We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
where
Now we find the moment of inertia by integrating from
to
The moment of inertia is
(from (-a/2) to
(a/2))
Answer:
ω = 0.1 rad/s
v = 0.002 m/s
Explanation:
The angular velcoity of the second hand of the clock can be found by:
ω = θ/t
where,
ω = Angular Speed
θ = Angular Displacement
t = time taken
Now, for one complete revolution of second hand of the clock:
θ = 2π rad
t = 60 s
Therefore,
ω = 2π rad/60 s
<u>ω = 0.1 rad/s</u>
Now, for the linear speed (V):
V = rω
where,
V = Linear Speed of Second Hand = ?
r = radius = length of second hand = 0.02 m
Therefore,
V = (0.02 m)(0.1 rad/s)
<u>V = 0.002 m/s</u>
1. Jumping off a building
2. Hitting a tennis ball
3. Swinging a bucket of water in a circle
<span>So we want to know why the does a bouncing ball rise to a lower height with each bounce. So lets say the ball is first on some height h. There it has potential energy Ep=m*g*h. Then as the ball starts falling to the ground the energy converts to kinetic energy Ek=(1/2)*m*v^2. When the ball falls to the ground, the kinetic energy transforms to elastic energy because the ball deforms as it hits the ground and some small quantity of heat. The heat goes to the air and to the ground so it gets removed from the system. So there is less energy in the system to be converted back to kinetic energy as the ball starts to rise in height again. Thats why the ball is not able to get bact to the same height as it started from. </span>
B is true, however that is not a way in which conserving and recycling paper helps manage energy sources...
