Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = 
L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Answer:
Linear and rotational Kinetic Energy + Gravitational potential energy
Explanation:
The ball rolls off a tall roof and starts falling.
Let us first consider the potential energy or more specifically gravitational potential energy (
; 
= mass of the ball, 
= acceleration due to gravity, 
= height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.
Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy (
; 
=moment of inertia of the ball & 
= angular velocity).
Finally comes the linear kinetic energy or simply, kinetic energy (
) which is caused due to the velocity 
of the ball.
Answer:
A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.
Explanation:
Assumes the shape and volume of its container
<span>particles can move past one another</span>
