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.
AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
Answer:
the velocity is 10 m/s
Explanation:
Using the expression for kinetic energy we have:
When a red giant has insufficient mass it pretty much sheds its outer layer to leave the center core know as the white dwarf
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J