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Degger [83]
3 years ago
12

Can you please help me

Physics
1 answer:
Tems11 [23]3 years ago
5 0
Number 1 is D and number 2 is A
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In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi
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Answer:

(a) 1,569.63 N

(b)  195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional  area (CSA) of small piston = A₁

Let the  cross-sectional  area (CSA) of the bigger piston  = A₂

Let the Force applied at the smaller piston  = F₁

Let the Force applied at the bigger piston  = F₂

The principle of hydraulic lift  assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a)  F₁=  F₂ xA₁ /A₂

F₂  = 13,300 N

A₁ = π r₁²

    =π x (0.0505)²

   =  0.008011 m²

A₂= π r₂²

    =π x (0.147)²

   =  0.06788 m²

Substituting into (1)

F₁ = 13,300 x  0.008011/0.06788

   = 1,569.6272

   ≈ 1,569.63 N

(b)   Air pressure = Force/Area

                            =  F₁/A₁

                            = 1,569.6272/ 0.008011

                            =   195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d)  F₂  =  F₁ x  A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

    =π x (0.079)²

   =  0.01960 m²

A₂= π r₂²

    =π x (0.353)²

   =  0.3914 m²

Substituting into (2)

F₂= 825 x  0.3914/ 0.01960

   = 16,474.7448

   ≈ 16,474.74 N

  = 1,647.47 kg

 

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