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1 answer:
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(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to
where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and is the variation of height. Substituting the numbers into the formula, we find
(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:
However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.
(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):
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