Ask Question
17 November, 12:07

What must be the contact area between a suction cup (completely exhausted) and a ceiling in order to support the weight of a 54.3 kg student? the acceleration of gravity is 9.8 m/s 2?

+3
Answers (1)
  1. 17 November, 13:01
    0
    But it all comes from the equation of pressure: P = F / S

    P = Pressure, in Pascal, [P]

    F = Force, in Newton [N]

    S = Surface, in squared meters [S]

    Atmospheric pressure * area = weight

    101325 pascal * area = 54.3 pascal * 9.8 N

    = 532.14 kg^2 converting this to m, it will be 0.0053 m^2
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “What must be the contact area between a suction cup (completely exhausted) and a ceiling in order to support the weight of a 54.3 kg ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers