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
143m/s if you just perhaps by what you know you'll figure it out
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
Explanation:
If you want to get speed, u have to divided distance over time
The lowest speed will lose
Answer:
0.15
Explanation:
Assuming the rope is horizontal, sum the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum the forces in the x direction:
∑F = ma
F − Nμ = ma
Substitute:
F − mgμ = ma
mgμ = F − ma
μ = (F − ma) / (mg)
Plug in values:
μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)
μ = 0.15
