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A 75.0-kg athlete performs a single-hand handstand. If the area of the hand in contact with the floor is 125 cm2, what pressure is exerted on the floor

Respuesta :

Eduard22sly

The pressure exerted on the floor by the athlete having a mass of 75 Kg is 58800 N/m²

How to determine the force

  • Mass (m) = 75 Kg
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Force (F) =?

F = mg

F = 75 × 9.8

F = 735 N

How to determine the pressure

  • Force (F) = 735 N
  • Area (A) = 125 cm² = 125×10¯⁴ m²
  • Pressure (P) =?

P = F / A

P = 735 / 125×10¯⁴

P = 58800 N/m²

Thus, the pressure exerted on the floor is 58800 N/m²

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