The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area of 0.05m2.<

Question

The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area of 0.05m2.
Critical Thinking

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Kim Cúc 4 years 2021-07-25T11:16:52+00:00 2 Answers 18 views 0

Answers ( )

  1. Bo
    0
    2021-07-25T11:18:17+00:00
    Reply

    Answer:

    50N

    Explanation:

    Pressure = Force/Area

    Where P1 = P2

    F1/A1 = F2/A2

    Given that F1 =? A1 = 0.01m^2

    F2= 250N A2= 0.05m^2

    F1/0.01m^2 = 250N/0.05m^2

    F1 × 0.05= 0.01 × 250

    F1 = 0.01×250/0.05

    F1 = 2.5/0.05

    F1 = 50N

  2. Nick
    0
    2021-07-25T11:18:48+00:00
    Reply

    Answer:

    Explanation:

    Given that,

    Small piston Hydraulic piston has an area

    A1 = 0.01m²

    If the force applied is 250N is applied to the small piston at an area of 0.05 m²

    Then,

    F2 = 250 N and A2 = 0.05m²

    Then, applying pascal principle,

    Pressure at small area = pressure are bigger area

    P1 = P2

    F1 / A1 = F2 / A2

    F1 / 0.01 = 250 / 0.05

    F1 / 0.01 = 5000

    Cross multiply

    F1 = 5000 × 0.01

    F1 = 50 N

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