stable equilibrium, if displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement.
unstable equilibrium, if displaced it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly.
neutral equilibrium, is when an equilibrium is independent of displacements from its original position.
Have a good day, hope this helps
Answer:
The two dogs sitting here are already poor and ignorant
Q1. The answer is 8.788 m/s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?
V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s
Q2. The answer is 9.22 s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V2 = 0 (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?
0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s
The mass of the second car is 1434.21 kg
<u>Explanation:</u>
Using law of conservation of momentum,
Given:
= 1090 kg
= 11 m/s
= 0
v = 4.75 m/s
We need to find 
When substituting the given values in the above equation, we get
