Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
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Answer: 0.790 g/cm3
Explanation:
The density of acetone is 790 Kg/m3.
To convert from Kg to g we multiply by 1000 (1 Kg = 1000 g)
To convert from m3 to cm3 we multiply by 10∧6
So, The density of acetone in (g/cm3) = (790 x 1000) / (10∧6) = 0.79 g/cm3
