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:
When you are running the most important force that you should understand is friction. Friction is a force that opposes movement between two objects, but for runners friction makes you faster. Friction gives you a better and more efficient way to use your energy into speed.
Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v
v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.
Explanation:
(a)
Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.
Using Snell's law as:
Where,
is the angle of incidence
is the angle of refraction = 90°
is the refractive index of the refraction medium
is the refractive index of the incidence medium
Thus,
The formula for the calculation of critical angle is:
Where,
is the critical angle
(b)
No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.
