Answer: 
Explanation:
Given
Length of plank is 1.6 m
Force
is applied on the left side of plank
Force
is applied 43 cm from the left end O.
Mass of the plank is 
for equilibrium
Net torque must be zero. Taking torque about left side of the plank

Net vertical force must be zero on the plank

Answer:
the property of liquid are
1 they can flow from one place to another if surface is slanted
2 it cannot be compressed
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:

where
m is the mass of the object
is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is

(b) 155 N
The impulse can also be rewritten as

where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
is the duration of the collision
In this situation, we have

So we can re-arrange the equation to find the magnitude of the average force:
