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:
a) Initial angular speed = 30 rad/s
b) Final angular speed = 70 rad/s
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here s = 400 radians
t = 8 s
a = 5 rad/s²
Substituting
400 = u x 8 + 0.5 x 5 x 8²
u = 30 rad/s
Initial angular speed = 30 rad/s
b) We have equation of motion v = u + at
Here u = 30 rad/s
t = 8 s
a = 5 rad/s²
Substituting
v = 30 + 5 x 8 = 70 rad/s
Final angular speed = 70 rad/s
It is number 3 because I know it is
It would destroy animals homes shelter etc. it also would make global warming go faster. Hope this helped :D
