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
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Answer:
x=22.57 m
Explanation:
Given that
35 m in W of S
angle = 40 degrees
25 m in east
From the diagram
The angle
From the triangle OAB
x=22.57 m
Therefore the answer of the above problem will be 22.57 m
Answer is B raindrops conduct electricity from clouds to the ground.
Lightning is formed when air, water droplets, and even ice crystals rub violently against each other inside a thundercloud, creating two opposite kinds of electrical charge
Answer:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' = 
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'
V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.
