Answer:
We say fictitious because the actual source of the centrifugal acceleration is somewhat indirect and the experience one has results from the unbalanced forces acting on the reference frame, not a force. Note, it is an acceleration not a force. For instance, imagine yourself on a swing.
Answer:
conservative
Explanation:
Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.
hope this helped!
The question is concerned with the regions found within California, which are the Coastal Region, Mountain Region, Central Valley Region, and the Desert Region.
The Coastal Region is located furthest to the west out of all of these regions. The Coastal Region is where the California meets the Pacific Ocean, and it has a somewhat moderate and constant climate throughout the year due to its location near the ocean.
Mechanical energy is the sum of kinetic energy and potential energy
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
How energy is conserved
Em₀ = ![E_{mf}](https://tex.z-dn.net/?f=E_%7Bmf%7D)
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)