Answer:
How fast is it going? 29.4 Meters per second
How far has it fallen? 44.1 meters
Explanation: Gravitional Acceleration: 9.8 meters per secnd squared!
Answer: c) increases
Explanation:
Pressure increases with decreasing height
Answer:
a)
& 
b) 
c) 
Explanation:
Given:
mass of the book, 
combined mass of the student and the skateboard, 
initial velocity of the book, 
angle of projection of the book from the horizontal, 
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

where:
final velcotiy of the student after throwing the book
b)



c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.



Answer:
Option A
Explanation:
From the question we are told that:
Mass 
Velocity 
Generally the equation for momentum for Ball A is mathematically given by
Initial Momentum



Final Momentum

Therefore

Generally the equation for momentum for Ball B is mathematically given by
Initial Momentum



Final Momentum

Therefore

Option A
Answer:
So when you shine a laser through a prism, there's nothing to be separated, and the light stays together.
Explanation: