A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
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Answer:
Approximately . (Assuming that
, and that the tabletop is level.)
Explanation:
Weight of the book:
.
If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, .
As a side note, the and
on this book are not equal- these two forces are equal in size but point in the opposite directions.
When the book is moving, the friction on it will be equal to
, the coefficient of kinetic friction, times
, the normal force that's acting on it.
That is:
.
Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that applied force. The net force on the book shall be:
.
Apply Newton's Second Law to find the acceleration of this book:
.
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
where v is final velocity which is zero at max height and u is it initial
hence
now we can find time in the 15 cm ascent
using quadratic formula
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
t=0.0409