Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction
Put the value into the formula
Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of
Coordinate of position of
Time = 2.00 s
We need to calculate the acceleration
For x coordinates
On differentiate w.r.to t
On differentiate again w.r.to t
The acceleration in x axis at 2 sec
For y coordinates
On differentiate w.r.to t
On differentiate again w.r.to t
The acceleration in y axis at 2 sec
The acceleration is
We need to calculate the net force
The magnitude of the force
Hence, The net force acting on this object is 180.89 N.