The force exerted by gravity is:
F = m g
F = 3300 kg * 9.8 m/s^2
F = 32,430 N
The force exerted due to the inclined plane is:
F tractor = 32,430 N * sin 14
<span>F tractor = 7,823.75 N (answer)</span>
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:
Work done, W = 2675.4 J
Given:
mass, m = 70.0 kg
height, H = 3.90 m
Solution:
According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:
Work done, W = mgH
where
g = acceleration due to gravity = ![9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J](https://tex.z-dn.net/?f=9.8%20m%2Fs%5E%7B2%7D%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%2070.0%5Ctimes%209.8%5Ctimes%203.90%20%3D%202675.4%20J)
closed and unsecure circuit
Answer:
I think it is meteorologist because they also study of the atmosphere, atmospheric phenomena, and atmospheric effects on our weather.