Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
The answer for this question, If I am correct, should be answer "D".
Answer:
Explanation: Determine the gravitational acceleration. ...
Decide whether the object has an initial velocity. ...
Choose how long the object is falling. ...
Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt
Answer:

Explanation:
We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

where
F is the net force on the object
m is its mass
a is its acceleration
In this problem:
F = 40 N is the force on the object
m = 2 kg is its mass
Therefore, the acceleration of the object is
