Question: 1. In a hydraulic system a 20.0-N force is applied to the small piston with cross-sectional area 25.0 cm2. What weight can be lifted by
2. If the diameter of the larger piston in Problem 1 is doubled, how is the weight able to be lifted changed?
3. If a dentist’s chair weighs 1600 N and is raised by a large piston with cross-sectional area of 75.0 cm2, what force must be exerted on a small piston of cross-sectional area 3.75 cm2 to lift the chair?
4. A hydraulic jack whose piston has a cross-sectional area of 115 cm2 supports a pickup truck weighing 1.20 x 104 N. Compressed air is used to apply a force on the second piston with cross-sectional area 25.0 cm2. How large must this force be to support the truck?
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1 F 2 F 1 A 2 A 1 200 N 500 cm 2 250 cm 2 400 N 2 Since the a... View full answer
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