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:
Explanation:
Given that,
Radius, r = 2 m
Velocity, v = 1 m/s
We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :
So, the magnitude of centripetal acceleration is 
.
the weight of the balloon is .030 * 10 = 0.3 N
the weight of the gas of volume v is 0.54*10 N
The lifting force of a volume of v m³ of displaced air is 1.29v N
so, we need
1.29*10*v = 0.3 + 0.54*10*v
or
1.29v = 0.03+0.54v
Explanation:
Below is an attachment containing the solution.
Correct order, from lowest potential energy to highest potential energy:
E - C - D - B - A
Explanation:
The gravitational potential energy of the car is given by:
where
m is the car's mass
g is the gravitational acceleration
h is the height of the car relative to the ground
In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.
