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.
When supernova explodes
because of the collapsing cores, the core itself will become a neutron star or
a white dwarf without undergoing any kind of explosive transformation. The white
dwarf stars can also become supernova if they orbit another star in a binary
system and steal material from their companion. In addition, the end state of a
star, whether it will explode as a supernova become a black hole.
intensity is the answer. i just got it on the test
Answer:
8.505 m
Explanation:
Let V1 and V2 be velocities of puck A and B respectively
Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s
Or
Vr=7.4 m/s
Distance=S= 18 m
Time =t=?
S=Vr×t
==> t=S/Vr
==> t= 18/7.4=2.43 sec
At this time both will strike together
<em><u>Distance by puck A</u></em>
<em>V1=3.5 m/s</em>
Time=t= 2.43 sec
Distance covered=d=?
d=V1×t=3.5×2.43=8.505 m
So, puck A will cover 8.505 meters before collision