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:
<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>
Explanation:
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately