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
Atmosphere
Atmospheric gas from prehistoric eras is found trapped in glaciers in the form of bubbles. These gas bubbles are the basis of studying ice cores as they provide us with accurate estimates of the conditions of past climates. The bubbles allow us to determine the composition of atmospheric air, such as the carbon dioxide and methane concentrations, as well as allow us to determine air temperatures in the past.
Energy Density = 1/2 × ε(0) × (V/d)^2
V = 100, d = 0.01, ε(0) = 8.85 x 10^-12
Answer:The mass of ball B is 10 kg.
Explanation;
Mass of ball A = 
Velocity of the ball A before collision:
Velocity of ball A after collision=
Mass of ball B= 
Velocity of the ball B before collision:
Velocity of ball B after collision=
The mass of ball B is 10 kg.
Answer:
25 mm = 0 deg C
200 mm = 100 deg C
200 - 25 = 175 = change in thread per 100 deg C
95 - 25 = 70 mm - change in thread from 0 deg C
70 / 175 * 100 = 40 deg C final temperature at 95 mm
= speed of sound in air = 343 m/s
= speed of sound in the rod =
= times interval between the hearing the sound twice = 0.12 s
= length of the rod
= time taken by the sound to travel through the rod
= time taken by the sound to travel to through air to the same point =