area/π)
=√(.258/π)=.286 m
diameter=2radius=2(.286)=.572 m=57.2 cm
**note some rounding error may be present Xp **
Review Part A A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m If her piston is 17.0 cm in diameter, what is the diameter of the football players' piston? Express your answer with the appropriate units. 106 cm Submit
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 0.800 m Part A If her piston is 16.0 cm in diameter, what is the diameter of the football players' piston? Express your answer with the appropriate units. IA Value cm Submit Request Answer
Review Part A A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m If her piston is 17.0 cm in diameter, what is the diameter of the football players' piston? Express your answer with the appropriate units. 49.3 cm Submit Incorrect; Try Again; 4 attempts remaining SWER 1: Deduction:-3% 0.49 m NSWER 2: Deduction:-3% 49.3 cm ANSWER 3: Deduction:-3% 87.4 cm ANSWER 4: Deduction:-3% 106 cm
The answer is E
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