Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
Answer:
A rocket taking off from earth which pushes gasses in one direction and the rocket in
the other
Answer:
4000J
Explanation:
Given parameters:
Weight of the man = 800N
Height of ladder = 5m
Unknown:
Gravitational potential energy gained = ?
Solution:
The gravitational potential energy is due to the position of a body.
Gravitational potential energy = weight x height
Now insert the parameters;
Gravitational potential energy = 800 x 5 = 4000J
Answer:
Explanation:
Given that height of the projectile as a function of time is
here we know that
h = 147 ft
so from above equation
now by solving above quadratic equation we know that
Answer:
mesa
Explanation:

A mesa is a flat-topped mountain or hill. It is a wide, flat, elevated landform with steep sides. ... Spanish explorers of the American southwest, where many mesas are found, used the word because the tops of mesas look like the tops of tables.
