Answer:
d = 90 ft
Explanation:
As we know that after each bounce it reaches to 4/5 times of initial height
so we can say

so the distance covered is given as

here we know that
h = 10 feet



Answer:
i'm sorry i'm not a physics student
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.
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find

2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
