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the diameter of the large piston of hydraulic press us 200 cm and the cross sectional area of the small piston is 5cm² if the f...Asked by Angeles Marie Ann Sambong
the diameter of the large piston of hydraulic press us 20 cm and the cross sectional area of the small piston is 5cm² if the force of 400N is applied to the small piston what is the resulting force exerted on the large piston?
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We can use the principle of Pascal's law to solve this problem. According to Pascal's law, the pressure applied to a confined fluid is transmitted uniformly in all directions.
Let's first find the pressure applied to the small piston:
Pressure = Force / Area
= 400 N / 5 cm²
= 80 N/cm²
Now, according to Pascal's law, this pressure is transmitted uniformly to all parts of the fluid and exerted on the large piston. Therefore, we can find the resulting force on the large piston by using the pressure and area of the large piston:
Force = Pressure x Area
= 80 N/cm² x (π x (20 cm/2)²)
= 80 N/cm² x π x 100 cm²
= 25,132 N
Therefore, the resulting force exerted on the large piston is 25,132 N.
Let's first find the pressure applied to the small piston:
Pressure = Force / Area
= 400 N / 5 cm²
= 80 N/cm²
Now, according to Pascal's law, this pressure is transmitted uniformly to all parts of the fluid and exerted on the large piston. Therefore, we can find the resulting force on the large piston by using the pressure and area of the large piston:
Force = Pressure x Area
= 80 N/cm² x (π x (20 cm/2)²)
= 80 N/cm² x π x 100 cm²
= 25,132 N
Therefore, the resulting force exerted on the large piston is 25,132 N.
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