Potential energy relative to the surface is
(mass) x (<span>acceleration due to gravity) x (height above the surface).
At 1.0 meter above the surface, that is
</span> (mass) x (<span>acceleration due to gravity) x (1.0 meter) .
The object's mass doesn't change, so the only thing that has any effect
on its potential energy at 1 meter above the surface is the acceleration
of gravity or, in other words, the surface of <em><u>what</u></em> ?
</span>
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
<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.
Therefore, your correct answer would be "all of the above".