Answer:
Explanation:
The height to which a ball will bounce depends on the height from which it is dropped, what the ball is made out of (and if it is inflated, what the pressure is), and what the surface it bounces from is made out of. The radius of the ball doesn't really matter, if you are measuring the height of the ball from the bottom of the ball to the ground.
A ball's gravitational potential energy is proportional to its height. At the bottom, just before the bounce, this energy is now all in the form of kinetic energy. After the bounce, the ball and the ground or floor have absorbed some of that energy and have become warmer and have made a noise. This energy lost in the bounce is a more or less constant fraction of the energy of the ball before the bounce. As the ball goes back up, kinetic energy (now a bit less) gets traded back for gravitational potential energy, and it will rise back to a height that is the original height times (1-fraction of energy lost). We'll call this number f. For a superball, f may be around 90% (0.9) or perhaps even bigger. For a steel ball on a thick steel plate, f is >0.95. For a properly inflated basketball, f is about 0.75. For a squash ball, f might be less than 0.5 or 0.25 - squash balls are not very bouncy. The steel ball on an unvarnished pine wood floor may not bounce at all, but rather make a dent, and so what the floor is made out of makes quite a lot of difference.
Answer:
Kindly find the graphs attached
Explanation:
For figure 1: There is a steady increase in the position of the object as time increases. This is because despite the negative acceleration (deceleration), the object continues to move and cover more ground as time goes by.
<em>The straight line graph is observed because the acceleration is constant and not varying.</em>
For Figure 2: The graph of velocity vs time will have an inverted nature. This is because since the object is decelerating, it is reducing in its velocity as time goes by (increases). <em>This is also in a straight line since the deceleration is constant.</em>
Velocity stands for Displacement w.r.t time
Or
Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g
k = 224 N/m
b) 
c) 
d) 
e) 
A = 0.682 m
Force =
=
F = 94.20 N
Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
