3. A hydraulic lift is used to jack a 950-kg car 45 cm off the floor. The diameter of the output piston is 23 cm, and the input force is 375 N. a. What is the area of the input piston? (10pts) b. What is the work done in lifting the car 45 cm? (10pts) c. If the input piston moves 15 cm in each stroke, how high does the car move up for each stroke? (10pts) d. How many strokes are required to jack the car up 45 cm? (10pts) e. Show that the energy is conserved. (10pts)

a. A input = 0.001669 m²

b. W= 4.193775 KJ

c. h output = 0.60357 cm

d. n = 74.55638 strokes

e. 4.1937 N = 4.1937 N

Explanation:

Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.

Parameters given:

Diameter of the output piston,   d =  0.23 m

Mass of the car,  m= 950 kg

Force applied at the input piston,  f = 375 N

Height, h  = 0.45 m

(a) Finding the area of the input piston:

First, we use Pascal's principle to find the area

(f ÷ A output)  ÷ (f ÷ A input)

Where A= area

            g = 9.81

          A output = πd² ÷ 4

(f ÷ (πd² ÷ 4)) = (f ÷ A input)

[(950 x 9.81)  ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input

9319.5 ÷ 0.0415 = 375 ÷ A input

A input = 0.001669 m²

(b) Finding the work done in lifting the car 45 cm

Work done, W = force, f x distance ( which in this case is height, h)

= (950 x 9.81) x 0.45

W= 4.193775 KJ

(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.

W output = W input

F x h output = F x h input

= (950 x 9.81) x h output = 375 x 0.15

h output =  0.0060357 m

h output = 0.60357 cm

(d) Finding the number of strokes that are required to jack the car up 45 cm

n = h ÷ h output

n = 45 ÷ 0.60357 cm

n = 74.55638 strokes

(e) How the energy is conserved

W output = W input

F x h output = F x h input x n

(950 x 9.81) x 0.45 = 375  x  0.15  x 74.55638

4.1937 N = 4.1937 N

explanation:

hdshusjshssh

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)