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
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2.
Answer:
A ferromagnetic material is a temporary magnet. The domains in a ferromagnetic material are randomly arranged. Under certain actions, the domains align in a particular direction and the material acts as a magnet. The actions that can cause alignment of domains in a ferromagnetic material are:
- rubbing the material against a magnet would cause the alignment of domains in the same direction as of the magnet.
- passing electricity around the material would generate magnetic field which would cause domains to align along the direction of the field.
- placing the material near a strong magnet would cause the alignment of domains in the direction of the field generated by the strong magnet.
Other actions like heating the material, placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.
The area of the sprinkles can be determined through the area of a circle that is pi * r^2 in which the given dimensions above are the radii, r. The second scenarios radius is only half of the original, that is 4 ft. In this case, we can compute the area of the second again. We calculate next the difference of two areas of circles.
I think it might be heat energy. light transforms into heat energy
