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)891.64 N
(b)0.7
Explanation:
Mass of crate 
Crate slows down in 
initial speed 
inclination 
From Work-Energy Principle
Work done by all the Forces is equal to change in Kinetic Energy
change in kinetic energy
(b)Coefficient of sliding friction
and 
<span>We can think this through intuitively. A frequency of 256 Hz means that the wave has 256 cycles each second. If the wavelength is 1.33 meters, then there are 256 of them each second. Therefore, we just need to multiply the wavelength by the frequency to find the speed of sound. (Note that the units Hz = 1 / s)
v = (frequency) x (wavelength)
v = (256 Hz) x (1.33 m)
v = 340.5 m/s
The speed of sound in the vicinity of the fork is 340.5 m/s</span>
Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s
