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:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0

From the figure, you can observe that a/b < 1, thus E < W
Answer:
195.168 m
Explanation:
To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse
Pythagorean Theorem:

Plug values in

Simplify

Add the two values

Take the square root of both sides

Answer:
Options B, A, D, C
Explanation:
When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.