Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
= I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
The answer is c. it requires no works
Answer:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
Answer:
4 hoop, disk, sphere
Explanation:
Because
We are given data that
Hoop, disk, sphere have Same mass and radius
So let
And Initial angular velocity, = 0
The Force on each be F
And Time = t
Also let
Radius of each = r
So let's find the inertia shall we!!
I1 = m r² /2
= 0.5 mr² the his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere wiil be
I3 = (2/5) mr²
= 0.4 mr²
So
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
So we can see that
ωf is inversely proportional to moment of inertia.
And so we take the
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
hoop, disk, sphere
Option - 4
#1. A. Waxing crescent.
#2. 1.
#3. C.
#4. C.
