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
Answer:
1min since there is no gravity on the moon so it will take time to drop on the moon.
Explanation:
Answer:
F = 0.00156[N]
Explanation:
We can solve this problem by using Newton's proposed universal gravitation law.

Where:
F = gravitational force between the moon and Ellen; units [Newtos] or [N]
G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1= Ellen's mass [kg]
m2= Moon's mass [kg]
r = distance from the moon to the earth [meters] or [m].
Data:
G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]
m1 = 47 [kg]
m2 = 7.35 * 10^22 [kg]
r = 3.84 * 10^8 [m]
![F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]](https://tex.z-dn.net/?f=F%3D6.67%2A10%5E%7B-11%7D%20%2A%20%5Cfrac%7B47%2A7.35%2A10%5E%7B22%7D%20%7D%7B%283.84%2A10%5E8%29%5E%7B2%7D%20%7D%5C%5C%20F%3D%200.00156%20%5BN%5D)
This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.