By definition we have the momentum is:
 P = m * v
 Where,
 m = mass
 v = speed
 Before the impact:
 P1 = (0.048) * (26) = 1.248 kg * m / s
 After the impact:
 P2 = (0.048) * (- 17) = -0.816 Kg * m / s.
 Then we have that deltaP is:
 deltaP = P2-P1
 deltaP = (- 0.816) - (1,248)
 deltaP = -2,064 kg * m / s.
 Then, by definition:
 deltaP = F * delta t
 Clearing F:
 F = (deltaP) / (delta t)
 Substituting the values
 F = (- 2.064) / (1/800) = - 1651.2N
 answer:
 the magnitude of the average force exerted on the superball by the sidewalk is 1651.2N
        
             
        
        
        
I think that chalk is a compound. it is made up of molecules containing one calcium atom ,one carbon atom and three oxygen atoms.
        
             
        
        
        
Answer:
 The coconut hit the ground after 1.28 s
Explanation:
Velocity: This can be defined as the rate of change of displacement. Or it is defined as speed in a specified direction. Velocity is a vector quantity. The S.I unit of velocity is m/s.
Using the equation of motion,
S = ut + 1/2gt²....................................... Equation 1
Where S = distance, u = initial velocity, t = time, g = acceleration due to gravity
<em>Given: s= 8 m, u = 0 m/s ( because it was drop from a height), </em>
<em>Constant :   g = +9.8 m/s²</em>
<em>Substituting these values into equation 1,</em>
<em>8 = 0×t + 1/2(9.8×t²)</em>
<em>8 = 4.9t²</em>
<em>t² = 8/4.9</em>
<em>t = √1.63</em>
<em>t = 1.276 s</em>
t = 1.28 s
Therefore the coconut hit the ground after 1.28 s
 
        
             
        
        
        
The motion of the buoy is a composition of two independent motions:
- a uniform motion on the horizontal axis, with constant speed vx=50 m/s
- an uniformly accelerated motion on the vertical axis, with constant acceleration 

Since we want to find the vertical displacement, we are only interested in the vertical motion.
The law of motion on the vertical direction is given by:

where
h is the initial height of the buoy

 is the initial vertical velocity of the buoy, which is zero
t is the time
We know that the buoy lands after t=21 seconds, this means that the vertical position at t=21 s is y(21 s)=0. If we substitute these data into the equation, we can find the value of h, the initial height of the buoy:


And this corresponds to the vertical displacement of the buoy.