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.
Answer:
A) The process that give us day and night.
Explanation:
From the answer choices provided the one that is not an example of this is the process that give us day and night. This is because, the day/night cycle occurs as the Earth spins on it's axis. This cause one side of the Earth to be facing the Sun, which is the side that is currently experiencing day time while the other side is experiencing night time. As the Earth spins the cycles repeat. This is not an example of revolving.
Answer:
48.16 %
Explanation:
coefficient of restitution = 0.72
let the incoming speed be = u
let the outgoing speed be = v
kinetic energy = 0.5 x mass x 
- incoming kinetic energy = 0.5 x m x
- coefficient of restitution =
0.72 =
v = 0.72u
therefore the outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5 x m x 
outgoing kinetic energy = 0.5184 (0.5 x m x )
recall that 0.5 x m x is our incoming kinetic energy, therefore
outgoing kinetic energy = 0.5184 x (incoming kinetic energy)
from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.
The energy lost would be 100 - 51.84 = 48.16 %
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
