Respuesta :

[tex]\begin{gathered} F_{input}=20N \\ F_{output}=100N \\ MA=? \\ MA=\frac{F_{output}}{F_{input}}=\frac{100N}{20N}=5 \\ MA=5 \\ The\text{ mechanical advantage is 5} \\ \\ Question\text{ 2\rparen} \\ W_{input}=380J \\ W_{output}=500J \\ \eta=? \\ \eta=\frac{W_{output}}{W_{input}}=\frac{500J}{380J}=1.316=131.6 \\ The\text{ efficiency is 131.6\%} \end{gathered}[/tex]

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